# Patent

## Title
Method for constructing an intelligent system processing uncertain causal relationship information

## Abstract

The present invention disclosed a method constructing an intelligent system processing uncertain causal relationship information. It can express, monitor and analyze the causal logic relationship among the different variables in complex systems directly, implicitly or in both way of them under the circumstance of unsure, dynamic, having a logic loop, lacking of statistical data, unclear evidence, mixture of discrete and continuous variables, incomplete knowledge, multi-resource of knowledge. It gave effective bases to solve the problems in the domain of production, monitoring, detection, diagnosis, prediction, et al.

## Background
RELATED APPLICATIONS
This application is a national stage application (under 35 U.S.C. §371) of PCT/CN2006/002070, filed Aug. 15, 2006.
FIELD OF THE INVENTION
This invention involves the intelligence technology processing information, in particular a method for processing the uncertain causality type information to help people make decisions through the computation of computers by means of the technology of knowledge representation and inference of intelligent system, i.e. a method for constructing an intelligent system for processing the uncertain causality information.
BACKGROUND OF THE INVENTION
The so called intelligent system in the modern artificial intelligence technology includes at least two necessary factors: one is the knowledge representation model; another is the inference method based on this knowledge representation model. The two factors are necessary for constructing an intelligent system, because the inference method is directly based on the knowledge representation model. Different knowledge representation model decides different inference method. Vice versa, choosing which knowledge representation model must consider not only the capability of the representation, the flexibility of using this representation model, and the difficulty of obtaining or learning the related data, but also how to build the reasonable and effective inference method based on the knowledge representation model. Such inference method must satisfy the natural law (in this invention, it is the basic law of probability theory), and have the capability and efficiency of dealing with various problems in real applications. In other words, the knowledge representation model and the inference method is an organic whole body. They together compose an intelligent system. Therefore, to construct an intelligent system must include a knowledge representation model and an inference method in accordance with this knowledge representation model.
According to the knowledge representation model and the inference method included in the intelligent system, the ordinary software engineers can develop the specific intelligent system software products by using various software development tools. By installing such a software product in a computer, this computer becomes a specific intelligent system device. As users, the domain engineers can input the data or information related to their problems to be solved by means of the functions provided by this device. In cope with the obtained information online, this device can perform the specific inference computation and provide the useful information to help solve the problems. If this intelligent system device including the knowledge of the specific application is installed as a component in the control system with closed loop, the intelligent automatic control can be realized.
The engineers in different domains may input different knowledge/information by means of the same functions provided by the same intelligent system device and use it in different domains. Therefore, the intelligent system can usually be used in many areas and have great commercial values.
Uncertain causality information is a particularly important type of information of various types of knowledge information to be dealt with by intelligent systems. This makes the research and development of the intelligent systems dealing with the uncertain causality information be an important development direction of the intelligent system technology, because such intelligent systems can be widely applied in the fault diagnosis of industrial systems, the prediction of disasters, the analyses of economy or finance, the risk prediction, the detection, the decision consultation, etc. So far, a lot of resources have been invested in this area by many countries in the world. For example, the relative program in the National Natural Science Foundation of China is named as “intelligent system and knowledge engineering”.
The online fault diagnosis of nuclear power plants is one of the examples of applying such intelligent systems dealing with the uncertain causality information. The main parameters and the component states related to the plant operation are collected in the control room through the data collection system, and are displayed in various instrument meters. The task of the operators is to check these data periodically, judge whether or not they are normal; when abnormal case or alarm appears, diagnose the root cause and take measures in time, so as to remove or control the fault. Usually, however, the number of important parameters is in the scale of a few hundreds, the amount of data is huge, the situations are complex, and the burden of the operators in the control room is heavy. These factors cause the nervous moods of the operators, leading to the difficulty of correctly diagnosing the root cause of the abnormal state and of taking the correct measures in time. This may result in a big loss.
The Three Mile Island accident of the US in 1979 is such a typical example. This accident is caused by an ordinary component failure. But the operators made an incorrect judgment to the abnormal signals and took an incorrect measure. Not only the fault was not removed or controlled, but the fault was enlarged, resulting in a serious accident, while the root cause is just a small failure. The core of the reactor was burned. The whole nuclear power plant was ruined down.
The astounding Chernobyl nuclear power plant accident happened during the post Soviet Union is also caused by the incorrect judgment of operators and incorrect measures. It causes a lot of death, wounded persons and a big loss of property. So far, the serious result has not disappeared yet.
The prediction to the flood is also an important engineering and technical problem related to intelligent systems. This problem usually deals with the comprehensive analysis and prediction to the possibility of the dangerous down river water level in the following days, the judgment to the degree of danger, and then providing the gist for decisions such as remove people, reinforce the bank, or even bomb the bank somewhere for flood discharge, etc., so as to reduce the loss of flood. It is a realistic technical scheme to apply the intelligent system to solve this problem, based on which the uncertain inference prediction can be made according to the uncertain parameters such as the water levels at different places of the valley, the weather forecast, etc., and the uncertain causalities among them.
The traditional method to deal with the causality information is the rule-based expert system. This methodology takes the rules in the type of IF-THEN to represent the causalities among the real things. For example, the Chinese patent (ZL90103328.6) named as “computer aided decision method” by TIC of the US is such a rule-based expert system. The rule-based expert systems are good in the cases without uncertainty. Its technical scheme involves mainly how to represent and organize the rules and facts, as well as how to invoke, match and eliminate these rules and facts in the inference. When uncertainties exist, the technologies dealing with uncertainties have to be applied.
The uncertainty (including dynamical cases) is currently the important research area in this field. This is because the uncertainty exists universally and is the most difficult technical problem. For this, the international academic community establishes the association of uncertainty in artificial intelligence (AUAI) and holds international conferences every year. So far, such conferences have been held more than 20 times (www.auai.org).
The certainty factor method presented by Shortliffe, the evidence reasoning method presented by Dempster-Shafer, the fuzzy logic method presented by Zahdy, etc., take the non-probabilistic parameters to measure the uncertainty. Although they have unique features, their applications are limited due to the limitations of the non-probabilistic parameters themselves and other causes.
Of the intelligent systems dealing with uncertainty, the intelligent systems that take graphs instead of only languages in dealing with the uncertain causality information are more and more welcome by users. This is because the graphs are intuitive to be understood, convenient in representation, etc., in which the neural network (NN) was and is still one of such methods. For example, the recent granted Chinese invention patents No. 01139043.3 named as “the structure based method for the construction and optimization of NN”, No. 03137640.1 named as “the NN based processing method for information pattern recognition” and No. 02139414.8 named as “the recognition method for the chaos signals and general noise”, etc., are such methods. NN imitates brain's neural network, adjusts the network structure and parameters by learning from a large amount of data, so as to obtain and represent the knowledge. After this, the states of the things in concern can be inferred according to the observed information. However, due to the lack of the enough research to the neural network of brain and the limitations of NN such as the black or gray box representation model that does not correspondingly and clearly reflect the logics among the things in concern, and the lack of the data to learn from, its applications mainly focused on the pattern recognition and some other areas. There are less applications dealing with uncertain causality information.
The Bayesian/Bayes/Belief Network (BN) presented by Judea Pearl, al. has been so far another good method to deal with the uncertain causality problems. Its feature is to use the direct acyclic graph (DAG) to represent the causalities among the things and use the conditional probability table (CPT) to represent the degree of the causality uncertainty. Then, based an the observed evidence E, the forward, backward or bidirectional probabilistic inference can be made. FIG. 40 shows two examples of BN, in which FIG. 40-1 is singly connected and FIG. 40-2 is multiply connected. The directed arcs represent the causalities. The static logic cycles (e.g. FIG. 41) are prohibited in BN, because otherwise, there is no solution. In general, BN is graphically intuitive, has clear physical meaning, based on the probability theory strictly, easy to utilize the statistical data, localized in the computation steps, self-consistent as a whole theory framework, etc. But, although its every computation step is localized, the probability distributions of the nodes in concern can only be obtained by the computation of bidirectional information propagation throughout the whole network. This increases the computation amount significantly. Moreover, it is difficult for BN to deal with the static logic cycles, the dynamical change of data and logic structures, and the lack, imprecision and incompleteness of data, etc. In spite of this, BN is still applied widely.
Because of the success of BN, Judea Pearl won the excellent research award issued once per two years by IJCAI in 1999, as well as many other awards (http://bayes.cs.ucla.edu/). So far, BN has become one of the popular intelligent systems.
Based on the advantages of BN such as the graphical representation and strictly based on probability theory, etc., a method named as the single-valued Dynamical Causality Diagram (DCD) was developed in 1994. FIG. 42 is an example of the single-valued DCD. This single-valued DCD includes AND gate, OR gate and the logical cycle. But every variable can only has one state for which the causalities can be represented, and another state for which no causality can be represented. Otherwise, the case becomes multi-valued. The DCD solves, in the single-valued cases, the problems of logic cycles, dynamics, representing the causalities with linkage intensities instead of CPTs so that the statistic dada can be less relied on, etc. The single-valued DCD method involves the disjoint operation of logic expressions, which is an NP hard difficulty. That is to say, as the scale of cases becomes large, there will be the combination explosion in the disjoint operation.
In 2001, a further method (briefly called the multi-valued DCD) was developed, which transfers the multi-valued cases into the single-valued cases. This method treats all the abnormal states of a variable as a polymeric state besides a normal state whose causalities are not represented, so as to transfer the multi-valued cases as the single-valued cases for computation. After this, the probability of the polymeric state is allocated among the abnormal states according to some proportion. However, the theory of computing the proportion is not well founded. It does not really solve the conflict between the independence of representing knowledge and the correlation resulted from the exclusion among the different states of a variable in a multi-valued DCD. Furthermore, it requires that every variable has a special normal state for which no causality should be represented. Therefore, it is not sound and cannot be widely applied. Moreover, the free mixture and transformation from each other between the explicit representation of the multi-valued DCD and the implicit CPT representation of BN, the fuzzy evidence, the free mixture and unified treatment of the discrete and continuous variables, the complex logic combinations, the lack or incompleteness of data, the dynamics, etc., have not been solved yet.
SUMMARY OF THE INVENTION
To solve the problems existing in the present intelligent systems, this invention presents a new technical scheme. That is the method to construct the intelligent system named as the Dynamical Uncertain Causality Graph (DUCG). The intelligent system presented in this invention provides not only the new model to represent the uncertain causality knowledge or information (mainly see sections §1-10), but also the inference method based on this new representation model (mainly see sections §11-22). By applying this intelligent system, people can easily represent various causality information among the real things encountered in practice, which is complex, multi-valued, uncertain and dynamical. According to the online received various data or information, the dynamical intelligent analysis for prediction, diagnosis, or both can be made, so as to provide people with the valuable information for the fault diagnoses of industrial systems, disaster prediction, financial/economical analyses, risk assessment, detection, decision consultation, etc.
In what follows, the technical terminology used in this invention specification is explained.
In this invention, the so called causality means the logic relations between any causes and consequences/effects, or the relations that can be formally represented as the causal logic relations. For example, in the weather forecast, the converging of cool air and the warm air may result in rain. The converging of cool air and warm air is the cause. The rain is the consequence. For another example, the leakage of a steam pipe will result in the low pressure in this steam pipe. The leakage is the cause. The low pressure is the consequence. For further example, the increase of bank interest rate will restrain the product price. The increase of bank interest rate is the cause. The reduction of product price is the effect. All these are the causalities.
The so called multi-valued means that the thing in concern may have more than one effective discrete value (e.g. no rain, small rain and large rain) or continuous value (e.g. the temperature of a stove). It corresponds to the single-valued causality case in which only true (e.g. rain) or false (e.g. no rain) are the two states that one thing can be in, and only the true state is in concern. Therefore, only the logic relations among the true states are represented (single-valued). DUCG does not have this single-valued limitation. Except being specially specified, the DUCG in this document is always multi-valued.
The so called uncertain means that there is uncertainty in the logic relations among things. In the above example, whether or not is the converging of the cool air and the warm air the cause of the rain? If yes, whether or not does it rain? Is the rain the large rain or small rain? What is the large rain or small rain? What is cool or warm? All of these are uncertain. In this invention, these uncertainties are represented by the parameters such as the functional intensity, the state membership, the probability or belief of event occurrence, the relationship and the conditional probability, etc.
The so called dynamical means that the constructed logic relations and data can be dynamically changed according to the observed information including the occurrence order of events, the known states of some event variables, the start time of some process. The parameters can be the functions of time. The computation process can also reflect the dynamical change of things, by combining the dynamical information.
The so called inference means to reason the states of the things in concern by the intelligent system according to the above described relations among things, based on the observed and dynamically changed evidence E (i.e. to compute the probabilities of the events in concern, conditioned on the known evidence E). This computation can be either forward (prediction), or backward (diagnosis), or the mixture of both, e.g. to find the cause or consequence of the abnormal power plant state when the plant parameters or signals are partially abnormal and partially normal.
The so called domain engineers are those who are rich of the domain knowledge, familiar with the situations of the applications, and then can provide the professional knowledge or information required for this intelligent system to solve the problems in concern. These personal are also the direct users of this intelligent system.
The so called belief is one of the two type probabilities. One is frequency-based. For example, in 100 experiments, 30 show success. Then, the probability of success is 0.3. The other one is belief-based, which is based on the belief of domain engineers. This belief comes from the accumulation of the past statistic data and knowledge in the mind of the domain engineers. For the example of an experiment never being done, there is no data available. But the domain engineers may did similar experiments somewhere else, or the parts of this experiment. Then the domain engineers may judge the success probability of this experiment as 0.3, i.e. Belief=0.3, by their subjective synthesis and analogy.
To solve the problems mentioned above, this invention provides the technical scheme as follows:
§1. A method for constructing an intelligent system for processing the uncertain causality information. This method represents the causalities among the things in the explicit representation mode, which includes the following steps:
(1) Establish a representation system about the various cause variables Vi and consequence variables Xn in concern with the problem to be solved. The features of this representation system are the follows: {circle around (1)} Let V represent two type variables B and X, i.e. V∈{B,X}, in which B is the basic variable that is only the cause variable and X is the consequence variable that can be also the cause variable of the other consequence variable; {circle around (2)} No matter the states of the variable are discrete or not, represent them all as the discrete or fuzzy discrete states, so as to be dealt with by using the same manner, that is, represent the different states of Vi and Xn as Vij and Xnk respectively, where i and n index variables while j and k index the discrete or fuzzy discrete states of the variables; {circle around (3)} When Vi or Xn is continuous, the membership of an arbitrary value ei of Vi or en of Xn, belonging to Vij or Xnk respectively, is mij(ei) or mnk(en) respectively, and they satisfy




  
    
      
        
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{circle around (4)} Vij and Xnk are treated as events, i.e., Vij represents the event that Vi is in its state j and Xnk represents the event that Xn is in its state k; meanwhile, if j≠j′ and k≠k′, Vij is exclusive with Vij′ and Xnk is exclusive with Xnk′; {circle around (5)} If i≠i′, Bij and Bij′ are independent events, and their occurrence probabilities bij are known and satisfy





  
    
      
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(2) For the consequence variable Xn, determine its direct cause variables Vi, i∈SEXn, SEXn is the index set of the {B,X} type direct variables of Xn in the explicit representation mode;
(3) The functional variable Fn;i is used to represent the causality between Vi, i∈SEXn, and Xn. Vi is the input or cause variable of Fn;i and Xn is the output or consequence variable of Fn;i, the features of Fn;i are follows: {circle around (1)} The causality uncertainty between Vi and Xn is represented by the occurrence probability fnk;ij of the specific value Fnk;ij of Fn;i. Fnk;ij is a random event representing the uncertain functional mechanism of Vij causing Xnk. fnk;ij is the probability contribution of Vij to Xnk; {circle around (2)} fnk;ij=(rn;i/rn)ank;ij, where rn;i is called the relationship between Vi and Xn, rn is the normalization factor and




  
    
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§2. The functional variable Fn;i described in §1 can be the conditional functional variable, and the conditional functional variable is used to represent the functional relation between the cause variable Vi and the consequence variable Xn conditioned on Cn;i, where the condition Cn;i has the following features:
(1) Cn;i has only two states: true or false, and its state can be found according to the observed information or the computation results;
(2) When Cn;i is true, the conditional functional variable becomes the functional variable;
(3) When Cn;i is false, the conditional functional variable is eliminated.
§3. The explicit representation mode in §1 includes also to extend V∈{B,X} to V∈{B,X,G}, where G is the logic gate variable, i.e. the cause variable to influence the consequence variable by the state logic combinations of a group of cause variables; suppose the input variables of logic gate variable Gi are Vh, then the logic gate Gi is constructed as follows:
(1) The logic combinations between the input variables Vh, V∈{B,X,G}, are represented by the truth value table of Gi in which each input row is a logic expression composed of the input variable states and corresponds to a unique state of Gi; different rows of the logic expressions are exclusive with each other, wherein if a logic expression is true, the corresponding state of Gi is true;
(2) The set of the states of Gi is equal to or less than the set of all state combinations of the input variables;
(3) When the set of the states of Gi is less than the set of all state combinations of the input variables, there is a remnant state of Gi, which corresponds uniquely to the remnant state combinations of the input variables, so that all the states of Gi including the remnant state are exclusive with each other and just cover all the state combinations of the input variables;
(4) When Gi is the direct cause variable of Xn, Gi functions to Xn through the functional or conditional functional variable Fn;i;
(5) If a logic gate has only one input variable, this logic gate can be ignored, i.e. the input variable of the logic gate can be taken as the input variable of the functional variable or conditional functional variable Fn;i with this logic gate as its input variable;
(6) When Gi is the direct cause variable of Xn, the relationship between Gi and Xn is rn;i; when calculating fnk;ij, the calculation to rn includes the relationship between Gi and Xn; when calculating Pr{Xnk}, the fnk;ij between Gi and Xn is included.
§4. The explicit mode in §1 or §3 includes also the following contents:
(1) Extend V∈{B,X} as V∈{B,X,D}, or extend V∈{B,X,G} as V∈{B,X,G,D}, in which D is the default event or variable, Dn can appear only with Xn and is an independent cause variable that has only one inevitable state;
(2) Dn becomes a direct cause variable of Xn through Fn;D, where Fn;D is the functional variable between Dn and Xn;
(3) The causality uncertainty between Dn and Xn is represented by the occurrence probability fnk;D of the specific value Fnk;D of Fn;D, where Fnk;D is a random event representing the functional mechanism of Dn to Xn, and fnk;D is the probability contribution of Dn to Xnk;
(4) fnk;D=(rn;D/rn)ank;D, where ank;D is the probability of the event that Dn causes Xn regardless of the other cause variables of Xn, and satisfies




  
    
      
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§5. The explicit representation mode in §4 includes also the following contents: When the default variable of Xn is more than one, they can be combined as one default variable Dn. Let g be the index distinguishing two or more default variables. Corresponding to the case of only one default variable, the variable Dn and the parameter rn;D, ank;D are represented as Dng, rn;Dg, ank;Dg respectively; after combining Dng as Dn, the parameters of Dn are calculated according to




  
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§6. A method for constructing the intelligent system for processing the uncertain causality information. This method uses the implicit mode to represent the uncertain causalities among things, which includes the following steps:
(1) Establish a representation system about the various cause variables and the consequence variables in concern with the problem to be solved according to the method described in §1 (1);
(2) For the consequence variable Xn, determine its direct cause variables Vi, i∈SIXn. SIXn is the index set of the direct cause variables of Xn in the implicit representation mode;
(3) The conditional probability table (CPT) is used to represent the causality between the consequence variable Xn and its direct cause variables Vi, i∈SIXn, the features include: {circle around (1)} When no cause variable will be eliminated, CPT is composed of only the conditional probabilities pnk;ij, where pnk;ij≡Pr{Xnk|j} and j indexes the state combination of the cause variables Vi, i∈SIXn; {circle around (2)} When part or even all cause variables may be eliminated, CPT is composed of three parameters: pnk;ij, qnk;ij and dn;j, satisfying pnk;ij=qnk;ij/dn;j, so that CPT can be reconstructed when some of its cause variables are eliminated, where qnk;ij and dn;j are the sample number and occurrence number of Xnk respectively, conditioned on the state combination indexed by j of the cause variables.
§7. The implicit representation mode in §6 includes also the following contents:
(1) In the implicit representation mode, the cause variables Vi, i∈SIXn, can be separated as several groups, every group uses the implicit representation mode to represent the uncertain causality to Xn according to §6;
(2) Give the relationship rXn between every group of direct cause variables to the consequence variable Xn;
(3) If some cause variables in the group are eliminated for any reason, the CPT of this group can be reconstructed as follows: Suppose the variable to be eliminated is Vi, before the elimination, there are several subgroups of the state combinations of the input variables indexed by j′; in subgroup j′, the states of all the variables are same except the states of Vi; denote the index set of the state combination j in subgroup j′ as Sij′. Then




  
    
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(4) Repeat (3) to deal with the case in which more than one cause variable is eliminated.
§8. A method for constructing the intelligent system for processing the uncertain causality information. This method uses the hybrid representation mode to represent the uncertain causality among things, which includes the following steps:
(1) Establish the representation system about the various cause variables Vi and the consequence variables Xn in concern with the problem to be solved according to the method in §1 (1).
(2) For consequence variable Xn, determine its direct cause variables Vi, i∈SEXn, V∈{B,X,D,G}, and the direct cause variables Vj′, i′∈SIXn, V∈{B,X}, SIXn may have many, i.e. there may be more than one group of direct cause variables in the implicit representation mode;
(3) Represent the causalities between the cause variables included in SEXn and Xn according to the explicit representation mode, and represent the causalities between the cause variables included in SIXng and Xn according to the implicit representation mode;
(4) For a group of cause variables Vi′, i′∈SIXn, in the implicit representation mode, give the corresponding relationship rXn, while in the explicit representation mode, rn is renewed as rn=rn+rXn, in which the right side rn is before the renewing;
(5) If the implicit representation mode has more than one group, they can be indexed by g and every group relationship can be denoted as rXng; then the calculation equation in above (4) becomes




  
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§9. A method for constructing the intelligent system for processing the uncertain causality information. It takes the following steps to synthetically represent the uncertain causality among the things in concern with the problem to be solved:
(1) Establish the representation system about the various cause variables Vi and consequence variables Xn in concern with the problem to be solved according to the method described in §1 (1);
(2) According to the specific cases of every consequence variable Xn, represent the uncertain causalities between Xn and its direct cause variables in either explicit mode, implicit mode or hybrid mode respectively; the representations above for all the consequence variables compose the original DUCG;
(3) The evidence E in concern with the original DUCG is received during the online application and is expressed as




  
    E
    =
    
      
        E
        *
      
      ⁢
      
        
          ∏
          h
        
        ⁢
        
          
        
        ⁢
        
          E
          h
        
      
    
  
  ,




where Eh is the evidence indicating the state of the {B,X} type variable, E* represents the other evidence; if Eh is a fuzzy state evidence, i.e. the state of the variable Vh in the original DUCG is known in a state probability distribution, or if Eh is a fuzzy continuous evidence, i.e. the specific value eh of the continuous variable Vh is known in the fuzzy area of different fuzzy states of Vh, V∈{B,X}, then add Eh as a virtual evidence variable into the original DUCG and represent the causality between Vh and Eh according to the explicit mode so that Eh becomes the consequence variable of the cause variable Vh; after finishing these steps, the original DUCG becomes the E conditional original DUCG.

§10. In step (3) of the method described in §9, the step of adding Eh as a virtual evidence variable into the original DUCG and representing the causality between Vh and Eh according to the explicit mode so that Eh becomes the consequence variable of the cause variable Vh includes the following features: Suppose mhj=mhj(eh) is the membership of Eh belonging to the fuzzy state j, or mhj is the probability of Xhj indicated by the fuzzy state evidence Eh, i.e., mhj=Pr{Vhj|Eh}, j∈SEh, SEh is the index set of state j in which mhj≠0 and includes at least two different indexes, while satisfying




  
    
      ∑
      
        j
        ∈
        
          S
          Eh
        
      
    
    ⁢
    
      m
      hj
    
  
  =
  
    1
    ⁢
    
      :
    
  




(1) As the virtual consequence variable of Vh, Eh has only one inevitable state, has only one direct cause variable Vh, and is not the cause variable of any other variable;
(2) The virtual functional variable from Vhj to Eh is FE;h and its specific value FE;hj is the virtual random event that Vhj causes Eh; the functional intensity parameter fE;hj of FE;hj may be given by domain engineers;
(3) If the domain engineers cannot give fE;hj, it can be calculated from




  
    
      f
      
        E
        ;
        hj
      
    
    =
    
      
        
          
            m
            hj
          
          ⁢
          
            v
            hk
          
        
        
          
            m
            hk
          
          ⁢
          
            v
            hj
          
        
      
      ⁢
      
        f
        
          E
          ;
          hk
        
      
    
  
  ,




where j≠k, j∈SEh, k∈SEh, vhj≡Pr{Vhj}, vhk≡Pr{Vhk}. Given fE;hk>0, fE;hj can be calculated.

§11. Based on the E conditional original DUCG described in §9, the following steps are taken to perform the inference, so as to provide the effective gist for solving the problems in concern:
(1) According to E, simplify the E conditional original DUCG to get the simplified DCUG;
(2) Transform the simplified DUCG to EDUCG or IDUCG, where if there is any logic cycle, transfer to only EDUCG;
(3) If transform to IDUCG, the BN method can be used to calculate the state probability distribution of the variables in concern conditioned on E;
(4) If transform to EDUCG, outspread the evidence events Eh included in E, which determine the states of the {B,X} type variables, and the events Hkj in concern. In the process of outspread, break the logic cycles;
(5) In the case of transforming to EDUCG, based on the outspreaded logic expressions of Eh and Hkj, further outspread




  
    
      ∏
      h
    
    ⁢
    
      
    
    ⁢
    
      
        E
        h
      
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        H
        kj
      
      ⁢
      
        
          ∏
          h
        
        ⁢
        
          
        
        ⁢
        
          E
          h
        
      
    
  
  ;




(6) In the case of transforming to EDUCG, calculate the state probability and the rank probability of the concerned event Hkj conditioned on E according to the following equations:
The state probability:




  
    h
    kj
    s
  
  =
  
    
      
        
          P
          r
        
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            E
          
          }
        
      
      
        Pr
        ⁢
        
          {
          E
          }
        
      
    
    .
  




The rank probability:




  
    h
    kj
    r
  
  =
  
    
      
        h
        kj
        s
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          h
          kj
          s
        
      
    
    =
    
      
        
          P
          r
        
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            E
          
          }
        
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              E
            
            }
          
        
      
    
  




Where S is the set of all the events in concern.

§12. In the inference computation steps described in §11, if the problem to be dealt with is about the process system, the following steps can be further applied in (5) and (6) in §11:
(1) Express the evidence set




  
    ∏
    h
  
  ⁢
  
    
  
  ⁢
  
    E
    h
  




indicating the states of the {B,X} type variables as E′E′′, in which





  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      
    
    ⁢
    
      E
      i
      ′
    
  




is the evidence set composed of the evidence events indicating the abnormal states of variables, and





  
    E
    ′′
  
  =
  
    
      ∏
      
        i
        ′
      
    
    ⁢
    
      
    
    ⁢
    
      E
      
        i
        ′
      
      ′′
    
  




is the evidence set composed of the evidence events indicating the normal states of variables;

(2) Outspread




  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      
    
    ⁢
    
      E
      i
      ′
    
  




and determine the possible solution set S conditioned on E, where every possible solution Hkj is an event in concern for the problem to be solved;

(3) Calculate two types of the state probability and rank probability of Hkj conditioned on E:
The state probability with incomplete information:




  
    
      h
      kj
      s
    
    =
    
      
        
          P
          r
        
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            
              E
              ′
            
          
          }
        
      
      
        Pr
        ⁢
        
          {
          
            E
            ′
          
          }
        
      
    
  
  ;




The state probability with complete information:




  
    
      h
      kj
      s
    
    =
    
      
        
          h
          kj
          
            s
            ′
          
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                |
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                |
                
                  E
                  ′
                
              
              }
            
          
        
      
      =
      
        
          
            h
            kj
            
              s
              ′
            
          
          ⁢
          Pr
          ⁢
          
            {
            
              
                E
                ′′
              
              |
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
              
            
            }
          
        
        
          
            ∑
            j
          
          ⁢
          
            
              h
              kj
              
                s
                ′
              
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                |
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
        
      
    
  
  ;




The rank probability with incomplete information:




  
    
      h
      kj
      
        r
        ′
      
    
    =
    
      
        
          h
          kj
          
            s
            ′
          
        
        
          
            ∑
            
              
                H
                kj
              
              ∈
              S
            
          
          ⁢
          
            h
            kj
            
              s
              ′
            
          
        
      
      =
      
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              
                E
                ′
              
            
            }
          
        
        
          
            ∑
            
              
                H
                kj
              
              ∈
              S
            
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
              
              }
            
          
        
      
    
  
  ;




The rank probability with complete information:




  
    h
    kj
    r
  
  =
  
    
      
        h
        kj
        s
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          h
          kj
          s
        
      
    
    =
    
      
        
          
            h
            kj
            
              r
              ′
            
          
          ⁢
          Pr
          ⁢
          
            {
            
              
                E
                ′′
              
              |
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
              
            
            }
          
        
        
          
            ∑
            j
          
          ⁢
          
            
              h
              kj
              
                r
                ′
              
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                |
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
        
      
      .
    
  




In which, if HkjE′ is null, Pr{E′′|HkjE′}≡0.

§13. In the inference computation method described in step (1) of §11, the step to simplify the E conditional original DUCG includes the following contents: suppose Vi is the direct cause variable of Xn, V∈{B,X,G,D}, then
(1) According to E, determine whether or not the condition Cn;i of the conditional functional variable Fn;i is valid: {circle around (1)} if yes, change the conditional functional variable as the functional variable; {circle around (2)} if not, eliminate this conditional functional variable; {circle around (3)} If cannot determine whether or not Cn;i is valid, keep the conditional functional variable until Cn;i can be determined;
(2) According to E, if Vih is not the cause of any state of Xn, when E shows that Vih is true, eliminate the functional or conditional functional variable Fn;i that is from Vi to Xn;
(3) According to E, if Xnk cannot be caused by any state of Vi, when E shows that Xnk is true, eliminate the functional or conditional functional variable from Vi to Xn;
(4) In the explicit mode of representation, if the X or G type variable without any cause or input appears, eliminate this variable along with the F type variables starting from this variable;
(5) If there is any group of isolated variables without any logic connection to the variables related to E, eliminate this group variables;
(6) If E shows that Xnk is true, while Xnk is not the cause of any other variable and Xn has no connection with the other variables related to E, denote the index set of the index n of such Xn as SEnk; When Vi and its logic connection variables Fn;i have no logic connection with the variables related to E except the variables indexed in SEnk, eliminate Xn, Vi and the functional or conditional functional variables Fn;i along with all other variables logically connected with Vi;
(7) If E shows that Xnk appears earlier than Vij, so that for sure Vij is not the cause of Xnk, eliminate the functional or conditional functional variables that are in the causality chains from Vi to Xn but are not related to the influence of other variables to Xn;
(8) Upon demand, the above steps can be in any order and can be repeated.
§14. In step (2) of the inference method described in §11, the step to transform the DUCG with implicit or hybrid representation mode conditioned on E as all in the explicit mode, i.e. EDUCG, includes the following steps:
(1) For the consequence variable Xn in the implicit or hybrid mode, for every group of SIXn type cause variables, introduce a default variable Dn and a default functional variable Fn;D. Dn becomes the cause variable of Xn by Fn;D;
(2) The calculation method for the parameters of Fn;D is: for every k, find the smallest pnk;j in the CPT composed of the variables whose subscripts belong to SIXn; the smallest pnk;j is denoted as pnk, i.e.




  
    
      
        p
        nk
      
      =
      
        
          min
          j
        
        ⁢
        
          {
          
            p
            
              nk
              ;
              j
            
          
          }
        
      
    
    ;
    
      
        if
        ⁢
        
          
        
        ⁢
        
          
            ∑
            k
          
          ⁢
          
            p
            nk
          
        
      
      =
      0
    
  
  ,




Dn and Fn;D are not existing; if





  
    
      
        ∑
        k
      
      ⁢
      
        p
        nk
      
    
    ≠
    0
  
  ,




calculate





  
    a
    
      nk
      ;
      D
    
  
  =
  
    
      p
      nk
    
    /
    
      
        ∑
        k
      
      ⁢
      
        p
        nk
      
    
  




and the relationship of Fn;D:





  
    
      r
      
        n
        ;
        D
      
    
    =
    
      
        r
        Xn
      
      ⁢
      
        
          ∑
          k
        
        ⁢
        
          p
          nk
        
      
    
  
  ;




(3) Introduce a virtual logic gate variable Gi, in which the cause variables of SIXn are the input variables of Gi, and the number of the states of Gi and the input rows of the truth value table of Gi equal to the number of the state combinations of the cause variables in SIXn, while each of the state combination of the input variables is an input row of the truth value table of Gi and also a state of the virtual logic gate;
(4) Introduce the virtual functional variable Fn;i, in which Gi is the input variable and Xn is the output variable, so that Gi becomes the direct cause variable of Xn;
(5) In the CPT of the cause variables in SIXn, if




  
    
      1
      -
      
        
          ∑
          k
        
        ⁢
        
          p
          nk
        
      
    
    ≠
    0
  
  ,




there is





  
    
      p
      
        nk
        ;
        j
      
    
    =
    
      
        (
        
          
            p
            
              nk
              ;
              j
            
          
          -
          
            p
            nk
          
        
        )
      
      /
      
        (
        
          1
          -
          
            
              ∑
              k
            
            ⁢
            
              p
              nk
            
          
        
        )
      
    
  
  ,




in which the pnk;j on the right side are the values before the calculation and that on the left side are the values after the calculation. dn;j can remains and qnk;j=pnkdn;j, ank;ij=pnk;j, where the pnk;j in the two equations are all values after the calculation; the relationship of Fn;i is





  
    
      r
      
        n
        ;
        i
      
    
    =
    
      
        r
        Xn
      
      (
      
        1
        -
        
          
            ∑
            k
          
          ⁢
          
            p
            nk
          
        
      
      )
    
  
  ;




(6) In the original CPT of the cause variables in SIXn, if




  
    
      1
      -
      
        
          ∑
          k
        
        ⁢
        
          p
          nk
        
      
    
    =
    0
  
  ,




the variables in SIXn are fused as Dn, ank;D=pnk and rn;D=rXn; there is no need to introduce the virtual logic gate Gi and its functional variable;

(7) When there is only one input variable in Gi, such Gi can be ignored, i.e. the virtual functional variable takes the input variable of Gi as its input variable directly;
(8) When the groups of the SIXn type variables are more than one group, repeat the above steps for every groups, and then combine the default variables resulted according to the method described in §5.
§15. In step (2) of §11, the following steps are involved to transform the DUCG conditioned on E in the explicit representation mode or in the more than one group implicit representation mode as the IDUCG in which all representations are in the implicit representation mode with only one group direct cause variables:
(1) If Cn;i is valid, change the conditional functional variable as the functional variable; if Cn;i is invalid, eliminate the conditional functional variable;
(2) For any representation of the uncertain causality between the consequence variable Xn and its direct cause variables, if it is in the hybrid or more than one group implicit representation mode, transform the representation mode for Xn to the explicit mode according to the method described in §14;
(3) After the above steps, take the state combinations of the {B,X} type cause variables of the consequence variable Xn as the conditions indexed by j, calculate the conditional probability of Xnk Pr{Xnk|j} according to the explicit mode, where the connections between the {B,X} type cause variables and Xn may be or may not be through logic gates; in the calculation, all contributions from different types of direct cause variables should be considered, i.e. when the direct cause variables are {X,B,G} types,




  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          |
          j
        
        }
      
    
    =
    
      
        ∑
        i
      
      ⁢
      
        f
        
          nk
          ;
          ih
        
      
    
  
  ;




when the direct cause variables are {X,B,G,D} types,





  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          |
          j
        
        }
      
    
    =
    
      
        
          ∑
          i
        
        ⁢
        
          f
          
            nk
            ;
            ih
          
        
      
      +
      
        f
        
          nk
          ;
          D
        
      
    
  
  ;




(4) The case of ank;ih=1 can be understood as that Xnk is true for sure, i.e. when the input variable i is in its state h, all the states, except k, of Xn cannot be true; if this applies, when ank;ih=1, Pr{Xnk|j}=1, meanwhile Pr{Xnk′|j}=0, where k≠k′;
(5) If ank;ih=1, k∈Sm, Sm is the index set of such states of Xn that ank;ih=1 and the number of such states is m, then Pr{Xnk|j}=1/m and Pr{Xnk′|j}=0, where k′∉Sm;
(6) If such calculated




  
    
      
        ∑
        k
      
      ⁢
      
        Pr
        ⁢
        
          {
          
            
              X
              nk
            
            |
            j
          
          }
        
      
    
    <
    1
  
  ,
  
    
      let
      ⁢
      
        
      
      ⁢
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          |
          j
        
        }
      
    
    =
    
      1
      -
      
        
          ∑
          
            k
            ≠
            n
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                X
                nk
              
              |
              j
            
            }
          
        
      
    
  
  ,




where η indexes the default state of Xn;

(7) If there is no default state η in (6), follow the normalization method below:




  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          |
          j
        
        }
      
    
    =
    
      Pr
      ⁢
      
        
          {
          
            
              X
              nk
            
            |
            j
          
          }
        
        /
        
          
            ∑
            k
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  X
                  nk
                
                |
                j
              
              }
            
          
        
      
    
  
  ,




where the Pr{Xnk|j} on the right side are the values before the normalization;

(8) After satisfying the normalization, Pr{Xnk|j} becomes the conditional probability pnk;nj in the standard implicit representation mode;
(9) Connect the {X,B} type direct cause variables of Xn through or not through logic gates with Xn according to the implicit representation mode, the DUCG conditioned on E is transformed as the IDUCG.
§16. According to the inference method described in §11 or §12, in order to outspread E, E′, HkjE or HkjE′, such method is involved that outspreads the evidence Eh indicating the states of the {B,X} type variables and the X type variables included in Hkj, and breaks the logic cycles during the outspread, which has the following features:
(1) When Eh indicates that Xn is in its state k, then Eh=Xnk; if Eh is the virtual consequence variable of Xn,




  
    
      E
      h
    
    =
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            E
            ;
            nk
          
        
        ⁢
        
          X
          nk
        
      
    
  
  ;




when Eh indicates that Bi is in its state j, then Eh=Bij. If Eh is the virtual consequence variable of Bi,





  
    
      E
      h
    
    =
    
      
        ∑
        j
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            E
            ;
            ij
          
        
        ⁢
        
          B
          ij
        
      
    
  
  ;




(2) Outspread Xnk according to




  
    
      X
      nk
    
    =
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            nk
            ;
            i
          
        
        ⁢
        
          V
          i
        
      
    
  
  ,




where Vi are the direct cause variables of Xn, i∈SEXn, V∈{X,B,G,D};

(3) When Vi is a logic gate, the input variables of Vi are outspreaded according to the truth value table of this logic gate; if the input variables are logic gates again, outspread these input variables in the same way;
(4) Consider every non-F type variable in the logic expression outspreaded from (2) and (3): {circle around (1)} If it is such an X type variable that has not appeared in the causality chain, repeat the logic outspread process described in (2) and (3); {circle around (2)} If it is a {B,D} type variable or such an X type variable that has appeared in the causality chain, no further outspread is needed;
(5) In (4) {circle around (2)} above, the X type variable that has appeared in the causality chain is called the repeated variable; in the dynamical case, the repeated variable is the same variable but is in the earlier moment; the probability distribution of this variable is known according to the computation or the observed evidence in the earlier moment; in the static case, the repeated variable as cause is treated as null, i.e. {circle around (1)} if the repeated variable as cause is connected to the consequence variable by only an F type variable without any logic gate, this F type variable is eliminated, meanwhile the relationship corresponding to this F type variable is eliminated from rn; {circle around (2)} if the repeated variable as cause is connected with the consequence variable by being an input variable of a logic gate in which the repeated variable is logically combined with other input variables, this repeated variable is eliminated from the input variables of the logic gate.
§17. In step (5) {circle around (2)} in §16, the following steps to eliminate an input variable of a logic gate is involved; suppose the variable to be eliminated from the logic gate is Vi, then,
(1) When the logic gate is a virtual logic gate, eliminate the direct cause variable Vi in the corresponding implicit mode first, reconstruct the conditional probability table according to the method described in §7 (3), and then transform the new implicit mode case to a new virtual logic gate and a new virtual functional variable according to the method described in §14; correspondingly, some new default variable, the virtual functional variable from the new logic gate and the default functional variable of the new default event may be introduced;
(2) When the logic gate is not a virtual logic gate, make the logic gate as the most simplified logic gate first; based on the most simplified logic gate, calculate the logic expression in every input row in the truth value table by treating any state of Vi as null, eliminate the input row along with the corresponding logic gate state when this row is calculated as null; the functional or conditional functional events with this logic gate state as their input events are also eliminated;
(3) If all the input variables of a non-virtual logic gate are eliminated, or all the input rows of the truth value table are eliminated, this logic gate becomes null;
(4) Repeat the above steps to treat the case when more than one input variables are eliminated.
§18. In §16, for outspreading E, E′HkjE or HkjE′, the following steps are involved.
(1) According to the steps to simplify DUCG, which is described in §13, and the method to outspread the X type variables and breaking logic cycles, which is described in §16, the input variables and the truth value table of the logic gate in EDUCG may change; after the change, make the expression in the truth value table of the logic gate as the exclusive expression; then, the logic gate is outspreaded according to the exclusive expressions of the input rows in the truth value table;
(2) In the logic AND operation of the same logic gate but with different input variables in different cases, along with the logic AND operation of the F type variables with such logic gate as input, let Gik denote the kth case of variable elimination of the logic gate Gi, Sik denote the index set of the input variables of Gik, Fkn;i denote the functional variable from Gik to Xn, where if k=0, Gik=Gi, Fkn;i=Fn;i and Sik=Si0. Si0 is the index set including all input variables of Gi; thus, {circle around (1)} GikGik′=Gih, Fkn;iFk′n;i=Fhn;i, in which Sih=SikSik′, Sih is the index set including those variables that are not only the input variables of Gik, but also the input variables of Gik′; {circle around (2)} when Sih=Si0, Gih=Gi; when Sih≠Si0, the truth value table of Gih is formed according to the change from Gi to Gih; {circle around (3)} when Sih≠Si0, Fhn;i is the remnant part of the Fn;i in which Fnk;ij are eliminated, where j∈Sih, Sih is the index set of the eliminated states of Gi compared with Gih;
(3) The result of the AND operation of different initiating events is null “0”;
(4) Given j≠j′, k≠k′, then Fnk;ijFnk′;ij=0, Fnk;ijFnk;ij′=0, Fnk;ijFnk′;ij′=0 and VijVi′j′=0, where V∈{B,X,G};
(5) If the logical outspread to the default state Xnη of Xn is necessary, while the direct cause variables of Xnη are not represented, outspread Xnη according to




  
    
      X
      
        n
        ⁢
        
          
        
        ⁢
        η
      
    
    =
    
      
        ∏
        
          k
          ≠
          η
        
      
      ⁢
      
        
      
      ⁢
      
        
          X
          _
        
        nk
      
    
  
  ;




(6) If Xnk, k≠η, does not have input or the input is null, Xnk=0;
(7) When the condition Cn;i of the conditional functional variable Fn;i becomes invalid during the outspread, Fn;i is eliminated.
§19. In §12 (2), to find the possible solution set S, the following steps are involved.
(1) Outspread




  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      
    
    ⁢
    
      E
      i
      ′
    
  




according to the steps described in §16, §17 and §18, so as to obtain the sum-of-product type logic expression composed of only the {B,D,F} type events, where “product” indicates the logic AND, “sum” indicates the logic OR, and a group of events ANDed together is an “item”;

(2) After Eliminating the {F;D} type events and other inevitable events in all items, further simplify the outspreaded expression by logically absorbing or combining the physically same items;
(3) After finishing the above steps, every item in the final outspreaded expression is composed of only the B type events and every item is a possible solution event; all these items compose the possible solution set S conditioned on E, in which the item with same B type variables is denoted as Hk, and the item with same B type variables but in different states is denoted as Hkj. Hkj is a possible solution.
§20. The method described in §11 can be extended to the dynamical case involving more than one time point, that is, transform the case that the process system dynamically changes according to time as the static cases at sequential discrete time points, and perform the computation for each time point; then, combine all the static computation results at different time points together so as to correspond the dynamical change of the process system, including the following steps.
(1) Classify the time as discrete time points t1, t2, . . . , tn; for each time point ti, collect the static evidence E(ti) at that time point; find all the possible solutions Hkj conditioned on E(ti), these possible solutions compose the static possible solution set S(ti) at time ti; more specifically, treat E(ti) as E, {circle around (1)} Construct the E(ti) conditional original DUCG according to the methods described in §9 (3) and §10; {circle around (2)} Simplify the E(ti) conditional original DUCG according to the method described in §13; {circle around (3)} transform the simplified DUCG as EDUCG according to the methods described in §14; {circle around (4)} Outspread




  
    E
    ⁡
    
      (
      
        t
        i
      
      )
    
  
  =
  
    
      ∏
      k
    
    ⁢
    
      
    
    ⁢
    
      
        E
        k
      
      ⁡
      
        (
        
          t
          i
        
        )
      
    
  




according to the method described in §16-19, then obtain the possible solution set Si at time ti;

(2) Calculate




  
    
      S
      ⁡
      
        (
        
          t
          n
        
        )
      
    
    =
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        
      
      ⁢
      
        S
        i
      
    
  
  ,




S(tn) is called the dynamical possible solution at time tn.

(3) Eliminate the other possible solutions included in EDUCG but not included in S(tn), further simplify the EDUCG according to the method described in §13;
(4) Based on the above simplified EDUCG, according to the method described in §12, calculate the static state probabilities with incomplete and complete information hkjs′(ti) and hkjs(ti) respectively, the static rank probabilities with incomplete and complete information hkjr′(ti) and hkjr(ti) respectively, of Hkj in S(tn), as well as the unconditional probability hkj(t0)=Pr{Hkj};
(5) Calculate the dynamical state and rank probabilities with incomplete and complete information of Hkj included in S(tn) as follows: {circle around (1)} The dynamical state probabilities with incomplete and complete information:



  
    
      
        h
        kj
        
          s
          ′
        
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              kj
              
                s
                ′
              
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          ∑
          j
        
        ⁢
        
          
        
        ⁢
        
          
            ∏
            
              i
              =
              1
            
            n
          
          ⁢
          
            
          
          ⁢
          
            
              
                h
                kj
                
                  s
                  ′
                
              
              ⁡
              
                (
                
                  t
                  i
                
                )
              
            
            /
            
              
                (
                
                  
                    h
                    kj
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                )
              
              
                n
                -
                1
              
            
          
        
      
    
  




  
    
      
        h
        kj
        s
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              kj
              s
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          ∑
          j
        
        ⁢
        
          
        
        ⁢
        
          
            ∏
            
              i
              =
              1
            
            n
          
          ⁢
          
            
          
          ⁢
          
            
              
                h
                kj
                s
              
              ⁡
              
                (
                
                  t
                  i
                
                )
              
            
            /
            
              
                (
                
                  
                    h
                    kj
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                )
              
              
                n
                -
                1
              
            
          
        
      
    
  



In which, when hkj(t0)=0, hkjs′(ti)/(hkj(t0))n-1=0 and hkjs(ti)/(hkj(t0))n-1=0; {circle around (2)} The dynamical rank probabilities with incomplete and complete information:




  
    
      
        h
        kj
        
          r
          ′
        
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              kj
              
                r
                ′
              
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            
              S
              ⁡
              
                (
                
                  t
                  n
                
                )
              
            
          
        
        ⁢
        
          
        
        ⁢
        
          
            ∏
            
              i
              =
              1
            
            n
          
          ⁢
          
            
          
          ⁢
          
            
              
                h
                kj
                
                  r
                  ′
                
              
              ⁡
              
                (
                
                  t
                  i
                
                )
              
            
            /
            
              
                (
                
                  
                    h
                    kj
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                )
              
              
                n
                -
                1
              
            
          
        
      
    
  




  
    
      
        h
        kj
        r
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              kj
              r
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            
              S
              ⁡
              
                (
                
                  t
                  n
                
                )
              
            
          
        
        ⁢
        
          
        
        ⁢
        
          
            ∏
            
              i
              =
              1
            
            n
          
          ⁢
          
            
          
          ⁢
          
            
              
                h
                kj
                r
              
              ⁡
              
                (
                
                  t
                  i
                
                )
              
            
            /
            
              
                (
                
                  
                    h
                    kj
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                )
              
              
                n
                -
                1
              
            
          
        
      
    
  



In which, when hkj(t0)=0, hkjr′(ti)/(hkj(t0))n-1=0 and hkjr(ti)/(hkj(t0))n-1=0.

§21. In step (3) of §11, when IDUCG includes the virtual consequence variable Eh, to solve the problem by treating IDUCG as BN includes the following steps:
(1) Introduce another event Ēh, where Pr{Eh}≡1 and Pr{Ēh}≡0;
(2) The CPT between the virtual consequence variable Eh and its only cause variable Vh is constructed as follows: {circle around (1)} When



  
    
      
        
          max
          j
        
        ⁢
        
          {
          
            f
            Ehj
          
          }
        
      
      ≤
      1
    
    ,
    
      
        Pr
        ⁢
        
          {
          
            
              E
              h
            
            ❘
            
              V
              hj
            
          
          }
        
      
      =
      
        f
        Ehj
      
    
    ,
    
      

    
    ⁢
    and
  




  
    
      
        Pr
        ⁢
        
          {
          
            
              
                E
                _
              
              h
            
            ❘
            
              V
              hj
            
          
          }
        
      
      ≡
      
        1
        -
        
          Pr
          ⁢
          
            {
            
              
                E
                h
              
              ❘
              
                V
                hj
              
            
            }
          
        
      
    
    ;
  



{circle around (2)} When




  
    
      
        
          
            max
            j
          
          ⁢
          
            {
            
              f
              Ehj
            
            }
          
        
        >
        1
      
      ,
      
        
          Pr
          ⁢
          
            {
            
              
                E
                h
              
              ❘
              
                V
                hj
              
            
            }
          
        
        =
        
          
            f
            Ehj
          
          /
          
            
              max
              j
            
            ⁢
            
              
                {
                
                  f
                  Ehj
                
                }
              
              ⁢
              
                
              
              ⁢
              and
            
          
        
      
    
    ⁢
    
      
    
  




  
    
      
    
    ⁢
    
      
        Pr
        ⁢
        
          {
          
            
              E
              h
            
            ❘
            
              V
              hj
            
          
          }
        
      
      =
      
        1
        -
        
          Pr
          ⁢
          
            
              {
              
                
                  E
                  h
                
                ❘
                
                  V
                  hj
                
              
              }
            
            .
          
        
      
    
  



§22. The case that there is spurious evidence or part of the DUCG is incorrect can be dealt with as follows:
When the possible solution set conditioned on E is null or all the members in this set is excluded so that it becomes null, there must be spurious evidence or the part of the DUCG is incorrect (imperfect). As to which evidence or part is wrong is unknown. In this case, we can eliminate Ei from E one by one, and perform the computation for each case of the elimination. The collection of all the computation results corresponding to the eliminations is then the computation result in the imperfect case. If such elimination of one by one cannot solve the problem, eliminate the evidence one group by one group, so as to obtain the result. But this means the increase of computation due to the combination explosion.



BRIEF DESCRIPTIONS OF THE DRAWINGS
FIG. 1 is the example of a simple explicit representation mode.
FIG. 2 is the example of the fuzzy discretization for the continuous variables.
FIG. 3 is the example of the explicit representation mode including a logic gate.
FIG. 4 is the example of the truth value table of a logic gate.
FIG. 5 is the truth value table of the logic gate in FIG. 4, which includes the remnant state.
FIG. 6 is the illustration of combining the default variables.
FIG. 7 is an example of transforming the explicit representation mode to the equivalent causality trees or forest.
FIG. 8 is an example of the standard implicit representation mode.
FIG. 9 is an example of the non-standard implicit representation mode.
FIG. 10 is the example of transforming the non-standard implicit representation mode in FIG. 9 to the standard implicit representation mode.
FIG. 11 is an example of the hybrid representation mode.
FIG. 12 is the explanation in terms of the explicit representation mode to the hybrid representation mode.
FIG. 13 is an example to treat the evidence event Ei as the virtual consequence event of Vi.
FIG. 14 is the new logic gate and new truth value table in the case after eliminating Vi in FIG. 4.
FIG. 15 is the truth value table of the logic gate including the remnant state in FIG. 14,
FIG. 16 is the truth value table in FIG. 4 in which the expressions are exclusive.
FIG. 17 is the exclusive truth value table of the logic gate including the remnant state shown in FIG. 5.
FIG. 18 is the exclusive truth value table of the logic gate shown in FIG. 5.
FIG. 19 is the brief flow chart of the DUCG intelligent system constructed based on this invention.
FIG. 20 is the brief figure showing the secondary loop system of a nuclear power plant.
FIG. 21 is the brief figure of the fault influence DUCG of the secondary loop system of the nuclear power plant shown in FIG. 20.
FIG. 22 is the DUCG resulted from FIG. 21, in which the output functional variables of the variables in normal state are eliminated.
FIG. 23 is the DUCG after eliminating the part without any connection with the abnormal state variables in FIG. 22.
FIG. 24 is the DUCG after eliminating the part connected with only the variables in normal state in FIG. 23.
FIG. 25 is the DUCG after eliminating the invalid conditional functional variables in FIG. 24.
FIG. 26 is the DUCG after eliminating the output functional variables starting from the normal state variables in FIG. 21.
FIG. 27 is the DUCG after eliminating the functional variables that cannot cause the known states of the consequence variables in FIG. 26.
FIG. 28 is the DUCG after eliminating the part without any connection with the abnormal state variables in FIG. 27.
FIG. 29 is the DUCG after eliminating the part directly connected with only the normal state variables in FIG. 28.
FIG. 30 is the DUCG after eliminating the functional variables inconsistent with the occurrence order of events in FIG. 29.
FIG. 31 is the DUCG after eliminating the exclusive or condition invalid variables and the isolated normal state variables in FIG. 30.
FIG. 32 is the simplified DUCG about the price influence of the agricultural products.
FIG. 33 is the transformed EDUCG from FIG. 32.
FIG. 34 is the simplified EDUCG based on FIG. 33.
FIG. 35 is the EDUCG adding E8 based on FIG. 34.
FIG. 36 is an illustration of the weighing factor function.
FIG. 37 is an illustration of the time distribution of the problem can be solved by this invention.
FIG. 38 is the illustration of the approximate calculation using the time differentiating method.
FIG. 39 is the illustration of the observation positions of the rain amount and the water level of a river.
FIG. 40 is the DUCG for predicting the river flood of the river shown in FIG. 39.
FIG. 41 is the illustration of a present technology, i.e. the singly and multiply connected BNs.
FIG. 42 is an example of the BN that has logic cycles and cannot be solved by the present technology.
FIG. 43 is an example of the single-valued DCD as a present technology.



THE DETAILED DESCRIPTION OF EMBODIMENTS OF THE INVENTION
In what follows, the methods in this invention will be explained in details. Examples 1-21 explain the methods in §1-21. Examples 22-24 are the real examples of synthetically applying this invention. These examples provide not only the detailed explanation to the methods included in the claims, but also other new contents.
Example 1
FIGS. 1 and 2 are idiographic examples about the method mentioned in §1, which is explained below.
§1.1. The consequence variable Xn can be drawn as  which indicates the consequence variable indexed by n (in this example, n=2,4), e.g. the temperature of a stove. Xn is a brief notation of the consequence variable (consequence/effect event variable). It can have either discrete or continuous state value. For example, the stove temperature is abnormally high, normal, abnormally low, or the stove temperature is 1800 C°, 2000 C°, etc. Xn may have more than one input or cause variable (in this example, the input or cause variables of X4 are B1 and X2) and more than one output variable (i.e. the direct downstream consequence variables). The continuous variable can be fuzzily discretized (see §1.8 for details). The discrete or fuzzy discrete state of Xn is denoted as Xnk that is a specific event (e.g. the stove temperature is abnormally high), i.e. the consequence variable indexed by n is in its state k. xnk≡Pr{Xnk}, which can be the function of time and satisfies




  
    
      ∑
      k
    
    ⁢
    
      
    
    ⁢
    
      x
      nk
    
  
  =
  1.




Xnk can be expressed in terms of matrix:



  
    
      X
      n
    
    =
    
      
        (
        
          
            
              
                X
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              
                X
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            
              ⋮
            
          
          
            
              
                X
                nk
              
            
          
          
            
              ⋮
            
          
          
            
              
                X
                nK
              
            
          
        
        )
      
      =
      
        
          (
          
            
              
                
                  X
                  
                    n
                    ⁢
                    
                      
                    
                    ⁢
                    1
                  
                
              
              
                
                  X
                  
                    n
                    ⁢
                    
                      
                    
                    ⁢
                    2
                  
                
              
              
                ...
              
              
                
                  X
                  nk
                
              
              
                ...
              
              
                
                  X
                  nK
                
              
            
          
          )
        
        T
      
    
  




  Or




  
    
      X
      n
    
    =
    
      
        (
        
          
            
              
                x
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              
                x
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            
              ⋮
            
          
          
            
              
                x
                nk
              
            
          
          
            
              ⋮
            
          
          
            
              
                x
                nK
              
            
          
        
        )
      
      =
      
        
          (
          
            
              
                
                  x
                  
                    n
                    ⁢
                    
                      
                    
                    ⁢
                    1
                  
                
              
              
                
                  x
                  
                    n
                    ⁢
                    
                      
                    
                    ⁢
                    2
                  
                
              
              
                ...
              
              
                
                  x
                  nk
                
              
              
                ...
              
              
                
                  x
                  nK
                
              
            
          
          )
        
        T
      
    
  



In which, K is the upper bound of k. It is seen that Xn represents not only the event matrix, but also the data matrix. Moreover, Xn can be expressed in the form of function:

Xn=Xn{V1, V2, . . . , VN}

In which, Xn{ } is the operator of Xn about its cause variables, V1, V2, . . . , VN are the direct cause or input variables, N is the number of input variables.

§1.2. The basic variable Bi can be drawn as  i.e. the basic variable indexed by i (in this example, i=1), e.g. the state of an electromagnetic valve. Bi is the brief notation of the basic event variable and can have several discrete or continuous state values. For example, the electromagnetic valve is blocked, leaking, closed, or the open degree is 70%, 80%, etc. Bi does not have any input, but has at least one output. The so called basic variable means such variables whose causes do not need be found and whose states are random independently or given online. The continuous basic variable can be fuzzily discretized (see §11.8 for details). The discrete or fuzzy discrete state of Bi is denoted as Bij that is a specific event (e.g. the electromagnetic valve is closed or the open degree is middle), i.e. the basic variable indexed by i is in its state j. The occurrence probability of this event bij≡Pr{Bij} is given by the domain engineers when they construct the representation model of the consequence variable Xn as well as its various cause variables Vi involved in the problem to be solved, and can be the function of time while satisfies




  
    
      ∑
      j
    
    ⁢
    
      
    
    ⁢
    
      b
      ij
    
  
  ≤
  1.




This invention does not require the data normalization or completeness, i.e. the case





  
    
      ∑
      j
    
    ⁢
    
      b
      ij
    
  
  <
  1




is allowed. This is for the convenience of users. For example, when the domain engineers give the probability parameters bij of the states of Bi, they need give only the abnormal state parameters, but not the normal state parameters.

bij can be replaced in some cases by the frequency (i.e. bij=λij occurrence times of Bij in the unit time interval. See §11.2 for details).
Bi can be expressed in terms of matrix:

Bi=(Bi1 Bi2 . . . Bij . . . BiJ)T 

Or

Bi=(bi1 Bi2 . . . bij . . . biJ)T 

In which, J is the upper bound of j. It is seen that Bi can represent both the event matrix and the data matrix.

§1.3. The cause variable Vi is used to generally denote the variables as causes. V∈{B,X} means that the cause variable can be either the B type variable or the X type variable. When V=X, the consequence variable Xi is also the cause variable of another consequence variable Xn. Moreover, V can also represent G and D type cause variables (see examples 3 and 4 for details).
§1.4. When the problem to be solved is about the continuously operating process system (e.g. the power plant, chemical system, etc), the basic variable can be further classified as the initiating event variable and the non-initiating event variable, and can be distinguished by different drawings or colors. For example, the initiating event variable can be drawn as  and the non-initiating event variable can be drawn as  The so called initiating event variable means that all the abnormal states of the variable are initiating events. The non-initiating event variable means that all the abnormal states of the variable are non-initiating events. The division between the initiating event and non-initiating event is the further information representation, which will make the computation process be simplified significantly. If the abnormal states of a variable are partially the initiating events and partially the non-initiating events, this variable is a hybrid variable that can be represented by another drawing and cannot be simply treated as the initiating variable or the non-initiating variable. Instead, it must be treated at the event level.
The so called non-initiating event indicates that the event itself cannot directly function and cannot affect the operation state of the system. Only when the other initiating event occurs, will it possibly function. For example, the event that the depressurizing valve of a power plant cannot open to depressurize the pressure when the abnormal high pressure appears is a non-initiating event. Here, if the abnormal high pressure does no appear, the non-initiating event will not function. While its function does not mean that it appears just at this moment. Instead, it may have being existed for a long time, e.g. the mistake of design or installation, the incorrect reinstallation after maintenance, the circuit short of the valve electromotor due to the damp in some moment before, etc. In general, the occurrence time may be much earlier than the function time.
In other words, the occurrence of the initiating event can be viewed as in the meantime (i.e., within a vary small time interval Δt) of the operation state change of the process system, while the occurrence of the non-initiating event can be viewed as within a relatively long time interval T (Δt<<T) before the operation state change of the process system. Based on this, it can be concluded that the probability of the simultaneous occurrence of two or more independent initiating events is a high order small value compared with the occurrence probability of one initiating event plus some (including none) non-initiating events. When the occurrence probability of the basic event is much less than 1, the simultaneous occurrence of two or more independent initiating events can be viewed as impossible. But the simultaneous occurrence of one initiating event ANDed with some non-initiating events is possible. For example, the failure rate λ of many types of equipments or components is usually ranging from 10−2-10−5/year, while the time interval Δt is usually measured in minutes or seconds. In comparison, T is usually counted in months. Thus, the probability of the initiating event that the equipment or component fails during Δt is b≈λΔt<<1, and (λΔt)n<<λΔt(λT)m, where n>1 and m≧0. It is seen that the probability of the simultaneous occurrence of two or more initiating events is a high order small value, and can be ignored. In other words, when two or more different initiating events are ANDed together, the result will be a null set.
§1.5. The functional variable as the brief name of the functional event variable can be represented with the directed arc →. The conditional functional variable as the brief name of the conditional functional event variable can be represented with the dashed directed arc  They start from the cause or input variable Vi (in this example, V1=B1 and V2=X2) and stop at the consequence or output variable Xn (X4 in this example), and are denoted as Fn;i (in this example, F4;1 is a functional variable and F4;2 is a conditional functional variable). The specific value of Fn;i is Fnk;ij that is a weighted random event indicating that the state j of the variable indexed by i causes the state k of the consequence variable indexed by n. The occurrence probability of Fnk;ij quantifies the uncertainty and the uncertain degree of the event that the cause event causes the consequence event. The difference between the conditional functional variable and the functional variable is only at that the former is conditionally valid (see §2.1 for details). Therefore, they can be generally called the functional variable.
The significance of introducing the functional event Fnk;ij is that in the real world, there must be a functional mechanism to realize that the cause event makes the consequence event occur. The detailed functional mechanism is usually unclear or very complicated. Meanwhile, what time the functional mechanism is realized is usually uncertain. That is, the causalities in the real world appear uncertain. In this invention, the introduction of the functional event is to represent this uncertain functional mechanism. It not only avoids the detailed explanation of the functional mechanism, but also fully represents the functional effect of this uncertain causal functional mechanism. With this representation, the functional mechanism does function means that the functional event does occur; the functional mechanism does not function means that the functional event does not occur. The probability of the functional event is the probability of the event that the functional mechanism does function.
When the cause variable and the consequence variable are all discrete or fuzzy discrete variables, the functional variable is corresponding to a matrix:




  
    F
    
      n
      ;
      i
    
  
  =
  
    
      (
      
        
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            ⋯
          
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                ij
              
            
          
          
            ⋯
          
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                iJ
              
            
          
        
        
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            ⋯
          
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                ij
              
            
          
          
            ⋯
          
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                iJ
              
            
          
        
        
          
            ⋮
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
        
        
          
            
              F
              
                nk
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              F
              
                nk
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            ⋯
          
          
            
              F
              
                nk
                ;
                ij
              
            
          
          
            ⋯
          
          
            
              F
              
                nk
                ;
                iJ
              
            
          
        
        
          
            ⋮
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
        
        
          
            
              F
              
                nK
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              F
              
                nK
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            ⋯
          
          
            
              F
              
                nK
                ;
                ij
              
            
          
          
            ⋯
          
          
            
              F
              
                nK
                ;
                iJ
              
            
          
        
      
      )
    
    =
    
      (
      
        
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                i
              
            
          
        
        
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                i
              
            
          
        
        
          
            ⋮
          
        
        
          
            
              F
              
                nk
                ;
                i
              
            
          
        
        
          
            ⋮
          
        
        
          
            
              F
              
                nK
                ;
                i
              
            
          
        
      
      )
    
  




In which, K is the upper bound of k, J is the upper bound of j, and

Fnk;i≡(Fnk;i1 Fnk;i2 . . . Fnk;ij . . . Fnk;iJ)

The rows of the matrix Fn;i correspond to the states of the consequence variable, while the columns correspond to the states of the cause variable. The element Fnk;ij of the matrix is the functional event that the state j of the cause variable i causes the state k of the consequence variable n. Define fnk;ij≡Pr{Fnk;ij}. fnk;ij is called the functional intensity, i.e. the probability contribution of Vij to Xnk by Fnk;ij.
For domain engineers, when they give the occurrence probability of Vij causing Xnk by Fnk;ij, they usually do not consider or are in difficulty to consider the functions of other cause variables influencing Xnk. In the case of not considering the other cause variables, the occurrence probability value of Xnk by Fnk;ij given by the domain engineers is called the original functional intensity and is expressed as ank;ij. In terms of matrix,




  
    A
    
      n
      ;
      i
    
  
  ≡
  
    (
    
      
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                1
              
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                1
              
            
          
        
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                1
              
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                2
              
            
          
        
        
          ⋯
        
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                1
              
              ;
              ij
            
          
        
        
          ⋯
        
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                1
              
              ;
              iJ
            
          
        
      
      
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                2
              
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                1
              
            
          
        
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                2
              
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                2
              
            
          
        
        
          ⋯
        
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                2
              
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                1
                ⁢
                j
              
            
          
        
        
          ⋯
        
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                2
              
              ;
              iJ
            
          
        
      
      
        
          ⋮
        
        
          
            
          
        
        
          
            
          
        
        
          
            
          
        
        
          
            
          
        
        
          
            
          
        
      
      
        
          
            a
            
              nk
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                1
              
            
          
        
        
          
            a
            
              nk
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                2
              
            
          
        
        
          ⋯
        
        
          
            a
            
              nk
              ;
              ij
            
          
        
        
          ⋯
        
        
          
            a
            
              nk
              ;
              iJ
            
          
        
      
      
        
          ⋮
        
        
          
            
          
        
        
          
            
          
        
        
          
            
          
        
        
          
            
          
        
        
          
            
          
        
      
      
        
          
            a
            
              nK
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                1
              
            
          
        
        
          
            a
            
              nK
              ;
              
                i
                ⁢
                
                  
                
                ⁢
                2
              
            
          
        
        
          ⋯
        
        
          
            a
            
              nK
              ;
              ij
            
          
        
        
          ⋯
        
        
          
            a
            
              nK
              ;
              iJ
            
          
        
      
    
    )
  




In which, ank;ij are given by domain engineers and can be the functions of time.

In this invention, ank;ij satisfy




  
    
      ∑
      k
    
    ⁢
    
      a
      
        nk
        ;
        ij
      
    
  
  ≤
  1.




Usually,





  
    ∑
    k
  
  ⁢
  
    a
    
      nk
      ;
      ij
    
  




equals to 1, i.e. the sum of data in any column of the matrix An;i sum up to 1. But in this invention, only ≦1 is required, i.e. the data completeness is not required. This is for the consideration that the domain engineers may not concern the occurrence probabilities of all the states of Xn caused by Vij, but only the occurrence probabilities of partial states of Xn. In this case, the domain engineers need only give the probabilities of the states in concern of Xn, but not all. Moreover, for some Vij, the domain engineers may not concern the causalities between Vij and Xn, and therefore do not need give the original functional intensity starting from Vij for any state of Xn. In other words, in this invention, the data and causalities can be incomplete, thus to provide grate convenience for domain engineers to freely and explicitly represent the causalities in the real world.

The above original functional intensity ank;ij is given without considering the other cause variables. But in the real world, the cause variables are not unrelated. For the example 1, when B1 causes X41 occur, it is impossible for X2 to cause X42, because X41 and X42 are exclusive. To solve the conflict between the independency of giving the original functional intensity ank;ij and the mutual exclusion (correlation) of different states of Xn, this invention defines that the probability or functional intensity of the event that every cause variable causes some state of the consequence variable is only a contribution to the state probability distribution of the consequence variable. As to in which state the consequence variable will be is decided randomly according to the probability distribution.
It is obvious that as the different states of a cause or consequence variable are exclusive, the corresponding functional events are exclusive. Moreover, since the probability contributions from different cause variables to a same state of a consequence variable are simply in the summation relation, the functional events from different cause variables to a same state of a consequence variable are exclusive with each, other in the sense of effect. In other words, all the functional events as the outputs to the states of a same consequence variable are exclusive with each other, while the logic AND of the functional events from different cause variables to a same state of a consequence variable are mutually independent.
Considering that the state probability distribution of the consequence variable should satisfy the normalization. This invention defines the functional intensity fnk;ij=(rn;i/rn)ank;ij, where




  
    r
    n
  
  =
  
    
      ∑
      i
    
    ⁢
    
      r
      
        n
        ;
        i
      
    
  




is the normalization factor and rn;i is the relationship of Fn;i. Thus,





  
    F
    
      n
      ;
      i
    
  
  =
  
    
      
        (
        
          
            r
            
              n
              ;
              i
            
          
          /
          
            r
            n
          
        
        )
      
      ⁢
      
        A
        
          n
          ;
          i
        
      
    
    =
    
      (
      
        
          
            
              f
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              f
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            ⋯
          
          
            
              f
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                ij
              
            
          
          
            ⋯
          
          
            
              f
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                iJ
              
            
          
        
        
          
            
              f
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              f
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            ⋯
          
          
            
              f
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                ij
              
            
          
          
            ⋯
          
          
            
              f
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                iJ
              
            
          
        
        
          
            ⋮
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
        
        
          
            
              f
              
                nk
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              f
              
                nk
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            ⋯
          
          
            
              f
              
                nk
                ;
                ij
              
            
          
          
            ⋯
          
          
            
              f
              
                nk
                ;
                iJ
              
            
          
        
        
          
            ⋮
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
          
            
              
            
          
        
        
          
            
              f
              
                nK
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              f
              
                nK
                ;
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            ⋯
          
          
            
              f
              
                nK
                ;
                ij
              
            
          
          
            ⋯
          
          
            
              f
              
                nK
                ;
                iJ
              
            
          
        
      
      )
    
  




That is to say, Fn;i represents not only the event matrix, but also the data matrix.

§1.6. Every functional variable Fn;i is associated with a factor rn;i (briefly called the relationship) representing the causal relation degree between the input variable Vi and the output variable Xn. In general, 0<rn;i≦1. When people are not sure for weather or not there exists causality between the cause variable and the consequence variable, the relationship can be used to represent the uncertain degree. Usually, when rn;i=1, the causality exists with 100% confidence; when rn;i=0, the causality does not exist, but in this case, the functional variable should not exist too. Therefore, rn;i≠0. The middle case can be represented by a number between 0 and 1. As the default case, the causality can be viewed as 100%, i.e. rn;i=1. rn;i can be the function of time.
In nature, the relationship represents the influence weight of the cause variable to the consequence variable, because rn;i appears always in the form of (rn;i/rn) in the computation process, where




  
    r
    n
  
  =
  
    
      ∑
      i
    
    ⁢
    
      
        r
        
          n
          ;
          i
        
      
      .
    
  




That is, rn;i always satisfy the normalization. Therefore, rn;i≦1 is no long the restriction must be satisfied. In fact, the method presented in this invention allows the case rn;i>1. Even more, it can be all rn;i<1, because in (rn;i/rn), no matter rn;i is positive or negative, the result is same.

The difference between the relationship and the functional intensity is: The relationship represents the direct causality correlation degree between the cause variable and the consequence variable or the weighing degree among the cause variables to influence the consequence variable in the sense of causality. It has nothing to do with the states of the cause variables and consequence variable. The functional intensity represents the probability distribution contribution over the different states of a consequence variable from the different states of a cause variable, in the precondition that the direct causality already exists.
§1.7. Given rn;i, the state probabilities of the consequence variable are calculated according to the following equations.




  
    Pr
    ⁢
    
      {
      
        X
        nk
      
      }
    
  
  =
  
    
      
        ∑
        i
      
      ⁢
      
        (
        
          
            (
            
              
                r
                
                  n
                  ;
                  i
                
              
              /
              
                r
                n
              
            
            )
          
          ⁢
          
            
              ∑
              j
            
            ⁢
            
              
                a
                
                  nk
                  ;
                  ij
                
              
              ⁢
              Pr
              ⁢
              
                {
                
                  V
                  ij
                
                }
              
            
          
        
        )
      
    
    =
    
      
        ∑
        
          i
          ,
          j
        
      
      ⁢
      
        
          f
          
            nk
            ;
            ij
          
        
        ⁢
        Pr
        ⁢
        
          {
          
            V
            ij
          
          }
        
      
    
  




It can be proved that the sum of the probabilities of all the states of Xn are normalized conditioned on





  
    
      ∑
      j
    
    ⁢
    
      a
      
        nk
        ;
        ij
      
    
  
  =
  
    
      1
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        
          ∑
          j
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              V
              ij
            
            }
          
        
      
    
    =
    1.
  




Proof:





  
    
      
        
          
            ∑
            k
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                X
                nk
              
              }
            
          
        
        =
        
          
            ∑
            k
          
          ⁢
          
            
              ∑
              i
            
            ⁢
            
              (
              
                
                  (
                  
                    
                      r
                      
                        n
                        ;
                        i
                      
                    
                    /
                    
                      r
                      n
                    
                  
                  )
                
                ⁢
                
                  
                    ∑
                    j
                  
                  ⁢
                  
                    Pr
                    ⁢
                    
                      {
                      
                        V
                        ij
                      
                      }
                    
                    ⁢
                    
                      a
                      
                        nk
                        ;
                        ij
                      
                    
                  
                
              
              )
            
          
        
      
    
  
  
    
      
        =
        
          
            ∑
            i
          
          ⁢
          
            (
            
              
                (
                
                  
                    r
                    
                      n
                      ;
                      i
                    
                  
                  /
                  
                    r
                    n
                  
                
                )
              
              ⁢
              
                
                  ∑
                  j
                
                ⁢
                
                  Pr
                  ⁢
                  
                    {
                    
                      V
                      ij
                    
                    }
                  
                  ⁢
                  
                    a
                    
                      nk
                      ;
                      ij
                    
                  
                
              
            
            )
          
        
      
    
  
  
    
      
        =
        
          
            
              ∑
              i
            
            ⁢
            
              (
              
                
                  r
                  
                    n
                    ;
                    i
                  
                
                /
                
                  r
                  n
                
              
              )
            
          
          =
          
            
              
                ∑
                i
              
              ⁢
              
                
                  r
                  
                    n
                    ;
                    i
                  
                
                /
                
                  r
                  n
                
              
            
            =
            1
          
        
      
    
  




▪

But, as this invention does not require the completeness, there may be




  
    
      
        ∑
        k
      
      ⁢
      
        a
        
          nk
          ;
          ij
        
      
    
    <
    
      1
      ⁢
      
        
      
      ⁢
      or
      ⁢
      
        
      
      ⁢
      
        
          ∑
          j
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              V
              ij
            
            }
          
        
      
    
    <
    1
  
  ,
  
    
  
  ⁢
  
    
      so
      ⁢
      
        
      
      ⁢
      that
      ⁢
      
        
          
        
        ⁢
        
          
        
      
      ⁢
      
        
          ∑
          k
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              X
              nk
            
            }
          
        
      
    
    <
    1.
  




Usually, there is a default state η of Xn, which represents such a state of Xn that its probability is not in concern. To satisfy the state probability normalization of Xn, there is





  
    
      
        Pr
        ⁢
        
          {
          
            X
            nη
          
          }
        
      
      ≡
      
        Pr
        ⁢
        
          {
          
            
              ∏
              
                k
                ≠
                η
              
            
            ⁢
            
              
            
            ⁢
            
              
                X
                —
              
              nk
            
          
          }
        
      
    
    =
    
      1
      -
      
        
          ∑
          
            k
            ≠
            η
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              X
              nk
            
            }
          
        
      
    
  
  ,
  
    
      
        ∑
        k
      
      ⁢
      
        Pr
        ⁢
        
          {
          
            X
            nk
          
          }
        
      
    
    =
    1
  




so that can be satisfied. The default state is usually the normal state of the variable, e.g. the stove temperature is normal, etc. Since people do not concern the probability of this state, people usually do not give the cause of this state (the cause is absent), nor give it as the cause of other variables. In the explicit representation mode, the inference computation process is usually not related to the calculation of the default state probability, and therefore the incompleteness does not usually influence the computation result. If the completeness is not met and the default state is not given, the sum of the calculated probabilities of different states may not be normalized (less than 1). In this case, the following normalization method can be applied (only when necessary).





  
    Pr
    ⁢
    
      {
      
        V
        ij
      
      }
    
  
  =
  
    Pr
    ⁢
    
      
        {
        
          V
          ij
        
        }
      
      /
      
        
          ∑
          j
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              V
              ij
            
            }
          
        
      
    
  




Although this compulsory normalization is not strict enough in theory, it reflects the limitation for people to know the real world, but not the limitation of the method itself. In contrast, it is because this method allows the limitation of people in knowing the real world, that this method has more advantages than the other methods.
§1.8. When the cause or the consequence variable is the continuous variable, the fuzzy method can be used to make it discrete. For example, suppose a variable represents the continuous value of a temperature. It can be discretized as three discrete states: low (e.g. lower than) 115° C.), normal (e.g. 107-134° C.) and high (e.g. more than 125° C.). There may be fuzzy area between states. The degree of a given temperature ei belonging to state j is quantified by the membership mij(ei). For any value ei of a given continuous variable,




  
    
      ∑
      j
    
    ⁢
    
      
        m
        ij
      
      ⁡
      
        (
        
          e
          i
        
        )
      
    
  
  =
  1.




FIG. 2 provides an example. mij(ei) can be briefly written as mij.

After the discretization of the continuous variables, the original functional intensities ank;ki between the cause variables and the consequence variable are the same as for the discrete variables, and can be given by domain engineers. When the continuous functional intensity is given by the functional intensity density function φn;i, φn;j can be transformed as φnk;ij:
{circle around (1)} When the cause variable and the consequence variable are all continuous, φm;i(ei,en). The meaning of φn;i(ei,en)den is the probability that Vi causes the value en of Xn within a very small interval den, given the value of Vi is ei. Since the incompleteness is allowed, for any ei,



  
    
      
        ∫
        
          e
          n
        
      
      ⁢
      
        
          
            φ
            
              n
              ;
              i
            
          
          ⁡
          
            (
            
              
                e
                i
              
              ,
              
                e
                n
              
            
            )
          
        
        ⁢
        
          
        
        ⁢
        
          d
          
            e
            n
          
        
      
    
    ≤
    1
  




  
    
      
        a
        
          nk
          ;
          i
        
      
      ⁡
      
        (
        
          e
          i
        
        )
      
    
    =
    
      
        ∫
        
          e
          n
        
      
      ⁢
      
        
          
            m
            nk
          
          ⁡
          
            (
            
              e
              n
            
            )
          
        
        ⁢
        
          
            φ
            
              n
              ;
              i
            
          
          ⁡
          
            (
            
              
                e
                i
              
              ,
              
                e
                n
              
            
            )
          
        
        ⁢
        
          
        
        ⁢
        
          d
          
            e
            n
          
        
      
    
  



In which, ank;i(ei) is the occurrence probability that Vi causes Xnk, given the value of Vi is ei. ank;i(ei) is the function of ei. What people want to know is the probability of Xn being in its fuzzy discrete state k in average when Vi is in the fuzzy discrete state j after the fuzzy discretization. Therefore, the average for ank;i(ei) weighted by mij(ei) should be done:





  
    
      
        
          a
          
            
              nk
              ;
              ij
            
            ⁢
            
              
            
          
        
        =
        
        ⁢
        
          
            
              ∫
              
                e
                i
              
            
            ⁢
            
              
                
                  m
                  ij
                
                ⁡
                
                  (
                  
                    e
                    i
                  
                  )
                
              
              ⁢
              
                
                  a
                  
                    nk
                    ;
                    i
                  
                
                ⁡
                
                  (
                  
                    e
                    i
                  
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  i
                
              
            
          
          
            
              ∫
              
                e
                i
              
            
            ⁢
            
              
                
                  m
                  ij
                
                ⁡
                
                  (
                  
                    e
                    i
                  
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  i
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              ∫
              
                e
                i
              
            
            ⁢
            
              
                ∫
                
                  e
                  n
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  
                    m
                    nk
                  
                  ⁡
                  
                    (
                    
                      e
                      n
                    
                    )
                  
                
                ⁢
                
                  
                    φ
                    
                      n
                      ;
                      i
                    
                  
                  ⁢
                  
                    
                  
                  (
                  
                    
                      e
                      i
                    
                    ,
                    
                      e
                      n
                    
                  
                  )
                
                ⁢
                
                  d
                  
                    e
                    n
                  
                
                ⁢
                
                  
                
                ⁢
                
                  d
                  
                    e
                    i
                  
                
              
            
          
          
            
              ∫
              
                e
                i
              
            
            ⁢
            
              
                
                  m
                  ij
                
                ⁡
                
                  (
                  
                    e
                    i
                  
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  i
                
              
            
          
        
      
    
  




As the same as the discrete variable, we have





  
    
      
        
          
            ∑
            k
          
          ⁢
          
            a
            
              nk
              ;
              ij
            
          
        
        =
        
        ⁢
        
          
            
              ∑
              k
            
            ⁢
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  ∫
                  
                    e
                    n
                  
                
                ⁢
                
                  
                    
                      m
                      ij
                    
                    ⁡
                    
                      (
                      
                        e
                        i
                      
                      )
                    
                  
                  ⁢
                  
                    
                      m
                      nk
                    
                    ⁡
                    
                      (
                      
                        e
                        n
                      
                      )
                    
                  
                  ⁢
                  
                    
                      φ
                      
                        n
                        ;
                        i
                      
                    
                    ⁢
                    
                      
                    
                    (
                    
                      
                        e
                        i
                      
                      ,
                      
                        e
                        n
                      
                    
                    )
                  
                  ⁢
                  
                    d
                    
                      e
                      n
                    
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    d
                    
                      e
                      i
                    
                  
                
              
            
          
          
            
              ∫
              
                e
                i
              
            
            ⁢
            
              
                
                  m
                  ij
                
                ⁡
                
                  (
                  
                    e
                    i
                  
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  i
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              ∫
              
                e
                i
              
            
            ⁢
            
              
                
                  m
                  ij
                
                ⁡
                
                  (
                  
                    e
                    i
                  
                  )
                
              
              ⁢
              
                
                  ∫
                  
                    e
                    n
                  
                
                ⁢
                
                  
                    
                      φ
                      
                        n
                        ;
                        i
                      
                    
                    ⁡
                    
                      (
                      
                        
                          e
                          i
                        
                        ,
                        
                          e
                          n
                        
                      
                      )
                    
                  
                  ⁢
                  
                    
                      ∑
                      k
                    
                    ⁢
                    
                      
                        
                          m
                          nk
                        
                        ⁡
                        
                          (
                          
                            e
                            n
                          
                          )
                        
                      
                      ⁢
                      
                        
                      
                      ⁢
                      
                        d
                        
                          e
                          n
                        
                      
                      ⁢
                      
                        
                      
                      ⁢
                      
                        d
                        
                          e
                          i
                        
                      
                    
                  
                
              
            
          
          
            
              ∫
              
                e
                i
              
            
            ⁢
            
              
                
                  m
                  ij
                
                ⁡
                
                  (
                  
                    e
                    i
                  
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  i
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  
                    ∫
                    
                      e
                      n
                    
                  
                  ⁢
                  
                    
                      
                        φ
                        
                          n
                          ;
                          i
                        
                      
                      ⁡
                      
                        (
                        
                          
                            e
                            i
                          
                          ,
                          
                            e
                            n
                          
                        
                        )
                      
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      d
                      
                        e
                        n
                      
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      d
                      
                        e
                        i
                      
                    
                  
                
              
            
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  d
                  
                    e
                    i
                  
                
              
            
          
          ≤
          
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  d
                  
                    e
                    i
                  
                
              
            
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  d
                  
                    e
                    i
                  
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        1
      
    
  




{circle around (2)} When the cause variable Vi is continuous but the consequence variable Xn is discrete, φn;i=φn;i(ei,Xnk), and




  
    a
    
      nk
      ;
      ij
    
  
  =
  
    
      
        ∫
        
          e
          i
        
      
      ⁢
      
        
          
            m
            ij
          
          ⁡
          
            (
            
              e
              i
            
            )
          
        
        ⁢
        
          
            φ
            
              n
              ;
              i
            
          
          ⁡
          
            (
            
              
                e
                i
              
              ,
              
                X
                nk
              
            
            )
          
        
        ⁢
        
          
        
        ⁢
        
          d
          
            e
            i
          
        
      
    
    
      
        ∫
        
          e
          i
        
      
      ⁢
      
        
          
            m
            ij
          
          ⁡
          
            (
            
              e
              i
            
            )
          
        
        ⁢
        
          
        
        ⁢
        
          d
          
            e
            i
          
        
      
    
  




In which, φn;i(ei,Xnk) is the functional intensity density function between the continuous cause variable Vi and the discrete state k of the consequence variable Xn. For any ej,





  
    
      ∑
      k
    
    ⁢
    
      
        φ
        
          n
          ;
          i
        
      
      ⁡
      
        (
        
          
            e
            i
          
          ,
          
            X
            nk
          
        
        )
      
    
  
  ≤
  1




Thus we have





  
    
      
        
          
            ∑
            k
          
          ⁢
          
            a
            
              nk
              ;
              ij
            
          
        
        =
        
        ⁢
        
          
            
              ∑
              k
            
            ⁢
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  
                    φ
                    
                      n
                      ;
                      i
                    
                  
                  ⁢
                  
                    
                  
                  (
                  
                    
                      e
                      i
                    
                    ,
                    
                      X
                      nk
                    
                  
                  )
                
                ⁢
                
                  
                
                ⁢
                
                  d
                  
                    e
                    i
                  
                
              
            
          
          
            
              ∫
              
                e
                i
              
            
            ⁢
            
              
                
                  m
                  ij
                
                ⁡
                
                  (
                  
                    e
                    i
                  
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  i
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  
                    ∑
                    k
                  
                  ⁢
                  
                    
                      
                        φ
                        
                          n
                          ;
                          i
                        
                      
                      ⁡
                      
                        (
                        
                          
                            e
                            i
                          
                          ,
                          
                            X
                            nk
                          
                        
                        )
                      
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      d
                      
                        e
                        i
                      
                    
                  
                
              
            
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  d
                  
                    e
                    i
                  
                
              
            
          
          ≤
          
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  d
                  
                    e
                    i
                  
                
              
            
            
              
                ∫
                
                  e
                  i
                
              
              ⁢
              
                
                  
                    m
                    ij
                  
                  ⁡
                  
                    (
                    
                      e
                      i
                    
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  d
                  
                    e
                    i
                  
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        1
      
    
  




{circle around (3)} When the cause variable Vi is discrete and the consequence variable Xn is continuous, φn;i=φn;i(Vij,en),




  
    a
    
      nk
      ;
      ij
    
  
  =
  
    
      ∫
      
        e
        n
      
    
    ⁢
    
      
        
          m
          nk
        
        ⁡
        
          (
          
            e
            n
          
          )
        
      
      ⁢
      
        
          φ
          
            n
            ;
            i
          
        
        ⁡
        
          (
          
            
              V
              ij
            
            ,
            
              e
              n
            
          
          )
        
      
      ⁢
      
        
      
      ⁢
      
        d
        
          e
          n
        
      
    
  




In which, φn;j(Vij,en) is the functional intensity density function between the discrete cause variable Vi and the continuous consequence variable Xn. For any Vij,





  
    
      ∫
      
        e
        n
      
    
    ⁢
    
      
        
          φ
          
            n
            ;
            i
          
        
        ⁡
        
          (
          
            
              V
              ij
            
            ,
            
              e
              n
            
          
          )
        
      
      ⁢
      
        
      
      ⁢
      
        d
        
          e
          n
        
      
    
  
  ≤
  1




Then we have





  
    
      
        
          
            ∑
            k
          
          ⁢
          
            a
            
              nk
              ;
              ij
            
          
        
        =
        
        ⁢
        
          
            ∑
            k
          
          ⁢
          
            
              ∫
              
                e
                n
              
            
            ⁢
            
              
                
                  m
                  nk
                
                ⁡
                
                  (
                  
                    e
                    n
                  
                  )
                
              
              ⁢
              
                
                  φ
                  
                    n
                    ;
                    i
                  
                
                ⁢
                
                  
                
                (
                
                  
                    V
                    ij
                  
                  ,
                  
                    e
                    n
                  
                
                )
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  n
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            ∫
            
              e
              n
            
          
          ⁢
          
            
              ∑
              k
            
            ⁢
            
              
                
                  m
                  nk
                
                ⁡
                
                  (
                  
                    e
                    n
                  
                  )
                
              
              ⁢
              
                
                  φ
                  
                    n
                    ;
                    i
                  
                
                ⁢
                
                  
                
                (
                
                  
                    V
                    ij
                  
                  ,
                  
                    e
                    n
                  
                
                )
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  n
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              ∫
              
                e
                n
              
            
            ⁢
            
              
                
                  φ
                  
                    n
                    ;
                    i
                  
                
                ⁢
                
                  
                
                (
                
                  
                    V
                    ij
                  
                  ,
                  
                    e
                    n
                  
                
                )
              
              ⁢
              
                
              
              ⁢
              
                d
                
                  e
                  n
                
              
            
          
          ≤
          1
        
      
    
  




Example 2
Still as shown in FIG. 1, the meaning of the conditional functional variable is as follows.
§2.1. The only difference between the functional variable (F4;1 in this example) and the conditional functional variable (F4;2 in this example) is that the conditional functional variable is added an validation condition Cn,i (C4;2 in this example), while the others remain same. The meaning of Cn,i is: when Cn,i is valid, Fn,i is valid; otherwise, Fn,i is eliminated. The function of the conditional functional variable is that when the condition Cn;i is not valid, break the causality between the input variable and the output variable. In FIG. 1, it appears as that the directed arc representing the functional variable is eliminated. Vise versa, if the condition is valid, the conditional functional variable becomes the functional variable. In FIG. 1, the dashed directed arc becomes the solid directed arc.
The reason of introducing the conditional functional variable is because the causalities between things are not always determinable in advance. Some of them have to be determined according to the online received evidence or the middle computation results. For example, suppose B12 represents the rupture of the U type pipes in a steam generator of a nuclear power plant, X2 represents the feed water flow rate of the steam generator, and X4 represents the water level of the steam generator. Only when B12 does not occur, will X2 become the cause of X4, i.e. C4;2= B12. But in advance, people do not know whether or not B12 occurs. Thus F4;2 is a conditional functional variable. When the online evidence or the computation process shows that C4;2 is valid, F4;2 becomes a functional variable. In accordance, the dashed directed arc becomes a solid directed arc; When the online evidence or the computation process shows that C4;2 is invalid, F4;2 is eliminated.
Example 3
FIGS. 3-5 are the examples about the logic gate, and are explained below.
§3.1. As shown in FIG. 3, a logic gate is drawn as  i.e. the logic gate indexed by i (in this example, i=4), which is used to represent any state logical combination of input variables. Gi has at least one input (three inputs X1, X2 and B3 in this example) and at least one output (two outputs X5 and X6 in this example). The input variables are connected with the logic gate by the directed arc→that is different from the functional variable→and the conditional functional variable  and are denoted as Ui;h (U4;1, U4;2 and U4;3 in this example), in which h is the input variable index (h=1,2,3 in this example). The elements of Ui;h are denoted as Uik;hj representing that the state j of the input variable h participates in the logic operation and results in the occurrence of the state k of the logic gate i. Uik;hj is either an inevitable event whose probability is always 1, or an impossible event whose probability is always 0 (i.e. the state j of the input variable h has nothing to do with the state k of the logic gate i). However, at least one element of Un;i is the inevitable event, otherwise Vh is not the input of logic gate Gi. The logic gate can also be written as in the form of function:

Gi=Gi{U1;1V1, Ui;2V2, . . . , Ui;NVN}

Or simply written as





  
    G
    i
  
  =
  
    
      
        G
        i
      
      ⁢
      
        {
        
          
            V
            1
          
          ,
          
            V
            2
          
          ,
          ...
          ⁢
          
            
          
          ,
          
            V
            N
          
        
        }
      
    
    =
    
      (
      
        
          
            
              G
              
                i
                ⁢
                
                  
                
                ⁢
                1
              
            
          
        
        
          
            
              G
              
                i
                ⁢
                
                  
                
                ⁢
                2
              
            
          
        
        
          
            ⋮
          
        
        
          
            
              G
              iK
            
          
        
      
      )
    
  




Where Gi{ } is the operator of the logic gate, V1, V2, . . . , VN are the input variables of Gi, Gij denotes the state j of Gi, and K is the number of the states of the logic gate. A variable Ui;h can be added in front of every corresponding input variable, so as to ease the investigation of the causality chain according to the subscripts. It can also be ignored. V∈{B,X,G} indicates that the logic gate can also be the input variable of another logic gate. Of course, usually the multiple logic gate representation is not necessary, because the multiple logic combination can be fully represented in one logic gate. Therefore, for simplicity, only the case of the single logic gate will be discussed below.

The states of the logic gate are discrete. The different states of the logic gate reflect the different effects of the state combinations of the input variables. Every logic gate has a corresponding truth value table specifying the corresponding relation between the different state combinations of the input variables and the states of the logic gate, i.e. specifying the specific meaning of the operator Gi{ }. For the example shown in FIG. 4, suppose the input variables of G4 are V1, V2 and V3, and they have two states each. The graphical illustration and the truth value table are given in FIG. 4. In the table, the second subscript indexes the different state of the variable. The serial number of the input row is to distinguish the logic expressions of the state combinations of input variables. The logic expressions indexed by the different serial numbers correspond to different states of the logic gate, and therefore are exclusive with each other. Gij denotes the state j of the logic gate i. The value “1” in the truth value table indicates that this state of the logic gate is true. The value “0” in the truth value table indicates not true. “+” denotes the logic OR (may not be exclusive). The conjunctive two sides of “+” are different logic items. The truth of any one of the items can result in the truth of the same state of the logic gate. The events in every logic item are in the AND relation. For the example above, the logic expression indexed by the serial number 1 is composed of three items. The truth of any one of the items can result in the truth of G41. The logic expression indexed by the serial number 2 is composed of two items. The truth of any one of the items can result in the truth of G42.
The state combination expressions of the input variables in the truth value table do not have to cover all the state combinations of the input variables. As shown in FIG. 4, V22V32 is not covered by the two expressions in the truth value table. That is, the domain engineers need only consider the state combinations of the input variables in concern, but not have to be all, i.e. the incomplete representation is allowed.
§3.2. To satisfy the normalization of the logic gate, all the uncovered state combinations of input variables can be treated as a new added expression that is called the remnant expression into the truth value table, and correspondingly a new state of the logic gate is added. For the example shown in FIG. 4, the truth value table of the logic gate covering all the state combinations is shown in FIG. 5.
It can be proved that the sum of all the states including the remnant state of the logic gate is normalized.
Proof:




  
    
      
        
          
            ∑
            j
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                G
                ij
              
              }
            
          
        
        =
        
        ⁢
        
          
            ∑
            j
          
          ⁢
          Pr
          ⁢
          
            {
            
              the
              ⁢
              
                
              
              ⁢
              logic
              ⁢
              
                
              
              ⁢
              expression
              ⁢
              
                
              
              ⁢
              corresponding
              ⁢
              
                
              
              ⁢
              to
              ⁢
              
                
              
              ⁢
              
                G
                ij
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                ∑
                j
              
              ⁢
              
                the
                ⁢
                
                  
                
                ⁢
                logic
                ⁢
                
                  
                
                ⁢
                expression
                ⁢
                
                  
                
                ⁢
                corresponding
                ⁢
                
                  
                
                ⁢
                to
                ⁢
                
                  
                
                ⁢
                
                  G
                  ij
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        Pr
        ⁢
        
          {
          
            all
            ⁢
            
              
                
              
              ⁢
              
                
              
            
            ⁢
            the
            ⁢
            
              
            
            ⁢
            state
            ⁢
            
              
            
            ⁢
            combinations
            ⁢
            
              
                
              
              ⁢
              
                
              
            
            ⁢
            of
            ⁢
            
              
            
            ⁢
            the
            ⁢
            
              
            
            ⁢
            input
            ⁢
            
              
            
            ⁢
            variables
          
          }
        
      
    
  
  
    
      
        =
        
        ⁢
        1
      
    
  




▪

When the logic relation of Gi is the complete combination of all the states of input variables,




  
    G
    i
  
  =
  
    
      
        G
        i
      
      ⁢
      
        {
        
          
            V
            1
          
          ,
          
            V
            2
          
          ,
          ...
          ⁢
          
            
          
          ,
          
            V
            N
          
        
        }
      
    
    =
    
      
        (
        
          
            
              
                G
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              
                G
                
                  i
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            
              ⋮
            
          
          
            
              
                G
                iK
              
            
          
        
        )
      
      =
      
        (
        
          
            
              
                V
                11
              
            
            
              
                V
                21
              
            
            
              ...
            
            
              
                V
                
                  N
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              
                V
                11
              
            
            
              
                V
                21
              
            
            
              ...
            
            
              
                V
                
                  N
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            
              
                
              
            
            
              ⋮
            
            
              
                
              
            
            
              
                
              
            
          
          
            
              
                V
                
                  1
                  ⁢
                  A
                
              
            
            
              
                V
                
                  2
                  ⁢
                  B
                
              
            
            
              ...
            
            
              
                V
                NZ
              
            
          
        
        )
      
    
  




The index A is the state number of V1, the index B is the state number of V2, . . . , the index Z is the state number of VN. This type logic gate is called the complete combination logic gate. For the example shown in FIG. 4, the complete combination logic gate is




  
    G
    4
  
  =
  
    
      
        G
        4
      
      ⁢
      
        {
        
          
            V
            1
          
          ,
          
            V
            2
          
          ,
          
            V
            3
          
        
        }
      
    
    =
    
      
        (
        
          
            
              
                G
                41
              
            
          
          
            
              
                G
                42
              
            
          
          
            
              
                G
                43
              
            
          
          
            
              
                G
                44
              
            
          
          
            
              
                G
                45
              
            
          
          
            
              
                G
                46
              
            
          
          
            
              
                G
                47
              
            
          
          
            
              
                G
                48
              
            
          
        
        )
      
      =
      
        (
        
          
            
              
                V
                11
              
            
            
              
                V
                21
              
            
            
              
                V
                
                  3
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
              
            
          
          
            
              
                V
                11
              
            
            
              
                V
                21
              
            
            
              
                V
                
                  3
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
              
            
          
          
            
              
                V
                11
              
            
            
              
                V
                22
              
            
            
              
                V
                31
              
            
          
          
            
              
                V
                11
              
            
            
              
                V
                22
              
            
            
              
                V
                32
              
            
          
          
            
              
                V
                12
              
            
            
              
                V
                21
              
            
            
              
                V
                31
              
            
          
          
            
              
                V
                12
              
            
            
              
                V
                21
              
            
            
              
                V
                32
              
            
          
          
            
              
                V
                12
              
            
            
              
                V
                22
              
            
            
              
                V
                31
              
            
          
          
            
              
                V
                12
              
            
            
              
                V
                22
              
            
            
              
                V
                32
              
            
          
        
        )
      
    
  




In fact, any logic gate can be transformed from the complete combination logic gate. For the example above, logic gate G4 can be expressed as





  
    
      
        
          G
          4
        
        =
        
        ⁢
        
          (
          
            
              
                
                  
                    
                      V
                      11
                    
                    ⁢
                    
                      V
                      21
                    
                  
                  +
                  
                    
                      V
                      11
                    
                    ⁢
                    
                      V
                      31
                    
                  
                  +
                  
                    
                      V
                      21
                    
                    ⁢
                    
                      V
                      31
                    
                  
                
              
            
            
              
                
                  
                    
                      V
                      12
                    
                    ⁢
                    
                      V
                      21
                    
                    ⁢
                    
                      V
                      32
                    
                  
                  +
                  
                    
                      V
                      12
                    
                    ⁢
                    
                      V
                      22
                    
                    ⁢
                    
                      V
                      31
                    
                  
                
              
            
          
          )
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            (
            
              
                
                  1
                
                
                  1
                
                
                  1
                
                
                  0
                
                
                  0
                
                
                  0
                
                
                  0
                
                
                  0
                
              
              
                
                  0
                
                
                  0
                
                
                  0
                
                
                  0
                
                
                  1
                
                
                  1
                
                
                  1
                
                
                  0
                
              
            
            )
          
          ⁢
          
            (
            
              
                
                  
                    V
                    11
                  
                
                
                  
                    V
                    21
                  
                
                
                  
                    V
                    
                      3
                      ⁢
                      
                        
                      
                      ⁢
                      1
                    
                  
                
              
              
                
                  
                    V
                    11
                  
                
                
                  
                    V
                    21
                  
                
                
                  
                    V
                    
                      3
                      ⁢
                      
                        
                      
                      ⁢
                      2
                    
                  
                
              
              
                
                  
                    V
                    11
                  
                
                
                  
                    V
                    22
                  
                
                
                  
                    V
                    31
                  
                
              
              
                
                  
                    V
                    11
                  
                
                
                  
                    V
                    22
                  
                
                
                  
                    V
                    32
                  
                
              
              
                
                  
                    V
                    12
                  
                
                
                  
                    V
                    21
                  
                
                
                  
                    V
                    31
                  
                
              
              
                
                  
                    V
                    12
                  
                
                
                  
                    V
                    21
                  
                
                
                  
                    V
                    32
                  
                
              
              
                
                  
                    V
                    12
                  
                
                
                  
                    V
                    22
                  
                
                
                  
                    V
                    31
                  
                
              
              
                
                  
                    V
                    12
                  
                
                
                  
                    V
                    22
                  
                
                
                  
                    V
                    32
                  
                
              
            
            )
          
        
      
    
  




In the matrix, “1” denotes the complete set and “0” denotes the null set.

If considering the remnant state of the logic gate, there is




  
    
      
        
          G
          4
        
        =
        
        ⁢
        
          (
          
            
              
                
                  
                    
                      V
                      11
                    
                    ⁢
                    
                      V
                      21
                    
                  
                  +
                  
                    
                      V
                      11
                    
                    ⁢
                    
                      V
                      31
                    
                  
                  +
                  
                    
                      V
                      21
                    
                    ⁢
                    
                      V
                      31
                    
                  
                
              
            
            
              
                
                  
                    
                      V
                      12
                    
                    ⁢
                    
                      V
                      21
                    
                    ⁢
                    
                      V
                      32
                    
                  
                  +
                  
                    
                      V
                      12
                    
                    ⁢
                    
                      V
                      22
                    
                    ⁢
                    
                      V
                      31
                    
                  
                
              
            
            
              
                
                  the
                  ⁢
                  
                    
                  
                  ⁢
                  remnant
                  ⁢
                  
                    
                  
                  ⁢
                  expression
                
              
            
          
          )
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            (
            
              
                
                  1
                
                
                  1
                
                
                  1
                
                
                  0
                
                
                  0
                
                
                  0
                
                
                  0
                
                
                  0
                
              
              
                
                  0
                
                
                  0
                
                
                  0
                
                
                  0
                
                
                  1
                
                
                  1
                
                
                  1
                
                
                  0
                
              
              
                
                  0
                
                
                  0
                
                
                  0
                
                
                  1
                
                
                  0
                
                
                  0
                
                
                  0
                
                
                  1
                
              
            
            )
          
          ⁢
          
            (
            
              
                
                  
                    
                      V
                      11
                    
                    ⁢
                    
                      V
                      21
                    
                    ⁢
                    
                      V
                      31
                    
                  
                
              
              
                
                  
                    
                      V
                      11
                    
                    ⁢
                    
                      V
                      21
                    
                    ⁢
                    
                      V
                      32
                    
                  
                
              
              
                
                  
                    
                      V
                      11
                    
                    ⁢
                    
                      V
                      22
                    
                    ⁢
                    
                      V
                      31
                    
                  
                
              
              
                
                  
                    
                      V
                      11
                    
                    ⁢
                    
                      V
                      22
                    
                    ⁢
                    
                      V
                      32
                    
                  
                
              
              
                
                  
                    
                      V
                      12
                    
                    ⁢
                    
                      V
                      21
                    
                    ⁢
                    
                      V
                      31
                    
                  
                
              
              
                
                  
                    
                      V
                      12
                    
                    ⁢
                    
                      V
                      21
                    
                    ⁢
                    
                      V
                      32
                    
                  
                
              
              
                
                  
                    
                      V
                      12
                    
                    ⁢
                    
                      V
                      22
                    
                    ⁢
                    
                      V
                      31
                    
                  
                
              
              
                
                  
                    
                      V
                      12
                    
                    ⁢
                    
                      V
                      22
                    
                    ⁢
                    
                      V
                      32
                    
                  
                
              
            
            )
          
        
      
    
  




Example 4
FIG. 6-2 is an example about the default variable that is explained below.
§4.1. The default event can be drawn as  i.e. the default event or the default basic event indexed by n (n=3 in this example). It does not have any cause variable, has only one output variable, and has only one state. Therefore, it is an inevitable event. Although the default event does not have more than one state, for convenience, the default event is also called the default variable or the default basic variable. One default event corresponds to one and only one consequence variable. For example, Dn corresponds to only Xn, i.e. Dn is the default event of Xn. The meaning is: Dn is a self-independent cause of the states of Xn. This cause functions with other causes to decide the state probability distribution of Xn. The difference between D and B is that D has only one state and the occurrence probability is always 1, while B has at least two states and the occurrence of each state depends on the state probability distribution of B.
§4.2. The functional variable with Dn as the direct cause variable of Xn is denoted as Fn;D (F3;D in FIG. 6-2). Its specific value is denoted as Fnk;D. The corresponding parameters are denoted as fnk;D, ank;D and rn;D respectively. The meanings and the conditions to be satisfied are the same as for fnk;ij, ank;ij and rn;j. In terms of matrix,




  
    
      F
      
        n
        ;
        D
      
    
    =
    
      (
      
        
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                D
              
            
          
        
        
          
            
              F
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                D
              
            
          
        
        
          
            ⋮
          
        
        
          
            
              F
              
                nk
                ;
                D
              
            
          
        
        
          
            ⋮
          
        
        
          
            
              F
              
                nK
                ;
                D
              
            
          
        
      
      )
    
  
  ,
  
    
      A
      
        n
        ;
        D
      
    
    =
    
      (
      
        
          
            
              a
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  1
                
                ;
                D
              
            
          
        
        
          
            
              a
              
                
                  n
                  ⁢
                  
                    
                  
                  ⁢
                  2
                
                ;
                D
              
            
          
        
        
          
            ⋮
          
        
        
          
            
              a
              
                nk
                ;
                D
              
            
          
        
        
          
            ⋮
          
        
        
          
            
              a
              
                nK
                ;
                D
              
            
          
        
      
      )
    
  
  ,
  
    
      F
      
        n
        ;
        D
      
    
    =
    
      
        
          (
          
            
              r
              
                n
                ;
                D
              
            
            /
            
              r
              n
            
          
          )
        
        ⁢
        
          A
          
            n
            ;
            D
          
        
      
      =
      
        (
        
          
            
              
                f
                
                  
                    n
                    ⁢
                    
                      
                    
                    ⁢
                    1
                  
                  ;
                  D
                
              
            
          
          
            
              
                f
                
                  
                    n
                    ⁢
                    
                      
                    
                    ⁢
                    2
                  
                  ;
                  D
                
              
            
          
          
            
              ⋮
            
          
          
            
              
                f
                
                  nk
                  ;
                  D
                
              
            
          
          
            
              ⋮
            
          
          
            
              
                f
                
                  nK
                  ;
                  D
                
              
            
          
        
        )
      
    
  




In which, K is the upper bound of k and





  
    r
    n
  
  =
  
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        r
        
          n
          ;
          i
        
      
    
    +
    
      
        r
        
          n
          ;
          D
        
      
      .
    
  




Correspondingly, because of adding the direct cause variable Dn,





  
    Pr
    ⁢
    
      {
      
        X
        nk
      
      }
    
  
  =
  
    
      
        ∑
        
          i
          ,
          j
        
      
      ⁢
      
        
      
      ⁢
      
        f
        
          nk
          ;
          ij
        
      
    
    +
    
      
        f
        
          nk
          ;
          D
        
      
      .
    
  




§4.3. In many cases, the default event D of the consequence variable corresponds to only one state of the consequence variable, or is only one of the causes of this default state of the consequence variable. The default state is given by the domain engineers in the process of constructing DUCG, and is a specific state of the various states of the consequence variable. Usually, it is the normal state. The so called normal state is the state that the variable is normally in. When people represent the causalities among things, they usually concern the abnormal states, but not the normal state. This results in that the domain engineers usually cannot give all the causes for all the states of the consequence variable. For the example of the secondary loop system of a nuclear power plant, the normal water pressure, the normal water temperature, etc, are the normal states of these variables. When the domain engineers represent the causalities among the states of various variables, they mainly concern the relations among the abnormal states, e.g. the influence of the high water temperature to the water pressure, etc, but not the causes of the normal states. Therefore, the normal states are usually the default states of the consequence variables. The default state is a special state and is indexed by η. For the default states without clearly describing the causes, the causes are or partially are the default events. The default events can be denoted clearly in DUCG, or be ignored because the default event Dn belongs always to Xn.
In general, DUCG allows represent only part of the causal relations among variables, but not necessarily all. Usually, the correctness of the computation based on the represented part of a DUCG is not influenced by the incompleteness of the representation, because in the explicit representation mode, the representation and computation of various variable states are independent. For the example of X3=F31B1+F32B2, suppose B1, B2 and X3 have only two states each. When the domain engineers give just a32,11, a32;12 and a32,22, but not a31;11, a31;12, a31;21, a31;22 and a32;21, this means that the domain engineers concern only X32, and X32 is only related to B11, B12 and B22. This does not influence the calculation to Pr{X32}, because X32=F32;11B11+F32;12B12+F32;22B22, i.e., Pr{X32}=f32;11b11+f32;12b12+f32;22b22.
Since DUCG allows




  
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        a
        
          nk
          ;
          ij
        
      
    
    <
    
      1
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        
          ∑
          j
        
        ⁢
        
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              V
              ij
            
            }
          
        
      
    
    <
    1
  
  ,




there must be





  
    
      ∑
      k
    
    ⁢
    
      
    
    ⁢
    
      Pr
      ⁢
      
        {
        
          X
          nk
        
        }
      
    
  
  <
  1.




In this case, to satisfy the normalization, define





  
    
      Pr
      ⁢
      
        {
        
          X
          
            n
            ⁢
            
              
            
            ⁢
            η
          
        
        }
      
    
    ≡
    
      Pr
      ⁢
      
        {
        
          
            ∏
            
              k
              ≠
              η
            
          
          ⁢
          
            
          
          ⁢
          
            
              X
              _
            
            nk
          
        
        }
      
    
  
  =
  
    1
    -
    
      
        ∑
        
          k
          ≠
          η
        
      
      ⁢
      
        
      
      ⁢
      
        Pr
        ⁢
        
          
            {
            
              X
              nk
            
            }
          
          .
        
      
    
  




Then we have





  
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        Pr
        ⁢
        
          {
          
            X
            nk
          
          }
        
      
    
    =
    1
  
  ,




in which η indexes the default state. If there is not the default state, it denotes the state without input. In the example above, η=1, Pr{X31}=1−Pr{X32}. However, usually people do not need know the probability of the default state or the state without input.

Example 5
FIG. 6 is an example combining two default variables as one default variable, which is explained below.
§5.1. As shown in FIG. 6-1, the case with more than one default variable can appear when constructing the original DUCG and transforming the multiple groups in the implicit representation mode to the explicit representation mode (see §14 for details). For Xn (n=3 in this example), all default variables Dnh (D31 and D32 in this example) are inevitable events. What influences the state probability distribution of Xn are only the functional variables Fn;Dh (F3;D1 and F3:D2 in this example) along with the parameters fhk;Dh, ank;Dh and rn;Dh. The same type parameters of different default variables are always in the summation relation during the computation. Therefore, it is not necessary for the different default variables to exist respectively. They can be combined as one default variable Dn in advance. FIG. 6-2 shows the result after the combination.
Additional specification: Beside the graphical representation shown in above figures, the explicit representation mode of DUCG can also be drawn graphically as in causality trees and causality forest.
The so called causality tree means to draw the cause variables and the functional variables of an event or event variable graphically, beginning with the consequence variable. On demand, the cause variable can be taken as the beginning point of drawing upstream cause variables and functional variables. This process can continue until all the leaves are basic variables. Such constructed tree type logical diagram is called the causality tree. For example, the explicit mode representation shown in FIG. 3 can be represented in terms of causality trees shown in FIG. 7, in which the meanings of X1, X2, B3, G4, X5 and X6 can be described in text.
Usually, multiple consequence variables should be represented by multiple causality trees, in which some variables may be repeated. The collection of all the related causality trees can be called the causality forest. The causality forest is fully equal to the corresponding DUCG in the explicit representation mode. They can be transformed from each other. Therefore, the causality forest is a transfiguration of the DUCG in the explicit representation mode. Different from the ordinary logic trees (e.g. the fault tree, event tree and decision tree, etc), there exists uncertainties in the causality trees, which are represented by the functional variables Fn;i and have multiple states.
Example 6
FIG. 8 is an example of the standard implicit representation mode described in §6, which is explained below.
§6.1. In the implicit representation mode, connect the consequence variable Xn (n=4 in this example) with its direct cause variables Vi (B1 and X2), i∈SIXn (i=1,2 in this example, i.e. SIXn={1,2}), by the directed arc→pointing to Xn, which is different from the functional variable→, the conditional functional variable  and the connection variable→that connects the logic gate and its input variables. Different from the explicit representation mode, the direct cause variables in the implicit representation mode are only the {B,X} type variables, but not the {G,D} type variables. The implicit representation mode takes the conditional probability table (CPT) to represent the uncertain causalities between the cause variables and the consequence variable. The so called conditional probability here is the probability of a state of Xn conditioned on the state combination j of its direct cause variables Vi, i∈SIXn, i.e. pnkj=Pr{Xnk|j}. For the example shown in FIG. 8, suppose B1 has three states, X2 and X4 have two states each. Then there are 6 state combinations of B1 and X2, in which the occurrence probability of the state k of Xn conditioned on the jth state combination (j=4 in this example) is denoted as pnk;j, where




  
    
      ∑
      k
    
    ⁢
    
      
    
    ⁢
    
      p
      
        nk
        ;
        j
      
    
  
  =
  1.




All the conditional probabilities are included in the conditional probability table. Thus, the causalities in FIG. 8 can be implicitly represented as shown in the following table:















State
Conditional Probabilities
Conditional Probabilities


j
Combination
corresponding to X41
corresponding to X42







1
B11X21
p41;1 = q41;1/d4;1
p42;1 = q42;1/d4;1


2
B11X22
p41;2 = q41;2/d4;2
p42;2 = q42;2/d4;2


3
B12X21
p41;3 = q41;3/d4;3
p42;3 = q42;3/d4;3


4
B12X22
p41;4 = q41;4/d4;4
p42;4 = q42;4/d4;4


5
B13X21
p41;5 = q41;5/d4;5
p42;5 = q42;5/d4;5


6
B13X22
p41;6 = q41;6/d4;6
p42;6 = q42;6/d4;6









In which, the meanings of q and d will be explained later.

The reason why this mode is called the implicit representation mode is because the logic combination relation among the cause variables and between the cause and consequence variables is not explicitly represented in the conditional probability table. Even more, some variables really having the causality with the consequence variable may not appear in the cause variables, while other variables not having causalities with the consequence variable may appear in the cause variables, although they do not have any function. All these are implicitly included in the conditional probability table.
§6.2. The conditional probabilities can be obtained from the statistic data. For the example above, suppose the number of samples of the state combination 5 is d4;5 (dn;j is called the number of samples), in which the number of samples including X41 is q41;5 (qnk;j is called the number of occurrence) and the number of samples including X42 is q42;5. Then the conditional probabilities p41;5=q41;5/d4;5 and p42;5=q42;5/d4;5. Obviously, p41;5+p42;5=1, because q41;5+q42;5=d4;5. Suppose the number of samples of the state combination 6 is d4;6, in which the number of samples including X41 is q41;6 and the number of samples including X42 is q42;5. Then the conditional probabilities p41;6=q41;6/d14;6 and p42;6=q42;6/d4;6. Similarly, p41;6+p42,6=1, because q41;6+q42;6=d4;6.
In the case that the direct cause variables will never be reduced, the conditional probability pnk;j can be given directly in the form of the calculation result of qnk;j/dn;j. If there is the case that the direct cause variables may be reduced, the conditional probability pnk;j should be given in the form of two parameters qnk;j and dn;j. This is because in this way, the new conditional probability table after one or more cause variables are eliminated can be calculated from the above calculation method.
Example 7
FIGS. 8-10 are the further examples of the implicit representation mode described in §7, which are explained below.
§7.1. In the case that the CPT is only composed of pnk;j, if the direct cause variable in the implicit representation mode is single-valued, this implicit representation mode is BN. As described in §6, the CPT can also be extended as being represented with pnk;j. qnk;j and dn;j satisfying pnk;j=qnk;j/dn;j. The purpose of this representation is that the CPT can be reconstructed dynamically in the case of dynamically eliminating the direct cause variables. Moreover, this invention allows more than one group of direct cause variables in the implicit representation mode, where every group has its own independent CPT. As shown in FIG. 9, the direct cause variables of X4 in the implicit representation mode are divided into two groups: {B1,X2} and {B3,X5}. The conditional probability table of the first group is CPT1. The conditional probability table of the second group is CPT2. CPT1 and CPT2 are composed independently. This case can be that the different engineers collect different data and form different CPT based on their different views and concerns. This is allowed in this invention.
In the case of more than one group, the relationship or the influence weight of every group to the consequence variable may be different. This needs be represented by a relationship parameter rXn;g. In FIG. 9, the relationship of the first group is rX4,1 and the relationship of the second group is rX4,2.
§7.2. No matter it is the standard or non-standard implicit representation mode, the direct cause variables may be dynamically reduced because of the following dynamical logic simplification or the operation to break the logic cycles during the computation, such that the CPT needs be reconstructed dynamically. As shown in FIG. 8, suppose X2 is no longer the cause of X4. Then the cause state combinations of X4 become three: B11, B12 and B13. The new conditional probability table corresponding to the original conditional probability table becomes

p41;1=q41;1/d4;1=(q41;1+q41;2)/(d4;1+d4;2), p42;1=q42;1/d4;1=(q42;1+q42;2)/(d4;1+d4;2)

p41;2=q41;2/d4;2=(q41;3+q41;4)/(d4;3+d4;4), p42;2=q42;2/d4;2=(q42;3+q42;4)/(d4;3+d4;4)

p41;3=q41;3/d4;3=(q41;5+q41;6)/(d4;5+d4;6), p42;3=q42;3/d4;3=(q42;5+q42;6)/(d4;5+d4;6)

In which, the right side on every second equator is the values before eliminating X2 and the left side on every second equator is the values after eliminating X2. Moreover, the numerators on the two sides of this equator are equal, and the denominators on the two sides of this equator are equal. The state combination index of the cause variable after eliminating X2 is denoted as j′. Obviously, the new pnk;j′, j′=1,2,3, are different from the original pnk;j, j=1,2,3,4,5,6.

§7.3. In the case of more than one group, the method of reconstructing CPT is same as for the single group case, because this reconstruction is only within the group.
The case of more than one group direct cause variables in the implicit representation mode can be transformed as the single group case by twice transformations. For the example, transform the case shown in FIG. 9 to the case shown in FIG. 10. The first transformation is to transform the case of more than one group in the implicit representation mode as the explicit representation mode (see example 14 for details). The second transformation is to transform the explicit representation mode to the single group implicit representation mode (see example 15 for details).
Example 8
FIGS. 11 and 12 are the examples of the hybrid representation mode described in §8, which are explained below.
§8.1. In this invention, the causalities between the same consequence variable and its direct cause variables can be represented partially by the implicit representation mode and partially by the explicit representation mode. That is the hybrid representation mode.
As shown in FIG. 11, X1 and B2 are involved in the standard implicit representation mode by using the conditional probability table to represent their functions to X4. X3 is in the explicit representation mode to function to X4, which may be added after collecting the statistic data of the conditional probabilities about X4 conditioned on X1 and B2. For example, suppose X1, B2 and X4 are same economic variables. The conditional probability table among them is obtained according to the past statistic data. X3 is a variable of the economic policies to be taken by the government, and is a newly added influence factor or cause variable to X4, which does not have any available statistic data except the belief of domain engineers. Then, the causality of X3 to X4 should be in the explicit representation mode. The real influence relation of the two representation modes to the consequence variable is as shown in FIG. 12. In which, G5 is a virtual logic gate, U5;1 and U5;2 are respectively the connection variables between G5 and {X1, B2} respectively, and are the inevitable event variables. To be distinguished from the ordinary logic gate, the virtual logic gate G5 is drawn in the dashed line.
The explicit functional relations represented in FIG. 12 among the various variables are actually the result of transforming the part in the implicit representation mode to be in the explicit representation mode. The explicit representation mode involves the relationship. Therefore, the relationship related to the implicit representation mode part in the hybrid representation mode needs also be given. As in this example, the domain engineers need give not only the conditional probability table about X1, B2 and X4, but also the whole relationship rX4 between X4 and {X1, B2}. In default, it can be defined that rX4=2 (i.e. the number of the cause variables in the implicit representation mode). Since the relationship is actually a relative weighing factor, the value can be larger than 1. If the case is in the non-standard implicit representation mode, then the relationship of the hth group is rn;h. All these relationships must be counted in the calculation of rn, i.e.




  
    
      r
      n
    
    =
    
      
        r
        n
      
      +
      
        
          ∑
          h
        
        ⁢
        
          
        
        ⁢
        
          r
          
            Xn
            ;
            h
          
        
      
    
  
  ,




in which rn on the right side is only the sum of the relationships in the explicit representation mode.

The calculation method of the truth value table and the functional intensity (f4k;5j in the above example) of a virtual logic gate (G5 in the above example) is presented in §14.
Example 9
FIG. 32 is an example of the comprehensive representation escribed in §9, which is explained below.
§9.1. As shown in FIG. 32, represent all the X type variables (X3, X4, X5, X6, X7, X8, X9 and X10 in this example) and their corresponding direct cause variables (also called the input variables) in the corresponding representation mode independently and respectively. For the X type variable, do not concern the output and represent only the input. Moreover, the representation reaches only the other X type variables or the {B,D} type variables. In this way, the representation is limited in a very small scale (e.g., the input variables of X6 are B1, B2, X3 and X5; the input variables of X7 are X4, X6 and B11; the input variable of X8 is X7). Every X type variable can be the input of other X type variables (e.g. X7 is the input variable of X8 in this example). Different X type variable may take the explicit, implicit or hybrid representation mode independently (For X6 and its input variables, the explicit representation mode is taken; for X7, X4 and X6, the standard implicit representation mode is taken; for X7 and B11, the explicit representation mode is taken, i.e. the hybrid representation mode is taken for X7; for X8 and its input variable, the standard implicit mode is taken). Then, simply connect all the X type variables represented in various representation modes together, the whole DUCG is obtained. In other words, no matter whether the explicit representation mode, the implicit representation mode or the hybrid representation mode is taken, the synthesis of all these representations is called the dynamical uncertain causality graph (DUCG). DUCG allows the logic cycles.
In summary, the construction steps of DUCG are: {circle around (1)} Initially decide the X type variables and the {B,D} type variables; {circle around (2)} Determine the modules according to the X type variables. Every module includes one X type variable. For every X type variable, decide its {X,B,D} type direct cause variables; {circle around (3)} According to the logic relation between the X type variable and its {X,B,D} type direct cause variables and the situation of the known data, represent the uncertain causalities among them by using either explicit, implicit or hybrid representation mode respectively; {circle around (4)} If it is necessary to increase, reduce or revise the X type variables or the {B,D} type variables, every module should use the new defined variables; {circle around (5)} Connect all the modules together to synthesize the whole original DUCG (e.g. X7 is the direct cause variable of X8, X4, X6 and B11 are the direct cause variables of X7. Then X4, X6 and B11 are indirectly the cause variables of X8). It is seen that the feature of constructing DUCG is to decompose the task of constructing a large DUCG as many small modules first, construct these modules respectively, and the whole DUCG can be synthesized by computer automatically by connecting them together. This feature reduces the difficulty of constructing DUCG greatly.
§9.2. When all the DUCG is fully or transformed as fully represented in the single group implicit representation mode (see §15) without logic cycle, this specific DUCG is the implicit dynamical uncertain causality graph (IDUCG). IDUCG can be transformed as BN according to the method described in §21, and be solved by the method of BN. If the DUCG is all represented in the implicit representation mode, but the logic cycles and more than one group are allowed, this IDUCG is called the general BN. Correspondingly, the DUCG without logic cycle and with only one group can be called the narrow BN. If the DUCG is transformed as all in the explicit representation mode (see §14 for details), this special DUCG is called the Explicit Dynamical Causality Graph (EDUCG). Obviously, the narrow BN or IDUCG is a special case of DUCG, and EDUCG is also a special case of DUCG. But any DUCG can be transformed as EDUCG and be dealt with, because EDCUG is always an applicable representation and computation method.
§9.3. It is not hard to see that in any conditional probability table in an implicit mode representation, the number of the conditional probabilities N=J1J2 . . . J1K, in which Ji is the number of the states of the input variable i, I is the number of the {B,X} type direct cause variables, and K is the number of the states of the consequence variable. If this group cause variables are represented with the explicit mode, the related data become the functional intensities and relationships, and the number of the data is N′=(J1+J2+ . . . +J1)K+1. Obviously, in many cases, N>>N′. That is, the explicit representation mode is usually more convenient than the implicit representation mode. In fact, in many cases, the conditional probabilities are difficult to be obtained from the statistics due to the lack of data. In this case, the uncertainty of the logical relations among things can only be represented by the belief of the domain engineers. If still use the implicit representation mode, it may be difficult to be implemented because of too many data to be given. For example, suppose Ji=5 (i=1,2, . . . , 1), I=5, K=5. Then N=56=15625. If the conditional probabilities are given by qnk;j and dn;j, then N=56×2=31250. Obviously, even for such a small case, the data are too many to be given by people, needless to say that these data are not explicit and are hard to be given by the domain engineers. If use the explicit representation mode, N′=53+5=130 that is much less than N. It is seen that the explicit representation mode is more suitable for the belief representation of the domain engineers. However, the conditional probabilities are obtained from the statistic data and are objective and reliable. Moreover, they do not need know the logic relations among the cause variables. On the other side, the functional intensities and the causal relationships are usually given by the domain engineers and are less objective. Moreover, they need know the logic relations among the cause variables. Therefore, the two representation modes have their own advantages and disadvantages, and are all necessary to be used independently or synthetically according to the specific situation.
§9.4. Add the fuzzy evidence Eh into the original DUCG as a virtual evidence variable. Let it be the consequence variable of Vh in the explicit representation mode. The method will be explained in example 10 in details.
§9.5. In the specific case of the process system, as the supplement to the original DUCG, a new type representation of the logic relations among things can be added, which is the relationship table of the B type basic events and the X type consequence events. This table is defined as below.
For every basic event or the state of the basic event variable, and for every consequence event or the state of the consequence variable, give the relationship Ψnk;ij such as that the former must cause the latter or must not, etc. For example, according to the knowledge of the domain engineers, given Bij, the relation between Bij and Xnk may be represented as: when Ψnk;ij=1, Xnk must be true; when Ψnk;ij=1, Xnk must not be true; when Ψnk;ij=0, whether or not Xnk is true is unknown. Etc.
Example 10
FIG. 13 is an illustration in which the evidence event Eh is taken as the virtual consequence event variable, which is explained below.
§10.1. All the online obtained evidence is expressed as E. E is composed of two type events: One is the group of evidence events with each being denoted as Eh, h is the index of such evidence events. They determine the states of the {B,X} type variables. Every Eh represents a specific evidence event such as “the flow rate is too high”, “the temperature is 185 C°”, “the pressure is normal”, “the alarm is on”, “the probabilities of the valve being blocked or not are 70% and 30% respectively”, etc. In this invention, this type events Eh can be classified as three types:
(1) The ordinary evidence: i.e. the evidence indicating the state of variable Vh certainly, where V∈{B,X}. For example, the temperature variable Vh is observed as 185 C°, while 185 C° is certainly in the high temperature area.
(2) The fuzzy continuous evidence: i.e. the evidence certainly indicating the value of the continuous variable Vh, and this value is just within the fuzzy area. For example, the observed temperature Vh is 185 C°, while 185 C° is just within the fuzzy area between the two fuzzily discretized states “normal” and “high” (see FIG. 2).
(3) The fuzzy state evidence: i.e. the evidence just indicating the state probability distribution of variable Vh, but not which state that Vh is really in. For example, suppose variable Vh represents the valve state (blocked or not). Before receiving any evidence, the probability distribution of whether or not the valve is blocked is given by the earlier statistic data. After receiving the evidence Eh, due to that the evidence is unclear or other reason, Eh does not indicate whether or not the valve is blocked, but indicates only that the probabilities of the two states of the valve are 70% and 30% respectively. Moreover, this probability distribution is usually not the same as the statistic data. Such evidence is the fuzzy state evidence.
The fuzzy continuous evidence and the fuzzy state evidence are all called the fuzzy evidence briefly. This invention treats the fuzzy evidence as the ordinary evidence by transforming the fuzzy evidence as the ordinary evidence. That is, treat Eh as the virtual consequence variable of Vh, while Eh has only one certainly true state. In this way, the case of the fuzzy evidence is transformed as the case of the ordinary evidence, and the method to deal with such case becomes the same as for the ordinary case.
The other evidence of E may not deal with the states of the {B,X} type variables, but include other useful information, such as the occurrence order of events, etc. This type evidence is denoted as E*. Its function is to simplify DUCG (see example 13 for details). Therefore,




  E
  =
  
    
      E
      *
    
    ⁢
    
      
        ∏
        h
      
      ⁢
      
        
      
      ⁢
      
        
          E
          h
        
        .
      
    
  




§10.2. The method to transform the fuzzy evidence as the ordinary evidence is as follows.
As shown in FIG. 13, treat Eh as the virtual consequence variable of Vh. Eh has only one inevitable state and has only one input Vh. Vh and Eh are connected by the virtual functional variable FE;h. Its virtual functional intensity fE;hj≡Pr{FE;hj} can be given by the domain engineers. But usually, fE:hj is hard to be given by the experiment or the domain engineers. In this case, the following calculation method can be applied.
Suppose the fuzzy continuous evidence Eh shows that the state membership of Vhj is mhj and the fuzzy state evidence Eh, shows that the state probability distribution of Vh is Pr{Vhj|Eh}=mhj. The fuzzy area related to Eh or the known state probability distribution deals with at least two states of Vh. The memberships of Eh belonging to these states or the probabilities of these states are all larger than 0. Meanwhile, the other states cannot be true. Denote Sm as the index set of those states of Vh for which mhj>0. Then,




  
    
      ∑
      
        j
        ∈
        
          S
          Eh
        
      
    
    ⁢
    
      
    
    ⁢
    
      m
      hj
    
  
  =
  
    
      1
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        
          ∑
          
            j
            ∈
            
              S
              Eh
            
          
        
        ⁢
        
          V
          hj
        
      
    
    =
    Ω
  




(the complete set). It can be proved that fE;hj can be calculated from the following equation.





  
    f
    
      E
      ;
      hj
    
  
  =
  
    
      
        
          m
          hj
        
        ⁢
        
          v
          hk
        
      
      
        
          m
          hk
        
        ⁢
        
          v
          hj
        
      
    
    ⁢
    
      f
      
        E
        ;
        hk
      
    
  




In which, j≠k, j∈SEh, k∈SEh, Vhj≡Pr{Vhj} and Vhk≡Pr{Vhk}. Given fE;hk, fE;hj can be calculated. For example, let fE;hj=1,





  
    f
    
      E
      ;
      hj
    
  
  =
  
    
      
        
          m
          hj
        
        ⁢
        
          v
          hk
        
      
      
        
          m
          hk
        
        ⁢
        
          v
          hj
        
      
    
    .
  




Proof:

(1) For the fuzzy state evidence, since Eh indicates the known probability distribution of a variable, we have




  
    
      
        
          m
          hj
        
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  V
                  hj
                
                ❘
                
                  E
                  h
                
              
              }
            
          
          =
          
            
              
                Pr
                ⁢
                
                  {
                  
                    
                      V
                      hj
                    
                    ⁢
                    
                      E
                      h
                    
                  
                  }
                
              
              
                Pr
                ⁢
                
                  {
                  
                    E
                    h
                  
                  }
                
              
            
            =
            
              
                
                  Pr
                  ⁢
                  
                    {
                    
                      
                        V
                        hj
                      
                      ⁢
                      
                        
                          ∑
                          
                            k
                            ⁢
                            
                              I
                              ^
                            
                            ⁢
                            
                              S
                              Eh
                            
                          
                        
                        ⁢
                        
                          
                            F
                            Ehk
                          
                          ⁢
                          
                            V
                            hk
                          
                        
                      
                    
                    }
                  
                
                ⁢
                
                  
                
              
              
                Pr
                ⁢
                
                  {
                  
                    
                      ∑
                      
                        k
                        ⁢
                        
                          I
                          ^
                        
                        ⁢
                        
                          S
                          Ehi
                        
                      
                    
                    ⁢
                    
                      
                        F
                        Ehk
                      
                      ⁢
                      
                        V
                        Ehk
                      
                    
                  
                  }
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              Pr
              ⁢
              
                {
                
                  
                    F
                    Ehj
                  
                  ⁢
                  
                    V
                    hj
                  
                
                }
              
            
            
              
                ∑
                
                  k
                  ∈
                  
                    S
                    Eh
                  
                
              
              ⁢
              
                
              
              ⁢
              
                Pr
                ⁢
                
                  {
                  
                    
                      F
                      Ehk
                    
                    ⁢
                    
                      V
                      hk
                    
                  
                  }
                
              
            
          
          =
          
            
              Pr
              ⁢
              
                {
                
                  F
                  Ehj
                
                }
              
              ⁢
              Pr
              ⁢
              
                {
                
                  V
                  hj
                
                }
              
            
            
              
                ∑
                
                  k
                  ∈
                  
                    S
                    Eh
                  
                
              
              ⁢
              
                
              
              ⁢
              
                Pr
                ⁢
                
                  {
                  
                    F
                    Ehk
                  
                  }
                
                ⁢
                Pr
                ⁢
                
                  {
                  
                    V
                    hk
                  
                  }
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              f
              Ehj
            
            ⁢
            
              v
              hj
            
          
          
            
              ∑
              
                k
                ∈
                
                  S
                  Eh
                
              
            
            ⁢
            
              
            
            ⁢
            
              
                f
                Ehk
              
              ⁢
              
                v
                hk
              
            
          
        
      
    
  




In which, vhj≡Pr{Vhj} and vhk≡Pr{Vhk}. The calculation method of vhj and vhk is: Transform the original DUCG as EDUCQ, outspread vhj as the logic expression composed of the {B,D,F} type events (see §16.1 for details), then take the values of the corresponding probability, frequency or the probability density into the expression, so as to get vhj and vhk. If there is not logic cycle, the original DUCG can also be transformed as IDCUG (see example 15 for details). In the case without evidence, IDUCG=BN (see §21) and the calculation method of BN can be used to calculate vhj and vhk.

(2) For the fuzzy continuous evidence, define ΔEh as the small interval including the value eh. The meaning of the operator Δ means: take the small interval including the value of the variable being operated. Δ→0 indicates the interval tends to be infinite small. According to the meaning of the fuzzy state membership of the continuous variable, we have




  
    
      
        
          m
          hj
        
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  V
                  hj
                
                ❘
                
                  E
                  h
                
              
              }
            
          
          =
          
            
              lim
              
                Δ
                →
                0
              
            
            ⁢
            
              Pr
              ⁢
              
                {
                
                  
                    V
                    hj
                  
                  ❘
                  
                    Δ
                    ⁢
                    
                      
                    
                    ⁢
                    
                      E
                      h
                    
                  
                
                }
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            lim
            
              Δ
              →
              0
            
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  
                    V
                    hj
                  
                  ⁢
                  Δ
                  ⁢
                  
                    
                  
                  ⁢
                  
                    E
                    h
                  
                
                }
              
            
            
              Pr
              ⁢
              
                {
                
                  Δ
                  ⁢
                  
                    
                  
                  ⁢
                  
                    E
                    h
                  
                
                }
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            lim
            
              Δ
              →
              0
            
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  
                    V
                    hj
                  
                  ⁢
                  
                    
                      ∑
                      
                        h
                        ∈
                        
                          S
                          Ei
                        
                      
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      
                        V
                        hk
                      
                      ⁢
                      Δ
                      ⁢
                      
                        
                      
                      ⁢
                      
                        F
                        Ehk
                      
                    
                  
                
                }
              
            
            
              Pr
              ⁢
              
                {
                
                  
                    ∑
                    
                      k
                      ∈
                      
                        S
                        Ehj
                      
                    
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      V
                      hk
                    
                    ⁢
                    Δ
                    ⁢
                    
                      
                    
                    ⁢
                    
                      F
                      Ehk
                    
                  
                
                }
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            lim
            
              Δ
              →
              0
            
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  
                    V
                    hj
                  
                  ⁢
                  Δ
                  ⁢
                  
                    
                  
                  ⁢
                  
                    F
                    Ehj
                  
                
                }
              
            
            
              Pr
              ⁢
              
                {
                
                  
                    ∑
                    
                      k
                      ∈
                      
                        S
                        Eh
                      
                    
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      V
                      hk
                    
                    ⁢
                    Δ
                    ⁢
                    
                      
                    
                    ⁢
                    
                      F
                      Ehk
                    
                  
                
                }
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            lim
            
              Δ
              →
              0
            
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  V
                  hj
                
                }
              
              ⁢
              Pr
              ⁢
              
                {
                
                  Δ
                  ⁢
                  
                    
                  
                  ⁢
                  
                    F
                    Ehj
                  
                
                }
              
            
            
              
                ∑
                
                  k
                  ∈
                  
                    S
                    Eh
                  
                
              
              ⁢
              
                
              
              ⁢
              
                Pr
                ⁢
                
                  {
                  
                    V
                    hk
                  
                  }
                
                ⁢
                Pr
                ⁢
                
                  {
                  
                    Δ
                    ⁢
                    
                      
                    
                    ⁢
                    
                      F
                      Ehk
                    
                  
                  }
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            lim
            
              Δ
              →
              0
            
          
          ⁢
          
            
              
                v
                hj
              
              ⁢
              
                
                  f
                  Ehj
                
                ⁡
                
                  (
                  
                    e
                    h
                  
                  )
                
              
              ⁢
              Δ
              ⁢
              
                
              
              ⁢
              
                e
                h
              
            
            
              
                ∑
                
                  k
                  ∈
                  
                    S
                    Eh
                  
                
              
              ⁢
              
                
              
              ⁢
              
                
                  v
                  hk
                
                ⁢
                
                  
                    f
                    Ehk
                  
                  ⁡
                  
                    (
                    
                      e
                      h
                    
                    )
                  
                
                ⁢
                Δ
                ⁢
                
                  
                
                ⁢
                
                  e
                  h
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              v
              hj
            
            ⁢
            
              f
              Ehj
            
          
          
            
              ∑
              
                k
                ∈
                
                  S
                  Eh
                
              
            
            ⁢
            
              
            
            ⁢
            
              
                v
                hk
              
              ⁢
              
                f
                Ehk
              
            
          
        
      
    
  




It is seen that the results of the fuzzy state evidence and the fuzzy continuous evidence are same, in which j∈SEh, mhj; and vhj are known. Thus we have





  
    
      
        v
        hj
      
      
        m
        hj
      
    
    ⁢
    
      f
      Ehj
    
  
  =
  
    
      ∑
      
        k
        ∈
        
          S
          Eh
        
      
    
    ⁢
    
      
        v
        hk
      
      ⁢
      
        
          f
          Ehk
        
        .
      
    
  




The right side of the equator is a constant independent of the left side. Therefore,





  
    
      
        
          v
          hj
        
        
          m
          hj
        
      
      ⁢
      
        f
        Ehj
      
    
    =
    
      
        
          v
          hk
        
        
          m
          hk
        
      
      ⁢
      
        f
        Ehk
      
    
  
  ,




≠k, and j,k∈SEH.

That is,




  
    f
    Ehj
  
  =
  
    
      
        
          m
          hj
        
        ⁢
        
          v
          hk
        
      
      
        
          m
          hk
        
        ⁢
        
          v
          hj
        
      
    
    ⁢
    
      f
      Ehk
    
  




▪

The functional intensity fE;hj calculated from the above method may be greater than 1, because it is actually the probability density. The functional intensity in the form of the probability density is the same for the calculation as in the form of probability, i.e. the probability density can be treated as the probability, because their significance is the relative magnitude. Whether or not the numerical value is greater than 1 as well as the dimension of the density does not affect the calculation result (see §11.2).
Example 11
This example illustrates the method presented in §11, which is explained below.
§11.1. The method to logically simplify the DUCG given E can be seen in details in example 13. The method to transform the simplified DUCG as EDUCG can be seen in details in example 14. The method to transform the simplified DUCG as IDUCG can be seen in details in example 15, in which the transformation is made only when the DUCG does not have any logic cycle. This is because in the case of logic cycle, there is no effective computation method at the present time. Therefore, when there is logic cycle in the simplified DUCG, it can only be transformed as EDUCG.
§11.2. The purpose of applying the intelligent system is to calculate the new probability distribution of the event in concern conditioned on E. Suppose the event in concern is Hkj. Usually, Hkj is composed of the {B,X,D} type events in the simplified DUCG. In which, Hk is the combination of the variables in concern, e.g. Bi, Xn, Bi, B1′, etc. They are distinguished by the index k. Hkj is the state j of Hk, e.g. Bij, Xnk, BijBi′j′, etc. In other words, Hkj is the state combination indexed by j of the variables included in Hk. Since D is the inevitable event, Hkj is in fact composed by only the {B,X} type events. The so called intelligent inference in the intelligent system is actually to calculate Pr{Hjk∥E}. In this invention, they are calculated according to the following equation.




  
    Pr
    ⁢
    
      {
      
        
          H
          kj
        
        ❘
        E
      
      }
    
  
  =
  
    
      Pr
      ⁢
      
        {
        
          
            H
            kj
          
          ⁢
          E
        
        }
      
    
    
      Pr
      ⁢
      
        {
        E
        }
      
    
  




This calculation result is called the state probability, because it is normalized according to the states of Hk.

Proof:





  
    
      
        
          
            ∑
            j
          
          ⁢
          
            
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ❘
                E
              
              }
            
          
        
        =
        
        ⁢
        
          
            
              ∑
              j
            
            ⁢
            
              
            
            ⁢
            
              Pr
              ⁢
              
                {
                
                  
                    H
                    kj
                  
                  ⁢
                  E
                
                }
              
            
          
          
            Pr
            ⁢
            
              {
              E
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              ∑
              j
            
            ⁢
            
              
            
            ⁢
            
              Pr
              ⁢
              
                {
                
                  
                    H
                    kj
                  
                  ⁢
                  E
                
                }
              
            
          
          
            Pr
            ⁢
            
              {
              
                E
                ⁢
                
                  
                    ∑
                    j
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    H
                    kj
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              ∑
              j
            
            ⁢
            
              
            
            ⁢
            
              Pr
              ⁢
              
                {
                
                  
                    H
                    kj
                  
                  ⁢
                  E
                
                }
              
            
          
          
            
              ∑
              j
            
            ⁢
            
              
            
            ⁢
            
              Pr
              ⁢
              
                {
                
                  
                    H
                    kj
                  
                  ⁢
                  E
                
                }
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        1
      
    
  




▪

The probability normalized according to the states is denoted as hkjs, i.e.




  
    h
    kj
    s
  
  =
  
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            E
          
          }
        
      
      
        Pr
        ⁢
        
          {
          E
          }
        
      
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            E
          
          }
        
      
      
        
          ∑
          j
        
        ⁢
        
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              E
            
            }
          
        
      
    
  




The normalization can also be based on the possible solution set S. In this case, the calculation result is called the rank probability denoted as hkjr. S is the set composed of all the possible solutions for the problem to be solved conditioned on E. These solutions are different events Hkj. Physically, they are exclusive with each other. For example, suppose Hkj=B1,2 and Hk′j′=B1,2B2,3. If just look the sets themselves, Hkj, must be true given Hk′j′, although the occurrence probabilities of them may be different. But physically, B1,2 and B1,2B2,3 may be totally different events. In this case, Hkj and Hk′j′ should be treated as two exclusive events (otherwise Hk′j′ should be absorbed by HO. After understanding the nature of the exclusion of the elements included in S, we have the following calculation equation.





  
    h
    kj
    r
  
  =
  
    
      
        h
        kj
        s
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          
        
        ⁢
        
          h
          kj
          s
        
      
    
    =
    
      
        
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ⁢
                E
              
              }
            
          
          
            Pr
            ⁢
            
              {
              E
              }
            
          
        
        /
        
          
            ∑
            
              
                H
                kj
              
              ∈
              S
            
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  
                    H
                    kj
                  
                  ⁢
                  E
                
                }
              
            
            
              Pr
              ⁢
              
                {
                E
                }
              
            
          
        
      
      =
      
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              E
            
            }
          
        
        
          
            ∑
            
              
                H
                kj
              
              ∈
              S
            
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ⁢
                E
              
              }
            
          
        
      
    
  




According to the above equation, it can be known that it does not matter what dimension of the parameters included in E is used, because the dimensions in both the nominator and denominator are the same. Thus the calculation result is always the number without dimension, i.e. the probability. In other words, the unconditional occurrence probability bij of Bij can be replaced by its occurrence rate λij. Meanwhile, the occurrence probability fnk;ij of Fnk;ij can be replaced by the probability density.
§11.3. To do the above calculation, E must be outspreaded. The events included in E can be classified as two types: One is the evidence event Eh determining the state of the {B,X} type variable. For example, suppose Xh is the temperature variable. If Eh shows that the temperature value is eh, then Eh is the evidence determining the state of the {B,X} type variable. The collection of this type evidence events is




  
    ∏
    h
  
  ⁢
  
    
  
  ⁢
  
    
      E
      h
    
    .
  




Another type includes the other evidence events, e.g. the occurrence order of events (e.g. Xnk occurs earlier than Xn′k′), etc. The set of these events is denoted as E*. Thus E is classified as





  E
  =
  
    
      E
      *
    
    ⁢
    
      
        ∏
        h
      
      ⁢
      
        
      
      ⁢
      
        
          E
          h
        
        .
      
    
  




The function of E* is to simplify the E conditional original DUCG (see example 13 for details). In the probability calculation based on the simplified DUCG,





  
    Pr
    ⁢
    
      {
      
        
          H
          kj
        
        ❘
        E
      
      }
    
  
  =
  
    
      Pr
      ⁢
      
        {
        
          
            H
            kj
          
          ❘
          
            
              E
              *
            
            ⁢
            
              
                ∏
                h
              
              ⁢
              
                
              
              ⁢
              
                E
                h
              
            
          
        
        }
      
    
    =
    
      Pr
      ⁢
      
        {
        
          
            H
            kj
          
          ❘
          
            
              ∏
              h
            
            ⁢
            
              
            
            ⁢
            
              E
              h
            
          
        
        }
      
    
  




This is because the information of E* has been used out in the simplified DUCG and is no longer related to Hkj and





  
    ∏
    h
  
  ⁢
  
    
  
  ⁢
  
    
      E
      h
    
    .
  




Therefore, to calculate hkjs and hkjr, we need only logically outspread and simplify




  
    
      ∏
      h
    
    ⁢
    
      
    
    ⁢
    
      
        E
        h
      
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        H
        kj
      
      ⁢
      
        
          ∏
          h
        
        ⁢
        
          
        
        ⁢
        
          E
          h
        
      
    
  
  ,




until the logic expression is compound of only the {B,X} type basic events and the F type functional events. The detailed method of the outspread and simplification is explained in example 16.

§11.4. As described above, the outspreaded and simplified expression of




  
    ∏
    h
  
  ⁢
  
    
  
  ⁢
  
    
      E
      h
    
    ⁢
    
      
    
    ⁢
    and
    ⁢
    
      
    
    ⁢
    
      H
      kj
    
    ⁢
    
      
        ∏
        h
      
      ⁢
      
        
      
      ⁢
      
        E
        h
      
    
  




are composed of only the {B,D} type basic events and the F type functional events.

These basic events and functional events are independent of each other. Moreover, the different items (i.e. the logic AND of the basic events and functional events) are exclusive with each other (this is because the direct cause variables in the explicit representation mode are in the simple probability summation relation, while the input logic expressions in the truth value table of a logic gate are also exclusive (see §18.1 for details)). Therefore, the parameters bij, fnk;ij and fnk;D of the basic events and the functional events can directly replace the corresponding events in the logic expression for the numerical calculation.

§11.5. It should be pointed out that the normalization factor rn used in fnk;ij and fhk;D should be calculated according to the actual direct cause variables of Xn in the dynamically outspreaded expression, but not the direct cause variables of Xn in the original DUCG. This is because the direct cause variables of Xn may change during the dynamical outspread. This principle applies in all numerical probability calculations of this invention.
Example 12
This example is about the method described in §12 and is explained below.
§12.1. As shown in §11.3, in the probability calculation, E is equivalent to




  
    
      ∏
      h
    
    ⁢
    
      
    
    ⁢
    
      E
      h
    
  
  ,




in which Eh is the evidence event indicating the state of the {B,X} type variable. In the specific case of process system,





  
    
      
        ∏
        h
      
      ⁢
      
        
      
      ⁢
      
        E
        h
      
    
    =
    
      
        E
        ′
      
      ⁢
      
        E
        ′′
      
    
  
  ,




in which





  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      
    
    ⁢
    
      E
      i
      ′
    
  




that is composed of the evidence Eh indicating the changed state of the {B,X} type variable, e.g. the temperature is high, etc;





  
    E
    ′′
  
  =
  
    
      ∏
      
        i
        ′
      
    
    ⁢
    
      
    
    ⁢
    
      E
      
        i
        ′
      
      ′′
    
  




is composed of the evidence E′h indicating the unchanged state of the {B,X} type variable, e.g. the pressure X′h′, is normal, etc. Logically outspread and simplify E′, until the expression is composed of only the {B,D,F} type events. The method of the outspread and simplification is described in §16. Moreover, according to the method described in §19, the possible solution set S conditioned on E may be further obtained. S is composed of the possible solution events Hkj conditioned on E. §12.2. According to the outspread expressions of E′ and HkjE′, calculate Pr{E′} and Pr{HkjE′}. Then, calculate the state probability hkjs′ of Hkj with incomplete information according to the following equation:





  
    h
    kj
    
      s
      ′
    
  
  =
  
    
      Pr
      ⁢
      
        {
        
          
            H
            kj
          
          ❘
          
            E
            ′
          
        
        }
      
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            
              E
              ′
            
          
          }
        
      
      
        Pr
        ⁢
        
          {
          
            E
            ′
          
          }
        
      
    
  




This is the probability normalized according to the states of Hkj with incomplete information, and is the probability distribution of Hk. It can be used to predict system faults. For example, when hkjs′=0.001≈Pr{Hkj}, the occurrence possibility of the fault Hkj is small. But when hkjs′=0.4>>Pr{Hkj}, the occurrence possibility of the fault Hkj is significantly increased, so that people should take measures to prevent or avoid this fault in time.

§12.3. According to




  
    h
    kj
    
      r
      ′
    
  
  =
  
    
      
        h
        kj
        
          s
          ′
        
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          
        
        ⁢
        
          h
          kj
          
            s
            ′
          
        
      
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            
              E
              ′
            
          
          }
        
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              
                E
                ′
              
            
            }
          
        
      
    
  




the rank probability hkjr′ of Hkj with incomplete information can be calculated. This probability of Hkj is conditioned on the incomplete information and normalized according to the possible solution set S. It is the rank of Hhk in S, and can be used to determine which is more possible when there is more than one possible solution. As the rank probability does not require the calculation to all states of Hkj, the default state of Hk is usually not involved. Moreover, since Pr{E′} is the same for all Hkj, it is possible to calculate only Pr{HkjE′}. In the case of the process system, the calculation to Pr{HkjE′} is usually simpler than Pr{E′}, because Hkj usually involves an initiating event and therefore the logic outspread can be simplified greatly, while E′ usually involves a lot of initiating events. This is because the result of the logic AND operation between different initiating events is null “0” (see §1.4).

The so called “incomplete information” means that the numerical calculation does not include E′′. However, when simplify IYUCG, the partial information of E′′ has been used, but not all. In the case without high requirement to the accuracy of calculation but with high requirement to the calculation speed, the calculation result with incomplete information can usually meet the requirement.
§12.4. The following method is usually applied in this invention to calculate the probability with the complete information.




  
    
      
        
          h
          kj
          s
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              E
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              
                
                  E
                  *
                
                ⁢
                
                  
                    ∏
                    h
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    E
                    h
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              
                
                  ∏
                  h
                
                ⁢
                
                  
                
                ⁢
                
                  E
                  h
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              
                
                  E
                  ′
                
                ⁢
                
                  E
                  ′′
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
                ⁢
                
                  E
                  ′′
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  E
                  ′
                
                ⁢
                
                  E
                  ′′
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
              
              }
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                ❘
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
          
            
              ∑
              j
            
            ⁢
            
              
            
            ⁢
            
              Pr
              ⁢
              
                {
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
                }
              
              ⁢
              Pr
              ⁢
              
                {
                
                  
                    E
                    ′′
                  
                  ❘
                  
                    
                      H
                      kj
                    
                    ⁢
                    
                      E
                      ′
                    
                  
                
                }
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                E
                ′
              
              }
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ❘
                
                  E
                  ′
                
              
              }
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                ❘
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                E
                ′
              
              }
            
            ⁢
            
              
                ∑
                j
              
              ⁢
              
                
              
              ⁢
              
                Pr
                ⁢
                
                  {
                  
                    
                      H
                      kj
                    
                    ⁢
                    
                      E
                      ′
                    
                  
                  }
                
                ⁢
                Pr
                ⁢
                
                  {
                  
                    
                      E
                      ′′
                    
                    ❘
                    
                      
                        H
                        kj
                      
                      ⁢
                      
                        E
                        ′
                      
                    
                  
                  }
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              h
              kj
              
                s
                ′
              
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                ❘
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
          
            
              ∑
              j
            
            ⁢
            
              
            
            ⁢
            
              
                h
                kj
                
                  s
                  ′
                
              
              ⁢
              Pr
              ⁢
              
                {
                
                  
                    E
                    ′′
                  
                  ❘
                  
                    
                      H
                      kj
                    
                    ⁢
                    
                      E
                      ′
                    
                  
                
                }
              
            
          
        
      
    
  




In which, if HkjE′=0, Pr{E′′|HkjE′}≡0. In this case, there must be hkjs=0.

Similar to the above, this invention calculates the rank probability of Hkj with the complete information according to the following calculation method.




  
    h
    kj
    r
  
  =
  
    
      
        h
        kj
        
          r
          ′
        
      
      ⁢
      Pr
      ⁢
      
        {
        
          
            E
            ′′
          
          ❘
          
            
              H
              kj
            
            ⁢
            
              E
              ′
            
          
        
        }
      
    
    
      
        ∑
        
          
            H
            kj
          
          ∈
          S
        
      
      ⁢
      
        
      
      ⁢
      
        
          h
          kj
          
            r
            ′
          
        
        ⁢
        Pr
        ⁢
        
          {
          
            
              E
              ′′
            
            ❘
            
              
                H
                kj
              
              ⁢
              
                E
                ′
              
            
          
          }
        
      
    
  




similarly, if HkjE′′0, Pr{E′′|HkjE′}≡0. In this case, there must be hkjr′=0.

The calculation method of Pr{E′′|HkjE′} is as follows: The first is to logically outspread E′′. The outspread continues until the events included in Hkj or E′ and the {B,D} type events. If the event appearing in the outspread is exclusive with or Hkj or E′, this event is null “0”. If the event appearing in the outspread is included in Hkj or E′, this event is the complete set “1”. Finally, the logic expression of E′′ will include only the {B,D,F} type events. HkjE′ can also be outspreaded as composed of the {B,D,F} type events, and thus be calculated according to the ordinary conditional probability calculation equation through the outspread of E′′HkjE′ and HkjE′ as the {B,D,F} type events. Thus,




  
    Pr
    ⁢
    
      {
      
        
          E
          ′′
        
        ❘
        
          
            H
            kj
          
          ⁢
          
            E
            ′
          
        
      
      }
    
  
  =
  
    
      Pr
      ⁢
      
        {
        
          
            E
            ′′
          
          ⁢
          
            H
            kj
          
          ⁢
          
            E
            ′
          
        
        }
      
    
    
      Pr
      ⁢
      
        {
        
          
            H
            kj
          
          ⁢
          
            E
            ′
          
        
        }
      
    
  




In general, the events in E′′ are the default events. Since DUCG allows the incomplete representation, the causes of the default events may not be represented in DUCG. In this case, the default events can be outspreaded according to the operation of AND after NOT of the non-default events (see example 22 for details).
From the above, the state and rank probabilities of Hkj with complete information can be calculated respectively.
Example 13
This Example is about the method presented in §13, which is explained below.
§13.1. The simplification of the original DUCG is based on the observed evidence E. First, E includes the observed information about the {X,B} type variables in the original DUCG, in which, some of them are the ordinary evidence and some of them are fuzzy evidence (see §10.1 for details). They are represented as




  
    ∏
    h
  
  ⁢
  
    
  
  ⁢
  
    
      E
      h
    
    .
  




More generally, E may include the other non-Eh type evidence E*. By means of the general E, the original DUCG can be simplified greatly, so that the subsequent computation can be reduced greatly.

§13.2. The detailed simplification method is explained below.
(1) By utilizing the relationship table between the basic events and the consequence events described in §12.1, some states of the B type basic event variables can be excluded (i.e. let them be null sets). If all the concerned states of a basic event variable are excluded, this variable is eliminated from the E conditional original DUCG. For example, when E includes Xnk, if the relationship between Bij and Xnk, is then Ψnk;ij=−1, then Bij=0. When E does not include Xnk but includes Xnh (h≠k), while the relationship between Bij and Xnk is Ψnk;ij=1, then Brij=0. If all meaningful states of Bi are null 0, Bi is eliminated, and the functional variable and the conditional functional variable with Bi as the input are also eliminated.
(2) According to E, determine whether or not the condition Cn;i of the conditional functional variable is valid, thus to determine whether the conditional variable becomes the functional variable or is eliminated. When only E cannot determine whether Cn;i is valid or not, keep it until the other information appears so that the validation can be determined. Since Cn;i is given in advance, the evidence determining whether or not Cn;i is valid should be collected consciously according to the need.
(3) Since DUCG does not require the completeness, there may be the case in which the partial states of some variables are the causes of a consequence variable, while the other states are not the causes. Suppose the partial states of V2 are the causes of X5. Then there must be some functional or conditional functional variables between V2 and X5 in the DUCG (for simplicity, except being specified specially, these functional and conditional functional variables are all called the directed arcs briefly). Suppose state V22 is not the cause of X5. When E shows that V2 is in its state V22, the directed arc from V2 to X5 can be eliminated.
(4) Suppose X53 cannot be caused by any state of V2. When E shows that X53 is true, the directed arc from V2 to X5 can be eliminated.
(5) If the consequence variable or the logic gate without input is generated, the consequence variable or the logic gate, as well as the corresponding directed arc starting from them (as causes) should be eliminated (the virtual logic gate is an exception, because once all the input variables of the virtual logic gate are eliminated, this virtual logic gate becomes a default variable).
(6) If there is any isolated part without any connection with the part related to E in the DUCG, this isolated part can be eliminated, because this part is useless for the computation of the new probability distributions of the variables remaining in DUCG conditioned on E.
(7) If E shows that X51 and X61 are true (η=1), while DUCG shows that X51 and X61 are not the causes of any other variable, X5 and X6 are also not the direct or indirect consequence variables of other variables related to E, meanwhile V2 along with its logic connection variables are not connected in any way with any variables related to E, then X5, X6, V2, the directed arcs F5;2 and F6;2 between V2 and X5 and X6, along with the variables connected with V2, can all be eliminated. This is because the eliminated part is not related to the part related to E and becomes an isolated part that is not related to the new probability distributions of the variables in DUCG conditioned on E.
(8) When Xnk is caused by Vij, Xnk can not appear earlier than Vij. Therefore, if E shows that Xnk appears earlier than Vij, which determines that Vij is impossible to be the cause of Xnk, the functional and conditional functional variables between Vi and Xn but not related to the influence of other variables to Xn are eliminated. The reason why without influence of other variables to Xn is required is because the states of other variables may appear earlier than the state of XB. For example, suppose X2 and X5 in DUCG are causes of each other. But E shows that the state of X5 appears earlier than the state of X2, then X2 cannot be the cause of X5. Therefore, the directed arc from X2 to X5 but not related to other causalities is eliminated.
(9) The above procedures can be implemented in any order on demand, and can be repeated at any time, so that the DUCG can be simplified greatly.
§13.3. It should be pointed out that as the time goes on, E may change dynamically. For the E at every time point, the simplification should be based on the original DUCG, so that the result of the simplification is consistent with the situation reflected by the E at that time point.
Example 14
FIGS. 11 and 12 are the illustration about the method described in §14, which is explained elow.
§14.1. The implicit representation mode or the hybrid representation mode includes at least one group direct cause variables in the implicit representation mode. Therefore, the transformation from the single group implicit representation mode to the explicit representation mode is the key to transform any implicit or hybrid representation mode to the explicit representation mode.
The single group implicit representation mode can be transformed to the explicit representation mode by adding a virtual logic gate. As shown in FIG. 11, the direct cause variables in the same group of implicit representation mode are B1 and X2. It can be transformed to the explicit representation mode as shown in FIG. 12. The default event Dn that is implicated in the CPT can be abstracted as an independent variable first (in FIG. 11, n=4). The calculation method of fnk;D is: for every k, seek the minimal pnk, denoted as pnk, i.e.




  
    p
    nk
  
  =
  
    
      min
      j
    
    ⁢
    
      
        {
        
          p
          
            nk
            ;
            j
          
        
        }
      
      .
    
  




If

pnk=0 for all k, Dn does not exist, because the state of Xn is completely related to the state combinations of the input variables, which means that there is no cause variable independent of the original input variables according to the CPT. The abstracted pnk can be viewed as the independent contribution of Dn to Xn, but cannot be treated as ank;D directly, because pnk is not normalized so far. After the normalization, we have





  
    a
    
      nk
      ;
      D
    
  
  =
  
    
      p
      nk
    
    /
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        
          p
          nk
        
        .
      
    
  




Meanwhile, the relationship of Fn;D is





  
    
      r
      
        n
        ;
        D
      
    
    =
    
      
        r
        Xn
      
      ⁢
      
        
          ∑
          k
        
        ⁢
        
          
        
        ⁢
        
          p
          nk
        
      
    
  
  ,




in which rXn is the whole relationship between the input variables and Xn in the implicit representation mode.





  
    ∑
    k
  
  ⁢
  
    
  
  ⁢
  
    p
    nk
  




is the proportion of Dn should have. If it is transformed as all in the single group implicit representation mode, rXn can be any value great than 0.

After the abstraction of Dn, the original CPT should be reconstructed: i.e. eliminate the part form the abstracted Dn and then perform the normalization:




  
    
      
        
          p
          
            nk
            ;
            j
          
        
        =
        
        ⁢
        
          
            (
            
              
                p
                
                  nk
                  ;
                  j
                
              
              -
              
                p
                nk
              
            
            )
          
          /
          
            
              ∑
              k
            
            ⁢
            
              
            
            ⁢
            
              (
              
                
                  p
                  
                    nk
                    ;
                    j
                  
                
                -
                
                  p
                  nk
                
              
              )
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            (
            
              
                p
                
                  nk
                  ;
                  j
                
              
              -
              
                p
                nk
              
            
            )
          
          /
          
            (
            
              1
              -
              
                
                  ∑
                  k
                
                ⁢
                
                  
                
                ⁢
                
                  p
                  nk
                
              
            
            )
          
        
      
    
  




In which, the right pnk;j is the value before the reconstruction and the left is the value after the reconstruction. The sample number dn;j remains same, while the occurrence number qnk;j=pnk;jdnj (pnk;j is the value after the reconstruction).





  
    
      
        If
        ⁢
        
          
        
        ⁢
        1
      
      -
      
        
          ∑
          k
        
        ⁢
        
          
        
        ⁢
        
          p
          nk
        
      
    
    =
    0
  
  ,




then the state combination of the cause variables in the implicit representation mode are not related to the state of Xn and depends only the influence of Dn. In this case, the cause variables in the implicit representation mode are not the cause of Xn, but only Dn is. Meanwhile, ank;D=Pnk and Rn;D=rXn.

After the above reconstruction, this group of cause variables in the implicit representation mode are taken as the input of a virtual logic gate (G5 in FIG. 12), and this logic gate is taken as the cause variable of the consequence variable (X4 in FIG. 12). They are connected through a virtual functional variable (F4;5 in FIG. 12). The states of the virtual logic gate are the state combinations of the cause variables. The virtual functional intensities are the reconstructed conditional probabilities, i.e. ank;ij=pnk;j, in which i indexes the virtual lgic gate (i=5 in example 12), j indexes the states of the virtual logic gate (j=1,2, . . . , 6 in example 12) and also indexes the state combinations of the cause variables, k indexes the state of the consequence variable (k=1,2 in example 12).
The relationship rn;j of Fn;j (F4;5 in example 12) is




  
    
      r
      
        n
        ;
        i
      
    
    =
    
      
        r
        Xn
      
      (
      
        1
        -
        
          
            ∑
            k
          
          ⁢
          
            
          
          ⁢
          
            p
            nk
          
        
      
      )
    
  
  ,




in which i indexes the virtual logic gate (i=5 in example 12). As mentioned earlier, rXn is the whole relationship between the direct cause variables and the consequence variable in the implicit representation mode.

When there is only one input variable of the virtual logic gate, this virtual logic gate can be ignored, i.e. take the input variable of the virtual logic gate directly as the input of the functional variable output from the virtual logic gate. The functional intensities are the new conditional probabilities after the reconstruction directly.
If the implicit representation mode involves more than one group, more than one default variable Dnh will be produced after transforming every group into the explicit representation mode. These default variables should be combined as one default variable Dn. The method is as described in §5.
Example 15
The example about the method described in §15, which is explained below.
§15.1. For every state combination of the direct cause variables, calculate the corresponding probabilities of the consequence variable according to the logic gates, functional intensities and relationships given in the explicit representation mode. The sum of these probabilities may be less than or equal to 1. If less than 1, the gap can be given to the default state, so that the sum is 1. This is the meaning of step (6) of §15. If there is no default state, there should be




  
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        a
        
          nk
          ;
          ij
        
      
    
    =
    1
  
  ,




because usually, in the case without default state, the complete probability data for every state combination between the consequence variable and the direct cause variables should be given, so that





  
    
      ∑
      k
    
    ⁢
    
      
    
    ⁢
    
      Pr
      ⁢
      
        {
        
          X
          nk
        
        }
      
    
  
  =
  1.




This is because after giving the state combination j of the direct cause variables,





  
    
      ∑
      k
    
    ⁢
    
      
    
    ⁢
    
      Pr
      ⁢
      
        {
        
          X
          nk
        
        }
      
    
  
  =
  
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        (
        
          
            (
            
              
                r
                
                  n
                  ;
                  i
                
              
              /
              
                r
                n
              
            
            )
          
          ⁢
          
            
              ∑
              k
            
            ⁢
            
              
            
            ⁢
            
              a
              
                nk
                ;
                ij
              
            
          
        
        )
      
    
    =
    
      
        
          ∑
          i
        
        ⁢
        
          
        
        ⁢
        
          
            r
            
              n
              ;
              i
            
          
          /
          
            r
            n
          
        
      
      =
      1.
    
  




If there is no default state, nor





  
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        a
        
          nk
          ;
          ij
        
      
    
    =
    1
  
  ,




these probabilities must be normalized, i.e. these probabilities are divided by the sum of them. This case is coming from the incompleteness of the representation, which is allowed in the explicit representation mode. But for the implicit representation mode, the conditional probability table must satisfy the normalization. Therefore, in the case of being transformed to the implicit representation mode, the normalization should be done. This is the meaning of step (7) of §15. Finally, the probabilities satisfying the normalization are then the conditional probabilities in the implicit representation mode.

§15.2. The probability contributions to the state probability of the consequence variable from the direct cause variables are in the simple summation relation, i.e. when there is more than one direct cause variable, it is impossible to define that a state of a direct cause variable certainly causes a state of a consequence variable. But when transforming it to the implicit representation mode, this limitation can be removed: The case that the functional intensity in the explicit representation mode ank;ij=1 is viewed as that the state k of the consequence variable n is certainly true, i.e. when the direct cause variable Vi is in its state j, the states except state k of the consequence variable n cannot be true. If this definition is chosen, when being transformed to the standard implicit representation mode, the state of the consequence variable corresponding to this functional intensity can be treated as certainly true (i.e. the conditional probability equals to 1), while the other states cannot be true. This is the content described in step (4). If there are m cases in which the functional intensity equals to 1 indicating that m different states of the consequence variable are certainly true, the m conditional probabilities equal to 1 should be normalized and the result is 1/m, while before the normalization these conditional probabilities equal to 1. This is because the different states of Xn cannot appear simultaneously, and therefore the compromise has to be made. Obviously, if we want to have the explicit representation mode and while choose to understand anj;ij=1 as that the state k of the consequence variable n must be true (this can be viewed as the non-standard explicit representation mode), we can transform this non-standard explicit representation mode as the single group implicit representation mode first, and then transform it back to the explicit representation mode as described in §1.
§15.3. With the method of transforming the explicit representation mode as the implicit representation mode, the correctness of the mutual transformations between the implicit representation mode and the explicit representation mode can be proved, i.e. transform the implicit representation mode to the explicit representation mode first, and then transform it back to the implicit representation mode. The proof is below:
For simplicity, consider only the case of single group implicit representation mode. Given the state combination j of the direct cause variables in the implicit representation mode, according to the calculation based on the transformed explicit representation mode, the probability contributions to Xnk come from two parts: the default variable Dn and the state j of the virtual logic gate Gi in the transformed explicit representation mode, i.e.,




  
    
      
        
          Pr
          ⁢
          
            {
            
              
                X
                nk
              
              ❘
              j
            
            }
          
        
        =
        
        ⁢
        
          
            f
            
              nk
              ;
              D
            
          
          +
          
            f
            
              nk
              ;
              ij
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              (
              
                
                  r
                  
                    n
                    ;
                    D
                  
                
                /
                
                  r
                  n
                
              
              )
            
            ⁢
            
              a
              
                nk
                ;
                D
              
            
          
          +
          
            
              (
              
                
                  r
                  
                    n
                    ;
                    i
                  
                
                /
                
                  r
                  n
                
              
              )
            
            ⁢
            
              a
              
                nk
                ;
                ij
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              (
              
                
                  r
                  Xn
                
                ⁢
                
                  
                    ∑
                    k
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      p
                      nk
                    
                    /
                    
                      r
                      n
                    
                  
                
              
              )
            
            ⁢
            
              
                p
                nk
              
              
                
                  ∑
                  k
                
                ⁢
                
                  
                
                ⁢
                
                  p
                  nk
                
              
            
          
          +
        
      
    
  
  
    
      
        
        ⁢
        
          
            (
            
              
                
                  r
                  Xn
                
                (
                
                  1
                  -
                  
                    
                      ∑
                      k
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      p
                      nk
                    
                  
                
                )
              
              /
              
                r
                n
              
            
            )
          
          ⁢
          
            
              
                p
                
                  nk
                  ;
                  j
                
              
              -
              
                p
                nk
              
            
            
              1
              -
              
                
                  ∑
                  k
                
                ⁢
                
                  
                
                ⁢
                
                  p
                  nk
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            (
            
              
                r
                Xn
              
              /
              
                r
                n
              
            
            )
          
          ⁢
          
            (
            
              
                p
                nk
              
              +
              
                p
                
                  nk
                  ;
                  j
                
              
              -
              
                p
                nk
              
            
            )
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              r
              Xn
            
            ⁢
            
              p
              
                nk
                ;
                j
              
            
          
          
            
              r
              
                n
                ;
                D
              
            
            +
            
              r
              
                n
                ;
                i
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              r
              Xn
            
            ⁢
            
              p
              
                nk
                ;
                j
              
            
          
          
            
              
                r
                Xn
              
              ⁢
              
                
                  ∑
                  k
                
                ⁢
                
                  
                
                ⁢
                
                  p
                  nk
                
              
            
            +
            
              
                r
                Xn
              
              (
              
                1
                -
                
                  
                    ∑
                    k
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    p
                    nk
                  
                
              
              )
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          p
          
            nk
            ;
            j
          
        
      
    
  




It is seen that the result is the conditional probability in the implicit representation mode.

According to the above transformation method, the DUCG in any representation mode can be transformed as all in the explicit representation mode (EDUCG), or transformed as all in the single group implicit representation mode (IDUCG). In which, when part of the DUCG is in the more than one group implicit representation mode, transform it as the explicit representation mode first according to the method described in §14, and then transform it as the single group implicit representation mode according to the method described in §15.
FIG. 33 is a specific example of transforming the DUCG in FIG. 32 as EDUCG.
Example 16
The illustration about the method presented in §16, which is explained below.
§16.1. As shown in §11.3, based on the simplified DUCG,




  
    Pr
    ⁢
    
      {
      
        
          H
          kj
        
        ❘
        E
      
      }
    
  
  =
  
    
      Pr
      ⁢
      
        {
        
          
            H
            kj
          
          ❘
          
            
              E
              *
            
            ⁢
            
              
                ∏
                h
              
              ⁢
              
                
              
              ⁢
              
                E
                h
              
            
          
        
        }
      
    
    =
    
      Pr
      ⁢
      
        {
        
          
            H
            kj
          
          ⁢
          
            
              ❘
              ∏
            
            h
          
          ⁢
          
            
          
          ⁢
          
            E
            h
          
        
        }
      
    
  




For the process system, there is (see §12)





  
    
      
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              E
            
            }
          
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              
                
                  ∏
                  h
                
                ⁢
                
                  
                
                ⁢
                
                  E
                  h
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              
                
                  E
                  ′
                
                ⁢
                
                  E
                  ′′
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              
                
                  ∏
                  i
                
                ⁢
                
                  
                
                ⁢
                
                  
                    E
                    i
                    ′
                  
                  ⁢
                  
                    
                      ∏
                      
                        i
                        ′
                      
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      E
                      
                        i
                        ′
                      
                    
                  
                
              
            
            }
          
        
      
    
  




In which,





  
    
      E
      ′
    
    =
    
      
        
          ∏
          i
        
        ⁢
        
          
        
        ⁢
        
          
            E
            i
            ′
          
          ⁢
          
            
          
          ⁢
          and
          ⁢
          
            
          
          ⁢
          
            E
            ′′
          
        
      
      =
      
        
          ∏
          
            i
            ′
          
        
        ⁢
        
          
        
        ⁢
        
          E
          
            i
            ′
          
          ′′
        
      
    
  
  ;




E′i and E′′i, are all the Eh type evidence events indicating the states of the {B,X} type variables.

If Eh is the ordinary evidence, Eh is a {B,X} type event. The B type events do not need the outspread. Only the X type events need the outspread. If Eh is the fuzzy evidence of Xn, as shown in §10, Eh=FE;nXn, where FE;n is the virtual functional variable from Xn to Eh. It is seen that Eh still come down to the outspread of Xn.
Moreover, Hkj is composed of the {B,X,D} type events, for which the {B,D} type events do not need the outspread. The only events needing the outspread are still the X type events.
According to the explicit representation mode,




  
    
      X
      nk
    
    =
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            nk
            ;
            i
          
        
        ⁢
        
          V
          i
        
      
    
  
  ,




or more generally





  
    
      X
      n
    
    =
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            n
            ;
            i
          
        
        ⁢
        
          V
          i
        
      
    
  
  ,




where V∈{X,B,G,D}.

It is obvious that when V=X, the outspread involves the further X type variables. During the outspread process, once there is logic cycle, the repeated variable on the cause or upstream side in the causality chain must be in the earlier moment, and its probability distribution must be known according to the earlier moment calculation. If the time is not involved, i.e. the system is viewed as in the same time slice, which is called the static case, the following principle is applied to break the logic cycle: the consequence cannot be the cause of itself in the same time slice. In other words, in the static outspread process, once a variable in the same causality chain is encountered repeatedly, this repeated variable is viewed as null. When all the input variables of a logic gate are eliminated, this logic gate and its output functional variables are all viewed as null.
The static case is the most common case for the intelligent system, because the dynamical case is usually simplified as the static case or is approximated by the static cases at a sequential time points. Then, the static case with logic cycle becomes normal. In the static case, the influence or the function of the cause variable to the consequence variable is immediate. In other words, the functional time is 0. This is not conflict with that the functional intensity is changeable dynamically with time (i.e. the probability value of Fnk;ij can dynamically change), because the different functional intensities at different time are all propagated to the consequence variable immediately. As the function is immediately done, when there is logic cycle, there must be the case that the same consequence is simultaneously the cause of itself. This is obviously inconsistent. Therefore, in the outspread process, once the repeated variable appears, this repeated variable (including the related functional or conditional functional variables) must be eliminated, i.e. be viewed as null. For example, suppose X1 is the cause of X2, the functional variable is F1;3; X2 is the cause of X3, the functional variable is F3;2; and X3 is the cause of X1, the functional variable is F3;1. Then, X3=F3;2X2{F2;1X1{F1;3X3}}. It can also be expressed directly with the variable, instead of with the function: X3=F3;2F2;1F1;3X3. The above is outspreaded with variables. The outspread can also be with events. In this example, suppose every variable has two states each. Then

X31=F31;21X21{F21;11X11{F11;31X31+F11;32}+F21;12X12{F12,31X31+F12;32X32}}+F31;22X22{F22;11X11{F11;31X31+F11;32X32}+F22;12X12{F11;31X31F11;32X32}}

Or be expressed with the event form instead of the function form:

X31=F31;21(F21;11(F11;31X31+F11;32X32)+F21;12(F12;31X31+F12;32X32))+F31;22(F22;11(F11;31X31+F11;32X32)+F22;12(F11;31X31+F11;32X32))

The outspread for X32 is similar.

In this case, no matter it is in the variable expression or in the event expression, the X3, X31 and X32 on the right side of the equator are all making the logic cycle to the X3 or X31 on the left side of the equator, and should all be viewed as null. But in this example, if eliminate X3 or X31 and X32, the right side of the equator becomes null. This means that the problem has no solution. Actually, besides that X1, X2 and X3 are cause and consequence of each other, there should be other variables, otherwise the problem does not have any reasonable physical meaning. Suppose B4 is the other cause of X2 and B4 has two states. Then

X3=F3;2(F2;1F1;3X3+F2;4B4)

Or

X31=F31;21(F21;11(F11;31X31+F11;32X32)+F21;12(F12;31X31+F12;32)+F21;41B41+F21;42B42)+F31;22(F22;11(F11;31X31+F11;32X32)+F22;12(F11;31X31+F11;32X32)+F22;41B41+F22;42B42)

Eliminating the repeated variables,




  
    
      X
      3
    
    =
    
      
        F
        
          3
          ;
          2
        
      
      ⁢
      
        F
        
          2
          ;
          4
        
      
      ⁢
      
        B
        4
      
    
  




  Or




  
    
      
        
          
            X
            31
          
          =
          
          ⁢
          
            
              
                F
                
                  31
                  ;
                  21
                
              
              ⁡
              
                (
                
                  
                    
                      F
                      
                        21
                        ;
                        41
                      
                    
                    ⁢
                    
                      B
                      41
                    
                  
                  +
                  
                    
                      F
                      
                        21
                        ;
                        42
                      
                    
                    ⁢
                    
                      B
                      42
                    
                  
                
                )
              
            
            +
            
              
                F
                
                  31
                  ;
                  22
                
              
              ⁡
              
                (
                
                  
                    
                      F
                      
                        22
                        ;
                        41
                      
                    
                    ⁢
                    
                      B
                      41
                    
                  
                  +
                  
                    
                      F
                      
                        22
                        ;
                        42
                      
                    
                    ⁢
                    
                      B
                      42
                    
                  
                
                )
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            
              
                F
                
                  31
                  ;
                  21
                
              
              ⁢
              
                F
                
                  21
                  ;
                  41
                
              
              ⁢
              
                B
                41
              
            
            +
            
              
                F
                
                  31
                  ;
                  21
                
              
              ⁢
              
                F
                
                  21
                  ;
                  42
                
              
              ⁢
              
                B
                42
              
            
            +
            
              
                F
                
                  31
                  ;
                  22
                
              
              ⁢
              
                F
                
                  22
                  ;
                  41
                
              
              ⁢
              
                B
                41
              
            
            +
          
        
      
    
    
      
        
          
          ⁢
          
            
              F
              
                31
                ;
                22
              
            
            ⁢
            
              F
              
                22
                ;
                42
              
            
            ⁢
            
              B
              42
            
          
        
      
    
  



This is the outspreaded logic expression of X3 or X31 in the form of “sum-of-products” composed of the {B,D,F} type variables or events without logic cycles.

The illustration involving breaking the logic cycles in the outspread expression is also shown in example 23.
§16.2. In the static cases, when there is the situation that the different input variables of a same logic gate are multiplied (the logic AND), there should be the fusion of the different input variables. This is because only in this way, can the logic gates multiplied be true simultaneously.
§16.3. Since there is the situation that the direct cause variables are reduced, in the numerical probability calculation, whether or not the direct cause variables in the exclusive logic outspread expression of the same consequence variable are reduced must be investigated. If yes, the calculation for the normalization factor rn included in the calculation for the functional intensity of the functional event fnk;ij or fnk;D should only be the sum of the really involved {B,X,D,G} type direct cause variable relationships, so as to ensure the normalization of the state probabilities of the consequence variable.
§16.4. When the causality expression between the consequence variable and its cause variables does not satisfy the normalization (i.e. incomplete), the default state is outspreaded according to the operation of AND after NOT of the non-default states; when the causes of the states of the non-default states are not given, these non-default states are treated as null; when the condition Cn;i of the conditional functional variable Fn;i appears invalid during the outspread process, Fn;i=0; otherwise Fn;i is treated as that Cn;i is valid.
The “AND after NOT” of the non-default states implies that all the non-default states are not true, where “not true” means “false” and “all” means “AND”. Suppose X1 has tree states, in which state 1 is the default state. Then X11= X12 X13. Suppose X12=F12;21V21+F12;32V32, V□{X,B,G,D}, then X12=( F12;21+ V21)( F12;32+ V32). Take V21 as the example. Suppose V2 has three states. If V21 is the default state, then V21=V22+V23; if V21 is the non-default state, outspread it according to the cause variables shown in the EDUCG directly.
Example 17
FIGS. 4, 14 and 15 are about the illustration for the method described in §17, which is explained below.
§17.1. The method to eliminate the input variables of the virtual logic gate is the same as to eliminate the direct cause variables in the single group implicit representation mode, i.e. eliminate the direct cause variables in the implicit representation mode corresponding to the virtual logic gate, and then reconstruct the CPT (see example 7 for details), and then transform the implicit representation mode after the elimination of input variables to the explicit representation mode according to the method described in §14, so as to produce new functional variables of the new logic gate and the new default event and its functional variable.
§17.2. In the situation of non-virtual logic gate, for the convenience of reconstruct the truth value table of the logic gate in the case of eliminating the input variables of the logic gate, the original logic gate or the original truth value table of the logic gate should be the most simplified. If the truth value table given by the domain engineers is not the most simplified, it should be transformed as the most simplified. The so called most simplified includes the row most simplified and the item most simplified. The so called row most simplified means that every row in the truth value table has specific different meaning; the so called item most simplified means that any variable in any item of every row cannot be eliminated, otherwise this item cannot result in the truth of the corresponding logic gate state.
To explain the row most simplified, suppose the logic expression of a row is V11V21. Moreover, suppose V31+V32=1 (the complete set). If divide V11V21 as two rows V11V21V31 and V11V21V32, which will add an extra logic gate state, whether or not these two rows or the two logic gate states have different meanings should be judged by the domain engineers. But it can also be judged and be most simplified just according to the expression. That is, consider the different groups of the functional variables and conditional functional variables as the output from the different logic gate states corresponding to the different rows. If these groups are exactly same, combine these different rows and the corresponding different logic gate states together (see §1.5 and §2). In this example, suppose V11V21V31 and V11V21V32 are two logic expressions of different rows. If the functional variables and the conditional functional variables as the output of the two logic gate states corresponding to the two rows are exactly the same, these two rows can be combined together as V11V21V31+V11V21V32, so as to satisfy the row most simplified.
To explain the item most simplified, consider the above example. Given V31+V32=1, eliminate V31 from the item V11V21V31 and eliminate V32 from the item V11V21V32 in the row V11V21V31+V11V21V32, there is no influence to the truth of the corresponding logic gate state, because V11V21V31+V11V21V32=V11V21. Therefore, once V11V21 is true. In general, to obtain the item most simplified, the twice complement operation can be applied to the logic expression: That is, when there is only one item in the input row, the engineers can judge whether or not the item is simplified and simplify it if not; when there are multiple items in one input raw of the logic expression, perform the twice complement operation, i.e. the logic expression in an input raw=the logic expression of this input raw. For the above example, according to the law of the complement operation and the rules of the absorption, exclusion and complete set operations,




  
    
      
        
          
            
              V
              11
            
            ⁢
            
              V
              21
            
            ⁢
            
              V
              31
            
          
          +
          
            
              V
              11
            
            ⁢
            
              V
              21
            
            ⁢
            
              V
              32
            
          
        
        =
        
        ⁢
        
          
            
              
                V
                11
              
              ⁢
              
                V
                21
              
              ⁢
              
                V
                31
              
            
            +
            
              
                V
                11
              
              ⁢
              
                V
                21
              
              ⁢
              
                V
                32
              
            
          
          
            _
            _
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              (
              
                
                  
                    V
                    _
                  
                  11
                
                +
                
                  
                    V
                    _
                  
                  21
                
                +
                
                  
                    V
                    _
                  
                  31
                
              
              )
            
            ⁢
            
              (
              
                
                  
                    V
                    _
                  
                  11
                
                +
                
                  
                    V
                    _
                  
                  21
                
                +
                
                  
                    V
                    _
                  
                  32
                
              
              )
            
          
          _
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                V
                _
              
              11
            
            +
            
              
                
                  V
                  _
                
                11
              
              ⁢
              
                
                  V
                  _
                
                21
              
            
            +
            
              
                
                  V
                  _
                
                11
              
              ⁢
              
                
                  V
                  _
                
                32
              
            
            +
            
              
                
                  V
                  _
                
                21
              
              ⁢
              
                
                  V
                  _
                
                11
              
            
            +
          
          _
        
      
    
  
  
    
      
        
        ⁢
        
          
            
              
                V
                _
              
              21
            
            +
            
              
                
                  V
                  _
                
                21
              
              ⁢
              
                
                  V
                  _
                
                32
              
            
            +
            
              
                
                  V
                  _
                
                31
              
              ⁢
              
                
                  V
                  _
                
                11
              
            
            +
            
              
                
                  V
                  _
                
                31
              
              ⁢
              
                
                  V
                  _
                
                21
              
            
            +
          
          _
        
      
    
  
  
    
      
        
        ⁢
        
          
            
              
                V
                _
              
              31
            
            ⁢
            
              
                V
                _
              
              32
            
          
          _
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                
                  V
                  _
                
                11
              
              +
              
                
                  V
                  _
                
                21
              
              +
              
                
                  
                    V
                    _
                  
                  31
                
                ⁢
                
                  V
                  _
                
              
            
            _
          
          32
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            V
            11
          
          ⁢
          
            
              V
              21
            
            ⁡
            
              (
              
                
                  V
                  31
                
                +
                
                  V
                  32
                
              
              )
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            V
            11
          
          ⁢
          
            V
            21
          
        
      
    
  




So that the most simplified expression is obtained. If V31+V32≠1, there is





  
    
      
        
          
            
              V
              11
            
            ⁢
            
              V
              21
            
            ⁢
            
              V
              31
            
          
          +
          
            
              V
              11
            
            ⁢
            
              V
              21
            
            ⁢
            
              V
              32
            
          
        
        =
        
        ⁢
        
          
            
              
                V
                11
              
              ⁢
              
                V
                21
              
              ⁢
              
                V
                31
              
            
            +
            
              
                V
                11
              
              ⁢
              
                V
                21
              
              ⁢
              
                V
                32
              
            
          
          
            _
            _
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            V
            11
          
          ⁢
          
            
              V
              21
            
            ⁡
            
              (
              
                
                  V
                  31
                
                +
                
                  V
                  32
                
              
              )
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              V
              11
            
            ⁢
            
              V
              21
            
            ⁢
            
              V
              31
            
          
          +
          
            
              V
              11
            
            ⁢
            
              V
              21
            
            ⁢
            
              V
              32
            
          
        
      
    
  




i.e. the original expression has already been the most simplified.

§17.3. If it appears the case in the online application that the input variable of a non-virtual logic gate no longer exists, let all state or related event of this input variable in the expressions composed of the states of the input variables in the most simplified truth value table become null 0, and then calculate these expressions in different rows. If the calculation result of the expression of a row becomes null, this raw is eliminated; if the calculation result is not null, this row remains and the calculated result becomes the new expression; if an input row becomes null after the calculation, the corresponding state of the logic gate is eliminated.
For the example shown in FIG. 4, according to the twice complement operation, it can be known that the expressions in rows no. 1 and no. 2 are the most simplified expressions. The proof is shown below:
For row no. 1:




  
    
      
        
          
            
              V
              11
            
            ⁢
            
              V
              21
            
          
          +
          
            
              V
              11
            
            ⁢
            
              V
              31
            
          
          +
          
            
              V
              21
            
            ⁢
            
              V
              31
            
          
        
        =
        
        ⁢
        
          
            
              
                V
                11
              
              ⁢
              
                V
                21
              
            
            +
            
              
                V
                11
              
              ⁢
              
                V
                31
              
            
            +
            
              
                V
                21
              
              ⁢
              
                V
                31
              
            
          
          
            _
            _
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              (
              
                
                  
                    V
                    _
                  
                  11
                
                +
                
                  
                    V
                    _
                  
                  21
                
              
              )
            
            ⁢
            
              (
              
                
                  
                    V
                    _
                  
                  11
                
                +
                
                  
                    V
                    _
                  
                  31
                
              
              )
            
            ⁢
            
              (
              
                
                  
                    V
                    _
                  
                  21
                
                +
                
                  
                    V
                    _
                  
                  31
                
              
              )
            
          
          _
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                
                  V
                  _
                
                11
              
              ⁢
              
                
                  V
                  _
                
                21
              
            
            +
            
              
                
                  V
                  _
                
                11
              
              ⁢
              
                
                  V
                  _
                
                31
              
            
            +
            
              
                
                  V
                  _
                
                21
              
              ⁢
              
                
                  V
                  _
                
                31
              
            
          
          _
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            (
            
              
                V
                11
              
              +
              
                V
                21
              
            
            )
          
          ⁢
          
            (
            
              
                V
                11
              
              +
              
                V
                31
              
            
            )
          
          ⁢
          
            (
            
              
                V
                21
              
              +
              
                V
                31
              
            
            )
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              V
              11
            
            ⁢
            
              V
              21
            
          
          +
          
            
              V
              11
            
            ⁢
            
              V
              31
            
          
          +
          
            
              V
              21
            
            ⁢
            
              V
              31
            
          
        
      
    
  




For row no. 2:




  
    
      
        
          
            
              V
              12
            
            ⁢
            
              V
              21
            
            ⁢
            
              V
              32
            
          
          +
          
            
              V
              12
            
            ⁢
            
              V
              22
            
            ⁢
            
              V
              31
            
          
        
        =
        
          
            
              
                
                  V
                  12
                
                ⁢
                
                  V
                  21
                
                ⁢
                
                  V
                  32
                
              
              +
              
                
                  V
                  12
                
                ⁢
                
                  V
                  22
                
                ⁢
                
                  V
                  31
                
              
            
            _
          
          _
        
      
    
  
  
    
      
        =
        
          
            
              (
              
                
                  
                    V
                    —
                  
                  12
                
                +
                
                  
                    V
                    —
                  
                  21
                
                +
                
                  
                    V
                    —
                  
                  32
                
              
              )
            
            +
            
              (
              
                
                  
                    V
                    —
                  
                  12
                
                +
                
                  
                    V
                    —
                  
                  22
                
                +
                
                  
                    V
                    —
                  
                  31
                
              
              )
            
          
          _
        
      
    
  
  
    
      
        =
        
          
            
              V
              12
            
            +
            
              
                V
                21
              
              ⁢
              
                V
                22
              
            
            +
            
              
                V
                21
              
              ⁢
              
                V
                31
              
            
            +
            
              
                V
                22
              
              ⁢
              
                V
                32
              
            
            +
            
              
                V
                31
              
              ⁢
              
                V
                32
              
            
          
          _
        
      
    
  
  
    
      
        =
        
          
            
              V
              12
            
            ⁡
            
              (
              
                
                  V
                  21
                
                +
                
                  V
                  22
                
              
              )
            
          
          ⁢
          
            (
            
              
                V
                21
              
              +
              
                V
                31
              
            
            )
          
          ⁢
          
            (
            
              
                V
                22
              
              +
              
                V
                32
              
            
            )
          
          ⁢
          
            (
            
              
                V
                31
              
              +
              
                V
                32
              
            
            )
          
        
      
    
  
  
    
      
        =
        
          
            
              V
              12
            
            ⁢
            
              V
              21
            
            ⁢
            
              V
              32
            
          
          +
          
            
              V
              12
            
            ⁢
            
              V
              22
            
            ⁢
            
              V
              31
            
          
        
      
    
  




Suppose that V1 is no longer the input variable of G4. Then V1j=0, and the logic expression of the state combination in row no. 1 becomes V11V21+V11V31+V21V31=V21V31. The result is not null and should remain; the logic expression of the state combination in row no. 2 becomes V12V21V32+V12V22V31=0. The result is null and this row should be eliminated. Since the or row no. 2 is eliminated, the corresponding logic gate state G42 is also eliminated. Finally, FIG. 4 is reconstructed as FIG. 14.

The above operation can be repeated so as to deal with the cases in which more than one input variable is eliminated. If all the input variables of a non-virtual logic gate are eliminated, this non-virtual logic gate along with its output variables is also eliminated.
The truth value table of FIG. 14 that includes the remnant state of the non-virtual logic gate is shown in FIG. 15.
Example 18
FIGS. 4, 5, 14, 16, 17 and 18 are about the illustration for the method described in §18, which is explained below.
§18.1. After determining the input variables and also the truth value table of a logic gate, if these are the non-exclusive items in a same expression, make them as the exclusive items, so that the probability calculation for the expression can be done directly.
For the example of the truth value table shown in FIG. 4, the corresponding exclusive result is shown in FIG. 16. If consider the remnant state, the truth value table shown in FIG. 5 is made exclusive as shown in FIG. 17. Similarly, the truth value table including the remnant state of the logic gate shown in FIG. 5 is made exclusive as shown in FIG. 18.
After the item exclusion, the outspread can be done according to the rules of the algebra calculation, and the probability values on the two sides of “+” can simply sum up. After considering the remnant state, the input and output states of the logic gate cover the whole set and the normalization of the logic gate is ensured. The adding of the remnant state is mainly for satisfying the state completeness of the system. In applications, it is usually not necessary.
§18.2. It should be pointed out that the notations Gij in both before and after eliminating V1 from the truth value table (FIGS. 4 and 14) or the exclusive truth value table (FIGS. 16 and 18) have different meanings, although they look the same. If they are ANDed together, the logic fusion of the input variables must be done, i.e. the input variables are fused as the input variables only related to the AND operation. In such a way, the two truth value tables are fused as one truth value table. This truth value table is based on the most simplified truth value tables of the logic gates.
It must be emphasized that if the input variables are reduced, the new truth value table must be reconstructed based on the most simplified truth value table. Only after this can the exclusion of the expressions, adding the remnant expression and its corresponding logic gate state, as well as the transformation to the complete combined logic gate be done.
Example 19
The detailed explanation about the method described in §19 is below.
§19.1. As mentioned before, in the case of process system, E′ is composed of the evidence indicating the state change of variables. E′ can be logically outspreaded and simplified, until the expression is composed of only the {B,D,F} type variables. The outspread method is described in §16-18. It should be emphasized that during the outspread process, once more than one initiating event is ANDed together, the result is null. Therefore, in every item (the group of events occurring simultaneously) of the final expression, there is one and only one initiating event, with one or none or more non-initiating events. In addition, there are some functional events or conditional functional events. By eliminating all the functional events and conditional functional events (i.e. let them be the complete set 1), there may be some repeated items. The logical absorption operation is necessary so that the items with the same real effect are combined and minimized. Thus, in the final expression, the set of one initiating event and some non-initiating events and default events in one item is a possible solution Hkj, where k is the index of the variable combination, e.g. Hk=Bn, Hk=Xn, Hk=BnBm or Hk=BnDnBm, etc; j is the state index of the variable state combination, e.g. Hkj=Bnh, HkjXnh; Hkj=BnhBmg, Hkj=BnhDnBmg, etc. In which, only one of Bnh and Bmg can be the initiating event, while the other one is the non-initiating event. The so called necessary absorption means the logic absorption of the items with same physical meaning. If the physical meanings of Bnh, BnhBmg and BnhDnBmg are different, Bnh, BnhBmg and BnhBmg are the exclusive possible solutions. If the physical meanings of them are same, the later two are absorbed by Bnh. After the above operation, the items in the expression of E′ are the members Hkj of the possible solution set SH, i.e. SH={Hkj}. Then, the selection of Hkj will be limited in only these members. In other words, the cause of the system state change is limited only in these possible solutions.
Example 20
The illustration about the method described in §20, which is explained below.
§20.1. In the dynamical cases, when the new information is added ceaselessly during the diagnosis process, the known information E(ti-1) at the earlier time point ti-1 and the known information E(ti) at the later time point ti may be different. They need be treated as E based on the original DUCG respectively and repeat the computation described in §12-19 to obtain, based on the original DUCG, the E conditional original DUCGs. But, according to §12-19, the earlier computed possible solution set S may be different from the later computed possible solution set Si. Only the intersection of them can satisfy both. Therefore, the intersection of all the possible solution sets before tn (including tn) is the dynamical possible solution set S(tn). Correspondingly, the possible solution set Si at different time point is called the static possible solution set. Recursively computed according to the sequential time points, the intersection becomes smaller and smaller, while the diagnosis becomes more and more accurate, so as to perform the dynamical logic operation and then the dynamical probability calculation.
Specifically, suppose the time points are sequentially t1, t2, . . . , ti, . . . , tn. The known information or evidence corresponding to these time points are sequentially E(ti), E(t2), . . . , E(ti), . . . , E(tn). The static possible solution sets are sequentially S1, S2, . . . , Si, . . . , Sn. The dynamical possible solution sets are sequentially S(t1), S(t2), . . . , S(ti), . . . , S(tn). Then




  
    S
    ⁡
    
      (
      
        t
        n
      
      )
    
  
  =
  
    
      ∏
      
        i
        =
        1
      
      n
    
    ⁢
    
      
    
    ⁢
    
      S
      i
    
  




Usually, when the information is rich enough, the dynamical possible solution set S(tn) may have only one member, i.e. the diagnostic result is unique and the rank probability calculation is not necessary.
Actually, S(tn) excludes some other possible solutions in the EDUCG. For the EDUCG shown in FIG. 30 of example 22, the possible fault variables are B79, B80, B81, B100, B101, B102 and B117. But S(t2) shows that only B102,2 is the possible solution. Therefore, B79, B80, B81, B100 and B101 are all eliminated from the EDUCG in FIG. 30. Furthermore, the so resulted isolated variables are also eliminated, which leads to the simplified EDUCG shown in FIG. 31.
§20.2. When it is required to calculate the dynamical state probability and the rank probability of Hkj included in the dynamical possible solution set SH(t), the first is to calculate the static state probability conditioned on the incomplete information Hkjs′(ti) and the static state probability conditioned on the complete information hkjs(ti), the static rank probability conditioned on the incomplete information hkjr′(ti) and the static rank probability conditioned on the complete information hkjr(ti), of Hkj at different time points according to the method presented in §11. Moreover, the unconditional probability Pr{Hkj} should also be calculated, which is briefly denoted as hkj(t0) where t0 represents the time without receiving any online information or evidence, i.e. hkj(t0)≡Pr{Hkj}. Note that hkj(t0) does not have the difference between incomplete information and complete information as well as the state probability and rank probability. When Hkj is the basic event Big, hkj(t0)=Pr{Hkj}=Pr{Big}=big.
Hkj can be either the member of S(tn) or the member of Sn. If calculate only the static probabilities of Hkj, the probabilities of Hkj calculated based on the members in S(tn) have include the partial dynamical information in fact, although they are still called the static values. The probabilities of Hkj calculated based on the members in Sn are according to only the information of that time point, and can be called the complete static values so as to be distinguished. To include all the dynamical information, the calculation should be done according to the method described in §20.3.
§20.3. In the dynamical case of more than one time point, the calculation method including all the information is as follows:




  
    
      
        
          
            h
            kj
          
          ⁡
          
            (
            t
            )
          
        
        ≡
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                
                  H
                  kj
                
                ❘
                
                  
                    ∏
                    
                      i
                      =
                      1
                    
                    n
                  
                  ⁢
                  
                    E
                    ⁡
                    
                      (
                      
                        t
                        i
                      
                      )
                    
                  
                
              
              ,
              
                
                  t
                  n
                
                ≤
                t
                ≤
                
                  t
                  
                    n
                    +
                    1
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              
                
                  ∏
                  
                    i
                    =
                    1
                  
                  n
                
                ⁢
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ⁢
                
                  
                    ∏
                    
                      i
                      =
                      1
                    
                    n
                  
                  ⁢
                  
                    E
                    ⁡
                    
                      (
                      
                        t
                        i
                      
                      )
                    
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  ∏
                  
                    i
                    =
                    1
                  
                  n
                
                ⁢
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                H
                kj
              
              }
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  
                    ∏
                    
                      i
                      =
                      1
                    
                    n
                  
                  ⁢
                  
                    E
                    ⁡
                    
                      (
                      
                        t
                        i
                      
                      )
                    
                  
                
                ❘
                
                  H
                  kj
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  ∏
                  
                    i
                    =
                    1
                  
                  n
                
                ⁢
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
              
              }
            
          
        
      
    
  




If




  
    ∏
    
      i
      =
      1
    
    n
  
  ⁢
  
    E
    ⁡
    
      (
      
        t
        i
      
      )
    
  




is indeed caused by Hkj, then given Hkj, the static evidence E(ti) are exclusive with each other, and the above equation becomes





  
    
      h
      kj
    
    ⁡
    
      (
      t
      )
    
  
  =
  
    
      Pr
      ⁢
      
        {
        
          H
          kj
        
        }
      
      ⁢
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                E
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              ❘
              
                H
                kj
              
            
            }
          
        
      
    
    
      Pr
      ⁢
      
        {
        
          
            ∏
            
              i
              =
              1
            
            n
          
          ⁢
          
            E
            ⁡
            
              (
              
                t
                i
              
              )
            
          
        
        }
      
    
  




In which, if hkj(t0)≡Pr{Hkj}=0, there is





  
    
      Pr
      ⁢
      
        {
        
          
            H
            kj
          
          ❘
          
            
              ∏
              
                i
                =
                1
              
              n
            
            ⁢
            
              E
              ⁡
              
                (
                
                  t
                  i
                
                )
              
            
          
        
        }
      
    
    =
    0
  
  ;




otherwise,





  
    
      
        
          Pr
          ⁢
          
            {
            
              
                E
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              ❘
              
                H
                kj
              
            
            }
          
        
        =
        
          
            Pr
            ⁢
            
              {
              
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
                ⁢
                
                  H
                  kj
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                H
                kj
              
              }
            
          
        
      
    
  
  
    
      
        =
        
          
            Pr
            ⁢
            
              {
              
                E
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              }
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ❘
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                H
                kj
              
              }
            
          
        
      
    
  
  
    
      
        =
        
          
            Pr
            ⁢
            
              {
              
                E
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              }
            
            ⁢
            
              
                h
                kj
              
              ⁡
              
                (
                
                  t
                  i
                
                )
              
            
          
          
            
              h
              kj
            
            ⁡
            
              (
              
                t
                0
              
              )
            
          
        
      
    
  




In which, hkj(ti) is the static value of Hkj at time ti. The ignorance of the superscript of hkj(ti) indicates that this value can be conditioned on either the incomplete information or the complete information, and can be either the state value or the rank value depending on the content of the conditioned evidence E (E′ or E′E′′). Therefore,





  
    
      
        
          
            h
            kj
          
          ⁡
          
            (
            t
            )
          
        
        =
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ❘
              
                
                  ∏
                  
                    i
                    =
                    1
                  
                  n
                
                ⁢
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
          
            Pr
            ⁢
            
              {
              
                H
                kj
              
              }
            
            ⁢
            
              
            
            ⁢
            
              
                ∏
                
                  i
                  =
                  1
                
                n
              
              ⁢
              
                Pr
                ⁢
                
                  {
                  
                    
                      E
                      ⁡
                      
                        (
                        
                          t
                          i
                        
                        )
                      
                    
                    ❘
                    
                      H
                      kj
                    
                  
                  }
                
              
            
          
          
            Pr
            ⁢
            
              {
              
                
                  ∏
                  
                    i
                    =
                    1
                  
                  n
                
                ⁢
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
          
            
              
                h
                kj
              
              ⁡
              
                (
                
                  t
                  0
                
                )
              
            
            ⁢
            
              
                ∏
                
                  i
                  =
                  1
                
                n
              
              ⁢
              
                
                  
                    h
                    kj
                  
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
                ⁢
                
                  
                    ∏
                    
                      i
                      =
                      1
                    
                    n
                  
                  ⁢
                  
                    Pr
                    ⁢
                    
                      {
                      
                        E
                        ⁡
                        
                          (
                          
                            t
                            i
                          
                          )
                        
                      
                      }
                    
                  
                
              
            
          
          
            Pr
            ⁢
            
              {
              
                
                  ∏
                  
                    i
                    =
                    1
                  
                  n
                
                ⁢
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      i
                    
                    )
                  
                
              
              }
            
            ⁢
            
              
                (
                
                  
                    h
                    kj
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                )
              
              n
            
          
        
      
    
  
  
    
      
        =
        
          
            
              
                ∏
                
                  i
                  =
                  1
                
                n
              
              ⁢
              
                Pr
                ⁢
                
                  {
                  
                    E
                    ⁡
                    
                      (
                      
                        t
                        i
                      
                      )
                    
                  
                  }
                
              
            
            
              Pr
              ⁢
              
                {
                
                  
                    ∏
                    
                      i
                      =
                      1
                    
                    n
                  
                  ⁢
                  
                    E
                    ⁡
                    
                      (
                      
                        t
                        i
                      
                      )
                    
                  
                
                }
              
            
          
          ⁢
          
            
              
                ∏
                
                  i
                  =
                  1
                
                n
              
              ⁢
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
            
            
              
                (
                
                  
                    h
                    kj
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                )
              
              
                n
                -
                1
              
            
          
        
      
    
  
  
    
      
        =
        
          α
          ⁢
          
            
              ∏
              
                i
                =
                1
              
              n
            
            ⁢
            
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              /
              
                
                  
                    h
                    kj
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                
                  n
                  -
                  1
                
              
            
          
        
      
    
  




In which,





  α
  =
  
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        Pr
        ⁢
        
          {
          
            E
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          }
        
      
    
    
      Pr
      ⁢
      
        {
        
          
            ∏
            
              i
              =
              1
            
            n
          
          ⁢
          
            E
            ⁡
            
              (
              
                t
                i
              
              )
            
          
        
        }
      
    
  




is a normalization constant independent of Hkj.

If when hkj(t0)=0, define hkj(ti)/(hkj(t0))n-1≡0, then no matter hkj(t0)=0 or not, we always have




  
    
      h
      kj
    
    ⁡
    
      (
      t
      )
    
  
  =
  
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        
          
            h
            kj
          
          ⁡
          
            (
            
              t
              i
            
            )
          
        
        /
        
          
            (
            
              
                h
                kj
              
              ⁡
              
                (
                
                  t
                  0
                
                )
              
            
            )
          
          
            n
            -
            1
          
        
      
    
    
      
        ∑
        j
      
      ⁢
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
            
              h
              kj
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
    
  




In the real application, the normalization can be made according to the requirement of the incomplete information, the complete information, the state probability and the rank probability:

(1) The dynamical incomplete information state probability:




  
    
      h
      kj
      
        s
        ′
      
    
    ⁡
    
      (
      t
      )
    
  
  =
  
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        
          
            h
            kj
            
              s
              ′
            
          
          ⁡
          
            (
            
              t
              i
            
            )
          
        
        /
        
          
            (
            
              
                h
                kj
              
              ⁡
              
                (
                
                  t
                  0
                
                )
              
            
            )
          
          
            n
            -
            1
          
        
      
    
    
      
        ∑
        j
      
      ⁢
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
            
              h
              kj
              
                s
                ′
              
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
    
  




In which, when hkj(t0)=0, hkjs′(ti)/(hkj(t0))n-1≡0.

(2) The dynamical incomplete information rank probability:




  
    
      h
      kj
      
        r
        ′
      
    
    ⁡
    
      (
      t
      )
    
  
  =
  
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        
          
            h
            kj
            
              r
              ′
            
          
          ⁡
          
            (
            
              t
              i
            
            )
          
        
        /
        
          
            (
            
              
                h
                kj
              
              ⁡
              
                (
                
                  t
                  0
                
                )
              
            
            )
          
          
            n
            -
            1
          
        
      
    
    
      
        ∑
        
          
            H
            kj
          
          ∈
          
            
              S
              H
            
            ⁡
            
              (
              t
              )
            
          
        
      
      ⁢
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
            
              h
              kj
              
                r
                ′
              
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
    
  




In which, when hkj(t0)=0, hkjr′(ti)/(hkj(t0))n-1≡0.

(3) The dynamical complete information state probability:




  
    
      h
      kj
      s
    
    ⁡
    
      (
      t
      )
    
  
  =
  
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        
          
            h
            kj
            s
          
          ⁡
          
            (
            
              t
              i
            
            )
          
        
        /
        
          
            (
            
              
                h
                kj
              
              ⁡
              
                (
                
                  t
                  0
                
                )
              
            
            )
          
          
            n
            -
            1
          
        
      
    
    
      
        ∑
        j
      
      ⁢
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
            
              h
              kj
              s
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
    
  




In which, when hkj(t0)=0, hkjs(ti)/(hkj(t0)n-1≡0.

(4) The dynamical complete information rank probability:




  
    
      h
      kj
      r
    
    ⁡
    
      (
      t
      )
    
  
  =
  
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        
          
            h
            kj
            r
          
          ⁡
          
            (
            
              t
              i
            
            )
          
        
        /
        
          
            (
            
              
                h
                kj
              
              ⁡
              
                (
                
                  t
                  0
                
                )
              
            
            )
          
          
            n
            -
            1
          
        
      
    
    
      
        ∑
        
          
            H
            kj
          
          ∈
          
            
              S
              H
            
            ⁡
            
              (
              t
              )
            
          
        
      
      ⁢
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
            
              h
              kj
              r
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
    
  




In which, when hkj(t0)=0, hkjr(ti)/(hkj(t0))n-1≡0.

The above equations are all come down to the static calculations to different time points. Obviously, there are other types of the state and rank probability calculations based on the situation what and how the evidence is conditioned on, e.g. the evidence at different time point is partially incomplete and partially complete, etc. They cannot be all listed here.
Example 21
The detailed explanation about the calculation method described in §21 is below.
§21.1. IDUCG is similar to BN. The only difference is that there may be the virtual consequence variable Eh with only one state in IDUCG. To transform IDUCG as BN and use the method of BN to do the calculation, a new event Ēh can be introduced in IDUCG, where Pr{Eh}=1 and Pr{Ēh}≡0. Thus IDUCG becomes BN. The construction method of the CPT between Eh and its only direct cause variable Vh is: {circle around (1)} When



  
    
      
        
          
            max
            j
          
          ⁢
          
            {
            
              f
              
                Eh
                ;
                j
              
            
            }
          
        
        ≤
        1
      
      ,
      
        
          Pr
          ⁢
          
            {
            
              
                E
                h
              
              ❘
              
                V
                hj
              
            
            }
          
        
        =
        
          f
          
            Eh
            ;
            j
          
        
      
    
    ⁢
    
      
    
  




  
    and
    ⁢
    
      
    
  




  
    
      Pr
      ⁢
      
        {
        
          
            
              E
              _
            
            h
          
          ❘
          
            V
            hj
          
        
        }
      
    
    =
    
      1
      -
      
        Pr
        ⁢
        
          
            {
            
              
                E
                h
              
              ❘
              
                V
                hj
              
            
            }
          
          .
        
      
    
  



{circle around (2)} When




  
    
      
        
          max
          j
        
        ⁢
        
          {
          
            f
            
              Eh
              ;
              j
            
          
          }
        
      
      >
      1
    
    ,
    
      
        Pr
        ⁢
        
          {
          
            
              E
              h
            
            ❘
            
              V
              Vj
            
          
          }
        
      
      =
      
        
          f
          
            Eb
            ;
            j
          
        
        /
        
          
            max
            j
          
          ⁢
          
            {
            
              f
              
                Eh
                ;
                j
              
            
            }
          
        
      
    
  




  and




  
    
      Pr
      ⁢
      
        {
        
          
            
              E
              _
            
            h
          
          ❘
          
            V
            hj
          
        
        }
      
    
    =
    
      1
      -
      
        Pr
        ⁢
        
          
            {
            
              
                E
                h
              
              ❘
              
                V
                hj
              
            
            }
          
          .
        
      
    
  



This is because the parameter fEh;j of FEh;j between Eh and Vh calculated according to the method described in §10 may be greater than 1 due to that the meaning of fEh;j is the probability density. The conditional probability in BN does not include the probability density. Therefore, the probability density must be transformed as the probability distribution (i.e. CPT). Then the method of BN can be applied. This method is actually to compress the values of fEh;j when they are greater than 1, while keeping their proportion, so as to satisfy the normalization. Since the values of fEh;j have only the relative meaning, this transformation is reasonable.

Example 22
The comprehensive illustration with the fault diagnosis of the secondary loop system of a nuclear power plant.
FIG. 20 shows the secondary loop system of a nuclear power plant imitated by the imitator located at Tsinghua University for training the operators of a real nuclear power plant. The high temperature and pressure water heated by the nuclear reactor goes through three steam generators SGA, SGB and SGC to heat the water in the secondary loop system, so that it becomes the steam gathering at the steam header and going through the TV and the GV control valves to push the turbine to generate the electricity. The tired steam goes through the hot well to be cooled, the condensate pumps, the low temperature heaters, the high temperature heaters, the drain pumps, and the main feed water pumps, then feeds back to the steam generators, so as to form the circulation. The operation parameters and the states of the components are collected through the online data collection system and are shown in the control room. The task of the operators is to check periodically these data and judge the states. When the abnormal state or the alarm appears, diagnose the cause and take correct measures to remove or control the fault. But it is usually very hard for the operators to diagnose the cause of the abnormal system state and take the correct measures in time. This may cause big loss.
Currently, using the artificial intelligence technology to diagnose the fault online for large complex systems has become one of the important research issues in the international academic community. But as the systems like nuclear power plants and space equipment are so expensive and their failure statistic data are so rare, it is hard to construct the fault diagnosis model by data learning and mining. The main available data are the experience and belief of the domain engineers. Plus the significant large scale, complexity and dynamics of such systems, many techniques including the neural network and the BN are hard to be applied.
The DUCG intelligent system presented in this invention can better fulfill this task. The features of DUCG is at that it does not rely much on the completeness and accuracy of the statistic data, can utilize the belief of the domain engineers to describe flexibly the fault causal relations among the various parameters and states of components (without the requirement of the completeness and accuracy), and can perform the real time and online monitoring to the secondary loop according to the online collected information related to the loop state. Once the abnormal signals appear (the evidence E′ is not null), the diagnosis process is started immediately to find the fault cause, so as to inform people to take correct control measures. Usually, the fault can be monitored even it is in the early stage and can be oriented rapidly, and furthermore the development state of the system can be predicted. Thus, the fault diagnosis and prediction of the system can be realized, so that people can take measures in advance to avoid the development of the fault and avoid the big economic loss.
The brief fault propagation DUCG of the secondary loop of a nuclear power plant is shown in FIG. 21. Since the faults of the nuclear power plant is very rare, there is no sufficient failure data available. The realistic way is to mainly rely on the belief of the domain engineers. Hence, this DUCG will use the explicit representation mode (see §1) only, i.e. EDUCG, so as to ease the belief representation of the domain engineers. In other words, FIG. 21 and all other figures of this example are actually EDUCGs. Therefore, the methods of §11 (4)-(6) and §12 can be applied. There is no need to transform the EDUCG.
The variables of the DUCG in FIG. 21 are shown in table 1 (the probability data are ignored). In which, the first state is the normal state, i.e. the default state. The other states are abnormal states. The abnormal state at time t1 is written in bold italic, and at time t2 is in bold only. The shading denotes the abnormally high. The underline denotes the abnormally low. The other state of the consequence variable is normal. All the states of the B type variables and the logic gates are unknown.
Moreover, the nuclear power plant is a typical process system. The B type variables in table 1 are all the initiating event variables, i.e. all the abnormal states of the B type variables are the initiating events. According to §1, the AND of more than one initiating event is null. In fact, if considering the details fully, this example should contain the non-initiating events, such as the depressurizing valve cannot restore after it opens for depressurizing, etc. But such cases are relatively rare and are hard to be dealt with. To save space, this case is not considered in this example.
As the purpose of this DUCG is the fault diagnosis, the DUCG in FIG. 21 needs only represent the causal relations of the fault influence. In other words, the normal state of any variable has no influence to other variables. Therefore, this DUCG is the DUCG without completeness. The property that DUCG does not require the completeness reduces the requirement to the domain engineers and the data, so that users can focus on the problems in concern only, without rigorous requirement to the completeness of data and model. This reduces the difficulty to construct DUCG greatly.
In more details, according to the steps described in §1, the various cause variables Vi, (B and X types) and the consequence variables Xn (X type only) in concern related to the fault diagnosis of the secondary loop system of this nuclear power plant are determined. According to the real situation of this example, the important X and B type variables are determined as shown in table 1, in which the continuous variables have been discretized. For the example of variable X23, the temperature in the feed water header is a continuous variable, which is discretized as three fuzzy discrete states: X23;1 (between 107-134° C.), X23;2 (>125 C°) and X23;3 (<115 C°). Obviously, there are two fuzzy areas: (107-115 C°) and (125-134 C°). When the real temperature value e23≦107 C°, m23;1=m23,2=0 and m23,3=1; When e23=110 C°, m23;1=m23,3=0.5 and m23,2=0; When 115 C°≦e23≦125 C, m23,1=1 and m23;2=m23,3=0; When e23≦130 C°, m23;1=m23,2=0.5 and m23,3=0; When e23≧134 C°, m23;1=m23,3=0 and m23,2=1. The membership mij of other temperature can be obtained from FIG. 2 easily. The parameters bij of the abnormal states of Bi (i≠1) are replaced with λij, and are given by the domain engineers according to the mean time between failures (MTBF) included in the component specifications along with other factors. For simplicity, these parameters are ignored in table 1. But the value ranges can be described as below: As the components of the nuclear power plant are all high reliable, the failure rates are generally in the range of 10−4-10−6/year, which is much less than 1. Therefore, the simultaneously occurrence of more than one initiating event can be treated as impossible event. Thus, we establish the causality representation framework about the cause variables and the consequence variables of the secondary loop of the nuclear power plant for the fault diagnosis.
Moreover, according to the above causal relations among variables, for every X type variable, determine its direct cause variables so as to form a module. For example, the module for X16 indicating the vacuum of the hot well and then representing the pressure of the hot well, has the direct cause variables X49 representing the state of the depressurizing control valve PCV400A, X50 representing the state of the depressurizing control valve PCV400AB, B119 representing the state of the vacuum pump VPA, and B120 representing the state of the vacuum pump VPB. As the influences of VPA and VPB to the pressure of the hot well are common and the influence extent depends on the state combination of both pumps, B119 and B120 function together to X16 through logic gate G18 (see §3). Based on the knowledge about this vacuum system, the truth value table of G18 is constructed as shown below:














No.
Expression of the state combination
G18,1
G18,2
G18,3







1
B119,1B120,1
1
0
0


2
B119,2B120,1 + B119,1B120,2
0
1
0









This is because when the two pumps operate normally (G18,1), the vacuum cannot be low; when one of the two pumps stops (G18,2), the pressure of the hot well may be high (i.e. the vacuum may be low); when two pumps stop simultaneously (G18,3), the pressure of the hot well may be very high (the vacuum may be very low), where G18,3 is the logic AND of two initiating events B119,2 B120,2 and should be null, which means that the state G18,3 is an impossible state (null). But in practice, VPA and VPB may stop simultaneously due to some common causes (e.g. the common power failure, etc), therefore, G18,3 is possible. In this case, B119 and B120 should be replaced with X119 and X120 that have their upstream cause variables including independent cause variables and the common cause variables. For simplicity, only the independent case will be considered in this example, and therefore G18,3 is a null event.

Finally, connect the cause variables with the consequence variable X16 through F16;49, F16;50 and F16,18, and determine their causal relations, the module of X16 with its direct cause variables being represented in the explicit representation mode is constructed. In which, X16,1, X49,1, X50,1 and G18,1 are the default states and are not in concern, because people concern only the fault causalities. Therefore, they are ignored in the representation of the functional variables Fn;i. In other words, people need only represent F16j;49k, F16j;50k and F16j;18k (j=2,3 and k=2,3). Moreover, considering that the influence of G18 to X16 is relatively large, let r16;18=1; As the influence of X49 and X50 to X16 is relatively small, let r16;49=r16;50=0.5. Then, r16=r16;49+r16;50+r16;18=2. The original functional intensity ank;ji of Fnk;ij is given by the domain engineers according to the statistic data or the beliefs. For example, a16,2;18,2=0.8; a16,3;18.2=0.1; a16,2;18,3=0.3; a16,3;18,3=0.7; etc. In which, a16,2;18,2+a16,3;18,2≦0.9<1. This is because whether or not the stop of any one of the two vacuum pumps results in the low vacuum is uncertain. Thus, f16,2;18,2=(1⁄2)0.8=0.4; f16,3;18,2=(1⁄2)0.1=0.05; f16,3;18,2=(1⁄2)0.3=0.15; f16,3;18,3=(1⁄2)0.7=0.35; and so on.
After all the X type variables in table 1 are represented in the explicit representation mode similar to that for X16, put all these modules together according to the method described in §9 (2), the DUCG shown in FIG. 21 is constructed. In this example, the default state of the X type variable corresponds to a default variable D that is the direct cause of the default state. However, since people do not concern the default states, all the D type variables are ignored in FIG. 21.
Moreover, F12;6 is a conditional functional variable (see §2). Its condition is C12;6= X9,2 B102,2 (the feed water flow rate in line C does not increase and there is no U type pipe rupture in the steam generator SGC), otherwise the steam flow rate X6 in line C is not the cause of the abnormal water level X12 in SGC.
For simplicity, the specific values of An;i and rn;i in table 1 and FIG. 21 are ignored.







TABLE 1





The variable description for FIG. 21



















X1
pressure of steam line A
normal
high
low


X2
pressure of steam line B
normal
high
low


X3
pressure of steam line C
normal
high
low


X4
flow rate of steam line A
normal
high
low


X5
flow rate of steam line B
normal
high
low


X6
flow rate of steam line C
normal
 
low


X7
flow rate of feed water line A
normal
high
low


X8
flow rate of feed water line B
normal
high
low


X9
flow rate of feed water line C
normal
high
low


X10
water level of steam generator A
normal
high
low


X11
water level of steam generator B
normal
high
low


X12
water level of steam generator C
normal
high
low


X13
pressure of main steam header
normal
high
low


X14
First stage pressure of turbine
normal
high
low


X15
water level of hot well
normal
high
low


X16
pressure of hot well
normal
high
very high


X17
level of water storage tank
normal
high
low


G18
state combination of vacuum pumps VPA and
normal
one pump
all fail



VPB

fail


X19
pressure of suction header of low pressure heater
normal
high
low


X20
pressure of main feed water pump suction header
normal
high
low


X21
discharging header pressure of main feed water
normal
high
low



pumps


X22
pressure of feed water header
normal
high
low


X23
temperature of feed water header
normal
high
low


X27
pressure of low pressure heater drain pump HDPA
normal
high
low


X28
pressure of low pressure heater drain pump HDPB
normal
high
low


X30
load of turbine
normal
high
low


X31
speed of turbine
normal
high
low


X32
pressure of lubricating oil
normal
high
low


X35
state signal of bypass valve V21
open
close


X36
state signal of bypass valve V22
open
close


X37
state signal of bypass valve V31
open
close


X38
state signal of output valve V1
open
close


X39
state signal of output valve V2
open
close


X40
state signal of motor valve V26
open
close


X41
state signal of motor valve V27
open
close


X42
state signal of motor valve V28
open
close


X43
state signal of valve V3
open
close


X44
state signal of valve V4
open
close


X45
state signal of main feed water pump MFPA
normal
fail high
fail low


X46
state signal of main feed water pump MFPB
normal
fail high
fail low


X47
state signal of condensation pump CPA
normal
fail high
fail low


X48
state signal of condensation pump CPB
normal
fail high
fail low


X49
state signal of depressurizing valve PCV400A
close
open


X50
state signal of depressurizing valve PCV400B
close
open


X51
state signal of depressurizing valve PCV400C
close
open


X52
state signal of feed water control valve FCV478
normal
high
low


X53
state signal of feed water control valve FCV488
normal
high
low


X54
state signal of feed water control valve FCV498
normal
high
low


X55
state signal of depressurizing valve PCV308A
close
open


X56
state signal of depressurizing valve PCV308B
close
open


X57
state signal of depressurizing valve PCV308C
close
open


X58
state signal of isolation valve V11 in steam line A
open
close


X59
state signal of isolation valve V12 in steam line B
open
close


X60
state signal of isolation valve V13 in steam line C
open
close


X61
state signal of turbine main control valve TV
normal
high
low


X62
state signal of turbine main control valve GV
normal
high
low


B63
condensation pump CPA
normal
fail stop


B64
condensation pump CPB
normal
fail stop


B65
control valve FCV478
normal
fail high
fail low


B66
control valve FCV488
normal
fail high
fail low


B67
control valve FCV498
normal
fail high
fail low


B68
feed water pipe line A
normal
leak


B69
feed water pipe line B
normal
leak


B70
feed water pipe line C
normal
leak


B71
turbine
normal
fail


B72
lubricating oil system
normal
fail


B73
high pressure heat drain pump HDPA
normal
fail stop


B74
low pressure heat drain pump HDPB
normal
fail stop


B75
load of turbine
normal
increase
decrease


B77
main feed water pump MFPA
normal
fail stop


B78
main feed water pump MFPB
normal
fail stop


B79
main steam isolation valve MSIV-A
normal
fail close


B80
main steam isolation valve MSIV-B
normal
fail close


B81
main steam isolation valve MSIV-C
normal
fail close


B82
valve PCV308A
close
fail open


B83
valve PCV308B
close
fail open


B84
valve PCV308C
close
fail open


B85
valve PCV400A
close
fail open


B86
valve PCV400B
close
fail open


B87
valve PCV400C
close
fail open


B88
leakage between hot well and condensation pumps
no
yes


B89
leakage in low pressure heaters
no
yes


B90
leakage of water storage tank
no
yes


B91
leakage of main steam header
no
yes


B92
leakage of steam line A
no
yes


B93
leakage of steam line B
no
yes


B94
leakage of steam line C
no
yes


B95
turbine valve TV
normal
fail high
fail low


B96
turbine valve GV
normal
fail high
fail low


B100
tubes in SGA
normal
leak


B101
tubes in SGB
normal
leak


B102
tubes in SGC
normal
leak


B103
valve V1
open
fail close


B104
valve V2
open
fail close


B105
valve V3
open
fail close


B106
valve V4
open
fail close


B107
valve V11
open
fail close


B108
valve V12
open
fail close


B109
valve V13
open
fail close


B110
valve V21
close
fail open


B111
valve V22
close
fail close


B112
valve V26
open
fail close


B113
valve V27
open
fail close


B114
valve V28
open
fail close


B115
valve V31
open
fail close


B116
valve LCV1900
normal
fail open
fail close


B117
valve LCV1910
normal
fail open
fail close


B119
vacuum pump VPA
normal
fail stop


B120
vacuum pump VPB
normal
fail stop









In the operation state of the nuclear power plant, this intelligent system monitors the states or the consequence variables online. These consequence variables are the system parameters of the nuclear power plant shown in the control room collected online and in time. Once the abnormal variable state appears (E′ is not null), the fault diagnosis program of the DUCG is started. Suppose t1 is the time that the first abnormal variable state appears. According to the experiment of the imitator of this nuclear power plant, the variable states are shown in table 1, in which the position of the italic letter indicates that only X6 is in the abnormal state at this moment (E′(t1)=X6,2: the steam flow rate in line C is high, which is marked by the italic and shading). The other consequence variables are in the normal states (these states compose E′′(t1)). The states of the basic variables are unknown. In this example, all the evidence is certain instead of fuzzy. The case of fuzzy evidence will be discussed later.
Since the abnormal signal is monitored, the fault diagnosis program of DUCG is started as follows:
1. From the DUCG in FIG. 21, eliminate all the consequence variables with normal state and their output functional variables influencing the other variables, so as to obtain the DUCG shown in FIG. 22. This is because the DUCG represents only the influences of the abnormal states to other variables, which is based on the requirement of modeling, and therefore the normal states have no influence to other variables.
2. Eliminate the isolated part without any connection with the abnormal variables in FIG. 22, so as to obtain the DUCG shown in FIG. 23. This is because the isolated part without any connection with the abnormal variables does not include any information related to the abnormal variables and their causes, and therefore is not related to the fault diagnosis.
3. Eliminate the part in FIG. 23 which is connected directly with only the normal consequence variable (X15 in this example), so as to obtain the DUCG shown in FIG. 24-1. This is because this part, blocked by the normal variables, does not have any information about the abnormal variables and their causes, and is therefore not related to the fault diagnosis.
4. The DUCG in FIG. 24-1 includes only one basic variable B102 (the U type pipe rupture in the steam generator SGC). In addition, its normal state does not influence the other variables. Therefore, only its failure state (state 2) is the possible state. Then the static possible solution set in this moment is SH(t1)={B102,2}. Since there is only one possible solution in SH(t3), we h102,2r′(t1)=h102,2r(t1)=1 without any calculation.
5. Since B102,2 has been found certainly, the condition C12;6 of F12;6 in FIG. 24-1 is not valid, and therefore F12;6 is eliminated so as to obtain FIG. 25-1.
6. To show the method, the static state probability of B102,2 with incomplete information is calculated as follows:




  
    
      
        
          
            h
            
              102
              ,
              2
            
            
              s
              ′
            
          
          ⁡
          
            (
            
              t
              1
            
            )
          
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ❘
                
                  
                    E
                    ′
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
              
              =
              
                
                  X
                  
                    6
                    ,
                    2
                  
                
                =
                
                  
                    F
                    
                      6
                      ,
                      2
                      ,
                      102
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ⁢
                
                  F
                  
                    6
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  B
                  
                    102
                    ,
                    2
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  F
                  
                    6
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  B
                  
                    102
                    ,
                    2
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  F
                  
                    6
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  B
                  
                    102
                    ,
                    2
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  F
                  
                    6
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  B
                  
                    102
                    ,
                    2
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        1
      
    
  




In which, B102,2B102,2=B102,2;

Similarly, the static state probability of B102,1 with incomplete information can be calculated as follows:




  
    
      
        
          
            h
            
              102
              ,
              1
            
            
              s
              ′
            
          
          ⁡
          
            (
            
              t
              1
            
            )
          
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                
                  B
                  
                    102
                    ,
                    1
                  
                
                ❘
                
                  
                    E
                    ′
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
              
              =
              
                
                  X
                  
                    6
                    ,
                    2
                  
                
                =
                
                  
                    F
                    
                      6
                      ,
                      2
                      ,
                      102
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  B
                  
                    102
                    ,
                    1
                  
                
                ⁢
                
                  F
                  
                    6
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  B
                  
                    102
                    ,
                    2
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  F
                  
                    6
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  B
                  
                    102
                    ,
                    2
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          0
          
            Pr
            ⁢
            
              {
              
                
                  F
                  
                    6
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  B
                  
                    102
                    ,
                    2
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        0
      
    
  




In which, B102,1B102,2=0.

Furthermore, by noting h102,1s′(t1)=0, the static state probability with complete information of B102,2 is:




  
    
      
        
          
            h
            
              102
              ,
              2
            
            s
          
          ⁡
          
            (
            
              t
              1
            
            )
          
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                B
                
                  102
                  ,
                  2
                
              
              ❘
              
                E
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                h
                
                  102
                  ,
                  2
                
                
                  s
                  ′
                
              
              ⁡
              
                (
                
                  t
                  1
                
                )
              
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  
                    E
                    ′′
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
                ❘
                
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                  ⁢
                  
                    X
                    
                      6
                      ,
                      2
                    
                  
                
              
              }
            
          
          
            
              ∑
              
                j
                =
                1
              
              2
            
            ⁢
            
              
                
                  h
                  
                    102
                    ,
                    j
                  
                  
                    s
                    ′
                  
                
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
              ⁢
              Pr
              ⁢
              
                {
                
                  
                    
                      E
                      ′′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                  ❘
                  
                    
                      B
                      
                        102
                        ,
                        j
                      
                    
                    ⁢
                    
                      X
                      
                        6
                        ,
                        2
                      
                    
                  
                
                }
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  
                    E
                    ′′
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
                ❘
                
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                  ⁢
                  
                    X
                    
                      6
                      ,
                      2
                    
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  
                    E
                    ′′
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
                ❘
                
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                  ⁢
                  
                    X
                    
                      6
                      ,
                      2
                    
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        1
      
    
  




In which, E′′(t1)=X3,1X12,1X15,1. In the same way, the static state probability with complete information of B102,1 is:





  
    
      
        
          
            h
            
              102
              ,
              1
            
            s
          
          ⁡
          
            (
            
              t
              1
            
            )
          
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                B
                
                  102
                  ,
                  1
                
              
              ❘
              
                E
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                h
                
                  102
                  ,
                  1
                
                
                  s
                  ′
                
              
              ⁡
              
                (
                
                  t
                  1
                
                )
              
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  
                    E
                    ′′
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
                ❘
                
                  
                    B
                    
                      102
                      ,
                      1
                    
                  
                  ⁢
                  
                    X
                    
                      6
                      ,
                      2
                    
                  
                
              
              }
            
          
          
            
              ∑
              
                j
                =
                1
              
              2
            
            ⁢
            
              
                
                  h
                  
                    102
                    ,
                    j
                  
                  
                    s
                    ′
                  
                
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
              ⁢
              Pr
              ⁢
              
                {
                
                  
                    
                      E
                      ′′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                  ❘
                  
                    
                      B
                      
                        102
                        ,
                        j
                      
                    
                    ⁢
                    
                      X
                      
                        6
                        ,
                        2
                      
                    
                  
                
                }
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          0
          
            Pr
            ⁢
            
              {
              
                
                  
                    E
                    ′′
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
                ❘
                
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                  ⁢
                  
                    X
                    
                      6
                      ,
                      2
                    
                  
                
              
              }
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        0
      
    
  




The reason why the result is obtained without calculating Pr{E′′(t1)|102;jX6,2} is because h102,1s′(t1)=0. But as an illustration of the method, suppose the condition C12;6 of F12;6 in FIG. 24-1 is valid. According to FIG. 24-1,




  
    
      
        
          Pr
          ⁢
          
            {
            
              
                
                  E
                  ′′
                
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
              ❘
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ⁢
                
                  X
                  
                    6
                    ,
                    2
                  
                
              
            
            }
          
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                
                  X
                  
                    3
                    ,
                    1
                  
                
                ⁢
                
                  X
                  
                    12
                    ,
                    1
                  
                
                ⁢
                
                  X
                  
                    15
                    ,
                    1
                  
                
              
              ❘
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ⁢
                
                  X
                  
                    6
                    ,
                    2
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                
                  
                    X
                    _
                  
                  
                    3
                    ,
                    2
                  
                
                ⁢
                
                  
                    X
                    _
                  
                  
                    3
                    ,
                    3
                  
                
                ⁢
                
                  
                    X
                    _
                  
                  
                    12
                    ,
                    2
                  
                
                ⁢
                
                  
                    X
                    _
                  
                  
                    12
                    ,
                    3
                  
                
                ⁢
                
                  
                    X
                    _
                  
                  
                    15
                    ,
                    2
                  
                
                ⁢
                
                  
                    X
                    _
                  
                  
                    15
                    ,
                    3
                  
                
              
              ❘
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ⁢
                
                  X
                  
                    6
                    ,
                    2
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                
                  
                    F
                    
                      3
                      ,
                      2
                      ,
                      102
                      ,
                      2
                    
                  
                  _
                
                ⁢
                
                  0
                  _
                
                ⁢
                
                  
                    (
                    
                      
                        
                          F
                          
                            12
                            ,
                            2
                            ,
                            102
                            ,
                            2
                          
                        
                        ⁢
                        
                          B
                          
                            102
                            ,
                            2
                          
                        
                      
                      +
                      
                        
                          F
                          
                            12
                            ,
                            2
                            ,
                            6
                            ,
                            2
                          
                        
                        ⁢
                        
                          X
                          
                            6
                            ,
                            2
                          
                        
                      
                    
                    )
                  
                  _
                
                ⁢
                
                  0
                  _
                
                ⁢
                
                  
                    
                      F
                      
                        15
                        ,
                        2
                        ,
                        102
                        ,
                        2
                      
                    
                    ⁢
                    
                      B
                      
                        102
                        ,
                        2
                      
                    
                  
                  _
                
                ⁢
                
                  0
                  _
                
              
              ❘
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ⁢
                
                  X
                  
                    6
                    ,
                    2
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                
                  (
                  
                    
                      
                        F
                        _
                      
                      
                        3
                        ,
                        2
                        ,
                        102
                        ,
                        2
                      
                    
                    +
                    
                      
                        B
                        _
                      
                      
                        102
                        ,
                        2
                      
                    
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      
                        F
                        _
                      
                      
                        12
                        ,
                        2
                        ,
                        102
                        ,
                        2
                      
                    
                    +
                    
                      
                        B
                        _
                      
                      
                        102
                        ,
                        2
                      
                    
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      
                        F
                        _
                      
                      
                        12
                        ,
                        2
                        ,
                        6
                        ,
                        2
                      
                    
                    +
                    
                      
                        X
                        _
                      
                      
                        6
                        ,
                        2
                      
                    
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      
                        F
                        _
                      
                      
                        15
                        ,
                        2
                        ,
                        102
                        ,
                        2
                      
                    
                    +
                    
                      
                        B
                        _
                      
                      
                        102
                        ,
                        2
                      
                    
                  
                  )
                
              
              ❘
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ⁢
                
                  X
                  
                    6
                    ,
                    2
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                
                  
                    F
                    _
                  
                  
                    3
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  
                    F
                    _
                  
                  
                    12
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
                ⁢
                
                  
                    F
                    _
                  
                  
                    12
                    ,
                    2
                    ,
                    6
                    ,
                    2
                  
                
                ⁢
                
                  
                    F
                    _
                  
                  
                    15
                    ,
                    2
                    ,
                    102
                    ,
                    2
                  
                
              
              ❘
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ⁢
                
                  X
                  
                    6
                    ,
                    2
                  
                
              
            
            }
          
        
      
    
  




Based on FIG. 24-1, we have

B102,2X6,2=B102,2F6,2,102,2B102,2=F6,2,102,2B102,2 

It is independent of F3,2,102,2 F12,2,102,2 F12,2,6,2 F15,2,102,2. Then

Pr{E′′(t1)|B102,2X6,2}=Pr{ F3,2,102,2 F12,2,102,2 F12,2,6,2 F15,2,102,2}=(1−f2,3,102,2)(1−f12,2,102,2)(1−f12,2,6,2)(1−f15,2,102,2)

Therefore, even if h102,1s′(t1)≠0, h102,2s(t1) can also be calculated according to the method illustrated above, in which, conditioned on B102,2X6,2, both B102,2 and X6,2 are impossible and should be null 0, while conditioned on B102,1X6,2, B102,2 is certainly true and is therefore the complete set 1.

Moreover, since B102,1X6,2=0, Pr{E′′(t1)|B102,1X6,2}=0.
Of course, the calculation in this example should be based on FIG. 25-1, because when B102,2 is found as the possible solution, C12;6 is determined as invalid. The reason why based on FIG. 24-1 is only for using more complex example to illustrate the method.
So far, B102,2 has been determined as the only fault. Its static state and rank probabilities with both incomplete and complete information at time t1 are all equal to 1. In fact, B102,2 is indeed the fault given in a training course at the nuclear power plant imitator for retraining the operators working in a nuclear power plant. In this experiment, it takes seven and half minutes for these operators to find this fault with the help of the teachers, while by DUCG, this fault can be certainly found at the first moment (t1) when the abnormal signal just appears.
Moreover, the DUCG in FIG. 25-1 predicts that X3, X12 and X15 will possibly be abnormal. By analyzing the data of Fnk;102,2(n∈{3,12,15}, k∈{1,2,3}), which are not shown in the table but the qualitative analysis can be done, the possible abnormal states of the three consequence variables would be X3,2, X12,2 and X15,2, because the contributions of B102,2 to the other states of these variables are 0, i.e. Fnk;102,2=0 where k≠2.
To explain the fuzzy evidence case, suppose E6 does not determine X6,2, but only shows that the probability or the membership of X6 being in its normal state is 0.3 and that in its abnormal high state is 0.7, i.e. m6,1=0.3 and m62=0.7. Then E6 is the fuzzy evidence. According to §9 (3) and §10, E6 is taken as the consequence variable of X6 and is added into the original DUCG. Meanwhile, this DUCG is simplified conditioned on B according to the same method above, so as to obtain the final result shown in FIG. 24-1. The internal process is same as the certain evidence case, because E6 excludes the influence of X6,3 and X6,1 does not have any influence to the process. The only difference is that between FIGS. 24-1 and 24-2. The others are the same. For saving space, the internal process is ignored. Because m6,1=0.3, m6,2=0.7 and v6,1=Pr{X6,1}≈1 (the unconditional probability of the normal state is usually very close to 1), v6,2=Pr{X6,2}=Pr{F6,2;102B102+F6,3X3}≈0 (The normal process should be outspread X3 and break the cycle based on FIG. 21 first, and then perform the calculation. But it is too trivial and therefore is ignored. Compared with v6,1=1, v6,2 is the unconditional probability or frequency of the abnormal state and can be approximated as 0), according to §10, let fE;6,2=1, we get




  
    f
    
      
        E
        ;
        6
      
      ,
      1
    
  
  =
  
    
      
        
          
            
              m
              
                6
                ,
                1
              
            
            ⁢
            
              v
              
                6
                ,
                2
              
            
          
          
            
              m
              
                6
                ,
                2
              
            
            ⁢
            
              v
              
                6
                ,
                1
              
            
          
        
        ⁢
        
          f
          
            
              E
              ;
              6
            
            ,
            2
          
        
      
      ≈
      
        
          0.3
          ×
          0
        
        
          0.7
          ×
          1
        
      
    
    =
    0.
  




In this way, E6 is added as a consequence variable of X6 into the DUCG.

In FIG. 24-2, the only abnormal variable is E6, i.e.




  
    
      
        
          
            E
            '
          
          ⁢
          
            (
            
              t
              1
            
            )
          
        
        =
        
        ⁢
        
          E
          6
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            F
            
              E
              ;
              6
            
          
          ⁢
          
            X
            6
          
        
      
    
  
  
    
      
        
          =
          
          ⁢
          
            
              
                F
                
                  
                    E
                    ;
                    6
                  
                  ,
                  1
                
              
              ⁢
              
                F
                
                  6
                  ,
                  
                    1
                    ;
                    D
                  
                
              
              ⁢
              
                D
                6
              
            
            +
            
              
                F
                
                  
                    E
                    ;
                    6
                  
                  ,
                  2
                
              
              ⁢
              
                F
                
                  6
                  ,
                  
                    2
                    ;
                    D
                  
                
              
              ⁢
              
                D
                6
              
            
            +
            
              
                F
                
                  
                    E
                    ;
                    6
                  
                  ,
                  1
                
              
              ⁢
              
                F
                
                  6
                  ,
                  
                    1
                    ;
                    102
                  
                  ,
                  1
                
              
              ⁢
              
                B
                
                  102
                  ,
                  1
                
              
            
            +
          
        
        ⁢
        
          
        
      
    
  
  
    
      
        
        ⁢
        
          
            
              F
              
                
                  E
                  ;
                  6
                
                ,
                1
              
            
            ⁢
            
              F
              
                6
                ,
                
                  1
                  ;
                  102
                
                ,
                2
              
            
            ⁢
            
              B
              
                102
                ,
                2
              
            
          
          +
          
            
              F
              
                
                  E
                  ;
                  6
                
                ,
                2
              
            
            ⁢
            
              F
              
                6
                ,
                
                  2
                  ;
                  102
                
                ,
                1
              
            
            ⁢
            
              B
              
                102
                ,
                1
              
            
          
          +
        
      
    
  
  
    
      
        
        ⁢
        
          
            F
            
              
                E
                ;
                6
              
              ,
              2
            
          
          ⁢
          
            F
            
              6
              ,
              
                1
                ;
                102
              
              ,
              2
            
          
          ⁢
          
            B
            
              102
              ,
              2
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              F
              
                
                  E
                  ;
                  6
                
                ,
                1
              
            
            ⁢
            
              F
              
                6
                ,
                
                  1
                  ;
                  D
                
              
            
            ⁢
            
              D
              6
            
          
          +
          
            
              F
              
                
                  E
                  ;
                  6
                
                ,
                1
              
            
            ⁢
            
              F
              
                6
                ,
                
                  1
                  ;
                  102
                
                ,
                2
              
            
            ⁢
            
              B
              
                102
                ,
                2
              
            
          
          +
          
            
              F
              
                
                  E
                  ;
                  6
                
                ,
                2
              
            
            ⁢
            
              F
              
                6
                ,
                
                  1
                  ;
                  102
                
                ,
                2
              
            
            ⁢
            
              B
              
                102
                ,
                2
              
            
          
        
      
    
  




This is because F6,2;D=F6,1;102,1=F6,2;102,1=0, i.e. D6 does not have any contribution to the abnormal state of X6, and the normal state of B102 does not have any contribution to X6, in which D6 and FE;6 is ignored in the figure. Moreover, since fE;6,1=0 which means FE;6,1=0. Then,

E′(t1)=FE;6,2F6,1;102,2B102,2 

Thus we know SH(t1)={B102,2}, which is the same as the case of the certain evidence. Then, given B102,2, the condition C12;6 of F12;6 in FIG. 24-2 is invalid so that F12;6 can be eliminated. That is the result shown in FIG. 25-2.

From the above discussion, in the case of the fuzzy evidence, the only difference from the certain evidence case is that there is an added event FE;6,2 in the expression of E′(t1). Then,



  
    
      
        
          
            
              h
              
                102
                ,
                2
              
              
                s
                ′
              
            
            ⁡
            
              (
              
                t
                1
              
              )
            
          
          =
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                  ❘
                  
                    
                      E
                      ′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                
                =
                
                  
                    
                      F
                      
                        E
                        ;
                        6
                      
                    
                    ⁢
                    
                      X
                      6
                    
                  
                  =
                  
                    
                      F
                      
                        
                          E
                          ;
                          6
                        
                        ,
                        2
                      
                    
                    ⁢
                    
                      F
                      
                        6
                        ,
                        
                          2
                          ;
                          102
                        
                        ,
                        2
                      
                    
                    ⁢
                    
                      B
                      
                        102
                        ,
                        2
                      
                    
                  
                
              
              }
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                  ⁢
                  
                    F
                    
                      
                        E
                        ;
                        6
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    F
                    
                      6
                      ,
                      
                        2
                        ;
                        102
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
                }
              
            
            
              Pr
              ⁢
              
                {
                
                  
                    F
                    
                      
                        E
                        ;
                        6
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    F
                    
                      6
                      ,
                      
                        2
                        ;
                        102
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
                }
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  
                    F
                    
                      
                        E
                        ;
                        6
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    F
                    
                      6
                      ,
                      
                        2
                        ;
                        102
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
                }
              
            
            
              Pr
              ⁢
              
                {
                
                  
                    F
                    
                      
                        E
                        ;
                        6
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    F
                    
                      6
                      ,
                      
                        2
                        ;
                        102
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
                }
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          1
        
      
    
  




  
    
      
        
          
            
              h
              
                102
                ,
                1
              
              
                s
                ′
              
            
            ⁡
            
              (
              
                t
                1
              
              )
            
          
          =
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  
                    B
                    
                      102
                      ,
                      1
                    
                  
                  ❘
                  
                    
                      E
                      ′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                
                =
                
                  
                    
                      F
                      
                        E
                        ;
                        6
                      
                    
                    ⁢
                    
                      X
                      6
                    
                  
                  =
                  
                    
                      F
                      
                        
                          E
                          ;
                          6
                        
                        ,
                        2
                      
                    
                    ⁢
                    
                      F
                      
                        6
                        ,
                        
                          2
                          ;
                          102
                        
                        ,
                        2
                      
                    
                    ⁢
                    
                      B
                      
                        102
                        ,
                        2
                      
                    
                  
                
              
              }
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  
                    B
                    
                      102
                      ,
                      1
                    
                  
                  ⁢
                  
                    F
                    
                      
                        E
                        ;
                        6
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    F
                    
                      6
                      ,
                      
                        2
                        ;
                        102
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
                }
              
            
            
              Pr
              ⁢
              
                {
                
                  
                    F
                    
                      
                        E
                        ;
                        6
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    F
                    
                      6
                      ,
                      
                        2
                        ;
                        102
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
                }
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            0
            
              Pr
              ⁢
              
                {
                
                  
                    F
                    
                      
                        E
                        ;
                        6
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    F
                    
                      6
                      ,
                      
                        2
                        ;
                        102
                      
                      ,
                      2
                    
                  
                  ⁢
                  
                    B
                    
                      102
                      ,
                      2
                    
                  
                
                }
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          0
        
      
    
  




  
    
      
        
          
            
              h
              
                102
                ,
                2
              
              s
            
            ⁡
            
              (
              
                t
                1
              
              )
            
          
          =
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  B
                  
                    102
                    ,
                    2
                  
                
                ❘
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
              
              }
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            
              
                
                  h
                  
                    102
                    ,
                    2
                  
                  
                    s
                    ′
                  
                
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
              ⁢
              Pr
              ⁢
              
                {
                
                  
                    
                      E
                      ′′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                  ❘
                  
                    
                      B
                      
                        102
                        ,
                        2
                      
                    
                    ⁢
                    
                      X
                      6
                    
                  
                
                }
              
            
            
              
                ∑
                
                  j
                  =
                  1
                
                2
              
              ⁢
              
                
                  
                    h
                    
                      102
                      ,
                      j
                    
                    
                      s
                      ′
                    
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
                ⁢
                Pr
                ⁢
                
                  {
                  
                    
                      
                        E
                        ′′
                      
                      ⁡
                      
                        (
                        
                          t
                          1
                        
                        )
                      
                    
                    ❘
                    
                      
                        B
                        
                          102
                          ,
                          j
                        
                      
                      ⁢
                      
                        X
                        6
                      
                    
                  
                  }
                
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            
              Pr
              ⁢
              
                {
                
                  
                    
                      E
                      ′′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                  ❘
                  
                    
                      B
                      
                        102
                        ,
                        2
                      
                    
                    ⁢
                    
                      X
                      6
                    
                  
                
                }
              
            
            
              Pr
              ⁢
              
                {
                
                  
                    
                      E
                      ′′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                  ❘
                  
                    
                      B
                      
                        102
                        ,
                        2
                      
                    
                    ⁢
                    
                      X
                      6
                    
                  
                
                }
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          1
        
      
    
  




  
    
      
        
          
            
              h
              
                102
                ,
                1
              
              s
            
            ⁡
            
              (
              
                t
                1
              
              )
            
          
          =
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  B
                  
                    102
                    ,
                    1
                  
                
                ❘
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
              
              }
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            
              
                
                  h
                  
                    102
                    ,
                    1
                  
                  
                    s
                    ′
                  
                
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
              ⁢
              Pr
              ⁢
              
                {
                
                  
                    
                      E
                      ′′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                  ❘
                  
                    
                      B
                      
                        102
                        ,
                        1
                      
                    
                    ⁢
                    
                      X
                      6
                    
                  
                
                }
              
            
            
              
                ∑
                
                  j
                  =
                  1
                
                2
              
              ⁢
              
                
                  
                    h
                    
                      102
                      ,
                      j
                    
                    
                      s
                      ′
                    
                  
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
                ⁢
                Pr
                ⁢
                
                  {
                  
                    
                      
                        E
                        ′′
                      
                      ⁡
                      
                        (
                        
                          t
                          1
                        
                        )
                      
                    
                    ❘
                    
                      
                        B
                        
                          102
                          ,
                          1
                        
                      
                      ⁢
                      
                        X
                        6
                      
                    
                  
                  }
                
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          
            0
            
              Pr
              ⁢
              
                {
                
                  
                    
                      E
                      ′′
                    
                    ⁡
                    
                      (
                      
                        t
                        1
                      
                      )
                    
                  
                  ❘
                  
                    
                      B
                      
                        102
                        ,
                        2
                      
                    
                    ⁢
                    
                      X
                      6
                    
                  
                
                }
              
            
          
        
      
    
    
      
        
          =
          
          ⁢
          0
        
      
    
  



Since the secondary loop of the nuclear power plant is a dynamical system, although the fault has been found at time t1, the consistence of the later diagnosis based on the new evidence with the earlier diagnosis is still important. Usually, after adding new evidence, the diagnosis should be more accurate. For this, we consider the signals monitored at time t2. The states of the consequence variables at t2 are still shown in table 1, in which the positions of the bold letters show that there are 9 abnormal variable states:
X1,2 (the pressure in steam line A is high, shaded letter in the table),
X2,2 (the pressure in steam line B is high, shaded letter in the table),
X3,2 (the pressure in steam line C is high, shaded letter in the table),
X4,3 (the flow rate of steam line A is low, underline letter in the table),
X5,3 (the flow rate of steam line B is low, underline letter in the table),
X6,2 (the flow rate of steam line C is high, shaded letter in the table),
X12,2 (the water level in the steam generator C is high, shaded letter in the table),
X13,2 (the pressure in the main steam header is high, shaded letter in the table),
X15,2 (the water level of the hot well is high, shaded letter in the table);
i.e. E′(t2)=X1,2X2,2X3,2X4,3X5,3X6,2X12,2X13,2X15,2. The other variable states are normal (marked as the italic letters in the figures). Repeat the fault diagnosis steps similar to those for time t1:
7. In FIG. 21, eliminate all the functional variables influencing the other variables from the consequence variables in the normal states, so as to obtain the DUCG shown in FIG. 26.
8. Consider the special functional variables in FIG. 26, i.e. the functional variables directly connecting the cause variables whose state is unknown with the consequence variables whose state is known. If no matter what state the cause variable is in, the contribution of this functional variable to the known states of the consequence variables is 0 or not given, this functional variable is eliminated, so as to obtain the DUCG shown in FIG. 27. For example, in the functional variable of B100 to X4,3, F4,3;100j=0, and in the functional variable of B101 to X5,3, F5,3;101j=0, eliminate F4;100 and F5;101. This is because the leakage of the U type pipes in the steam generator cannot cause the low steam flow rate, meanwhile the normal states (j=1) of the basic variables do not have any influence to the other variables. For another example, whether or not the steam pipe leaks cannot cause the high water level of the steam generator or the high pressure of the steam pipe. Therefore, F12;94, F3;94, F1;92 and F2;93 are eliminated.
9. Eliminate the isolated part without connection with the consequence variables whose states are abnormal, so as to obtain the DUCG shown in FIG. 28.
10. Eliminate the part directly connected with only the state normal variables in FIG. 28, so as to obtain the DUCG shown in FIG. 29.
11. Eliminate the functional variables inconsistent with the time order. Because X6,2 appeared at time t1, X3,2 appearing at time t2 cannot be the cause of X6,2. Therefore, F6;3 is eliminate to obtain the DUCG shown in FIG. 30.
12. Outspread E′(t2) so as to obtain SH(t2). Only for this, the functional variables, the conditional functional variables and the states of variables can be ignored. After the ignorance, the variables in the expression can be absorbed with other. The repeated variables on the cause side are treated as null. Then

X1=B79+B100+X13=B79+B100+B80+B101+B81+B102 

X2=B80+B101+X13=B80+B101+B79+B100+B81+B102 

X3=B81+B102+X13+X12=B81+B102+B79+B100+B80+B101 

X4=X1=B79+B100+B80+B101B81+B102 

X4=X2=B80+B101+B79+B100+B81+B102 

X6=B102 

X12=B81+X6=B81+B102 

X13=X1+X2+X3=B79+B100+B80+B101+B81+B102 

X15=B100+B101+B102+B117 

Thus we know E′(t2)=X1X2X3X4X5X6X12X13X15=B102. This is because X6 includes only the initiating event variable B102, meanwhile all the other consequence variables have the items including B102. According to the rule that the AND of the different initiating events is null, only B102 in the outspreaded expression of E′(t2) remains. Furthermore, because B102,1 does not have any output, only B102,2 can be true, i.e. SH(t2)={B102,2}. Conditioned on both E(t1) and E(t2), the dynamical possible solution set SH(t) should be the intersection of SH(t1) and SH(t2):

SH(t)=SH(t1)SH(t2)=SH(t1)=SH(t2)={B102,2}

In fact, the DUCG of either FIG. 29 or FIG. 30 deals with only 7 basic variables. The normal states of these variables do not have any influence to other variables. Therefore, only their fault states can be the possible states, i.e. the most possible scope of SH(t2) can only be:

SH(t2)+{B100,2,B101,2,B79,2,B80,2,B81,2,B117,2, B117,3}

In this case, we still have

SH(t)=SH(t1)SH(t2)={B102,2}·{B100,2,B101,2,B102,2,B79,2,B80,2,B81,2,B117,2, B117,3}={B102,2}

In which, the operator “·” denotes the logic AND.

Since both the static and dynamical possible solution sets SH(t1) and SH(t2) have only one possible solution, there is no need to calculate the rank probability, the fault cause B102,2 is found accurately and the rank probability equals 1. In this example, as the fault has been found at time t1 uniquely, the diagnosis accuracy cannot be increased by the new information at t2. Suppose the elements in SH(t1) are not only one, then the elements in SH(t), (t≧t2), must be less than or equal to the elements in SH(t1) (but should not be null, otherwise the DUCG has defects or spurious signals), i.e. as the amount of information increases, the accuracy of diagnosis increases gradually.
13. According to




  
    
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The dynamical state probability with incomplete information B102,2 is





  
    
      
        
          
            h
            
              102
              ,
              2
            
            
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              ′
            
          
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            (
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            )
          
        
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          ⁢
          
            {
            
              
                B
                
                  102
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        =
        
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                h
                
                  102
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                (
                
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                )
              
            
            ⁢
            
              
                
                  h
                  
                    102
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                  102
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                  (
                  
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                    h
                    
                      102
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                  b
                  
                    102
                    ,
                    1
                  
                
              
            
            +
            
              
                
                  h
                  
                    102
                    ,
                    2
                  
                  
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                    ′
                  
                
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
              ⁢
              
                
                  
                    h
                    
                      102
                      ,
                      2
                    
                    
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                  ⁡
                  
                    (
                    
                      t
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                    )
                  
                
                /
                
                  b
                  
                    102
                    ,
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        =
        
        ⁢
        
          
            
              
                h
                
                  102
                  ,
                  2
                
                
                  s
                  ′
                
              
              ⁡
              
                (
                
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                  1
                
                )
              
            
            ⁢
            
              
                
                  h
                  
                    102
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              /
              
                b
                
                  102
                  ,
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                h
                
                  102
                  ,
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                  s
                  ′
                
              
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                (
                
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                )
              
            
            ⁢
            
              
                
                  h
                  
                    102
                    ,
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                    ′
                  
                
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                  (
                  
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              /
              
                b
                
                  102
                  ,
                  2
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        1
      
    
  




In which,

h102,1(t0)=b102,1 

h102,2(t0)=b102,2 

h102,1s′(t1)=0

h102,2s′(t1)=1

In the same way,





  
    
      
        
          
            h
            
              102
              ,
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              ′
            
          
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            (
            t
            )
          
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                B
                
                  102
                  ,
                  1
                
              
              ❘
              
                
                  
                    E
                    ′
                  
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                    (
                    
                      t
                      1
                    
                    )
                  
                
                ⁢
                
                  
                    E
                    ′
                  
                  ⁡
                  
                    (
                    
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                      2
                    
                    )
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
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                h
                
                  102
                  ,
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                  h
                  
                    102
                    ,
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                  (
                  
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              /
              
                b
                
                  102
                  ,
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                  h
                  
                    102
                    ,
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                  (
                  
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              ⁢
              
                
                  
                    h
                    
                      102
                      ,
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                    (
                    
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                /
                
                  b
                  
                    102
                    ,
                    1
                  
                
              
            
            +
            
              
                
                  h
                  
                    102
                    ,
                    2
                  
                  
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                  (
                  
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                  )
                
              
              ⁢
              
                
                  
                    h
                    
                      102
                      ,
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                    (
                    
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                      2
                    
                    )
                  
                
                /
                
                  b
                  
                    102
                    ,
                    2
                  
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          0
          
            
              
                h
                
                  102
                  ,
                  2
                
                
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                (
                
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                    102
                    ,
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                    s
                    ′
                  
                
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                  (
                  
                    t
                    2
                  
                  )
                
              
              /
              
                b
                
                  102
                  ,
                  2
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        0
      
    
  




According to





  
    
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      kj
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            h
            kj
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            (
            
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              i
            
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                h
                kj
              
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            n
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                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
    
  




The dynamical state probability with complete information of B102,2 is





  
    
      
        
          
            h
            
              102
              ,
              2
            
            s
          
          ⁡
          
            (
            t
            )
          
        
        =
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                B
                
                  102
                  ,
                  2
                
              
              ❘
              
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      1
                    
                    )
                  
                
                ⁢
                
                  E
                  ⁡
                  
                    (
                    
                      t
                      2
                    
                    )
                  
                
              
            
            }
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                h
                
                  102
                  ,
                  2
                
                s
              
              ⁡
              
                (
                
                  t
                  1
                
                )
              
            
            ⁢
            
              
                
                  h
                  
                    102
                    ,
                    2
                  
                  s
                
                ⁡
                
                  (
                  
                    t
                    2
                  
                  )
                
              
              /
              
                b
                
                  102
                  ,
                  2
                
              
            
          
          
            
              
                
                  h
                  
                    102
                    ,
                    1
                  
                  s
                
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
              ⁢
              
                
                  
                    h
                    
                      102
                      ,
                      1
                    
                    s
                  
                  ⁡
                  
                    (
                    
                      t
                      2
                    
                    )
                  
                
                /
                
                  b
                  
                    102
                    ,
                    1
                  
                
              
            
            +
            
              
                
                  h
                  
                    102
                    ,
                    2
                  
                  s
                
                ⁡
                
                  (
                  
                    t
                    1
                  
                  )
                
              
              ⁢
              
                
                  
                    h
                    
                      102
                      ,
                      2
                    
                    s
                  
                  ⁡
                  
                    (
                    
                      t
                      2
                    
                    )
                  
                
                /
                
                  b
                  
                    102
                    ,
                    2
                  
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              
                h
                
                  102
                  ,
                  2
                
                s
              
              ⁡
              
                (
                
                  t
                  1
                
                )
              
            
            ⁢
            
              
                
                  h
                  
                    102
                    ,
                    2
                  
                  s
                
                ⁡
                
                  (
                  
                    t
                    2
                  
                  )
                
              
              /
              
                b
                
                  102
                  ,
                  2
                
              
            
          
          
            
              
                h
                
                  102
                  ,
                  2
                
                s
              
              ⁡
              
                (
                
                  t
                  1
                
                )
              
            
            ⁢
            
              
                
                  h
                  
                    102
                    ,
                    2
                  
                  s
                
                ⁡
                
                  (
                  
                    t
                    2
                  
                  )
                
              
              /
              
                b
                
                  102
                  ,
                  2
                
              
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        1
      
    
  




Because in which,

h102,1s(t1)=h102,1s(t2)=0

In the same way, we can get h102,1s(t)=0.

14. From FIG. 30, eliminate the basic variables and their functional variables that have been determined should not be included. As X17 becomes an isolated normal variable due to the elimination of B117 and is irrelevant to the fault diagnosis, it is also eliminated. Meanwhile, C12;6 is invalid due to the determination of B102,2, which results in the elimination of F12,6. Then we get the DUCG shown in FIG. 31. This DUCG predicts that X10, X11, X49, X50, X51, X55, X56 and X57 might be abnormal. But, until the shut down of the reactor, these 8 variables do not appear abnormal. The reason is that the steam pressure increase resulted in by B102,2 is not large enough to open the depressurizing valves represented by the later 6 variables and the decrease of the stream flow rates in lines A and B is very small, such that the ordinary automatic control of the feed water flow rate maintains the normal water levels of the steam generators. But the predicted X3,2, X12,2 and X15,2 in step 4 at time t1 do appear at time t2, which proves the prediction accuracy of DUCG.
Example 23
The application illustration for predicting the effects of the economic policies.
A simple DUCG modeling the price influence of the agricultural products is as shown in FIG. 32. The construction method is similar to that for example 1: The first step is to determine the B and X type variables in the domain. These variables are described as shown in FIG. 32. In this example, all variables have three discrete or fuzzily discretized states indexed as “1, 2, 3” respectively. The method of fuzzy discretization is similar to that of fuzzily discretizing X16 in example 1. For every X type variable, determine its direct cause variables. For example, the direct cause variables of X7 are X4, X6 and B11. In which, as there are historical statistic data available, the causal relations between X4 and X6 to X7 are better to be represented in the implicit representation mode. B11 represents the proposed economic policies and there is no historical data available. Only the belief of the domain engineers is available. Therefore, the explicit representation mode is suitable. In other words, the causal relations between X7 and its direct cause variables are represented in the hybrid representation mode. The direct cause variable of X8 is only X7. Since there are the historical statistic data, the implicit representation mode can be used. The direct cause variables of X3, X4, X5, X6, X9 and X10 are also represented respectively as shown in FIG. 32, and are all in the explicit representation mode due to the data, etc. Put all the representations for the X type variables together, we get the DUCG as shown in FIG. 32. In which, the state probability parameters, the original functional intensities of the functional variables, the conditional probability tables involved in the implicit representation mode, the relationships, etc, are given by the domain engineers according to the statistic data or their belief. For example,




  
    
      B
      1
    
    =
    
      
        B
        2
      
      =
      
        (
        
          
            
              0.3
            
          
          
            
              0.4
            
          
          
            
              0.3
            
          
        
        )
      
    
  
  ,
  
    

  
  ⁢
  
    
      B
      11
    
    =
    
      (
      
        
          
            1
          
        
        
          
            0
          
        
        
          
            0
          
        
      
      )
    
  




In which, b11,1=1 is because we want to predict the effect of the proposed economic policy 1. Similarly, we can predict the effects of the other proposed economic policies. Moreover, in addition to the implicit representation mode,





  
    (
    
      
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                1
              
              ,
              ij
            
          
        
      
      
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                2
              
              ;
              ij
            
          
        
      
      
        
          
            a
            
              
                n
                ⁢
                
                  
                
                ⁢
                3
              
              ;
              ij
            
          
        
      
    
    )
  
  =
  
    (
    
      
        
          0.2
        
      
      
        
          0.3
        
      
      
        
          0.5
        
      
    
    )
  




Therefore,





  
    
      
        
          A
          
            3
            ;
            7
          
        
        =
        
        ⁢
        
          
            (
            
              
                
                  
                    a
                    
                      3
                      ,
                      
                        1
                        ;
                        7
                      
                      ,
                      1
                    
                  
                
                
                  
                    a
                    
                      3
                      ,
                      
                        1
                        ;
                        7
                      
                      ,
                      2
                    
                  
                
                
                  
                    a
                    
                      3
                      ,
                      
                        1
                        ;
                        7
                      
                      ,
                      3
                    
                  
                
              
              
                
                  
                    a
                    
                      3
                      ,
                      
                        2
                        ;
                        7
                      
                      ,
                      1
                    
                  
                
                
                  
                    a
                    
                      3
                      ,
                      
                        2
                        ;
                        7
                      
                      ,
                      2
                    
                  
                
                
                  
                    a
                    
                      3
                      ,
                      
                        2
                        ;
                        7
                      
                      ,
                      3
                    
                  
                
              
              
                
                  
                    a
                    
                      3
                      ,
                      
                        3
                        ;
                        7
                      
                      ,
                      1
                    
                  
                
                
                  
                    a
                    
                      3
                      ,
                      
                        3
                        ;
                        72
                      
                    
                  
                
                
                  
                    a
                    
                      3
                      ,
                      
                        3
                        ;
                        7
                      
                      ,
                      3
                    
                  
                
              
            
            )
          
          =
        
      
    
  
  
    
      
        
          A
          
            7
            ;
            11
          
        
        ≡
        
        ⁢
        
          (
          
            
              
                
                  a
                  
                    7
                    ,
                    
                      1
                      ;
                      11
                    
                    ,
                    1
                  
                
              
              
                
                  a
                  
                    7
                    ,
                    
                      1
                      ;
                      11
                    
                    ,
                    2
                  
                
              
              
                
                  a
                  
                    7
                    ,
                    
                      1
                      ;
                      11
                    
                    ,
                    3
                  
                
              
            
            
              
                
                  a
                  
                    7
                    ,
                    
                      2
                      ;
                      11
                    
                    ,
                    1
                  
                
              
              
                
                  a
                  
                    7
                    ,
                    
                      2
                      ;
                      11
                    
                    ,
                    2
                  
                
              
              
                
                  a
                  
                    7
                    ,
                    
                      2
                      ;
                      11
                    
                    ,
                    3
                  
                
              
            
            
              
                
                  a
                  
                    7
                    ,
                    
                      3
                      ;
                      11
                    
                    ,
                    1
                  
                
              
              
                
                  a
                  
                    7
                    ,
                    
                      3
                      ;
                      11
                    
                    ,
                    2
                  
                
              
              
                
                  a
                  
                    7
                    ,
                    
                      3
                      ;
                      11
                    
                    ,
                    3
                  
                
              
            
          
          )
        
      
    
  
  
    
      
        =
        
        ⁢
        
          (
          
            
              
                0.2
              
              
                0.2
              
              
                0.2
              
            
            
              
                0.3
              
              
                0.3
              
              
                0.3
              
            
            
              
                0.5
              
              
                0.5
              
              
                0.5
              
            
          
          )
        
      
    
  




Etc. All the relationships of the functional intensities are the default value 1. The causal relations of X4 and X6 to X7 are represented in the implicit representation mode. The conditional probability table CPT is:








TABLE 2







The conditional probability table of X4 and X6 to X7












Expression of






state


No. j
combination
q71;j/d7;j = p71;j
q72;j/d7;j = p72;j
q73;j/d7;j = p73;j





1
X41X61
50/100 = 0.5
50/100 = 0.5
0/100 = 0


2
X41X62
40/100 = 0.4
40/100 = 0.4
20/100 = 0.2


3
X41X63
100/200 = 0.5
50/200 = 0.25
50/200 = 0.25


4
X42X61
150/250 = 0.6
0/250 = 0
100/250 = 0.4


5
X42X62
0/100 = 0
40/100 = 0.4
60/100 = 0.6


6
X42X63
20/200 = 0.1
100/200 = 0.5
80/200 = 0.4


7
X43X61
40/100 = 0.4
40/100 = 0.4
20/100 = 0.2


8
X43X62
10/50 = 0.2
20/50 = 0.4
20/50 = 0.4


9
X43X63
20/80 = 0.25
40/80 = 0.5
20/80 = 0.25









The causal relation of X7 to X8 is also represented in the implicit representation mode. The conditional probability table is:







TABLE 3







The conditional probability table of X7 to X8












Expression of






state


No. j
combination
q81;j/d8;j = p81;j
q82;j/d8;j = p82;j
q83;j/d8;j = p83;j





1
X71
50/100 = 0.5
20/100 = 0.2
30/100 = 0.3


2
X72
60/80 = 0.75
8/80 = 0.1
12/80 = 0.15


3
X73
150/300 = 0.5
120/300 = 0.4
30/300 = 0.1









The DUCG in FIG. 32 is in the hybrid representation mode, and has the logic cycles. Therefore, it must be transformed as EDUCG for getting the solution, i.e. transform the hybrid representation mode of X7 and the implicit representation mode of X8 as the explicit representation mode. To do this, the logic gates G12 and G13 as well as the default events D7 and D8 are added. The transformed EDUCG is as shown in FIG. 33.
Since




  
    
      p
      
        7
        ⁢
        k
      
    
    =
    
      
        
          min
          j
        
        ⁢
        
          {
          
            p
            
              
                7
                ⁢
                k
              
              ;
              j
            
          
          }
        
      
      =
      0
    
  
  ,




k=1,2,3, D7 does not exist. Then





  
    A
    
      7
      ,
      12
    
  
  =
  
    (
    
      
        
          0.5
        
        
          0.4
        
        
          0.5
        
        
          0.6
        
        
          0
        
        
          0.1
        
        
          0.4
        
        
          0.2
        
        
          0.25
        
      
      
        
          0.5
        
        
          0.4
        
        
          0.25
        
        
          0
        
        
          0.4
        
        
          0.5
        
        
          0.4
        
        
          0.4
        
        
          0.5
        
      
      
        
          0
        
        
          0.2
        
        
          0.25
        
        
          0.4
        
        
          0.6
        
        
          0.4
        
        
          0.2
        
        
          0.4
        
        
          0.25
        
      
    
    )
  




Meanwhile, G13 can be ignored because it has only one input variable. Then we have FIG. 34. The numbers with the directed arcs are the relationships (default case is 1), in which, according to




  
    
      a
      
        
          8
          ⁢
          k
        
        ;
        D
      
    
    =
    
      
        
          p
          
            8
            ⁢
            k
          
        
        /
        
          
            ∑
            k
          
          ⁢
          
            
              p
              
                8
                ⁢
                k
              
            
            ⁢
            
              
            
            ⁢
            and
            ⁢
            
              
            
            ⁢
            
              p
              
                8
                ⁢
                k
              
            
          
        
      
      =
      
        
          min
          j
        
        ⁢
        
          {
          
            p
            
              
                8
                ⁢
                k
              
              ;
              j
            
          
          }
        
      
    
  
  ,




the original functional intensities of F8;D are calculated as a81;D= 5/7, a82;D= 1/7 and a83;D= 1/7.

After the transformation,

    
    
        rX8=r8;7+1=2, where r8;7=1 is given y domain engineers;
    
    






  
    
      r
      
        8
        ;
        D
      
    
    =
    
      
        
          r
          
            X
            ⁢
            
              
            
            ⁢
            8
          
        
        ⁢
        
          
            ∑
            k
          
          ⁢
          
            p
            
              8
              ⁢
              k
            
          
        
      
      =
      
        
          2
          ×
          0.7
        
        =
        1.4
      
    
  
  ,








  
    
      r
      
        8
        ;
        7
      
    
    =
    
      
        r
        
          8
          ;
          13
        
      
      =
      
        
          
            r
            
              X
              ⁢
              
                
              
              ⁢
              8
            
          
          (
          
            1
            -
            
              
                ∑
                k
              
              ⁢
              
                p
                
                  8
                  ⁢
                  k
                
              
            
          
          )
        
        =
        
          
            2
            ×
            0.3
          
          =
          0.6
        
      
    
  
  ,




where r8;7 is produced after ignore G13.

According to




  
    
      
        
          a
          
            
              8
              ⁢
              k
            
            ;
            
              7
              ⁢
              j
            
          
        
        =
        
        ⁢
        
          p
          
            
              8
              ⁢
              k
            
            ;
            j
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            (
            
              
                p
                
                  
                    8
                    ⁢
                    k
                  
                  ;
                  j
                
              
              -
              
                p
                
                  8
                  ⁢
                  k
                
              
            
            )
          
          /
          
            
              ∑
              k
            
            ⁢
            
              (
              
                
                  p
                  
                    
                      8
                      ⁢
                      k
                    
                    ;
                    j
                  
                
                -
                
                  p
                  
                    8
                    ⁢
                    k
                  
                
              
              )
            
          
        
      
    
  
  
    
      
        
          =
          
          ⁢
          
            
              (
              
                
                  p
                  
                    
                      8
                      ⁢
                      k
                    
                    ;
                    j
                  
                
                -
                
                  p
                  
                    8
                    ⁢
                    k
                  
                
              
              )
            
            /
            
              (
              
                1
                -
                
                  
                    ∑
                    k
                  
                  ⁢
                  
                    p
                    
                      8
                      ⁢
                      k
                    
                  
                
              
              )
            
          
        
        ,
      
    
  




the values of a8k;7j after the reconstruction are calculated as shown in the following table:








TABLE 4







The new conditional probability table of X7 to X8













q83;j′/d8;j′ =


j′
q81;j′/d8;j′ = p81;j′ = a81;7j′
q82;j′/d8;j′ = p82;j′ = a82;7j′
p83;j′ = a83;7j′





1
0/100 = 0
33.33/100 = 0.3333
66.67/100 = 0.6667


2
66.67/80 = 0.8333
0/80 = 0
13.33/80 = 0.1667


3
0/300 = 0
300/300 = 1
0/300 = 0









Then we have





  
    A
    
      8
      ;
      7
    
  
  =
  
    
      
        (
        
          
            
              0
            
            
              0.8333
            
            
              0
            
          
          
            
              0.3333
            
            
              0
            
            
              1.
            
          
          
            
              0.6667
            
            
              0.1667
            
            
              0
            
          
        
        )
      
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        A
        
          8
          ;
          D
        
      
    
    =
    
      (
      
        
          
            
              5
              7
            
          
        
        
          
            
              1
              7
            
          
        
        
          
            
              1
              7
            
          
        
      
      )
    
  




After the initial logic outspread of the consequence variables and logic gate in the above EDUCG, we have

X3=F3;7X7 

X4=F4;8X8 

X5=F5;9X9 

X6=F6;1B1+F6;2B2+F6;3X3+F6;5X5 

X7=F7;11B11+F7;12G12 

X8=F8;7X7+F8;DD8 

X9=F9;10X10 

X10=F10;7X7+F10;11B11 

G12=G12{U12;4X4,U12;6X6}  (2.1)

After the further outspread to X3, we have





  
    
      
        
          X
          3
        
        =
        
          
            
              F
              
                3
                ;
                7
              
            
            ⁡
            
              (
              
                
                  
                    F
                    
                      7
                      ;
                      11
                    
                  
                  ⁢
                  
                    B
                    11
                  
                
                +
                
                  
                    F
                    
                      7
                      ;
                      12
                    
                  
                  ⁢
                  
                    G
                    12
                  
                  ⁢
                  
                    {
                    
                      
                        
                          U
                          
                            12
                            ;
                            4
                          
                        
                        ⁢
                        
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁡
                          
                            (
                            
                              
                                
                                  F
                                  
                                    8
                                    ;
                                    7
                                  
                                
                                ⁢
                                
                                  X
                                  7
                                
                              
                              +
                              
                                
                                  F
                                  
                                    8
                                    ;
                                    D
                                  
                                
                                ⁢
                                
                                  D
                                  8
                                
                              
                            
                            )
                          
                        
                      
                      ,
                      
                        
                          U
                          
                            12
                            ;
                            6
                          
                        
                        ⁡
                        
                          (
                          
                            
                              
                                F
                                
                                  6
                                  ;
                                  1
                                
                              
                              ⁢
                              
                                B
                                1
                              
                            
                            +
                            
                              
                                F
                                
                                  6
                                  ;
                                  2
                                
                              
                              ⁢
                              
                                B
                                2
                              
                            
                            +
                            
                              
                                F
                                
                                  6
                                  ;
                                  3
                                
                              
                              ⁢
                              
                                X
                                3
                              
                            
                            +
                            
                              
                                F
                                
                                  6
                                  ;
                                  5
                                
                              
                              ⁢
                              
                                F
                                
                                  5
                                  ;
                                  9
                                
                              
                              ⁢
                              
                                
                                  F
                                  
                                    9
                                    ;
                                    10
                                  
                                
                                ⁡
                                
                                  (
                                  
                                    
                                      
                                        F
                                        
                                          10
                                          ;
                                          11
                                        
                                      
                                      ⁢
                                      
                                        B
                                        11
                                      
                                    
                                    +
                                    
                                      
                                        F
                                        
                                          10
                                          ;
                                          7
                                        
                                      
                                      ⁢
                                      
                                        X
                                        7
                                      
                                    
                                  
                                  )
                                
                              
                            
                          
                          )
                        
                      
                    
                    }
                  
                
              
              )
            
          
          =
          
            
              
                F
                
                  3
                  ,
                  7
                
              
              ⁢
              
                F
                
                  7
                  ,
                  11
                
              
              ⁢
              
                B
                11
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      U
                      
                        12
                        ;
                        4
                      
                    
                    ⁢
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      U
                      
                        12
                        ;
                        6
                      
                    
                    ⁢
                    
                      F
                      
                        6
                        ;
                        1
                      
                    
                    ⁢
                    
                      B
                      1
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      U
                      
                        12
                        ;
                        4
                      
                    
                    ⁢
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      U
                      
                        12
                        ;
                        6
                      
                    
                    ⁢
                    
                      F
                      
                        6
                        ;
                        2
                      
                    
                    ⁢
                    
                      B
                      2
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      U
                      
                        12
                        ;
                        4
                      
                    
                    ⁢
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      U
                      
                        12
                        ;
                        6
                      
                    
                    ⁢
                    
                      F
                      
                        6
                        ;
                        3
                      
                    
                    ⁢
                    
                      X
                      3
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      U
                      
                        12
                        ;
                        4
                      
                    
                    ⁢
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      U
                      
                        12
                        ;
                        6
                      
                    
                    ⁢
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁢
                    
                      F
                      
                        5
                        ;
                        9
                      
                    
                    ⁢
                    
                      F
                      
                        9
                        ;
                        10
                      
                    
                    ⁢
                    
                      F
                      
                        10
                        ;
                        11
                      
                    
                    ⁢
                    
                      B
                      11
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      U
                      
                        12
                        ;
                        4
                      
                    
                    ⁢
                    
                      
                        F
                        
                          4
                          ;
                          8
                        
                      
                      (
                      
                        
                          
                            F
                            
                              8
                              ;
                              7
                            
                          
                          ⁢
                          
                            X
                            7
                          
                        
                        ,
                        
                          
                            U
                            
                              12
                              ;
                              6
                            
                          
                          ⁢
                          
                            F
                            
                              6
                              ;
                              5
                            
                          
                          ⁢
                          
                            F
                            
                              5
                              ;
                              9
                            
                          
                          ⁢
                          
                            F
                            
                              9
                              ;
                              10
                            
                          
                          ⁢
                          
                            F
                            
                              10
                              ;
                              7
                            
                          
                          ⁢
                          
                            X
                            7
                          
                        
                      
                      }
                    
                  
                  +
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          
                            U
                            
                              12
                              ;
                              4
                            
                          
                          ⁢
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁢
                          
                            F
                            
                              8
                              ;
                              D
                            
                          
                          ⁢
                          
                            D
                            8
                          
                        
                        ,
                        
                          
                            U
                            
                              12
                              ;
                              6
                            
                          
                          ⁢
                          
                            F
                            
                              6
                              ;
                              1
                            
                          
                          ⁢
                          
                            B
                            1
                          
                        
                      
                      }
                    
                  
                  +
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          
                            U
                            
                              12
                              ;
                              4
                            
                          
                          ⁢
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁢
                          
                            F
                            
                              8
                              ;
                              D
                            
                          
                          ⁢
                          
                            D
                            8
                          
                        
                        ,
                        
                          
                            U
                            
                              12
                              ;
                              6
                            
                          
                          ⁢
                          
                            F
                            
                              6
                              ;
                              2
                            
                          
                          ⁢
                          
                            B
                            2
                          
                        
                      
                      }
                    
                  
                  +
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          
                            U
                            
                              12
                              ;
                              4
                            
                          
                          ⁢
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁢
                          
                            F
                            
                              8
                              ;
                              D
                            
                          
                          ⁢
                          
                            D
                            8
                          
                        
                        ,
                        
                          
                            U
                            
                              12
                              ;
                              6
                            
                          
                          ⁢
                          
                            F
                            
                              6
                              ;
                              3
                            
                          
                          ⁢
                          
                            X
                            3
                          
                        
                      
                      }
                    
                  
                  +
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          
                            U
                            
                              12
                              ;
                              4
                            
                          
                          ⁢
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁢
                          
                            F
                            
                              8
                              ;
                              D
                            
                          
                          ⁢
                          
                            D
                            8
                          
                        
                        ,
                        
                          
                            U
                            
                              12
                              ;
                              6
                            
                          
                          ⁢
                          
                            F
                            
                              6
                              ;
                              5
                            
                          
                          ⁢
                          
                            F
                            
                              5
                              ;
                              9
                            
                          
                          ⁢
                          
                            F
                            
                              9
                              ;
                              10
                            
                          
                          ⁢
                          
                            F
                            
                              10
                              ;
                              11
                            
                          
                          ⁢
                          
                            B
                            11
                          
                        
                      
                      }
                    
                  
                  +
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          
                            U
                            
                              12
                              ;
                              4
                            
                          
                          ⁢
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁢
                          
                            F
                            
                              8
                              ;
                              D
                            
                          
                          ⁢
                          
                            D
                            8
                          
                        
                        ,
                        
                          
                            U
                            
                              12
                              ;
                              6
                            
                          
                          ⁢
                          
                            F
                            
                              6
                              ;
                              5
                            
                          
                          ⁢
                          
                            F
                            
                              5
                              ;
                              9
                            
                          
                          ⁢
                          
                            F
                            
                              9
                              ;
                              10
                            
                          
                          ⁢
                          
                            F
                            
                              10
                              ;
                              7
                            
                          
                          ⁢
                          
                            X
                            7
                          
                        
                      
                      }
                    
                  
                
              
            
          
        
      
    
    
      
        (
        2.2
        )
      
    
  




In this equation, the outspread follows the sequence of consequence to cause from lett to right, in which X3 and X7 are compose the logic cycles. According to the rule to break the logic cycle, the consequence cannot be its own cause. Therefore, in the static case, the X3 and X7 on the right side of the above equation should be treated as null. In the dynamical case, the X3 and X7 must be the values in the earlier moment, Moreover, D8 and Un;i are the inevitable events and can be ignored, so as to get the following equation:





  
    
      
        
          X
          3
        
        =
        
          
            
              F
              
                3
                ;
                7
              
            
            ⁡
            
              (
              
                
                  
                    F
                    
                      7
                      ;
                      11
                    
                  
                  ⁢
                  
                    B
                    11
                  
                
                +
                
                  
                    F
                    
                      7
                      ;
                      12
                    
                  
                  ⁢
                  
                    G
                    12
                  
                  ⁢
                  
                    {
                    
                      
                        
                          F
                          
                            4
                            ;
                            8
                          
                        
                        ⁡
                        
                          (
                          
                            
                              
                                F
                                
                                  8
                                  ;
                                  7
                                
                              
                              ⁢
                              
                                X
                                7
                              
                            
                            +
                            
                              F
                              
                                8
                                ;
                                D
                              
                            
                          
                          )
                        
                      
                      ,
                      
                        (
                        
                          
                            
                              F
                              
                                6
                                ;
                                1
                              
                            
                            ⁢
                            
                              B
                              1
                            
                          
                          +
                          
                            
                              F
                              
                                6
                                ;
                                2
                              
                            
                            ⁢
                            
                              B
                              2
                            
                          
                          +
                          
                            
                              F
                              
                                6
                                ;
                                3
                              
                            
                            ⁢
                            
                              X
                              3
                            
                          
                          +
                          
                            
                              F
                              
                                6
                                ;
                                5
                              
                            
                            ⁢
                            
                              F
                              
                                5
                                ;
                                9
                              
                            
                            ⁢
                            
                              
                                F
                                
                                  9
                                  ;
                                  10
                                
                              
                              ⁡
                              
                                (
                                
                                  
                                    
                                      F
                                      
                                        10
                                        ;
                                        11
                                      
                                    
                                    ⁢
                                    
                                      B
                                      11
                                    
                                  
                                  +
                                  
                                    
                                      F
                                      
                                        10
                                        ;
                                        7
                                      
                                    
                                    ⁢
                                    
                                      X
                                      7
                                    
                                  
                                
                                )
                              
                            
                          
                        
                        )
                      
                    
                    }
                  
                
              
              )
            
          
          =
          
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  11
                
              
              ⁢
              
                B
                11
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        1
                      
                    
                    ⁢
                    
                      B
                      1
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        2
                      
                    
                    ⁢
                    
                      B
                      2
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        3
                      
                    
                    ⁢
                    
                      X
                      3
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁢
                    
                      F
                      
                        5
                        ;
                        9
                      
                    
                    ⁢
                    
                      F
                      
                        9
                        ;
                        10
                      
                    
                    ⁢
                    
                      F
                      
                        10
                        ;
                        11
                      
                    
                    ⁢
                    
                      B
                      11
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁢
                    
                      F
                      
                        5
                        ;
                        9
                      
                    
                    ⁢
                    
                      F
                      
                        9
                        ;
                        10
                      
                    
                    ⁢
                    
                      F
                      
                        10
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        D
                      
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        1
                      
                    
                    ⁢
                    
                      B
                      1
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        D
                      
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        2
                      
                    
                    ⁢
                    
                      B
                      2
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        D
                      
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        3
                      
                    
                    ⁢
                    
                      X
                      3
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        D
                      
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁢
                    
                      F
                      
                        5
                        ;
                        9
                      
                    
                    ⁢
                    
                      F
                      
                        9
                        ;
                        10
                      
                    
                    ⁢
                    
                      F
                      
                        10
                        ;
                        11
                      
                    
                    ⁢
                    
                      B
                      11
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  3
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        D
                      
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁢
                    
                      F
                      
                        5
                        ;
                        9
                      
                    
                    ⁢
                    
                      F
                      
                        9
                        ;
                        10
                      
                    
                    ⁢
                    
                      F
                      
                        10
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                
                }
              
            
          
        
      
    
    
      
        (
        2.3
        )
      
    
  




In the above equation, the probabilities are in the summation relation. For every item, if all the variables are independent of each other (if the DUCG is singly connected, the variables in any item are independent of each other), then the data matrixes can be used to do the probability calculation. If there is repeated variable, e.g. the sixth item in the above equation,

F3;7F7;12G12{F4;8F8;7X7,F6;5F5;9F9;10F10;7X7}

the outspread should be based on the event level. After removing the repeated events, the probability calculation can be performed.

The data matrix calculation can also be applied. But the calculation rules are different from the ordinary matrix calculation, so as to combine the event matrix calculation and the data matrix calculation together. For this example, take X7 out of { } in the above equation, we get

G12{F4;8F8;7X7,F6;5F5;9F9;10F10;7X7}=G12{(F4;8F8;7F6;5F5;9F9;10F10;7)X7}

In which, if there is no change in the items in { }, the data matrix can be used to do the calculation. The result must be the data matrix with same number of rows but possible different number of columns, while the calculated matrixes are divided by “,”. These matrixes are fused according to the complete combination mode. The fused matrix has the same columns, but the number of rows equal to the multiplication of the numbers of the rows of the matrixes before the fusion. The new matrix is then calculated with the data matrixes of the common taken out variables. The method of fusion is: the elements in the first row of the first matrix are multiplied with the corresponding elements of all the rows (suppose there are J rows) respectively of the second matrix, so as to get the first J rows of the resulted matrix; then the elements of the second row of the first matrix are multiplied with the corresponding elements of the J rows respectively of the second matrix, so as to get the second J rows of the resulted matrix; . . . ; the elements of the last row of the first matrix are multiplied with the corresponding elements of the J rows respectively of the J rows of the second matrix, so as to get the last J rows of the resulted matrix. Thus the first two matrixes are fused as one matrix. Taking this matrix as the first matrix, we can perform the same calculation to the following matrixes, until all matrixes are fused as one matrix. Then the resulted matrix can be calculated with the data matrixes of the common taken out variables.

Although this calculation method is a new invention, it is only a method of mathematics and therefore out of the claims, nor explained in details. However, it is easy for the professionals to see the general law related to this method.
At the event level, the outspread can be done. And then the numerical calculation can be performed after removing the repeated events:




  
    
      
        
          x
          
            3
            ⁢
            k
          
        
        =
        
          
            
              ∑
              
                j
                ,
                i
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                f
                
                  
                    7
                    ⁢
                    j
                  
                  ;
                  
                    11
                    ⁢
                    i
                  
                
              
              ⁢
              
                b
                
                  11
                  ⁢
                  i
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          8
                          ⁢
                          m
                        
                        ;
                        
                          7
                          ⁢
                          h
                        
                      
                    
                    ⁢
                    
                      
                        x
                        
                          7
                          ⁢
                          h
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            1
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      b
                      
                        1
                        ⁢
                        g
                      
                    
                  
                  )
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
              
            
            ⁢
            
              
                f
                
                  
                    k
                    ⁢
                    
                      
                    
                    ⁢
                    3
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          8
                          ⁢
                          m
                        
                        ;
                        
                          7
                          ⁢
                          h
                        
                      
                    
                    ⁢
                    
                      
                        x
                        
                          7
                          ⁢
                          h
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            2
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      b
                      
                        2
                        ⁢
                        g
                      
                    
                  
                  )
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          8
                          ⁢
                          m
                        
                        ;
                        
                          7
                          ⁢
                          h
                        
                      
                    
                    ⁢
                    
                      
                        x
                        
                          7
                          ⁢
                          h
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            3
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      x
                      
                        3
                        ⁢
                        g
                      
                    
                  
                  )
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
                ,
                z
                ,
                η
                ,
                β
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          8
                          ⁢
                          m
                        
                        ;
                        
                          7
                          ⁢
                          h
                        
                      
                    
                    ⁢
                    
                      
                        x
                        
                          7
                          ⁢
                          h
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            5
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          5
                          ⁢
                          g
                        
                        ;
                        
                          9
                          ⁢
                          z
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          9
                          ⁢
                          z
                        
                        ;
                        
                          10
                          ⁢
                          
                            
                          
                          ⁢
                          η
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          10
                          ⁢
                          η
                        
                        ;
                        
                          11
                          ⁢
                          β
                        
                      
                    
                    ⁢
                    
                      b
                      
                        11
                        ⁢
                        β
                      
                    
                  
                  )
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
                ,
                z
                ,
                n
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      
                        f
                        
                          
                            8
                            ⁢
                            m
                          
                          ;
                          
                            7
                            ⁢
                            b
                          
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            5
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          5
                          ⁢
                          g
                        
                        ;
                        
                          9
                          ⁢
                          z
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          9
                          ⁢
                          z
                        
                        ;
                        
                          10
                          ⁢
                          η
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          10
                          ⁢
                          η
                        
                        ;
                        
                          7
                          ⁢
                          h
                        
                      
                    
                  
                  )
                
              
              ⁢
              
                x
                
                  7
                  ⁢
                  h
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      
                        f
                        
                          8
                          ;
                          Dh
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            1
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      b
                      
                        1
                        ⁢
                        g
                      
                    
                  
                  )
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      
                        f
                        
                          8
                          ;
                          Dh
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            2
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      b
                      
                        2
                        ⁢
                        g
                      
                    
                  
                  )
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      
                        f
                        
                          8
                          ;
                          Dh
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            3
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      x
                      
                        3
                        ⁢
                        g
                      
                    
                  
                  )
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
                ,
                z
                ,
                η
                ,
                β
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      
                        f
                        
                          8
                          ;
                          Dh
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            5
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          5
                          ⁢
                          g
                        
                        ;
                        
                          9
                          ⁢
                          z
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          9
                          ⁢
                          z
                        
                        ;
                        
                          10
                          ⁢
                          η
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          10
                          ⁢
                          η
                        
                        ;
                        
                          11
                          ⁢
                          β
                        
                      
                    
                    ⁢
                    
                      b
                      
                        11
                        ⁢
                        β
                      
                    
                  
                  )
                
              
            
          
          +
          
            
              ∑
              
                j
                ,
                i
                ,
                n
                ,
                m
                ,
                h
                ,
                y
                ,
                g
                ,
                z
                ,
                η
                ,
                β
              
            
            ⁢
            
              
                f
                
                  
                    3
                    ⁢
                    k
                  
                  ;
                  
                    7
                    ⁢
                    j
                  
                
              
              ⁢
              
                
                  f
                  
                    
                      7
                      ⁢
                      j
                    
                    ;
                    
                      12
                      ⁢
                      i
                    
                  
                
                ⁡
                
                  (
                  
                    
                      f
                      
                        
                          4
                          ⁢
                          n
                        
                        ;
                        
                          8
                          ⁢
                          m
                        
                      
                    
                    ⁢
                    
                      
                        f
                        
                          8
                          ;
                          Dh
                        
                      
                      ·
                      
                        f
                        
                          
                            6
                            ⁢
                            y
                          
                          ;
                          
                            5
                            ⁢
                            g
                          
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          5
                          ⁢
                          g
                        
                        ;
                        
                          9
                          ⁢
                          z
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          9
                          ⁢
                          z
                        
                        ;
                        
                          10
                          ⁢
                          η
                        
                      
                    
                    ⁢
                    
                      f
                      
                        
                          10
                          ⁢
                          η
                        
                        ;
                        
                          7
                          ⁢
                          β
                        
                      
                    
                    ⁢
                    
                      x
                      
                        7
                        ⁢
                        β
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        (
        2.4
        )
      
    
  




In which every row and every item in the row are simply in the probability summation relation. All variables in an item are independent of each other, and are simply in the probability multiplication relation. The variables X3g and X7h are the values of the earlier moment, and are therefore independent of the variables already appeared in the items. Moreover, the operator “·” indicates the numerical multiplication, and meanwhile indicates the parallel in terms of time and causality.

In the same way,




  
    
      
        
          X
          8
        
        =
        
          
            
              F
              
                8
                ;
                D
              
            
            +
            
              (
              
                
                  F
                  
                    8
                    ;
                    7
                  
                
                ⁡
                
                  (
                  
                    
                      
                        F
                        
                          7
                          ;
                          11
                        
                      
                      ⁢
                      
                        B
                        11
                      
                    
                    +
                    
                      
                        F
                        
                          7
                          ;
                          12
                        
                      
                      ⁢
                      
                        G
                        12
                      
                      ⁢
                      
                        {
                        
                          
                            
                              F
                              
                                4
                                ;
                                8
                              
                            
                            ⁢
                            
                              X
                              8
                            
                          
                          ,
                          
                            (
                            
                              
                                
                                  F
                                  
                                    6
                                    ;
                                    1
                                  
                                
                                ⁢
                                
                                  B
                                  1
                                
                              
                              +
                              
                                
                                  F
                                  
                                    6
                                    ;
                                    2
                                  
                                
                                ⁢
                                
                                  B
                                  2
                                
                              
                              +
                              
                                
                                  F
                                  
                                    6
                                    ;
                                    3
                                  
                                
                                ⁢
                                
                                  F
                                  
                                    3
                                    ;
                                    7
                                  
                                
                                ⁢
                                
                                  X
                                  7
                                
                              
                              +
                              
                                
                                  F
                                  
                                    6
                                    ;
                                    5
                                  
                                
                                ⁢
                                
                                  F
                                  
                                    5
                                    ;
                                    9
                                  
                                
                                ⁢
                                
                                  
                                    F
                                    
                                      9
                                      ;
                                      10
                                    
                                  
                                  ⁡
                                  
                                    (
                                    
                                      
                                        
                                          F
                                          
                                            10
                                            ;
                                            7
                                          
                                        
                                        ⁢
                                        
                                          X
                                          7
                                        
                                      
                                      +
                                      
                                        
                                          F
                                          
                                            10
                                            ;
                                            11
                                          
                                        
                                        ⁢
                                        
                                          B
                                          11
                                        
                                      
                                    
                                    )
                                  
                                
                              
                            
                            )
                          
                        
                        }
                      
                    
                  
                  )
                
              
              )
            
          
          =
          
            
              F
              
                8
                ;
                D
              
            
            +
            
              
                F
                
                  8
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  11
                
              
              ⁢
              
                B
                11
              
            
            +
            
              
                F
                
                  8
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      X
                      8
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        1
                      
                    
                    ⁢
                    
                      B
                      1
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  8
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      X
                      8
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        2
                      
                    
                    ⁢
                    
                      B
                      2
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  8
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      X
                      8
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        3
                      
                    
                    ⁢
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  8
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      X
                      8
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁢
                    
                      F
                      
                        5
                        ;
                        9
                      
                    
                    ⁢
                    
                      F
                      
                        9
                        ;
                        10
                      
                    
                    ⁢
                    
                      F
                      
                        10
                        ;
                        7
                      
                    
                    ⁢
                    
                      X
                      7
                    
                  
                
                }
              
            
            +
            
              
                F
                
                  8
                  ;
                  7
                
              
              ⁢
              
                F
                
                  7
                  ;
                  12
                
              
              ⁢
              
                G
                12
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      X
                      8
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁢
                    
                      F
                      
                        5
                        ;
                        9
                      
                    
                    ⁢
                    
                      F
                      
                        9
                        ;
                        10
                      
                    
                    ⁢
                    
                      F
                      
                        10
                        ;
                        11
                      
                    
                    ⁢
                    
                      B
                      11
                    
                  
                
                }
              
            
          
        
      
    
    
      
        (
        2.5
        )
      
    
  




In which, X7 and X8 on the right side must be the values in the earlier moment.

1. The Static Case
In the static case, the cause cannot be the consequence. Therefore, X3 and X7 in equation (2.3) are all viewed as null. Then, equation (2.3) becomes

X3=F3;7F7;11B11+F3;7F7;12G12{F4;8F8;D,F6;1B1}+F3;7F7;12G12{F4;8F8;D,F6;2B2}+F3;7F7;12G12{F4;8F8;D,F6;5F5;9F9;10F10;11B11}  (2.6)

In which, all variables in any item are independent of each other, and therefore the data matrixes can be used in the calculation directly.

In the above equation, X3 as one of the input variables of X6 is eliminated due to breaking cycle, leading to r6=3. X7 as one of the input variables of X8 is eliminated due to breaking cycle, leading to r8=1.4. X7 as one of the input variables of X10 is eliminated due to breaking cycle, leading to r10=1.
Since r3;7/r3=1 and r10;11/r10=1, we know F3;7=A3;7 and F10;11=A10;11. Moreover, since r7;11/r7=1⁄3, we have




  
    
      
        
          
            
              
                F
                
                  7
                  ;
                  11
                
              
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  A
                  
                    7
                    ;
                    11
                  
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        Furthermore
        ,
      
    
    
      
        
      
    
  
  
    
      
        
          
            
              
                
                  F
                  
                    3
                    ;
                    7
                  
                
                ⁢
                
                  F
                  
                    7
                    ;
                    11
                  
                
                ⁢
                
                  B
                  11
                
              
              =
              
              ⁢
              
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
                ⁢
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      
                        1
                      
                    
                    
                      
                        0
                      
                    
                    
                      
                        0
                      
                    
                  
                  )
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        
          
            
              
                
                  F
                  
                    4
                    ;
                    8
                  
                
                ⁢
                
                  F
                  
                    8
                    ;
                    D
                  
                
              
              =
              
              ⁢
              
                
                  
                    A
                    
                      4
                      ;
                      8
                    
                  
                  ⁡
                  
                    (
                    
                      
                        r
                        
                          8
                          ;
                          D
                        
                      
                      /
                      
                        r
                        8
                      
                    
                    )
                  
                
                ⁢
                
                  A
                  
                    8
                    ;
                    D
                  
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
                ⁢
                
                  1.4
                  1.4
                
                ⁢
                
                  (
                  
                    
                      
                        
                          5
                          7
                        
                      
                    
                    
                      
                        
                          1
                          7
                        
                      
                    
                    
                      
                        
                          1
                          7
                        
                      
                    
                  
                  )
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                (
                
                  
                    
                      0.2
                    
                  
                  
                    
                      0.3
                    
                  
                  
                    
                      0.5
                    
                  
                
                )
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        
          
            
              
                
                  F
                  
                    6
                    ;
                    1
                  
                
                ⁢
                
                  B
                  1
                
              
              =
              
              ⁢
              
                
                  F
                  
                    6
                    ;
                    2
                  
                
                ⁢
                
                  B
                  2
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  (
                  
                    
                      r
                      
                        6
                        ;
                        2
                      
                    
                    /
                    
                      r
                      6
                    
                  
                  )
                
                ⁢
                
                  A
                  
                    6
                    ;
                    2
                  
                
                ⁢
                
                  B
                  2
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      
                        0.3
                      
                    
                    
                      
                        0.4
                      
                    
                    
                      
                        0.3
                      
                    
                  
                  )
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        
          
            
              
                
                  G
                  12
                
                ⁢
                
                  {
                  
                    
                      
                        F
                        
                          4
                          ;
                          8
                        
                      
                      ⁢
                      
                        F
                        
                          8
                          ;
                          D
                        
                      
                    
                    ,
                    
                      
                        F
                        
                          6
                          ;
                          1
                        
                      
                      ⁢
                      
                        B
                        1
                      
                    
                  
                  }
                
              
              =
              
              ⁢
              
                
                  G
                  12
                
                ⁢
                
                  {
                  
                    
                      
                        F
                        
                          4
                          ;
                          8
                        
                      
                      ⁢
                      
                        F
                        
                          8
                          ;
                          D
                        
                      
                    
                    ,
                    
                      
                        F
                        
                          6
                          ;
                          2
                        
                      
                      ⁢
                      
                        B
                        2
                      
                    
                  
                  }
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  
                    (
                    
                      
                        
                          0.2
                        
                      
                      
                        
                          0.3
                        
                      
                      
                        
                          0.5
                        
                      
                    
                    )
                  
                  ·
                  
                    1
                    3
                  
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                    
                  
                  )
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.04
                      
                    
                    
                      
                        0.06
                      
                    
                    
                      
                        0.1
                      
                    
                    
                      
                        0.06
                      
                    
                    
                      
                        0.09
                      
                    
                    
                      
                        0.15
                      
                    
                    
                      
                        0.1
                      
                    
                    
                      
                        0.15
                      
                    
                    
                      
                        0.25
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        
          
            
              
                
                  F
                  
                    3
                    ;
                    7
                  
                
                ⁢
                
                  F
                  
                    7
                    ;
                    12
                  
                
                ⁢
                
                  G
                  12
                
                ⁢
                
                  {
                  
                    
                      
                        F
                        
                          4
                          ;
                          8
                        
                      
                      ⁢
                      
                        F
                        
                          8
                          ;
                          D
                        
                      
                    
                    ,
                    
                      
                        F
                        
                          6
                          ;
                          1
                        
                      
                      ⁢
                      
                        B
                        1
                      
                    
                  
                  }
                
              
              =
              
              ⁢
              
                
                  F
                  
                    3
                    ;
                    7
                  
                
                ⁢
                
                  F
                  
                    7
                    ;
                    12
                  
                
                ⁢
                
                  G
                  12
                
                ⁢
                
                  {
                  
                    
                      
                        F
                        
                          4
                          ;
                          8
                        
                      
                      ⁢
                      
                        F
                        
                          8
                          ;
                          D
                        
                      
                    
                    ,
                    
                      
                        F
                        
                          6
                          ;
                          2
                        
                      
                      ⁢
                      
                        B
                        2
                      
                    
                  
                  }
                
              
            
          
        
        
          
            
              =
               
              ⁢
              
                
                  
                    A
                    
                      3
                      ;
                      7
                    
                  
                  ⁡
                  
                    (
                    
                      
                        r
                        
                          7
                          ;
                          12
                        
                      
                      /
                      
                        r
                        7
                      
                    
                    )
                  
                
                ⁢
                
                  A
                  
                    7
                    ;
                    12
                  
                
                ⁢
                
                  G
                  12
                
              
            
          
        
        
          
            
              
              ⁢
              
                {
                
                  
                    
                      F
                      
                        4
                        ;
                        8
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        D
                      
                    
                  
                  ,
                  
                    
                      F
                      
                        6
                        ;
                        2
                      
                    
                    ⁢
                    
                      B
                      2
                    
                  
                
                }
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
                ⁢
                
                  2
                  3
                
              
            
          
        
        
          
            
              
              ⁢
              
                
                  (
                  
                    
                      
                        0.5
                      
                      
                        0.4
                      
                      
                        0.5
                      
                      
                        0.6
                      
                      
                        0
                      
                      
                        0.1
                      
                      
                        0.4
                      
                      
                        0.2
                      
                      
                        0.25
                      
                    
                    
                      
                        0.5
                      
                      
                        0.4
                      
                      
                        0.25
                      
                      
                        0
                      
                      
                        0.4
                      
                      
                        0.5
                      
                      
                        0.4
                      
                      
                        0.4
                      
                      
                        0.5
                      
                    
                    
                      
                        0
                      
                      
                        0.2
                      
                      
                        0.25
                      
                      
                        0.4
                      
                      
                        0.6
                      
                      
                        0.4
                      
                      
                        0.2
                      
                      
                        0.4
                      
                      
                        0.25
                      
                    
                  
                  )
                
                ⁢
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.04
                      
                    
                    
                      
                        0.06
                      
                    
                    
                      
                        0.1
                      
                    
                    
                      
                        0.06
                      
                    
                    
                      
                        0.09
                      
                    
                    
                      
                        0.15
                      
                    
                    
                      
                        0.1
                      
                    
                    
                      
                        0.15
                      
                    
                    
                      
                        0.25
                      
                    
                  
                  )
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  2
                  9
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        
          
            
              
                
                  F
                  
                    6
                    ;
                    5
                  
                
                ⁢
                
                  F
                  
                    5
                    ;
                    9
                  
                
                ⁢
                
                  F
                  
                    9
                    ;
                    10
                  
                
                ⁢
                
                  F
                  
                    10
                    ;
                    11
                  
                
                ⁢
                
                  B
                  11
                
              
              =
              
              ⁢
              
                
                  (
                  
                    
                      r
                      
                        6
                        ;
                        5
                      
                    
                    /
                    
                      r
                      6
                    
                  
                  )
                
                ⁢
                
                  A
                  
                    6
                    ;
                    5
                  
                
                ⁢
                
                  A
                  
                    5
                    ;
                    9
                  
                
                ⁢
                
                  A
                  
                    9
                    ;
                    10
                  
                
                ⁢
                
                  A
                  
                    10
                    ;
                    11
                  
                
                ⁢
                
                  B
                  11
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  
                    (
                    
                      
                        
                          0.2
                        
                        
                          0.2
                        
                        
                          0.2
                        
                      
                      
                        
                          0.3
                        
                        
                          0.3
                        
                        
                          0.3
                        
                      
                      
                        
                          0.5
                        
                        
                          0.5
                        
                        
                          0.5
                        
                      
                    
                    )
                  
                  4
                
                ⁢
                
                  (
                  
                    
                      
                        1
                      
                    
                    
                      
                        0
                      
                    
                    
                      
                        0
                      
                    
                  
                  )
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        
          
            
              
                
                  G
                  12
                
                ⁢
                
                  {
                  
                    
                      
                        F
                        
                          4
                          ;
                          8
                        
                      
                      ⁢
                      
                        F
                        
                          8
                          ;
                          D
                        
                      
                    
                    ,
                    
                      
                        F
                        
                          6
                          ;
                          5
                        
                      
                      ⁢
                      
                        F
                        
                          5
                          ;
                          9
                        
                      
                      ⁢
                      
                        F
                        
                          9
                          ;
                          10
                        
                      
                      ⁢
                      
                        F
                        
                          10
                          ;
                          11
                        
                      
                      ⁢
                      
                        B
                        11
                      
                    
                  
                  }
                
              
              =
              
              ⁢
              
                
                  
                    (
                    
                      
                        
                          0.2
                        
                      
                      
                        
                          0.3
                        
                      
                      
                        
                          0.5
                        
                      
                    
                    )
                  
                  ·
                  
                    1
                    3
                  
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                    
                  
                  )
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.04
                      
                    
                    
                      
                        0.06
                      
                    
                    
                      
                        0.1
                      
                    
                    
                      
                        0.06
                      
                    
                    
                      
                        0.09
                      
                    
                    
                      
                        0.15
                      
                    
                    
                      
                        0.1
                      
                    
                    
                      
                        0.15
                      
                    
                    
                      
                        0.25
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        
          
            
              
                
                  F
                  
                    3
                    ;
                    7
                  
                
                ⁢
                
                  F
                  
                    7
                    ;
                    12
                  
                
                ⁢
                
                  G
                  12
                
                ⁢
                
                  {
                  
                    
                      
                        F
                        
                          4
                          ;
                          8
                        
                      
                      ⁢
                      
                        F
                        
                          8
                          ;
                          D
                        
                      
                    
                    ,
                    
                      
                        F
                        
                          6
                          ;
                          5
                        
                      
                      ⁢
                      
                        F
                        
                          5
                          ;
                          9
                        
                      
                      ⁢
                      
                        F
                        
                          9
                          ;
                          10
                        
                      
                      ⁢
                      
                        F
                        
                          10
                          ;
                          11
                        
                      
                      ⁢
                      
                        B
                        11
                      
                    
                  
                  }
                
              
              =
              
              ⁢
              
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
                ⁢
                
                  2
                  3
                
              
            
          
        
        
          
            
              
              ⁢
              
                
                  (
                  
                    
                      
                        0.5
                      
                      
                        0.4
                      
                      
                        0.5
                      
                      
                        0.6
                      
                      
                        0
                      
                      
                        0.1
                      
                      
                        0.4
                      
                      
                        0.2
                      
                      
                        0.25
                      
                    
                    
                      
                        0.5
                      
                      
                        0.4
                      
                      
                        0.25
                      
                      
                        0
                      
                      
                        0.4
                      
                      
                        0.5
                      
                      
                        0.4
                      
                      
                        0.4
                      
                      
                        0.5
                      
                    
                    
                      
                        0
                      
                      
                        0.2
                      
                      
                        0.25
                      
                      
                        0.4
                      
                      
                        0.6
                      
                      
                        0.4
                      
                      
                        0.2
                      
                      
                        0.4
                      
                      
                        0.25
                      
                    
                  
                  )
                
                ⁢
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.04
                      
                    
                    
                      
                        0.06
                      
                    
                    
                      
                        0.1
                      
                    
                    
                      
                        0.06
                      
                    
                    
                      
                        0.09
                      
                    
                    
                      
                        0.15
                      
                    
                    
                      
                        0.1
                      
                    
                    
                      
                        0.15
                      
                    
                    
                      
                        0.25
                      
                    
                  
                  )
                
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  2
                  9
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                    
                  
                  )
                
              
            
          
        
      
    
    
      
        
      
    
  
  
    
      
        Finally
        ,
      
    
    
      □
    
  
  
    
      
        
          X
          3
        
        =
        
          
            
              
                1
                3
              
              ⁢
              
                (
                
                  
                    
                      0.2
                    
                  
                  
                    
                      0.3
                    
                  
                  
                    
                      0.5
                    
                  
                
                )
              
            
            +
            
              
                2
                9
              
              ⁢
              
                (
                
                  
                    
                      0.2
                    
                  
                  
                    
                      0.3
                    
                  
                  
                    
                      0.5
                    
                  
                
                )
              
            
            +
            
              
                2
                9
              
              ⁢
              
                (
                
                  
                    
                      0.2
                    
                  
                  
                    
                      0.3
                    
                  
                  
                    
                      0.5
                    
                  
                
                )
              
            
            +
            
              
                2
                9
              
              ⁢
              
                (
                
                  
                    
                      0.2
                    
                  
                  
                    
                      0.3
                    
                  
                  
                    
                      0.5
                    
                  
                
                )
              
            
          
          =
          
            (
            
              
                
                  0.2
                
              
              
                
                  0.3
                
              
              
                
                  0.5
                
              
            
            )
          
        
      
    
    
      □
    
  




In the same way, in equation (2.5), X7 and X8 are eliminated due to breaking cycles. Then





  
    
      
        
          
            
              
                X
                8
              
              =
              
              ⁢
              
                
                  F
                  
                    8
                    ;
                    D
                  
                
                +
                
                  
                    F
                    
                      8
                      ;
                      7
                    
                  
                  ⁢
                  
                    F
                    
                      7
                      ;
                      11
                    
                  
                  ⁢
                  
                    B
                    11
                  
                
                +
                
                  
                    F
                    
                      8
                      ;
                      7
                    
                  
                  ⁢
                  
                    F
                    
                      7
                      ;
                      11
                    
                  
                  ⁢
                  
                    B
                    11
                  
                
                +
                
                  
                    F
                    
                      8
                      ;
                      7
                    
                  
                  ⁢
                  
                    F
                    
                      7
                      ;
                      12
                    
                  
                
              
            
          
        
        
          
            
              
              ⁢
              
                (
                
                  
                    
                      
                        
                          
                            G
                            12
                          
                          ⁢
                          
                            {
                            
                              
                                F
                                
                                  6
                                  ;
                                  1
                                
                              
                              ⁢
                              
                                B
                                1
                              
                            
                            }
                          
                        
                        +
                        
                          
                            G
                            12
                          
                          ⁢
                          
                            {
                            
                              
                                F
                                
                                  6
                                  ;
                                  2
                                
                              
                              ⁢
                              
                                B
                                2
                              
                            
                            }
                          
                        
                        +
                      
                    
                  
                  
                    
                      
                        
                          G
                          12
                        
                        ⁢
                        
                          {
                          
                            
                              F
                              
                                6
                                ;
                                5
                              
                            
                            ⁢
                            
                              F
                              
                                5
                                ;
                                9
                              
                            
                            ⁢
                            
                              F
                              
                                9
                                ;
                                10
                              
                            
                            ⁢
                            
                              F
                              
                                10
                                ;
                                11
                              
                            
                            ⁢
                            
                              B
                              11
                            
                          
                          }
                        
                      
                    
                  
                
                )
              
            
          
        
        
          
            
              =
              
              ⁢
              
                
                  F
                  
                    8
                    ;
                    D
                  
                
                +
                
                  
                    F
                    
                      8
                      ;
                      7
                    
                  
                  ⁢
                  
                    F
                    
                      7
                      ;
                      11
                    
                  
                  ⁢
                  
                    B
                    11
                  
                
                +
                
                  
                    F
                    
                      8
                      ;
                      7
                    
                  
                  ⁢
                  
                    F
                    
                      7
                      ;
                      12
                    
                  
                
              
            
          
        
        
          
            
              
              ⁢
              
                (
                
                  
                    
                      F
                      
                        6
                        ;
                        1
                      
                    
                    ⁢
                    
                      B
                      1
                    
                  
                  +
                  
                    
                      F
                      
                        6
                        ;
                        2
                      
                    
                    ⁢
                    
                      B
                      2
                    
                  
                  +
                  
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁢
                    
                      F
                      
                        5
                        ;
                        9
                      
                    
                    ⁢
                    
                      F
                      
                        9
                        ;
                        10
                      
                    
                    ⁢
                    
                      F
                      
                        10
                        ;
                        11
                      
                    
                    ⁢
                    
                      B
                      11
                    
                  
                
                )
              
            
          
        
      
    
    
      
        (
        2.7
        )
      
    
  




In the above equation, X4 as the input of the virtual logic gate G12 is eliminated due to breaking the cycle. There is only X6 as the input of G12. Therefore, G12 can be ignored, and correspondingly, A7;12 should be recalculated according to

a7k;12,1=p7k;1=(q7k;1+q7k;4+q7k;7)/(d7;1+d7;4+d7;7)

a7k;12,2=p7k;2=(q7k;2+q7k;5+q7k;8)/(d7;2+d7;5+d7;8)

a7k;12,3=p7k;1=(q7k;3+q7k;6+q7k;9)/(d7;3+d7;6+d7;9)

In which, the values on the right side of the equator are those in table 2. Thus,





  
    
      
        
          A
          
            7
            ;
            12
          
        
        =
        
        ⁢
        
          (
          
            
              
                
                  
                    50
                    +
                    150
                    +
                    40
                  
                  
                    100
                    +
                    250
                    +
                    100
                  
                
              
              
                
                  
                    40
                    +
                    0
                    +
                    10
                  
                  
                    100
                    +
                    100
                    +
                    50
                  
                
              
              
                
                  
                    100
                    +
                    20
                    +
                    20
                  
                  
                    200
                    +
                    200
                    +
                    80
                  
                
              
            
            
              
                
                  
                    50
                    +
                    0
                    +
                    40
                  
                  
                    100
                    +
                    250
                    +
                    100
                  
                
              
              
                
                  
                    40
                    +
                    40
                    +
                    20
                  
                  
                    100
                    +
                    100
                    +
                    50
                  
                
              
              
                
                  
                    50
                    +
                    100
                    +
                    40
                  
                  
                    200
                    +
                    200
                    +
                    80
                  
                
              
            
            
              
                
                  
                    0
                    +
                    100
                    +
                    20
                  
                  
                    100
                    +
                    250
                    +
                    100
                  
                
              
              
                
                  
                    20
                    +
                    60
                    +
                    20
                  
                  
                    100
                    +
                    100
                    +
                    50
                  
                
              
              
                
                  
                    50
                    +
                    80
                    +
                    20
                  
                  
                    200
                    +
                    200
                    +
                    80
                  
                
              
            
          
          )
        
      
    
  
  
    
      
        =
        
        ⁢
        
          (
          
            
              
                0.53333
              
              
                0.2
              
              
                029167
              
            
            
              
                0.2
              
              
                0.4
              
              
                0.39583
              
            
            
              
                0.26667
              
              
                0.4
              
              
                0.3125
              
            
          
          )
        
      
    
  




Moreover, r7;12=1 and r7=2. Therefore, F7:12=(r7;12/r7)A7;12=0.5A7;12.





  
    X
    8
  
  =
  
    
      
        F
        
          8
          ;
          D
        
      
      +
      
        
          F
          
            8
            ;
            7
          
        
        ⁢
        
          F
          
            7
            ;
            11
          
        
        ⁢
        
          B
          11
        
      
      +
      
        
          F
          
            8
            ;
            7
          
        
        ⁢
        
          
            F
            
              7
              ;
              12
            
          
          ⁡
          
            (
            
              
                
                  F
                  
                    6
                    ;
                    1
                  
                
                ⁢
                
                  B
                  1
                
              
              +
              
                
                  F
                  
                    6
                    ;
                    2
                  
                
                ⁢
                
                  B
                  2
                
              
              +
              
                
                  F
                  
                    6
                    ;
                    5
                  
                
                ⁢
                
                  F
                  
                    5
                    ;
                    9
                  
                
                ⁢
                
                  F
                  
                    9
                    ;
                    10
                  
                
                ⁢
                
                  F
                  
                    10
                    ;
                    11
                  
                
                ⁢
                
                  B
                  11
                
              
            
            )
          
        
      
    
    =
    
      
        
          
            1.4
            2
          
          ⁢
          
            (
            
              
                
                  
                    5
                    7
                  
                
              
              
                
                  
                    1
                    7
                  
                
              
              
                
                  
                    1
                    7
                  
                
              
            
            )
          
        
        +
        
          
            0.6
            2
          
          ⁢
          
            (
            
              
                
                  0
                
                
                  0.8333
                
                
                  0
                
              
              
                
                  0.333
                
                
                  0
                
                
                  1
                
              
              
                
                  0.6667
                
                
                  0.1667
                
                
                  0
                
              
            
            )
          
          ⁢
          
            1
            2
          
          ⁢
          
            (
            
              
                
                  0.2
                
                
                  0.2
                
                
                  0.2
                
              
              
                
                  0.3
                
                
                  0.3
                
                
                  0.3
                
              
              
                
                  0.5
                
                
                  0.5
                
                
                  0.5
                
              
            
            )
          
          ⁢
          
            (
            
              
                
                  1
                
              
              
                
                  0
                
              
              
                
                  0
                
              
            
            )
          
        
        +
        
          0.6
          ⁢
          
            (
            
              
                
                  0
                
                
                  0.8333
                
                
                  0
                
              
              
                
                  0.3333
                
                
                  0
                
                
                  1
                
              
              
                
                  0.6667
                
                
                  0.1667
                
                
                  0
                
              
            
            )
          
          ⁢
          
            1
            2
          
          ⁢
          
            (
            
              
                
                  0.53333
                
                
                  0.2
                
                
                  0.29167
                
              
              
                
                  0.2
                
                
                  0.4
                
                
                  0.39583
                
              
              
                
                  0.26667
                
                
                  0.4
                
                
                  0.3125
                
              
            
            )
          
          ×
          
            (
            
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      
                        0.3
                      
                    
                    
                      
                        0.4
                      
                    
                    
                      
                        0.3
                      
                    
                  
                  )
                
              
              +
              
                
                  1
                  3
                
                ⁢
                
                  (
                  
                    
                      
                        0.2
                      
                      
                        0.2
                      
                      
                        0.2
                      
                    
                    
                      
                        0.3
                      
                      
                        0.3
                      
                      
                        0.3
                      
                    
                    
                      
                        0.5
                      
                      
                        0.5
                      
                      
                        0.5
                      
                    
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      
                        0.3
                      
                    
                    
                      
                        0.4
                      
                    
                    
                      
                        0.3
                      
                    
                  
                  )
                
              
              +
              
                
                  1
                  3
                
                ⁢
                
                  
                    (
                    
                      
                        
                          0.2
                        
                        
                          0.2
                        
                        
                          0.2
                        
                      
                      
                        
                          0.3
                        
                        
                          0.3
                        
                        
                          0.3
                        
                      
                      
                        
                          0.5
                        
                        
                          0.5
                        
                        
                          0.5
                        
                      
                    
                    )
                  
                  4
                
                ⁢
                
                  (
                  
                    
                      
                        1
                      
                    
                    
                      
                        0
                      
                    
                    
                      
                        0
                      
                    
                  
                  )
                
              
            
            )
          
        
      
      =
      
        (
        
          
            
              0.58223
            
          
          
            
              0.25
            
          
          
            
              0.16767
            
          
        
        )
      
    
  




If the evidence E8={X8=(0.6 0.4 0)T} is observed, i.e., the membership of X8 being in X8,1 is m8;1=0.6, the membership of X8 being in X8,2 is m8;2=0.4, the membership of X8 being in X8,3 is m8;3=0, the EDUCG after adding the evidence E8 becomes FIG. 35, in which F8;E is the functional event variable from X8 to E8.
According to the above calculation, v81=Pr{X81}=0.58223 and v82=Pr{X82}=0.25. Let f8:2=1, we have




  
    
      f
      
        8
        ;
        1
      
    
    =
    
      
        
          
            
              m
              
                8
                ;
                1
              
            
            ⁢
            
              v
              82
            
          
          
            
              m
              
                8
                ;
                2
              
            
            ⁢
            
              v
              81
            
          
        
        ⁢
        
          f
          
            8
            ;
            2
          
        
      
      =
      
        
          
            0.6
            ×
            0.25
          
          
            0.4
            ×
            0.58223
          
        
        =
        0.6441
      
    
  
  ,




i.e. F8;E=(0.6441 1 0). Then, Pr{X3|E8}=α3Pr{X3E8}, where α3 is the state normalization factor of X3.

By treating E8 as ordinary evidence, it can be outspreaded as




  
    
      
        
          E
          8
        
        =
        
        ⁢
        
          
            F
            
              8
              ;
              E
            
          
          ⁢
          
            X
            8
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            F
            
              8
              ;
              E
            
          
          ⁡
          
            (
            
              
                F
                
                  8
                  ;
                  D
                
              
              +
              
                
                  F
                  
                    8
                    ;
                    7
                  
                
                ⁢
                
                  F
                  
                    7
                    ;
                    11
                  
                
                ⁢
                
                  B
                  11
                
              
              +
              
                
                  F
                  
                    8
                    ;
                    7
                  
                
                ⁢
                
                  
                    F
                    
                      7
                      ;
                      12
                    
                  
                  ⁡
                  
                    (
                    
                      
                        
                          G
                          12
                        
                        ⁢
                        
                          {
                          
                            
                              F
                              
                                6
                                ;
                                1
                              
                            
                            ⁢
                            
                              B
                              1
                            
                          
                          }
                        
                      
                      +
                      
                        
                          G
                          12
                        
                        ⁢
                        
                          {
                          
                            
                              F
                              
                                6
                                ;
                                2
                              
                            
                            ⁢
                            
                              B
                              2
                            
                          
                          }
                        
                      
                      +
                      
                        
                          G
                          12
                        
                        ⁢
                        
                          {
                          
                            
                              F
                              
                                6
                                ;
                                5
                              
                            
                            ⁢
                            
                              F
                              
                                5
                                ;
                                9
                              
                            
                            ⁢
                            
                              F
                              
                                9
                                ;
                                10
                              
                            
                            ⁢
                            
                              F
                              
                                10
                                ;
                                11
                              
                            
                            ⁢
                            
                              B
                              11
                            
                          
                          }
                        
                      
                    
                    )
                  
                
              
            
            )
          
        
      
    
  




Then,





  
    
      
        
          
            X
            3
          
          ⁢
          
            E
            8
          
        
        =
        
          
            (
            
              
                
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        11
                      
                    
                    ⁢
                    
                      B
                      11
                    
                  
                
              
              
                
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁢
                          
                            F
                            
                              8
                              ;
                              D
                            
                          
                        
                        ,
                        
                          
                            F
                            
                              6
                              ;
                              1
                            
                          
                          ⁢
                          
                            B
                            1
                          
                        
                      
                      }
                    
                  
                
              
              
                
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁢
                          
                            F
                            
                              8
                              ;
                              D
                            
                          
                        
                        ,
                        
                          
                            F
                            
                              6
                              ;
                              2
                            
                          
                          ⁢
                          
                            B
                            2
                          
                        
                      
                      }
                    
                  
                
              
              
                
                  
                    
                      F
                      
                        3
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          
                            F
                            
                              4
                              ;
                              8
                            
                          
                          ⁢
                          
                            F
                            
                              8
                              ;
                              D
                            
                          
                        
                        ,
                        
                          
                            F
                            
                              6
                              ;
                              5
                            
                          
                          ⁢
                          
                            F
                            
                              5
                              ;
                              9
                            
                          
                          ⁢
                          
                            F
                            
                              9
                              ;
                              10
                            
                          
                          ⁢
                          
                            F
                            
                              10
                              ;
                              11
                            
                          
                          ⁢
                          
                            B
                            11
                          
                        
                      
                      }
                    
                  
                
              
            
            )
          
          *
          
            (
            
              
                
                  
                    
                      F
                      
                        8
                        ;
                        E
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        D
                      
                    
                  
                
              
              
                
                  
                    
                      F
                      
                        8
                        ;
                        E
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        11
                      
                    
                    ⁢
                    
                      B
                      11
                    
                  
                
              
              
                
                  
                    
                      F
                      
                        8
                        ;
                        E
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          F
                          
                            6
                            ;
                            1
                          
                        
                        ⁢
                        
                          B
                          1
                        
                      
                      }
                    
                  
                
              
              
                
                  
                    
                      F
                      
                        8
                        ;
                        B
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          F
                          
                            6
                            ;
                            2
                          
                        
                        ⁢
                        
                          B
                          2
                        
                      
                      }
                    
                  
                
              
              
                
                  
                    
                      F
                      
                        8
                        ;
                        E
                      
                    
                    ⁢
                    
                      F
                      
                        8
                        ;
                        7
                      
                    
                    ⁢
                    
                      F
                      
                        7
                        ;
                        12
                      
                    
                    ⁢
                    
                      G
                      12
                    
                    ⁢
                    
                      {
                      
                        
                          F
                          
                            6
                            ;
                            5
                          
                        
                        ⁢
                        
                          F
                          
                            5
                            ;
                            9
                          
                        
                        ⁢
                        
                          F
                          
                            9
                            ;
                            10
                          
                        
                        ⁢
                        
                          F
                          
                            10
                            ;
                            11
                          
                        
                        ⁢
                        
                          B
                          11
                        
                      
                      }
                    
                  
                
              
            
            )
          
        
      
    
    
      
        (
        2.8
        )
      
    
  




The operator “*” means that all elements of the two matrixes are multiplied with each other crossly and then sum up.

In the event level of outspread, according to the rules of simplification: given j≠j′, k≠k′ and V∈{X,B}, there are

VijVij=Vij 

VijVij′=0

Fnk;ijFnk′;ij=0

Fnk;ijFnk;ij′=0

Fnk;ijFnk′;ij′=0

We can obtain the final vector matrix of equation (2.8), in which every element is a logic expression of the sum-of-products, where sum means exclusive and product means item. All the events in any item are independent of each other. Their probabilities can be used directly to calculate the expression.

Some calculation skill may be employed to perform the outspread of the data matrixes in equation (2.8):

X3E8=F3;7F7;11B11F8;EF8;D+F3;7(F8;EF8;7F7;11B11)+F3;7F7;11B11F8;EF8;7F7;12F6;1B1+F3;7F7;11B11FE;8F8;7F7;12F6;2B2+F3;7F7;11(F8;EF8;7F7;12F6;5F5;9F9;10F10;11B11)+F3;7F7;12G12{F4;8,F6;1B1}(F8;EF8;D)+F3;7F7;12G12{F4;8F8;D,F6;1B1}F8;EF8;7F7;11B11+ . . . +F3;7(F8;EF8;7F7;12G12{F4;8F8;D,F6;5F5;9F9;10F10;11B11})  (2.9)

There are 4×5=20 items in equation (2.9), in which the second item is resulted from the multiplication of F3;7F7;11B11 and F8;EF8;7F7;11B11. Since F7;11B11 is the common item of both, it is taken out and put after the operator . The non-common items including F8;E are put before  The two sides before and after  are put in ( ). The two sides are calculated respectively. The result before  is a row vector. The result after  is a column vector. They have the same number of elements. However,  means only the multiplication of the corresponding elements but not the summation. Such multiplied items compose a new column vector. For example, suppose the vectors before and after  are respectively

(θ1θ2 . . . θn) and (π1 π2 . . . πn)T 

Then,

(θ1 θ2 . . . θn)(π1 π2 . . . πn)T=(θ1π1 θ2π2 . . . θnπn)T 

The sixth item F3;7F7;12G12{F4;8,F6;1B1}(F8;EF8;D) in (2.9) is the multiplication of F3;7F7;12G12{F4;8F8;D,F6;1B1} and F8;EF8;E. Since F8;D is the common item, it is taken out after  with F8;E in ( ). After F8;D is taken out, the left item before “,” in “F4;8,F6;1B1” is not a column vector, but a matrix with multiple rows and columns. The calculation of every column vector and F6;1B1 is done according to the special operator “·”. The result is a row vector. This operator is defined in the form of example as follow: suppose




  
    F
    =
    
      
        
          (
          
            
              
                
                  f
                  11
                
              
              
                
                  f
                  12
                
              
            
            
              
                
                  f
                  21
                
              
              
                
                  f
                  22
                
              
            
          
          )
        
        ⁢
        
          
        
        ⁢
        and
        ⁢
        
          
        
        ⁢
        B
      
      =
      
        (
        
          
            
              
                b
                1
              
            
          
          
            
              
                b
                2
              
            
          
        
        )
      
    
  




  then




  
    
      F
      ·
      B
    
    =
    
      
        
          (
          
            
              
                
                  f
                  11
                
              
              
                
                  f
                  12
                
              
            
            
              
                
                  f
                  21
                
              
              
                
                  f
                  22
                
              
            
          
          )
        
        ·
        
          (
          
            
              
                
                  b
                  1
                
              
            
            
              
                
                  b
                  2
                
              
            
          
          )
        
      
      =
      
        (
        
          
            
              
                
                  f
                  11
                
                ⁢
                
                  b
                  1
                
              
            
            
              
                
                  f
                  12
                
                ⁢
                
                  b
                  1
                
              
            
          
          
            
              
                
                  f
                  11
                
                ⁢
                
                  b
                  2
                
              
            
            
              
                
                  f
                  12
                
                ⁢
                
                  b
                  2
                
              
            
          
          
            
              
                
                  f
                  21
                
                ⁢
                
                  b
                  1
                
              
            
            
              
                
                  f
                  22
                
                ⁢
                
                  b
                  1
                
              
            
          
          
            
              
                
                  f
                  21
                
                ⁢
                
                  b
                  2
                
              
            
            
              
                
                  f
                  22
                
                ⁢
                
                  b
                  2
                
              
            
          
        
        )
      
    
  



The 20th item in (2.9) is F3;7(F8;EF8;7F7;12G12{F4;8F8;D,F6;5F5;9F9;10F10;11B11}), which is the multiplication of the following two items:

F3;7F7;12G12{F4;8F8;D,F6;5F5;9F9;10F10;11B11} and F8;EF8;7F7;12G12{F6;5F5;9F9;10F10;11B11}

The common item F7;12G12{F4;8F8;D,F6;5F5;9F9;10F10,11B11} of both is taken out and put after  in ( ), while before  is F8;EF8;7 (note that the intersection of G12{F4;8F8;E,F6;5F5;9F9;10F10;11B11} and G12{F6;5F5;9F9;10F10;11B11} is former).

In this way, the value of X3E8 can be calculated. The result must be a column vector. Normalize this vector (i.e. every element is multiplied with the normalization factor α that is the reciprocal of the sum of all the elements in this vector), we have the result of Pr{X3|E8}.
As mentioned earlier, the calculation method that combines the event matrix operation and the data matrix together by outspreading equation (2.8) in terms of data matrix is also an innovation. However, since it is only a method of mathematics, it is not included in the claims in this invention.
2. The Dynamical Case
The problem in the dynamical case is: predict the probability distributions of the variables being in various possible states after a period of time T from taking some economic policy at time t=t1. The purpose is to evaluate, according to these predictions, the effects of these different economic policies, so as to provide the gist for the economic policy decision. In this case, during the period of time 0≦t≦tl, the states of all the variables are known, including the states in the fuzzy areas between two states. For example, the membership of the storage state variable being in state 1 is 0.4, while being in state 2 is 0.6, etc. After t>t1, these variables influence each other, and there is some delay in these influences, while the basic event variables change dynamically and independently. All these changes affect each other and influence the dynamical changes of these variables synthetically. In which, the economic policy variable is controlled by people.
In the dynamical case, there is time delay in the F type variables, and the information can circulate in the same chain. Then the F, X and B type variables are all the functions of time. Within the period of time T, the cause variables will influence the consequence variables sustainably and changeably. Then, equation (1.1) can be briefly written as:




  
    
      
        
          
            
              
                X
                3
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      3
                      ;
                      7
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      3
                      ;
                      7
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    X
                    7
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            3
                            ;
                            7
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
              
                X
                4
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      4
                      ;
                      8
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      4
                      ;
                      8
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    X
                    8
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            4
                            ;
                            8
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
              
                X
                5
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      5
                      ;
                      9
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      5
                      ;
                      9
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    X
                    9
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            5
                            ;
                            9
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
              
                X
                6
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    
                      W
                      
                        6
                        ;
                        1
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      F
                      
                        6
                        ;
                        1
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        ,
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      B
                      1
                    
                    ⁡
                    
                      (
                      t
                      )
                    
                  
                  ⁢
                  
                    
                      d
                      t
                    
                    /
                    
                      
                        ∫
                        0
                        T
                      
                      ⁢
                      
                        
                          
                            W
                            
                              3
                              ;
                              7
                            
                          
                          ⁡
                          
                            (
                            
                              T
                              -
                              t
                            
                            )
                          
                        
                        ⁢
                        
                          d
                          t
                        
                      
                    
                  
                
              
              +
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    
                      W
                      
                        6
                        ;
                        2
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      F
                      
                        6
                        ;
                        2
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        ,
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      B
                      2
                    
                    ⁡
                    
                      (
                      t
                      )
                    
                  
                  ⁢
                  
                    
                      d
                      t
                    
                    /
                    
                      
                        ∫
                        0
                        T
                      
                      ⁢
                      
                        
                          
                            W
                            
                              6
                              ;
                              2
                            
                          
                          ⁡
                          
                            (
                            
                              T
                              -
                              t
                            
                            )
                          
                        
                        ⁢
                        
                          d
                          t
                        
                      
                    
                  
                
              
              +
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    
                      W
                      
                        6
                        ;
                        3
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      F
                      
                        6
                        ;
                        3
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        ,
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      X
                      3
                    
                    ⁡
                    
                      (
                      t
                      )
                    
                  
                  ⁢
                  
                    
                      d
                      t
                    
                    /
                    
                      
                        ∫
                        0
                        T
                      
                      ⁢
                      
                        
                          
                            W
                            
                              6
                              ;
                              3
                            
                          
                          ⁡
                          
                            (
                            
                              T
                              -
                              t
                            
                            )
                          
                        
                        ⁢
                        
                          d
                          t
                        
                      
                    
                  
                
              
              +
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    
                      W
                      
                        6
                        ;
                        5
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      F
                      
                        6
                        ;
                        5
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        ,
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      X
                      5
                    
                    ⁡
                    
                      (
                      t
                      )
                    
                  
                  ⁢
                  
                    
                      d
                      t
                    
                    /
                    
                      
                        ∫
                        0
                        T
                      
                      ⁢
                      
                        
                          
                            W
                            
                              6
                              ;
                              5
                            
                          
                          ⁡
                          
                            (
                            
                              T
                              -
                              t
                            
                            )
                          
                        
                        ⁢
                        
                          d
                          t
                        
                      
                    
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
              X
              7
            
            ⁡
            
              (
              T
              )
            
          
          =
          
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      7
                      ;
                      11
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      7
                      ;
                      11
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    B
                    11
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            7
                            ;
                            11
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
            +
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      7
                      ;
                      12
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      7
                      ;
                      12
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                ⁢
                
                  G
                  12
                
                ⁢
                
                  {
                  
                    
                      
                        U
                        
                          12
                          ;
                          4
                        
                      
                      ⁢
                      
                        
                          X
                          4
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                    
                    ,
                    
                      
                        U
                        
                          12
                          ;
                          7
                        
                      
                      ⁢
                      
                        
                          X
                          6
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                    
                  
                  }
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            7
                            ;
                            12
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
          
        
        ⁢
        
          

        
        ⁢
        
          
            
              X
              8
            
            ⁡
            
              (
              T
              )
            
          
          =
          
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      8
                      ;
                      7
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      8
                      ;
                      7
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    X
                    7
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            8
                            ;
                            7
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
            +
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      8
                      ;
                      D
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      8
                      ;
                      D
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                ⁢
                
                  D
                  8
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            8
                            ;
                            D
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
          
        
        ⁢
        
          

        
        ⁢
        
          
            
              X
              9
            
            ⁡
            
              (
              T
              )
            
          
          =
          
            
              ∫
              0
              T
            
            ⁢
            
              
                
                  W
                  
                    9
                    ;
                    10
                  
                
                ⁡
                
                  (
                  
                    T
                    -
                    t
                  
                  )
                
              
              ⁢
              
                
                  F
                  
                    9
                    ;
                    10
                  
                
                ⁡
                
                  (
                  
                    T
                    ,
                    t
                  
                  )
                
              
              ⁢
              
                
                  X
                  10
                
                ⁡
                
                  (
                  t
                  )
                
              
              ⁢
              
                
                  d
                  t
                
                /
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        W
                        
                          9
                          ;
                          10
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
            
          
        
        ⁢
        
          

        
        ⁢
        
          
            
              X
              10
            
            ⁡
            
              (
              T
              )
            
          
          =
          
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      10
                      ;
                      7
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      10
                      ;
                      7
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    X
                    7
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            10
                            ;
                            7
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
            +
            
              
                ∫
                0
                T
              
              ⁢
              
                
                  
                    W
                    
                      10
                      ;
                      11
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    F
                    
                      10
                      ;
                      11
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    B
                    11
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ⁢
                
                  
                    d
                    t
                  
                  /
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          W
                          
                            10
                            ;
                            11
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
              
            
          
        
      
    
    
      
        (
        2.10
        )
      
    
  




In which, U and D are inevitable events and are independent of time, wn;i(T−t) is the weighing factor. Usually, as the time of the cause variable states is getting more and more close to the time T, the weight becomes larger. For example,





  
    
      
        {
        
          
            
              
                
                  
                    
                      w
                      
                        n
                        ;
                        i
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  =
                  
                    
                      
                        τ
                        
                          n
                          ;
                          i
                        
                        2
                      
                      -
                      
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                        2
                      
                    
                  
                
                ,
              
            
            
              
                t
                ≥
                
                  T
                  -
                  
                    τ
                    
                      n
                      ;
                      i
                    
                  
                
              
            
          
          
            
              
                
                  
                    
                      w
                      
                        n
                        ;
                        i
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  =
                  0
                
                ,
              
            
            
              
                t
                <
                
                  T
                  -
                  
                    τ
                    
                      n
                      ;
                      i
                    
                  
                
              
            
          
        
      
    
    
      
        (
        2.11
        )
      
    
  




The curve of this factor is shown in FIG. 36, in which τn;i denotes the longest time for the cause variable to function and is usually given by the domain engineers. The meaning is: before τn;i, the influence of the cause variable to the consequence variable is ignorable compared with the later influence. In the denominator,





  
    ∫
    0
    T
  
  ⁢
  
    
      
        w
        
          n
          ;
          i
        
      
      ⁡
      
        (
        
          T
          -
          t
        
        )
      
    
    ⁢
    
      d
      t
    
  




is to satisfy the probability normalization.

Fn;i(T,t) is given by the domain engineers according to the real situation and their domain knowledge. For example,




  
    
      
        {
        
          
            
              
                
                  
                    F
                    
                      n
                      ;
                      i
                    
                  
                  ⁡
                  
                    (
                    
                      T
                      ,
                      t
                    
                    )
                  
                
                =
                
                  (
                  
                    
                      
                        
                          X
                          n
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ⁢
                      
                        e
                        
                          
                            -
                            
                              (
                              
                                T
                                -
                                t
                              
                              )
                            
                          
                          ⁢
                          
                            
                              Γ
                              
                                n
                                ;
                                i
                              
                            
                            ⁡
                            
                              (
                              t
                              )
                            
                          
                        
                      
                    
                    +
                    
                      
                        
                          X
                          n
                        
                        ⁡
                        
                          (
                          
                            ∞
                            ij
                          
                          )
                        
                      
                      ⁢
                      
                        (
                        
                          1
                          -
                          
                            e
                            
                              
                                -
                                
                                  (
                                  
                                    T
                                    -
                                    t
                                  
                                  )
                                
                              
                              ⁢
                              
                                
                                  Γ
                                  
                                    n
                                    ;
                                    i
                                  
                                
                                ⁡
                                
                                  (
                                  t
                                  )
                                
                              
                            
                          
                        
                        )
                      
                    
                  
                  )
                
              
            
          
          
            
              
                
                  
                    Γ
                    
                      n
                      ;
                      i
                    
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                =
                
                  
                    Φ
                    
                      n
                      ;
                      i
                    
                  
                  ⁢
                  
                    {
                    
                      
                        
                          X
                          n
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ,
                      
                        
                          X
                          n
                        
                        ⁡
                        
                          (
                          
                            ∞
                            i
                          
                          )
                        
                      
                      ,
                      ...
                    
                    ⁢
                    
                      
                    
                    }
                  
                
              
            
          
        
      
    
    
      
        (
        2.12
        )
      
    
  




In which, Xn(t) is the state probability distribution of Xn at time t, Xn(∞i) is the functional intensity or probability contribution of Xi to the state probability distribution of Xn, after an infinite long time (i.e., the influence of time delay has disappeared completely); Γn;i(t)>0 is the time delay factor of Xi functioning to influence the state probability distribution of Xn; Φn;i is given by the domain engineers. It determines what factors affect Γn;i(t) and how. Usually, Γn;i(t) depends on Xn(t) and Xn(∞i). For example,





  
    
      
        
          
            Γ
            
              n
              ;
              i
            
          
          ⁡
          
            (
            t
            )
          
        
        =
        
          
            
              Φ
              
                n
                ;
                i
              
            
            ⁢
            
              {
              
                
                  
                    X
                    n
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ,
                
                  
                    X
                    n
                  
                  ⁡
                  
                    (
                    
                      ∞
                      i
                    
                    )
                  
                
              
              }
            
          
          =
          
            
              Λ
              
                n
                ;
                i
              
            
            ⁢
            
              
                ∑
                k
              
              ⁢
              
                
                  (
                  
                    
                      
                        X
                        n
                      
                      ⁡
                      
                        (
                        t
                        )
                      
                    
                    -
                    
                      
                        X
                        n
                      
                      ⁡
                      
                        (
                        
                          ∞
                          i
                        
                        )
                      
                    
                  
                  )
                
                2
              
            
          
        
      
    
    
      
        (
        2.13
        )
      
    
  




In which, Λn;i>0 is a constant independent of time. The meaning of this equation is: Γn;i(t) is proportional to the difference square between Xn(t) and Xn(∞i). The larger the difference is, the more rapid and obvious of the causality function appear.

In the example above, when T=t, the function of Xi to Xn has not appeared yet, and Fn;i(t,t)=Xn(t). When the time is long enough, the function of Xi to Xn has appeared fully, and Fn;i(∞,t)=Xn(∞i).
Since the cause variable can also vary according to time, the final value of Xn(T) should be the weighted average functional intensity as the contribution to the state probability distribution from the cause variable to the consequence variable at time T over the time interval [0,T].
Equations (2.10-2.13) are the brief expressions in terms of matrixes. The precise expressions should be:




  
    
      
        
          
        
        ⁢
        
          
            
              x
              
                3
                ⁢
                k
              
            
            =
            
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    ∑
                    j
                  
                  ⁢
                  
                    
                      
                        w
                        
                          3
                          ;
                          
                            7
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        f
                        
                          
                            3
                            ⁢
                            k
                          
                          ;
                          
                            7
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          ,
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        x
                        
                          7
                          ⁢
                          j
                        
                      
                      ⁡
                      
                        (
                        t
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    
                      w
                      
                        3
                        ;
                        
                          7
                          ⁢
                          j
                        
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    d
                    t
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
          
          ⁢
          
            
              
                x
                
                  4
                  ⁢
                  k
                
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    ∑
                    j
                  
                  ⁢
                  
                    
                      
                        w
                        
                          4
                          ;
                          
                            8
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        f
                        
                          
                            4
                            ⁢
                            k
                          
                          ;
                          
                            8
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          ,
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        x
                        
                          8
                          ⁢
                          j
                        
                      
                      ⁡
                      
                        (
                        t
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    
                      w
                      
                        4
                        ;
                        
                          8
                          ⁢
                          j
                        
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    d
                    t
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
          
          ⁢
          
            
              
                x
                
                  5
                  ⁢
                  k
                
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    ∑
                    j
                  
                  ⁢
                  
                    
                      
                        w
                        
                          5
                          ;
                          
                            9
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        f
                        
                          
                            5
                            ⁢
                            k
                          
                          ;
                          
                            9
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          ,
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        x
                        
                          9
                          ⁢
                          j
                        
                      
                      ⁡
                      
                        (
                        t
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    
                      w
                      
                        5
                        ;
                        
                          9
                          ⁢
                          j
                        
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    d
                    t
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
              
                x
                
                  6
                  ⁢
                  k
                
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      ∑
                      j
                    
                    ⁢
                    
                      
                        
                          w
                          
                            6
                            ;
                            
                              1
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          f
                          
                            
                              6
                              ⁢
                              k
                            
                            ;
                            
                              1
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            ,
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          b
                          
                            1
                            ⁢
                            j
                          
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          6
                          ;
                          
                            1
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              +
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      ∑
                      j
                    
                    ⁢
                    
                      
                        
                          w
                          
                            6
                            ;
                            
                              2
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          f
                          
                            
                              6
                              ⁢
                              k
                            
                            ;
                            
                              2
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            ,
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          b
                          
                            2
                            ⁢
                            j
                          
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          6
                          ;
                          
                            2
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              +
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      ∑
                      j
                    
                    ⁢
                    
                      
                        
                          w
                          
                            6
                            ;
                            
                              3
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          f
                          
                            
                              6
                              ⁢
                              k
                            
                            ;
                            
                              3
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            ,
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          x
                          
                            3
                            ⁢
                            j
                          
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
                
                  
                    
                      ∫
                      0
                      T
                    
                    ⁢
                    
                      
                        
                          w
                          
                            6
                            ;
                            
                              3
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                  +
                
              
              +
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      ∑
                      j
                    
                    ⁢
                    
                      
                        
                          w
                          
                            6
                            ;
                            
                              5
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          f
                          
                            
                              6
                              ⁢
                              k
                            
                            ;
                            
                              5
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            ,
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          x
                          
                            5
                            ⁢
                            j
                          
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          6
                          ;
                          
                            5
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
              
                x
                
                  7
                  ⁢
                  k
                
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      ∑
                      j
                    
                    ⁢
                    
                      
                        
                          w
                          
                            7
                            ;
                            
                              11
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          f
                          
                            
                              7
                              ⁢
                              k
                            
                            ;
                            
                              11
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            ,
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          b
                          
                            11
                            ⁢
                            j
                          
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          7
                          ;
                          
                            11
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              +
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      ∑
                      
                        
                          (
                          
                            n
                            ,
                            h
                          
                          )
                        
                        ->
                        j
                      
                    
                    ⁢
                    
                      
                        
                          w
                          
                            7
                            ;
                            
                              12
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          f
                          
                            
                              7
                              ⁢
                              k
                            
                            ;
                            
                              12
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            ,
                            t
                          
                          )
                        
                      
                      ⁢
                      Pr
                      ⁢
                      
                        {
                        
                          
                            
                              X
                              
                                4
                                ⁢
                                n
                              
                            
                            ⁡
                            
                              (
                              t
                              )
                            
                          
                          ⁢
                          
                            
                              X
                              
                                6
                                ⁢
                                h
                              
                            
                            ⁡
                            
                              (
                              t
                              )
                            
                          
                        
                        }
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          7
                          ;
                          
                            12
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
              
                x
                
                  8
                  ⁢
                  k
                
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      ∑
                      j
                    
                    ⁢
                    
                      
                        
                          w
                          
                            8
                            ;
                            
                              7
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          f
                          
                            
                              8
                              ⁢
                              k
                            
                            ;
                            
                              7
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            ,
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          x
                          
                            7
                            ⁢
                            j
                          
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          8
                          ;
                          
                            7
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              +
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          8
                          ;
                          D
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        f
                        
                          
                            8
                            ⁢
                            k
                          
                          ;
                          D
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          ,
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          8
                          ;
                          D
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
          
          ⁢
          
            
              
                x
                
                  9
                  ⁢
                  k
                
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    ∑
                    j
                  
                  ⁢
                  
                    
                      
                        w
                        
                          9
                          ;
                          
                            10
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        f
                        
                          
                            9
                            ⁢
                            k
                          
                          ;
                          
                            10
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          ,
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        x
                        
                          10
                          ⁢
                          j
                        
                      
                      ⁡
                      
                        (
                        t
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              
                
                  ∫
                  0
                  T
                
                ⁢
                
                  
                    
                      w
                      
                        9
                        ;
                        
                          10
                          ⁢
                          j
                        
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    d
                    t
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          
            
              
                x
                
                  10
                  ⁢
                  k
                
              
              ⁡
              
                (
                T
                )
              
            
            =
            
              
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      ∑
                      j
                    
                    ⁢
                    
                      
                        
                          w
                          
                            10
                            ;
                            
                              7
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          f
                          
                            
                              10
                              ⁢
                              k
                            
                            ;
                            
                              7
                              ⁢
                              j
                            
                          
                        
                        ⁡
                        
                          (
                          
                            T
                            ,
                            t
                          
                          )
                        
                      
                      ⁢
                      
                        
                          x
                          
                            7
                            ⁢
                            j
                          
                        
                        ⁡
                        
                          (
                          t
                          )
                        
                      
                      ⁢
                      
                        d
                        t
                      
                    
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          10
                          ;
                          
                            7
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
              +
              
                
                  
                    
                      
                        ∫
                        0
                        T
                      
                      ⁢
                      
                        
                          ∑
                          j
                        
                        ⁢
                        
                          
                            
                              w
                              
                                10
                                ;
                                
                                  11
                                  ⁢
                                  j
                                
                              
                            
                            ⁡
                            
                              (
                              
                                T
                                -
                                t
                              
                              )
                            
                          
                          ⁢
                          
                            
                              f
                              
                                
                                  10
                                  ⁢
                                  k
                                
                                ;
                                
                                  11
                                  ⁢
                                  j
                                
                              
                            
                            ⁡
                            
                              (
                              
                                T
                                ,
                                t
                              
                              )
                            
                          
                          ⁢
                          
                            
                              b
                              
                                11
                                ⁢
                                j
                              
                            
                            ⁡
                            
                              (
                              t
                              )
                            
                          
                        
                      
                    
                    )
                  
                  ⁢
                  
                    d
                    t
                  
                
                
                  
                    ∫
                    0
                    T
                  
                  ⁢
                  
                    
                      
                        w
                        
                          10
                          ;
                          
                            11
                            ⁢
                            j
                          
                        
                      
                      ⁡
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      t
                    
                  
                
              
            
          
        
      
    
    
      
        (
        2.14
        )
      
    
  




In the above equations, the inevitable events of U and D have ignored. Moreover, {n,h}→j means that the state combination of n and h of the two variables corresponds to the state j of the logic gate. Wn;ij is the element of Wn;i:





  
    
      
        {
        
          
            
              
                
                  
                    
                      w
                      
                        n
                        ;
                        ij
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  =
                  
                    
                      
                        τ
                        
                          n
                          ;
                          ij
                        
                        2
                      
                      -
                      
                        
                          (
                          
                            T
                            -
                            t
                          
                          )
                        
                        2
                      
                    
                  
                
                ,
              
            
            
              
                t
                ≥
                
                  T
                  -
                  
                    τ
                    
                      n
                      ;
                      ij
                    
                  
                
              
            
          
          
            
              
                
                  
                    
                      w
                      
                        n
                        ;
                        ij
                      
                    
                    ⁡
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  =
                  0
                
                ,
              
            
            
              
                t
                <
                
                  T
                  -
                  
                    τ
                    
                      n
                      ;
                      ij
                    
                  
                
              
            
          
        
      
    
    
      
        (
        2.15
        )
      
    
  




Correspondingly,





  
    
      
         
        
          {
          
            
              
                
                  
                    
                      
                        
                          
                            f
                            
                              nk
                              ;
                              ij
                            
                          
                          ⁡
                          
                            (
                            
                              T
                              ,
                              t
                            
                            )
                          
                        
                        =
                        
                        ⁢
                        
                          
                            (
                            
                              
                                r
                                
                                  n
                                  ;
                                  i
                                
                              
                              /
                              
                                r
                                n
                              
                            
                            )
                          
                          ⁢
                          
                            
                              a
                              
                                nk
                                ;
                                ij
                              
                            
                            ⁡
                            
                              (
                              
                                T
                                ,
                                t
                              
                              )
                            
                          
                        
                      
                    
                  
                  
                    
                      
                        =
                        
                        ⁢
                        
                          
                            (
                            
                              
                                r
                                
                                  n
                                  ;
                                  i
                                
                              
                              /
                              
                                r
                                n
                              
                            
                            )
                          
                          ⁢
                          
                            (
                            
                              
                                
                                  
                                    x
                                    nk
                                  
                                  ⁡
                                  
                                    (
                                    t
                                    )
                                  
                                
                                ⁢
                                
                                  e
                                  
                                    
                                      -
                                      
                                        (
                                        
                                          T
                                          -
                                          t
                                        
                                        )
                                      
                                    
                                    ⁢
                                    
                                      
                                        ω
                                        
                                          n
                                          ;
                                          ij
                                        
                                      
                                      ⁡
                                      
                                        (
                                        t
                                        )
                                      
                                    
                                  
                                
                              
                              +
                              
                                
                                  
                                    x
                                    nk
                                  
                                  ⁡
                                  
                                    (
                                    
                                      ∞
                                      ij
                                    
                                    )
                                  
                                
                                ⁢
                                
                                  (
                                  
                                    1
                                    -
                                    
                                      e
                                      
                                        
                                          -
                                          
                                            (
                                            
                                              T
                                              -
                                              t
                                            
                                            )
                                          
                                        
                                        ⁢
                                        
                                          
                                            ω
                                            
                                              n
                                              ,
                                              ij
                                            
                                          
                                          ⁡
                                          
                                            (
                                            t
                                            )
                                          
                                        
                                      
                                    
                                  
                                  )
                                
                              
                            
                            )
                          
                        
                      
                    
                  
                
              
            
            
              
                
                  
                    
                      ω
                      
                        n
                        ;
                        ij
                      
                    
                    ⁡
                    
                      (
                      t
                      )
                    
                  
                  =
                  
                    
                      φ
                      
                        n
                        ;
                        ij
                      
                    
                    ⁢
                    
                      {
                      
                        
                          
                            X
                            k
                          
                          ⁡
                          
                            (
                            t
                            )
                          
                        
                        ,
                        
                          
                            X
                            n
                          
                          ⁡
                          
                            (
                            
                              ∞
                              ij
                            
                            )
                          
                        
                        ,
                        ...
                      
                      ⁢
                      
                        
                      
                      }
                    
                  
                
              
            
          
        
      
    
    
      
        (
        2.16
        )
      
    
  




For example,





  
    
      
        
          
            ω
            
              n
              ;
              ij
            
          
          ⁡
          
            (
            t
            )
          
        
        =
        
          
            
              φ
              
                n
                ;
                ij
              
            
            [
            
              
                
                  X
                  n
                
                ⁡
                
                  (
                  t
                  )
                
              
              ,
              
                
                  X
                  n
                
                ⁡
                
                  (
                  
                    ∞
                    ij
                  
                  )
                
              
            
            }
          
          =
          
            
              β
              
                n
                ;
                ij
              
            
            ⁢
            
              
                ∑
                k
              
              ⁢
              
                
                  (
                  
                    
                      
                        x
                        nk
                      
                      ⁡
                      
                        (
                        
                          ∞
                          ij
                        
                        )
                      
                    
                    -
                    
                      
                        x
                        nk
                      
                      ⁡
                      
                        (
                        t
                        )
                      
                    
                  
                  )
                
                2
              
            
          
        
      
    
    
      
        (
        2.17
        )
      
    
  




In which, βn;ij>0 is a number depending on the state of the cause variable Xij but not the state k of the consequence variable Xn and the time, and is given by the domain engineers. The larger the βn;ij is, the less the time delay is. xnk(∞ij) is the functional intensity or the probability contribution of Xij to Xnk after passing the time delay or the influence has become stable. Obviously, the original functional intensity ank;ij(T,t) satisfies the probability normalization.

Proof:

Since the state probability normalization of Xn,




  
    
      ∑
      k
    
    ⁢
    
      
        x
        nk
      
      ⁡
      
        (
        t
        )
      
    
  
  =
  
    
      1
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        
          ∑
          k
        
        ⁢
        
          
            x
            nk
          
          ⁡
          
            (
            
              ∞
              ij
            
            )
          
        
      
    
    =
    1.
  




Therefore,





  
    
      
        
          
            ∑
            k
          
          ⁢
          
            
              a
              
                nk
                ;
                ij
              
            
            ⁡
            
              (
              
                T
                ,
                t
              
              )
            
          
        
        =
        
        ⁢
        
          
            ∑
            k
          
          ⁢
          
            (
            
              
                
                  
                    x
                    nk
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
                ⁢
                
                  e
                  
                    
                      -
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        ω
                        
                          n
                          ;
                          ij
                        
                      
                      ⁡
                      
                        (
                        t
                        )
                      
                    
                  
                
              
              +
              
                
                  
                    x
                    nk
                  
                  ⁡
                  
                    (
                    
                      ∞
                      ij
                    
                    )
                  
                
                ⁢
                
                  (
                  
                    1
                    -
                    
                      e
                      
                        
                          -
                          
                            (
                            
                              T
                              -
                              t
                            
                            )
                          
                        
                        ⁢
                        
                          
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                              n
                              ;
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                          ⁡
                          
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                            t
                            )
                          
                        
                      
                    
                  
                  )
                
              
            
            )
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              e
              
                
                  -
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    ω
                    
                      n
                      ;
                      ij
                    
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
              
            
            ⁢
            
              
                ∑
                k
              
              ⁢
              
                
                  x
                  nk
                
                ⁡
                
                  (
                  t
                  )
                
              
            
          
          +
          
            (
            
              1
              -
              
                e
                
                  
                    -
                    
                      (
                      
                        T
                        -
                        t
                      
                      )
                    
                  
                  ⁢
                  
                    
                      ω
                      
                        n
                        ;
                        ij
                      
                    
                    ⁡
                    
                      (
                      t
                      )
                    
                  
                
              
            
            )
          
        
      
    
  
  
    
      
        
        ⁢
        
          
            ∑
            k
          
          ⁢
          
            
              x
              nk
            
            ⁡
            
              (
              
                ∞
                ij
              
              )
            
          
        
      
    
  
  
    
      
        =
        
        ⁢
        
          
            
              e
              
                
                  -
                  
                    (
                    
                      T
                      -
                      t
                    
                    )
                  
                
                ⁢
                
                  
                    ω
                    
                      n
                      ;
                      ij
                    
                  
                  ⁡
                  
                    (
                    t
                    )
                  
                
              
            
            +
            
              (
              
                1
                -
                
                  e
                  
                    
                      -
                      
                        (
                        
                          T
                          -
                          t
                        
                        )
                      
                    
                    ⁢
                    
                      
                        ω
                        
                          n
                          ;
                          ij
                        
                      
                      ⁡
                      
                        (
                        t
                        )
                      
                    
                  
                
              
              )
            
          
          =
          1
        
      
    
  




That is, ank;ij(T,t) satisfies the normalization at any time ▪

The problem to be solved in this example is: given the change of the state probability distribution according to the time during the time interval [0,t1] while t1≧τn;i, i.e. the influence of the cause variable to the consequence variable before t=0 can be ignored, calculate the state probability distributions of the variables at time T (T>t1) (see FIG. 37).
How to calculate equations (2.14) is a pure mathematical calculation problem. Many numerical calculation methods can be applied. For example, the differential method can be applied, which is to divide the time interval [0,T] into many small time intervals Δti (as shown in FIG. 38), then recursively perform the calculation one small interval after another small interval. The smaller the classification of Δti is, the more accurate the calculation is.
It should be pointed out that equation (2.14 deals with the calculation of Pr{X4;hj(t)X6;gj(t)}, where X4hj and X6gj may not be independent of each other. Therefore, Pr{X4;hj(t)X6;gj(t)}=x4hj(t)x6gj(t) may not be valid, but can be applied in the approximate calculation. If the accurate calculation is required, we must further outspread X4nX6h as the consequence event according to the method shown in equation (1.2) (the detailed calculation is ignored here).
Example 24
The application example for predicting the flood level.
The present flood prediction method is mainly based on the dynamical mathematical-physical model composed of the water amount, the water quality, the shape of river bed, the weather, etc, in the valley, and then performing the deterministic numerical calculation in the mainframe computers. But it has been proved by fact that the accurate and complete mathematical-physical model is hard to be constructed, nor the accurate and complete data to be obtained, while the result of calculation relies strongly on these accuracy and completeness of the mathematical-physical model and data. It is often that these conditions cannot be satisfied such that this method is hard to be applied in practice.
With DUCG, people can utilize the historical statistic data and the domain engineer's belief to represent the uncertain causalities of various variables and time, and then the dynamical calculation can be performed according to the collected various evidence, so as to predict by what time the biggest flood peek will appear and how large its degree will be, and in such a way to avoid the problem of relying too much on the accuracy and completeness of applying the mathematical-physical model and data.
Suppose a river along with its rain amount and water level monitor positions are as shown in FIG. 39. The positions of the B type variables representing the rain amount are the rain amount monitor positions, and the positions of the X type variables representing the water level are the water level monitor positions. According to the method described in §1, the first step is to determine the cause and consequence variables related to the problem to be solved about the flood water level prediction. Obviously, in FIG. 39, the Bi and Xn are the cause and consequence variables related to the problem to be solved, in which all variables are continuous and need be fuzzily discretized. The method to fuzzily discretizing the continuous variables is the same as for X16 in example 1. According to the relations of the upriver and downriver of this river, it is easy to determine the direct cause variables of the X type variables. For example, X6 represents the water level at the converging position of two sub-rivers of the upriver. Obviously, this water level is affected by the upriver water levels X5 and X11, the local rain amount B7 and the downriver water level X13. Therefore, these variables are the direct cause variables of X6. As the problem to be solved is a dynamical prediction type problem, the explicit representation mode is more suitable. Thus, we connect the direct cause variables to X6 by F6;i, where i=5,11,7,13. Put all X type variables along with their direct cause variables modeled in the explicit representation mode together, we get the DUCG for predicting the flood of this river as shown in FIG. 40, in which the basic variables Bi represent the predicted rain amount in a following few days in the corresponding areas, where bij are functions dynamically changing with time. The consequence variables Xn represent the water levels of the river flood at the corresponding positions in a following few days. The numbers are the relationships of the functional variables. The default values are 1. In this figure, the bidirectional arcs represent two functional variables in opposite directions. This is because the downriver water level affects the upriver flood speed. Therefore, the influences of the upriver and downriver are bidirectional.
DUCG is not like the mathematical-physical model to perform the deterministic calculation, the result is the accurate values, while affected strongly by the accuracy and completeness of the data, but performs the uncertain calculation according to the probability distribution type statistic data and belief, while the result is the probability distributions of various possible states. In this example, the problem to be solved by DUCG is: conditioned on the known information E such as the observed rain amount and water levels at different positions during the past period of time, according to the weather forecast, predict the flood water level X22 at some city in the following few days, so as to decide whether or not to dynamite the bank somewhere to release the flood and remove people. Obviously, all the variables in the figure (including the functional variables) are the functions of time. According to E, the probability distributions bij(t) of the basic variables about the weather changes varying according to time in a following few days are given by the domain engineers, meanwhile, the functional variables can also be given by the domain engineers according to the statistic data or belief, in which the probability parameters ank;ij and fnk;ij of the elements Fnk;ij of the functional variables are similar to the functions of time in equations (2.15-2.17). Of course, rn;i and rn can also be the functions of time. Then, people can calculate the probability distribution of the flood water level X22 in a following few days according to the DWG. shown in FIG. 40 and conditioned on the known evidence E, so as to make decision.
As the calculation of this type is huge and complex, the calculation of this example is ignored and only the method is illustrated.
This invention is not limited to the specific ways and application examples described in the above specification.
INDUSTRIAL APPLICABILITY
This invention can be applied in industry.

## Claims

1. A method for constructing an intelligent system for processing the uncertain causality information, the method includes: representing the causalities among the things in the explicit representation mode, specifically including the following steps:
(1) Establish, by at least one processor, a representation system about the various cause variables Vi and consequence variables Xn in concern with the problem to be solved, wherein: {circle around (1)} Let V represent two type variables B and X, i.e. V∈{B,X}, in which B is the basic variable that is only the cause variable and X is the consequence variable that can be also the cause variable of the other consequence variables; {circle around (2)} No matter the states of the variable Vi or Xn are discrete or not, represent them all as the discrete or fuzzy discrete states, so as to be dealt with by using the same manner, that is, represent the different states of Vi and Xn as Vij and Xnk respectively, where i and n index variables while j and k index the discrete or fuzzy discrete states of the variables; {circle around (3)} When Vi or Xn is continuous, the membership of an arbitrary value ei of Vi or en of Xn, belonging to Vij or Xnk respectively, is mij(ei) or mnk(en) respectively, and they satisfy





  
    
      
        ∑
        j
      
      ⁢
      
        
      
      ⁢
      
        
          m
          ij
        
        ⁡
        
          (
          
            e
            i
          
          )
        
      
    
    =
    
      
        1
        ⁢
        
          
        
        ⁢
        and
        ⁢
        
          
        
        ⁢
        
          
            ∑
            j
          
          ⁢
          
            
          
          ⁢
          
            
              m
              ij
            
            ⁡
            
              (
              
                e
                i
              
              )
            
          
        
      
      =
      1
    
  
  ;




{circle around (4)} Vij and Xnk are treated as events, i.e., Vij represents the event that Vi is in its state j and Xnk represents the event that Xn is in its state k; meanwhile, if j≠j′ and k≠k′, Vij is exclusive with Vij′ and Xnk is exclusive with Xnk′; {circle around (5)} If i≠i′, Bij and Bij′ are independent events, and their occurrence probabilities bij satisfies




  
    
      
        ∑
        j
      
      ⁢
      
        
      
      ⁢
      
        b
        ij
      
    
    ≤
    1
  
  ;



(2) For the consequence variable Xn, determine, by the at least one processor, its direct cause variables Vi, i∈SEXn, SEXn is the index set of the {B,X} type direct variables of Xn in the explicit representation mode;
(3) The functional variable Fn;i is used, by the at least one processor, to represent the causality between Vi, i∈SEXn, and Xn. Vi is the input or cause variable of Fn;i and Xn is the output or consequence variable of Fn;i, wherein: {circle around (1)} The causality uncertainty between Vi and Xn is represented by the occurrence probability fnk;ij of the specific value Fnk;ij of Fn;i. Fnk;ij is a random event representing the uncertain functional mechanism of Vij causing Xnk. fnk;ij is the probability contribution of Vij to Xnk; {circle around (2)} fnk;ij=(rn;i/rn)ank;ij, where rn;i is called the relationship between Vi and Xn, rn is the normalization factor and





  
    
      r
      n
    
    =
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        r
        
          n
          ;
          i
        
      
    
  
  ,
  
    a
    
      nk
      ;
      ij
    
  




is the probability of the event that Vij causes Xnk regardless of any other cause variables and ank;ij and rn can be the function of time; {circle around (3)} ank;ij satisfies




  
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        a
        
          nk
          ;
          ij
        
      
    
    ≤
    1
  
  ;
  
    
      
        ◯
        ⁢
        4
      
      ⁢
      
        
      
      ⁢
      Pr
      ⁢
      
        {
        
          X
          nk
        
        }
      
    
    =
    
      
        ∑
        
          i
          ,
          j
        
      
      ⁢
      
        
      
      ⁢
      
        
          f
          
            nk
            ;
            ij
          
        
        ⁢
        Pr
        ⁢
        
          
            {
            
              V
              ij
            
            }
          
          .

2. The method according to claim 1, wherein the functional variable Fn;i in the explicit representation mode can be the conditional functional variable, the conditional functional variable is used to represent the functional relation between the cause variable Vi and the consequence variable Xn conditioned on Cn;i, wherein:
(1) Cn;i has only two states: true or false, and its state can be found according to the observed information or the computation results;
(2) When Cn;i is true, the conditional functional variable becomes the functional variable;
(3) When Cn;i is false, the conditional functional variable is eliminated.

3. The method according to claim 1, wherein the explicit representation mode includes also extending V∈{B,X} to V∈{B,X,G} in the explicit representation mode, where G is the logic gate variable, i.e. the cause variable to influence the consequence variable by the state logic combinations of a group of cause variables, suppose the input variables of logic gate variable Gi are Vh, then the logic gate Gi is constructed, by the at least one processor, by the following steps:
(31) The logic combinations between the input variables Vh, V∈{B,X,G}, are represented by the truth value table of Gi in which each input row is a logic expression composed of the input variable states and corresponds to a unique state of Gi, different rows of the logic expressions are exclusive with each other, wherein if a logic expression is true, the corresponding state of Gi is true;
(32) The set of the states of Gi is equal to or less than the set of all state combinations of the input variables;
(33) When the set of the states of Gi is less than the set of all state combinations of the input variables, there is a remnant state of Gi, which corresponds uniquely to the remnant state combinations of the input variables, so that all the states of Gi including the remnant state are exclusive with each other and just cover all the state combinations of the input variables;
(34) When Gi is the direct cause variable of Xn, Gi functions to Xn through the functional or conditional functional variable Fn;i;
(35) If a logic gate has only one input variable, this logic gate can be ignored, i.e. the input variable of the logic gate can be taken as the input variable of the functional variable or conditional functional variable Fn;i with this logic gate as its input variable;
(36) When Gi is the direct cause variable of Xn, the relationship between Gi and Xn is rn;i; when calculating fnk;ij, the calculation to rn includes the relationship between Gi and Xn; when calculating Pr{Xnk}, the fnk;ij between Gi and Xn is included.

4. The method according to claim 3, wherein further including:
(41) Extend, by at least one processor, V∈{B,X} as V∈{B,X,D}, or extend V∈{B,X,G} as V∈{B,X,G,D}, in which D is the default event or variable, Dn can appear only with Xn and is an independent cause variable that has only one inevitable state;
(42) Dn, becomes a direct cause variable of Xn through Fn;D, where Fn;D is the functional variable between Dn, and Xn;
(43) The causality uncertainty between Dn and Xn is represented by the occurrence probability fnk;D of the specific value Fnk;D of Fn;D, where Fnk;D is a random event representing the functional mechanism of Dn to Xn, and fnk;D is the probability contribution of Dn to Xnk;
(44) fnk;D=(rn;D/rn)ank;D, where ank;D is the probability of the event that Dn causes Xn regardless of the other cause variables of Xn, and satisfies





  
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        a
        
          nk
          ;
          D
        
      
    
    ≤
    1
  
  ,




rn;D is the relationship between Dn and Xn; after adding Dn,




  
    r
    n
  
  =
  
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        r
        
          n
          ;
          i
        
      
    
    +
    
      
        r
        
          n
          ;
          D
        
      
      ·
    
  




ank;D and rn;D be the function of time;
(45) The original





  
    Pr
    ⁢
    
      {
      
        X
        nk
      
      }
    
  
  =
  
    
      ∑
      
        i
        ,
        j
      
    
    ⁢
    
      
    
    ⁢
    
      
        f
        
          nk
          ;
          ij
        
      
      ⁢
      Pr
      ⁢
      
        {
        
          V
          ij
        
        }
      
    
  




is replaced by




  
    Pr
    ⁢
    
      {
      
        X
        nk
      
      }
    
  
  =
  
    
      
        ∑
        
          i
          ,
          j
        
      
      ⁢
      
        
      
      ⁢
      
        
          f
          
            nk
            ;
            ij
          
        
        ⁢
        Pr
        ⁢
        
          {
          
            V
            ij
          
          }
        
      
    
    +
    
      
        f
        
          nk
          ;
          D
        
      
      .

5. The method according to claim 4, wherein, further including: when the default variable of Xn is more than one, they can be combined as one default variable Dn; let g be the index distinguishing two or more default variables, Corresponding to the case of only one default variable, the variable Dn and the parameters rn;D, ank;D are represented as Dng, rn;Dg, ank;Dg respectively; after combining Dng as Dn, the parameters of Dn are calculated according to




  
    r
    
      n
      ;
      D
    
  
  =
  
    
      
        ∑
        g
      
      ⁢
      
        
      
      ⁢
      
        
          r
          
            n
            ;
            Dg
          
        
        ⁢
        
          
        
        ⁢
        and
        ⁢
        
          
        
        ⁢
        
          a
          
            nk
            ;
            D
          
        
      
    
    =
    
      
        ∑
        g
      
      ⁢
      
        
      
      ⁢
      
        
          a
          
            nk
            ;
            Dg
          
        
        .

6. The method according to claim 1, wherein the method further includes: using the implicit mode to represent the uncertain causalities among things, specifically including the following steps, performed by the at least one processor:
(4) The conditional probability table (CPT) is used to represent the causality between the consequence variable Xn and its direct cause variables Vi, i∈SIXn, wherein: {circle around (1)} When no cause variable will be eliminated, CPT is composed of only the conditional probabilities pnk;ij, where pnk;ij≡Pr{Xnk|j} and j indexes the state combination of the cause variables Vi, i∈SIXn; {circle around (2)} When part or even all cause variables may be eliminated, CPT is composed of three parameters: pnk;ij, qnk;ij and dn;j, satisfying pnk;ij=qnk;ij/dn;j, so that CPT can be reconstructed when some of its cause variables are eliminated, where qnk;ij and qn;nj are the sample number and occurrence number of Xnk respectively, conditioned on the state combination indexed by j of the cause variables.

7. The method according to claim 6, wherein the said step (4) further including:
(71) In the implicit representation mode, the cause variables Vi, i∈SIXn, can be separated as several groups, every group uses the implicit representation mode to represent the uncertain causality to Xn;
(72) Give the relationship rXn between every group of direct cause variables to the consequence variable Xn;
(73) If some cause variables in the group are eliminated for any reason, the CPT of this group can be reconstructed as follows: Suppose the variable to be eliminated is Vi, before the elimination, there are several subgroups of the state combinations of the input variables indexed by j′; in subgroup j′, the states of all the variables are same except the states of Vi; denote the index set of the state combination j in subgroup j′ as Sij′, then





  
    
      q
      
        nk
        ;
        
          j
          ′
        
      
    
    =
    
      
        ∑
        
          j
          ∈
          
            S
            
              ij
              ′
            
          
        
      
      ⁢
      
        
      
      ⁢
      
        q
        
          nk
          ;
          j
        
      
    
  
  ,
  
    
      d
      
        nj
        ′
      
    
    =
    
      
        ∑
        
          j
          ⁢
          
            I
            ^
          
          ⁢
          
            
          
          ⁢
          
            S
            
              ij
              ′
            
          
        
      
      ⁢
      
        
      
      ⁢
      
        d
        nj
      
    
  
  ,
  
    
      p
      
        nk
        ;
        
          j
          ′
        
      
    
    =
    
      
        q
        
          nk
          ;
          
            j
            ′
          
        
      
      /
      
        d
        
          n
          ;
          
            j
            ′
          
        
      
    
  



In which j′ is the new index of the remnant state combinations after the elimination of Vi;
(74) Repeat (73) to deal with the case in which more than one cause variable is eliminated.

8. The method according to claim 6, wherein further including the following steps by the at least one processor:
(5) For a group of cause variables Vi′, i′∈SIXn, in the implicit representation mode, give the corresponding relationship rXn, while in the explicit representation mode, rn is renewed as rn=rn+rXn, in which the right side rn is before the renewing;
(6) If the implicit representation mode has more than one group, they can be indexed by g and every group relationship can be denoted as rXng; then the calculation equation in above (5) becomes





  
    r
    n
  
  =
  
    
      r
      n
    
    +
    
      
        ∑
        h
      
      ⁢
      
        
      
      ⁢
      
        
          r
          Xng
        
        .

9. The method according to claim 8, further including:
(10) According to the specific cases of every consequence variable Xn, the representations above for all the consequence variables compose the original DUCG;
(11) The evidence E in concern with the original DUCG is received during the online application and is expressed as





  
    E
    =
    
      
        E
        *
      
      ⁢
      
        
          ∏
          h
        
        ⁢
        
          
        
        ⁢
        
          E
          h
        
      
    
  
  ,




where Eh is the evidence indicating the state of the {B,X} type variable, E* represents the other evidence; if Eh is a fuzzy state evidence, i.e. the state of the variable Vh in the original DUCG is known in a state probability distribution, or if Eh is a fuzzy continuous evidence, i.e. the specific value eh, of the continuous variable Vh is known in the fuzzy area of different fuzzy states of Vh, V∈{B,X}, then add Eh as a virtual evidence variable into the original DUCG and represent the causality between Vh and Eh according to the explicit mode so that Eh becomes the consequence variable of the cause variable Vh; after finishing these steps, the original DUCG becomes the E conditional original DUCG.

10. The method according to claim 9, wherein, the said step (11) including: adding Eh as a virtual evidence variable into the original DUCG, and further including the following steps: Suppose mhj=mhj(eh) is the membership of Eh belonging to the fuzzy state j, or mhj is the probability of Xhj indicated by the fuzzy state evidence Eh, i.e., mhj=Pr{Vhj|Eh}, j∈SEh, SEh is the index set of state j in which mhj≠0 and includes at least two different indexes, while satisfying




  
    
      ∑
      
        j
        ∈
        
          S
          Eh
        
      
    
    ⁢
    
      
    
    ⁢
    
      m
      hj
    
  
  =
  
    1
    :
  



(101) As the virtual consequence variable of Vh, Eh has only one inevitable state, has only one direct cause variable Vh, and is not the cause variable of any other variable;
(102) The virtual functional variable from Vhj to Eh is FE;h and its specific value FE;hj is the virtual random event that Vhj causes Eh; the functional intensity parameter fE;hj of FE;hj may be given by domain engineers;
(103) If the domain engineers cannot give fE;hj, it can be calculated from





  
    
      f
      
        E
        ;
        hj
      
    
    =
    
      
        
          
            m
            hj
          
          ⁢
          
            v
            hk
          
        
        
          
            m
            hk
          
          ⁢
          
            v
            hj
          
        
      
      ⁢
      
        f
        
          E
          ;
          hk
        
      
    
  
  ;




where j≠k′, j∈SEh, k∈SEh, vhj≡Pr{Vhj}, vhk≡Pr{Vhk} Given fE;hk>0, fE;hj can be calculated.

11. The method according to claim 10, wherein, further including the following steps performed by the at least one processor, to simplify the E conditional original DUCG: suppose Vi is the direct cause variable of Xn, V∈{B,X,G,D}, then
(111) According to E, determine whether or not the condition Cn;i of the conditional functional variable Fn;i is valid: {circle around (1)} if yes, change the conditional functional variable as the functional variable; {circle around (2)} if not, eliminate this conditional functional variable; {circle around (3)} if cannot determine whether or not Cn;i is valid, keep the conditional functional variable until Cn;i can be determined;
(112) According to E, if Vnk is not the cause of any state of Xn when E shows that Vih is true, eliminate the functional or conditional functional variable Fn;i that is from Vi to Xn;
(113) According to E, if Xnk cannot be caused by any state of Vi when E shows that Xnk is true, eliminate the functional or conditional functional variable from Vi to Xn;
(114) In the explicit mode of representation, if the X or G type variable without any cause or input appears, eliminate this variable along with the F type variables starting from this variable;
(115) If there is any group of isolated variables without any logic connection to the variables related to E, eliminate this group variables;
(116) If E shows that Xnk is true, while Xnk is not the cause of any other variable and Xn has no connection with the other variables related to E, denote the index set of the index n of such Xn as SEnk; When Vi and its logic connection variables Fn;i have no logic connection with the variables related to E except the variables indexed in SEnk, eliminate Xn, Vi and the functional or conditional functional variables Fn;i along with all other variables logically connected with Vi;
(117) If E shows that Xnk appears earlier than Vij, so that for sure Vij is not the cause of Xnk, eliminate the functional or conditional functional variables that are in the causality chains from Vi to Xn but are not related to the influence of other variables to Xn;
(118) Upon demand, the above steps can be in any order and can be repeated.

12. The method according to claim 11, wherein, further including the following steps performed by the at least one processor, to transform the DUCG with implicit or hybrid representation mode conditioned on E as all in the explicit mode, i.e. EDUCG:
(123) For the consequence variable Xn in the implicit or hybrid mode, for every group of SIXn type cause variables, introduce a virtual logic gate variable Gi, in which the cause variables of SIXn are the input variables of Gi, and the number of the states of Gi and the input rows of the truth value table of Gi equal to the number of the state combinations of the cause variables in SIXn, while each of the state combination of the input variables is an input row of the truth value table of Gi and also a state of the virtual logic gate;
(124) Introduce the virtual functional variable Fn;i, in which Gi is the input variable and Xn is the output variable, so that Gi becomes the direct cause variable of Xn;
(125) In the CPT of the cause variables in SIXn, ank;ij=pnk;j,; the relationship of Fn;i is; rni=rXn;
(126) When there is only one input variable in Gi, such Gi can be ignored, i.e. the virtual functional variable takes the input variable of Gi as its input variable directly;
(127) When the groups of the SIXn type variables are more than one group, repeat the above steps for every groups.

13. The method according to claim 11, wherein, further including the following steps performed by the at least one processor, to transform the DUCG conditioned on E in the explicit representation mode or in the more than one group implicit representation mode as the IDUCG in which all representations are in the implicit representation mode with only one group direct cause variables:
(131) If Cn;i is valid, change the conditional functional variable as the functional variable; If Cn;i is invalid, eliminate the conditional functional variable;
(132) For any representation of the uncertain causality between the consequence variable Xn and its direct cause variables, if it is in the hybrid or more than one group implicit representation mode, transform the representation mode for Xn to the explicit mode;
(133) After the above steps, take the state combinations of the {B,X} type cause variables of the consequence variable Xn as the conditions indexed by j, calculate the conditional probability of Xnk Pr{Xnk|j} according to the explicit mode, where the connections between the {B,X} type cause variables and Xn may be or may not be through logic gates; in the calculation, all contributions from different types of direct cause variables should be considered, i.e. when the direct cause variables are {X,B,G} types,





  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          ❘
          j
        
        }
      
    
    =
    
      
        ∑
        i
      
      ⁢
      
        f
        
          nk
          ;
          ih
        
      
    
  
  ;




when the direct cause variables are {X,B,G,D} types,




  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          ❘
          j
        
        }
      
    
    =
    
      
        
          ∑
          i
        
        ⁢
        
          f
          
            nk
            ;
            ih
          
        
      
      +
      
        f
        
          nk
          ;
          D
        
      
    
  
  ;



(134) The case of ank;ih=1 can be understood as that Xnk is true for sure, i.e. when the input variable i is in its state h, all the states, except k, of Xn cannot be true; if this applies, when ank;ih=1, Pr{Xnk|j}=1, meanwhile Pr{Xnk′|j}=0, where k≠k′;
(135) If ank;ih=1, k∈Sm, Sm, is the index set of such states of Xn that ank;ih=1 and the number of such states is m, then Pr{Xnk|j}=1/m and Pr{Xnk′,|j}=0, where k′ Sm;
(136) If such calculated





  
    
      
        ∑
        k
      
      ⁢
      
        Pr
        (
        
          
            X
            nk
          
          ❘
          j
        
        }
      
    
    <
    1
  
  ,
  
    
      let
      ⁢
      
        
      
      ⁢
      
        Pr
        (
        
          
            X
            nk
          
          ❘
          j
        
        }
      
    
    =
    
      1
      -
      
        
          ∑
          
            k
            ≠
            η
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                X
                nk
              
              ❘
              j
            
            }
          
        
      
    
  
  ,




where η indexes the default state of Xn;
(137) If there is no default state η in said step (136), the normalization method is used as follows:





  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          ❘
          j
        
        }
      
    
    =
    
      Pr
      ⁢
      
        
          {
          
            
              X
              nk
            
            ❘
            j
          
          }
        
        /
        
          
            ∑
            k
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  X
                  nk
                
                ❘
                j
              
              }
            
          
        
      
    
  
  ,




where the Pr{Xnk|j} on the right side are the values before the normalization;
(138) After satisfying the normalization, Pr{Xnk|j} becomes the conditional probability pnk;nj in the standard implicit representation mode;
(139) Connect the {X,B} type direct cause variables of Xn through or not through logic gates with Xn according to the implicit representation mode, the DUCG conditioned on E is transformed as the IDUCG.

14. The method according to claim 11, wherein, further including the following steps performed by the at least one processor:
(144) outspread the evidence events Eh, included in E, which determine the states of the {B,X} type variables, and the events Hkj in concern, and in the process of outspread, break the logic cycles;
(145) based on the outspreaded logic expressions of Eh and Hkj, further outspread





  
    
      ∏
      h
    
    ⁢
    
      
    
    ⁢
    
      
        E
        h
      
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        H
        kj
      
      ⁢
      
        
          ∏
          h
        
        ⁢
        
          
        
        ⁢
        
          E
          h
        
      
    
  
  ;



(146) calculate the state probability and the rank probability of the concerned event Hkj conditioned on E according to the following equations:
The state probability:






  
    
      h
      kj
      s
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            E
          
          }
        
      
      
        Pr
        ⁢
        
          {
          E
          }
        
      
    
  
  ;




The rank probability:






  
    h
    kj
    r
  
  =
  
    
      
        h
        kj
        s
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          h
          kj
          s
        
      
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            E
          
          }
        
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              E
            
            }
          
        
      
    
  



Where S is the set of all the events in concern.

15. The method according to claim 14, wherein, the said step (145) including:
(151) Express the evidence set





  
    ∏
    h
  
  ⁢
  
    
  
  ⁢
  
    E
    h
  




indicating the states of the {B,X} type variables as E′E′′, in which




  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      
    
    ⁢
    
      E
      i
      ′
    
  




is the evidence set composed of the evidence events indicating the abnormal states of variables, and




  
    E
    ′′
  
  =
  
    
      ∏
      
        i
        ′
      
    
    ⁢
    
      
    
    ⁢
    
      E
      
        i
        ′
      
      ′′
    
  




is the evidence set composed of the evidence events indicating the normal states of variables;
(152) Outspread





  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      
    
    ⁢
    
      E
      i
      ′
    
  




and determine the possible solution set S conditioned on E, where every possible solution Hkj is an event in concern for the problem to be solved;
And further the said step (146) including:
(153) Calculate two types of the state probability and rank probability of Hkj conditioned on E:
The state probability with incomplete information:





  
    
      h
      kj
      
        s
        ′
      
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            
              E
              ′
            
          
          }
        
      
      
        Pr
        ⁢
        
          {
          
            E
            ′
          
          }
        
      
    
  
  ;



The state probability with complete information:





  
    
      h
      kj
      s
    
    =
    
      
        
          h
          kj
          
            s
            ′
          
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                ❘
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                ❘
                
                  E
                  ′
                
              
              }
            
          
        
      
      =
      
        
          
            h
            kj
            
              s
              ′
            
          
          ⁢
          Pr
          ⁢
          
            {
            
              
                E
                ′′
              
              ❘
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
              
            
            }
          
        
        
          
            ∑
            j
          
          ⁢
          
            
              h
              kj
              
                s
                ′
              
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                ❘
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
        
      
    
  
  ;



The rank probability with incomplete information:





  
    
      h
      kj
      
        r
        ′
      
    
    =
    
      
        
          h
          kj
          
            s
            ′
          
        
        
          
            ∑
            
              
                H
                kj
              
              ∈
              S
            
          
          ⁢
          
            h
            kj
            
              s
              ′
            
          
        
      
      =
      
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              
                E
                ′
              
            
            }
          
        
        
          
            ∑
            
              
                H
                kj
              
              ∈
              S
            
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
              
              }
            
          
        
      
    
  
  ;



The rank probability with complete information:





  
    h
    
      k
      ⁢
      
        
      
      ⁢
      j
    
    r
  
  =
  
    
      
        h
        
          k
          ⁢
          
            
          
          ⁢
          j
        
        
          r
          ′
        
      
      ⁢
      Pr
      ⁢
      
        {
        
          
            E
            ′′
          
          ⁢
          
            
            
              
                H
                
                  k
                  ⁢
                  
                    
                  
                  ⁢
                  j
                
              
              ⁢
              
                E
                ′
              
            
            }
          
        
      
    
    
      
        ∑
        
          
            H
            
              k
              ⁢
              
                
              
              ⁢
              j
            
          
          ∈
          S
        
      
      ⁢
      
        
      
      ⁢
      
        
          h
          
            k
            ⁢
            
              
            
            ⁢
            j
          
          
            r
            ′
          
        
        ⁢
        Pr
        ⁢
        
          {
          
            
              E
              ′′
            
            ⁢
            
              
              
                
                  H
                  
                    k
                    ⁢
                    
                      
                    
                    ⁢
                    j
                  
                
                ⁢
                
                  E
                  ′
                
              
              }
            
          
        
      
    
  



In which, if HkjE′ is null, Pr{E′′|HkjE′}≡0.

16. The method according to claim 13, wherein, further including the following steps performed by the at least one processor,
(140) use the BN method to calculate the state probability distribution of the variables in concern conditioned on E.

17. The method according to claim 15, wherein, further including the following steps performed by the at least one processor, to outspread E, E′, HkjE or HkjE′, and to outspread the evidence Eh indicating the states of the {B,X} type variables and the X type variables included in Hkj, and breaks the logic cycles during the outspread:
(171) When Eh indicates that Xn is in its state k, then Eh=Xnk; if Eh is the virtual consequence variable of Xn,





  
    
      E
      h
    
    =
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            E
            ;
            
              n
              ⁢
              
                
              
              ⁢
              k
            
          
        
        ⁢
        
          X
          
            n
            ⁢
            
              
            
            ⁢
            k
          
        
      
    
  
  ;




when Eh indicates that Bi is in its state j, then Eh=Bij; if Eh is the virtual consequence variable of Bi,




  
    
      E
      h
    
    =
    
      
        ∑
        j
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            E
            ;
            
              i
              ⁢
              
                
              
              ⁢
              j
            
          
        
        ⁢
        
          B
          
            i
            ⁢
            
              
            
            ⁢
            j
          
        
      
    
  
  ;



(172) Outspread Xnk according to





  
    
      X
      
        n
        ⁢
        
          
        
        ⁢
        k
      
    
    =
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            
              n
              ⁢
              
                
              
              ⁢
              k
            
            ;
            i
          
        
        ⁢
        
          V
          i
        
      
    
  
  ,




where Vi are the direct cause variables of Xn, i∈SEXn, V∈{X,B,G,D};
(173) When Vi is a logic gate, the input variables of Vi are outspreaded according to the truth value table of this logic gate; if the input variables are logic gates again, outspread these input variables in the same way;
(174) Consider every non-F type variable in the logic expression outspreaded from (172) and (173): {circle around (1)} If it is such an X type variable that has not appeared in the causality chain, repeat the logic outspread process described in (172) and (173);
{circle around (2)} If it is a {B,D} type variable or such an X type variable that has appeared in the causality chain, no further outspread is needed;
(175) In the said step (174), {circle around (2)} the X type variable that has appeared in the causality chain is called the repeated variable; in the dynamical case, the repeated variable is the same variable but is in the near earlier moment; the probability distribution of this variable is known according to the computation or the observed evidence in the earlier moment; in the static case, the repeated variable as cause is treated as null, i.e. {circle around (1)} if the repeated variable as cause is connected to the consequence variable by only an F type variable without any logic gate, this F type variable is eliminated, meanwhile the relationship corresponding to this F type variable is eliminated from rn; {circle around (2)} if the repeated variable as cause is connected with the consequence variable by being an input variable of a logic gate in which the repeated variable is logically combined with other input variables, this repeated variable is eliminated from the input variables of the logic gate.

18. The method according to claim 17, wherein, the said step (175) {circle around (2)} including the following steps performed by the at least one processor, to eliminate an input variable of a logic gate is involved: suppose the variable to be eliminated from the logic gate is Vi, then,
(181) When the logic gate is a virtual logic gate, eliminate the direct cause variable Vi in the corresponding implicit mode first, reconstruct the conditional probability table and then transform the new implicit mode case to a new virtual logic gate and a new virtual functional variable; correspondingly, the new virtual functional variable may be introduced;
(182) When the logic gate is not a virtual logic gate, make the logic gate as the most simplified logic gate first; based on the most simplified logic gate, calculate the logic expression in every input row in the truth value table by treating any state of Vi as null, eliminate the input row along with the corresponding logic gate state when this row is calculated as null; the functional or conditional functional events with this logic gate state as their input events are also eliminated;
(183) If all the input variables of a non-virtual logic gate are eliminated, or all the input rows of the truth value table are eliminated, this logic gate becomes null;
(184) Repeat the above steps to treat the case when more than one input variables are eliminated.

19. The method according to claim 17, wherein further including the following steps performed by the at least one processor, to outspread E, E′ HkjE′ or HkjE′:
(191) to simplify DUCG and to outspread the X type variables for breaking logic cycles, it may change the input variables and the truth value table of the logic gate in EDUCG; after the change, make the expression in the truth value table of the logic gate as the exclusive expression; then, the logic gate is outspreaded according to the exclusive expressions of the input rows in the truth value table;
(192) The result of the AND operation of different initiating events is null “0”;
(193) If the logical outspread to the default state Xnη of Xn is necessary, while the direct cause variables of Xnη are not represented, outspread Xnη according to





  
    
      X
      
        n
        ⁢
        
          
        
        ⁢
        η
      
    
    =
    
      
        ∏
        
          k
          ≠
          η
        
      
      ⁢
      
        
      
      ⁢
      
        
          X
          _
        
        
          n
          ⁢
          
            
          
          ⁢
          k
        
      
    
  
  ;



(194) If Xnk, k≠η, does not have input or the input is null, Xnk=0;
(195) When the condition Cn;i of the conditional functional variable Fn;i becomes invalid during the outspread, Fn;i is eliminated.

20. The method according to claim 15, wherein, the said step (152) further including the following steps performed by the at least one processor, to find the possible solution set S:
(201) Outspread





  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      
    
    ⁢
    
      E
      i
      ′
    
  




so as to obtain the sum-of-product type logic expression composed of only the {B,D,F} type events, where “product” indicates the logic AND, “sum” indicates the logic OR, and a group of events ANDed together is an “item”;
(202) After Eliminating the {F,D} type events and other inevitable events in all items, further simplify the outspreaded expression by logically absorbing or combining the physically same items;
(203) After finishing the above steps, every item in the final outspreaded expression is composed of only the B type events and every item is a possible solution event; all these items compose the possible solution set S conditioned on E, in which the item with same B type variables is denoted as Hk, and the item with same B type variables but in different states is denoted as Hkj. Hkj is a possible solution.

21. The method according to claim 15, wherein, further including the following steps performed by the at least one processor, to extend the method to include the dynamical case involving more than one time point, that is, transform the case that the process system dynamically changes according to time as the static cases at sequential discrete time points, and perform the computation for each time point; then combine all the static computation results at different time points together so as to correspond the dynamical change of the process system:
(211) Classify the time as discrete time points t1, t2, . . . , tn for each time point ti, collect the static evidence E(ti) at that time point; find all the possible solutions Hkj conditioned on E(ti), these possible solutions compose the static possible solution set S(ti) at time ti; wherein: treat E(ti) as E, {circle around (1)} Construct the E(ti) conditional original DUCG; {circle around (2)} Simplify the E(ti) conditional original DUCG; {circle around (3)} transform the simplified DUCG as EDUCG; {circle around (4)} Outspread





  
    
      E
      ⁡
      
        (
        
          t
          i
        
        )
      
    
    =
    
      
        ∏
        k
      
      ⁢
      
        
      
      ⁢
      
        
          E
          k
        
        ⁡
        
          (
          
            t
            i
          
          )
        
      
    
  
  ,




then obtain the possible solution set Si at time ti;
(212) Calculate





  
    
      S
      ⁡
      
        (
        
          t
          n
        
        )
      
    
    =
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        
      
      ⁢
      
        S
        i
      
    
  
  ,




S(tn) is called the dynamical possible solution at time tn;
(213) Eliminate the other possible solutions included in EDUCG but not included in S(tn), further simplify the EDUCG;
(214) Based on the above simplified EDUCG, calculate the static state probabilities with incomplete and complete information hkjs′(ti) and hkjs(ti) respectively, the static rank probabilities with incomplete and complete information hkjr′(ti) and hkjr(ti) respectively, of Hkj in S(tn), as well as the unconditional probability hkj(t0)=Pr{Hkj};
(215) Calculate the dynamical state and rank probabilities with incomplete and complete information of Hkj included in S(tn) as follows:
{circle around (1)} The dynamical state probabilities with incomplete and complete information:




  
    
      
        h
        
          k
          ⁢
          
            
          
          ⁢
          j
        
        
          s
          ′
        
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              
                k
                ⁢
                
                  
                
                ⁢
                j
              
              
                s
                ′
              
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  
                    k
                    ⁢
                    
                      
                    
                    ⁢
                    j
                  
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          ∑
          j
        
        ⁢
        
          
        
        ⁢
        
          
            ∏
            
              i
              =
              1
            
            n
          
          ⁢
          
            
          
          ⁢
          
            
              
                h
                
                  k
                  ⁢
                  
                    
                  
                  ⁢
                  j
                
                
                  s
                  ′
                
              
              ⁡
              
                (
                
                  t
                  i
                
                )
              
            
            /
            
              
                (
                
                  
                    h
                    
                      k
                      ⁢
                      
                        
                      
                      ⁢
                      j
                    
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                )
              
              
                n
                -
                1
              
            
          
        
      
    
  




  
    
      
        h
        
          k
          ⁢
          
            
          
          ⁢
          j
        
        s
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              
                k
                ⁢
                
                  
                
                ⁢
                j
              
              s
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  
                    k
                    ⁢
                    
                      
                    
                    ⁢
                    j
                  
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          ∑
          j
        
        ⁢
        
          
        
        ⁢
        
          
            ∏
            
              i
              =
              1
            
            n
          
          ⁢
          
            
          
          ⁢
          
            
              
                h
                
                  k
                  ⁢
                  
                    
                  
                  ⁢
                  j
                
                s
              
              ⁡
              
                (
                
                  t
                  i
                
                )
              
            
            /
            
              
                (
                
                  
                    h
                    
                      k
                      ⁢
                      
                        
                      
                      ⁢
                      j
                    
                  
                  ⁡
                  
                    (
                    
                      t
                      0
                    
                    )
                  
                
                )
              
              
                n
                -
                1
              
            
          
        
      
    
  


In which, when hkj(t0)=0, hkjs′(ti)/(hkj(t0))n-1=0 and hkjs(ti)/(hkj(t0))n-1=0;
{circle around (1)}The dynamical rank probabilities with incomplete and complete information:




  
    
      
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In which, when hkj(t0)=0, hkjr′(t0))n-1=0 and hkjr(ti)/(hkj(t0))n-1=0.

22. A method for constructing an intelligent system for processing the uncertain causality information, the method includes: representing the causalities among the things in the implicit representation mode, specifically including the following steps performed by the at least one processor:
(1) Establish a representation system about the various cause variables Vi and consequence variables Xn in concern with the problem to be solved, wherein: {circle around (1)} Let V represent two type variables B and X, i.e. V∈{B,X}, in which B is the basic variable that is only the cause variable and X is the consequence variable that can be also the cause variable of the other consequence variables; {circle around (2)} No matter the states of the variable Vi or Xn, are discrete or not, represent them all as the discrete or fuzzy discrete states, so as to be dealt with by using the same manner, that is, represent the different states of Vi and Xn as Vij and Xnk respectively, where i and n index variables while j and k index the discrete or fuzzy discrete states of the variables; {circle around (3)}When Vi or Xn, is continuous, the membership of an arbitrary value ei of Vi or en of Xn, belonging to Vij or Xnk respectively, is mij(ei) or mnk(en) respectively, and they satisfy





  
    
      
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{circle around (4)} Vij and Xnk are treated as events, i.e., Vij represents the event that Vi is in its state j and Xnk represents the event that Xn is in its state k; meanwhile, if j≠j′ and k≠k′, Vij is exclusive with Vij′ and Xnk is exclusive with Xnk′; {circle around (5)} If i≠i′, Bij and Bij′ are independent events, and their occurrence probabilities bij satisfies




  
    
      
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(2) For the consequence variable Xn, determine its direct cause variables Vi, i∈SEXn, SEXn is the index set of the {B,X} type direct variables of Xn in the explicit representation mode;
(3) The conditional probability table (CPT) is used to represent the causality between the consequence variable Xn and its direct cause variables Vi, i∈SIXn, wherein: {circle around (1)} When no cause variable will be eliminated, CPT is composed of only the conditional probabilities pnk;ij, where pnk;ij≡Pr{Xnk|j} and j indexes the state combination of the cause variables Vi, i∈SIXn; {circle around (2)} When part or even all cause variables may be eliminated, CPT is composed of three parameters: pnk;ij; qnk;ij and dn;j, satisfying pnk;ij=qnk;ij/dn;j, so that CPT can be reconstructed when some of its cause variables are eliminated, where qnk;ij and qn;nj are the sample number and occurrence number of Xnk respectively, conditioned on the state combination indexed by j of the cause variables.

23. The method according to claim 22, wherein the said step (3) including the following steps performed by the at least one processor:
(231) In the implicit representation mode, the cause variables Vi, i∈SIXn, can be separated as several groups, every group uses the implicit representation mode to represent the uncertain causality to Xn;
(232) Give the relationship rXn between every group of direct cause variables to the consequence variable Xn;
(233) If some cause variables in the group are eliminated for any reason, the CPT of this group can be reconstructed as follows: Suppose the variable to be eliminated is Vi, before the elimination, there are several subgroups of the state combinations of the input variables indexed by j′; in subgroup j′, the states of all the variables are same except the states of Vi; denote the index set of the state combination j in subgroup j′ as Sij′, then





  
    
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In which j′ is the new index of the remnant state combinations after the elimination of Vi;
(234) Repeat (233) to deal with the case in which more than one cause variable is eliminated.

24. The method according to claim 22, wherein the method further includes: representing the causalities among the things in the explicit representation mode, specifically including the following steps performed by the at least one processor:
(4) The functional variable Fn;i is used to represent the causality between Vi, i∈SEXn, and Xn. Vi is the input or cause variable of Fn;i and Xn, is the output or consequence variable of Fn;i, wherein: {circle around (1)} The causality uncertainty between Vi and Xn is represented by the occurrence probability fnk:ij of the specific value Fnk;ij of Fn;i; Fnk;ij is a random event representing the uncertain functional mechanism of Vij causing Xnk. Fnk;ij is the probability contribution of Vij to Xnk; {circle around (2)} fnk;ij=(rn;i/rn)ank;ji, where rn;i is called the relationship between Vi and Xn, rn is the normalization factor and





  
    
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ank;ij is the probability of the event that Vij causes Xnk regardless of any other cause variables and ank;ij and rn can be the function of time; {circle around (3)} ank;ij satisfies




  
    
      
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25. The method according to claim 24, wherein the functional variable Fn;i in the explicit representation mode can be the conditional functional variable, the conditional functional variable is used to represent the functional relation between the cause variable Vi and the consequence variable Xn conditioned on Cn;i, wherein:
(1) Cn;i has only two states: true or false, and its state can be found according to the observed information or the computation results;
(2) When Cn;i is true, the conditional functional variable becomes the functional variable;
(3) When Cn;i is false, the conditional functional variable is eliminated.

26. A non-transitory computer-readable medium containing executable instructions that, when executed by a machine, cause the machine to implement a method for constructing an intelligent system for processing the uncertain causality information, the method includes: representing the causalities among the things in the explicit representation mode, specifically including the following steps;
(1) Establish a representation system about the various cause variables Vi and consequence variables Xn in concern with the problem to be solved, wherein: {circle around (1)} Let V represent two type variables B and X, i.e. V∈{B,X}, in which B is the basic variable that is only the cause variable and X is the consequence variable that can be also the cause variable of the other consequence variables; {circle around (2)} No matter the states of the variable Vi or Xn are discrete or not, represent them all as the discrete or fuzzy discrete states, so as to be dealt with by using the same manner, that is, represent the different states of Vi and Xn as and Xnk respectively, where i and n index variables while j and k index the discrete or fuzzy discrete states of the variables; {circle around (3)} When Vi or Xn is continuous, the membership of an arbitrary value ei of Vi or en of Xn, belonging to Vij or Xnk respectively, is mij(ei) or mnk(en) respectively, and they satisfy





  
    
      
        
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{circle around (4)} Vij and Xnk are treated as events, i.e., represents the event that Vi is in its state j and Xnk represents the event that Xn is in its state k; meanwhile, if j≠j′ and k≠k′, Vij is exclusive with Vij′ and Xnk is exclusive with Xnk′; {circle around (5)} If i≠i′ and Bij′ are independent events, and their occurrence probabilities bij satisfies




  
    
      
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(2) For the consequence variable Xn, determine its direct cause variables Vi, i∈SEXn, SEXn is the index set of the {B,X} type direct variables of Xn in the explicit representation mode;
(3) The functional variable Fn;i is used to represent the causality between Vi, i∈SEXn, and Xn. Vi is the input or cause variable of Fn;i and Xn is the output or consequence variable of Fn;i, wherein: The causality uncertainty between Vi and Xn is represented by the occurrence probability fnk;ij of the specific value Fnk;ij of Fn;i. Fnk;ij is a random event representing the uncertain functional mechanism of Vij causing Xnk. fnk;ij is the probability contribution of Vij to Xnk; {circle around (2)} fnk;ij=(rn;i/rn)ank;ij, where rn;i is called the relationship between Vi and Xn, rn is the normalization factor and





  
    
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ank;ij is the probability of the event that Vij causes Xnk regardless of any other cause variables and ank;ij and rn can be the function of time; {circle around (3)} ank;ij satisfies




  
    
      
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27. The medium according to claim 26, wherein the functional variable Fn;i in the explicit representation mode can be the conditional functional variable, the conditional functional variable is used to represent the functional relation between the cause variable Vi and the consequence variable Xn conditioned on Cn;i, wherein:
(1) Cn;i has only two states: true or false, and its state can be found according to the observed information or the computation results;
(2) When Cn;i is true, the conditional functional variable becomes the functional variable;
(3) When Cn;i is false, the conditional functional variable is eliminated.

28. The medium according to claim 26, wherein the explicit representation mode includes also extending V∈{B,X} to V∈{B,X,G} in the explicit representation mode, where G is the logic gate variable, i.e. the cause variable to influence the consequence variable by the state logic combinations of a group of cause variables, suppose the input variables of logic gate variable Gi are Vh, then the logic gate Gi is constructed by the following steps:
(31) The logic combinations between the input variables Vh, V∈{B,X,G}, are represented by the truth value table of Gi in which each input row is a logic expression composed of the input variable states and corresponds to a unique state of Gi, different rows of the logic expressions are exclusive with each other, wherein if a logic expression is true, the corresponding state of Gi is true;
(32) The set of the states of Gi is equal to or less than the set of all state combinations of the input variables;
(33) When the set of Gi the states of is less than the set of all state combinations of the input variables, there is a remnant state of Gi, which corresponds uniquely to the remnant state combinations of the input variables, so that all the states of Gi including the remnant state are exclusive with each other and just cover all the state combinations of the input variables;
(34) When Gi is the direct cause variable of Xn, Gi functions to Xn through the functional or conditional functional variable Fn;i;
(35) If a logic gate has only one input variable, this logic gate can be ignored, i.e. the input variable of the logic gate can be taken as the input variable of the functional variable or conditional functional variable Fn;i with this logic gate as its input variable;
(36) When Gi is the direct cause variable of Xn, the relationship between Gi and Xn is rn;i; when calculating fnk;ij, the calculation to rn includes the relationship between Gi and Xn; when calculating Pr{Xnk}, the fnk;ji between Gi and Xn is included.

29. The medium according to claim 28, wherein further including:
(41) Extend V∈{B,X} as V∈{B,X,D}, or extend V∈E {B,X,G} as V∈{B,X,G,D}, in which D is the default event or variable, Dn can appear only with Xn and is an independent cause variable that has only one inevitable state;
(42) Dn becomes a direct cause variable of Xn through Fn;D, where Fn;D is the functional variable between Dn, and Xn;
(43) The causality uncertainty between Dn and Xn is represented by the occurrence probability fnk;D of the specific value Fnk;D of Fn;D, where Fnk;D is a random event representing the functional mechanism of Dn to Xn, and fnk;D is the probability contribution of Dn to Xnk;
(44) fnk;D=(rn;D/rn)ank;D, where ank;D is the probability of the event that Dn causes Xn regardless of the other cause variables of Xn, and satisfies





  
    
      
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(45) The original





  
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30. The medium according to claim 29, wherein, further including: when the default variable of Xn is more than one, they can be combined as one default variable Dn; let g be the index distinguishing two or more default variables, Corresponding to the case of only one default variable, the variable Dn and the parameters rn;D, ank;D are represented as Dng, rn;Dg, ank;Dg respectively; after combining Dng as Dn, the parameters of Dn are calculated according to




  
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31. The medium according to claim 26, wherein the method further includes: using the implicit mode to represent the uncertain causalities among things, specifically including the following step:
(4) The conditional probability table (CPT) is used to represent the causality between the consequence variable Xn, and its direct cause variables Vi, i∈SIXn, wherein: {circle around (1)} When no cause variable will be eliminated, CPT is composed of only the conditional probabilities pnk;ij, where pnk;ij≡Pr{Xnk|j} and j indexes the state combination of the cause variables Vi, i∈SIXn; {circle around (2)} When part or even all cause variables may be eliminated, CPT is composed of three parameters: pnk;ij, qnk;ij and dn;j, satisfying pnk;ij=qnk;ij/dn;j, so that CPT can be reconstructed when some of its cause variables are eliminated, where qnk;ij and qn;nj are the sample number and occurrence number of Xnk respectively, conditioned on the state combination indexed by j of the cause variables.

32. The medium according to claim 31, wherein the said step (4) further including:
(71) In the implicit representation mode, the cause variables Vi, i∈SIXn, can be separated as several groups, every group uses the implicit representation mode to represent the uncertain causality to Xn;
(72) Give the relationship rXn between every group of direct cause variables to the consequence variable Xn;
(73) If some cause variables in the group are eliminated for any reason, the CPT of this group can be reconstructed as follows: Suppose the variable to be eliminated is Vi, before the elimination, there are several subgroups of the state combinations of the input variables indexed by j′; in subgroup j′, the states of all the variables are same except the states of Vi; denote the index set of the state combination j in subgroup j′ as Sij, then





  
    
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            j
            '
          
        
      
    
  



In which j′ is the new index of the remnant state combinations after the elimination of Vi;
(74) Repeat (73) to deal with the case in which more than one cause variable is eliminated.

33. The medium according to claim 31, wherein further including the following steps:
(5) For a group of cause variables Vi, i′∈SIXn, in the implicit representation mode, give the corresponding relationship rn=rn+rXn, while in the explicit representation mode, rn is renewed as rn=rn+rXn, in which the right side rn is before the renewing;
(6) If the implicit representation mode has more than one group, they can be indexed by g and every group relationship can be denoted as rXng; then the calculation equation in above (5) becomes





  
    r
    n
  
  =
  
    
      r
      n
    
    +
    
      
        ∑
        h
      
      ⁢
      
        
          r
          Xng
        
        ·

34. The medium according to claim 33, further including:
(10) According to the specific cases of every consequence variable Xn the representations above for all the consequence variables compose the original DUCG;
(11) The evidence E in concern with the original DUCG is received during the online application and is expressed as





  
    E
    =
    
      
        E
        *
      
      ⁢
      
        
          ∏
          h
        
        ⁢
        
          
        
        ⁢
        
          E
          h
        
      
    
  
  ,




where Eh is the evidence indicating the state of the {B,X} type variable, E* represents the other evidence; if Eh is a fuzzy state evidence, i.e. the state of the variable Vh in the original DUCG is known in a state probability distribution, or if Eh is a fuzzy continuous evidence, i.e. the specific value eh of the continuous variable Vh is known in the fuzzy area of different fuzzy states of Vh, V∈{B,X}, then add Eh as a virtual evidence variable into the original DUCG and represent the causality between Vh and Eh according to the explicit mode so that Eh becomes the consequence variable of the cause variable Vh; after finishing these steps, the original DUCG becomes the E conditional original DUCG.

35. The medium according to claim 34, wherein, the said step (11) including: adding Eh as a virtual evidence variable into the original DUCG, and further including the following steps: Suppose mhj=mhj(eh) is the membership of Eh belonging to the fuzzy state j, or mhj is the probability of Xhj indicated by the fuzzy state evidence Eh, i.e., mhj=Pr{Vhj|Eh}, j∈SEh, SEh is the index set of state j in which mhj≠0 and includes at least two different indexes, while satisfying




  
    
      ∑
      
        j
        ∈
        
          S
          Eh
        
      
    
    ⁢
    
      m
      hj
    
  
  =
  
    1
    ⁢
    
      :
    
  



(101) As the virtual consequence variable of Vh, Eh has only one inevitable state, has only one direct cause variable Vh, and is not the cause variable of any other variable;
(102) The virtual functional variable from Vhj to Eh is FE;h and its specific value FE;hj is the virtual random event that Vhj causes Eh; the functional intensity parameter fE;hj of FE;hj may be given by domain engineers;
(103) If the domain engineers cannot give fE;hj, it can be calculated from





  
    
      f
      
        E
        ;
        hj
      
    
    =
    
      
        
          
            m
            hj
          
          ⁢
          
            v
            hk
          
        
        
          
            m
            hk
          
          ⁢
          
            v
            hj
          
        
      
      ⁢
      
        f
        
          E
          ;
          hk
        
      
    
  
  ,




where j≠k, j∈SEh, k∈SEh, vhj≡Pr{Vhj}, vhk≡Pr{Vhk} Given fE;hk>0, fE;hj can be calculated.

36. The medium according to claim 35, wherein, further including the following steps to simplify the E conditional original DUCG: suppose Vi is the direct cause variable of Xn, V∈{B,X,G,D}, then (111) According to E, determine whether or not the condition Cn;i of the conditional functional variable Fn;i is valid: {circle around (1)} if yes, change the conditional functional variable as the functional variable; {circle around (2)} if not, eliminate this conditional functional variable; {circle around (3)} if cannot determine whether or not Cn;i is valid, keep the conditional functional variable until Cn;i can be determined;
(112) According to E, if Vih, is not the cause of any state of Xn, when E shows that Vih is true, eliminate the functional or conditional functional variable Fn;i that is from Vi to Xn;
(113) According to E, if Xnk cannot be caused by any state of Vi, when E shows that Xnk is true, eliminate the functional or conditional functional variable from Vi to Xn;
(114) In the explicit mode of representation, if the X or G type variable without any cause or input appears, eliminate this variable along with the F type variables starting from this variable;
(115) If there is any group of isolated variables without any logic connection to the variables related to E, eliminate this group variables;
(116) If E shows that Xnk is true, while Xnk is not the cause of any other variable and Xn has no connection with the other variables related to E, denote the index set of the index n of such Xn as SEnk; When Vi and its logic connection variables Fn;i have no logic connection with the variables related to E except the variables indexed in SEnk, eliminate Xn, Vi and the functional or conditional functional variables Fn;i along with all other variables logically connected with Vi;
(117) If E shows that Xnk appears earlier than Vij, so that for sure Vij is not the cause of Xnk, eliminate the functional or conditional functional variables that are in the causality chains from Vi to Xn but are not related to the influence of other variables to Xn;
(118) Upon demand, the above steps can be in any order and can be repeated.

37. The medium according to claim 36, wherein, further including the following steps to transform the DUCG with implicit or hybrid representation mode conditioned on E as all in the explicit mode, i.e. EDUCG:
(123) For the consequence variable Xn in the implicit or hybrid mode, for every group of SIXn type cause variables, introduce a virtual logic gate variable Gi, in which the cause variables of SIXn are the input variables of Gi, and the number of the states of Gi and the input rows of the truth value table of Gi equal to the number of the state combinations of the cause variables in SIXn, while each of the state combination of the input variables is an input row of the truth value table of Gi and also a state of the virtual logic gate;
(124) Introduce the virtual functional variable Fn;i, in which Gi is the input variable and Xn, is the output variable, so that Gi becomes the direct cause variable of Xn;
(125) In the CPT of the cause variables in SIXn, ank;ij=pnk;j; the relationship of Fn;i is; rni=rXn;
(126) When there is only one input variable in Gi, such Gi, can be ignored, i.e. the virtual functional variable takes the input variable of Gi as its input variable directly;
(127) When the groups of the SIXn type variables are more than one group, repeat the above steps for every groups.

38. The medium according to claim 36, wherein, further including the following steps to transform the DUCG conditioned on E in the explicit representation mode or in the more than one group implicit representation mode as the IDUCG in which all representations are in the implicit representation mode with only one group direct cause variables:
(131) If Cn;i is valid, change the conditional functional variable as the functional variable; If Cn;i is invalid, eliminate the conditional functional variable;
(132) For any representation of the uncertain causality between the consequence variable Xn and its direct cause variables, if it is in the hybrid or more than one group implicit representation mode, transform the representation mode for Xn to the explicit mode;
(133) After the above steps, take the state combinations of the {B,X} type cause variables of the consequence variable Xn as the conditions indexed by j, calculate the conditional probability of XnkPr{Xnk|j} according to the explicit mode, where the connections between the {B,X} type cause variables and Xn may be or may not be through logic gates; in the calculation, all contributions from different types of direct cause variables should be considered, i.e. when the direct cause variables are {X,B,G} types,





  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          |
          j
        
        }
      
    
    =
    
      
        ∑
        i
      
      ⁢
      
        f
        
          nk
          ;
          ih
        
      
    
  
  ;




when the direct cause variables are {X,B,G,D} types,




  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          |
          j
        
        }
      
    
    =
    
      
        
          ∑
          i
        
        ⁢
        
          f
          
            nk
            ;
            ih
          
        
      
      +
      
        f
        
          nk
          ;
          D
        
      
    
  
  ;



(134) The case of ank;ih=1 can be understood as that Xnk is true for sure, i.e. when the input variable i is in its state h, all the states, except k, of Xn cannot be true; if this applies, when ank;ih=1, Pr{Xnk|j}=1, meanwhile Pr{Xnk′|j}=0, where k≠k′;
(135) If ank;ih=1, k∈Sm, Sm is the index set of such states of Xn that ank;ih=1 and the number of such states is m, then Pr{Xnk|j}=1/m and Pr{Xnk|j}=0, where k′∉Sm;
(136) If such calculated





  
    
      
        ∑
        k
      
      ⁢
      
        Pr
        ⁢
        
          {
          
            
              X
              nk
            
            |
            j
          
          }
        
      
    
    <
    1
  
  ,
  
    
  
  ⁢
  
    
      let
      ⁢
      
        
      
      ⁢
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          |
          j
        
        }
      
    
    =
    
      1
      -
      
        
          ∑
          
            k
            ≠
            η
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                X
                nk
              
              |
              j
            
            }
          
        
      
    
  
  ,




where η indexes the default state of Xn;
(137) If there is no default state η in said step (136), the normalization method is used
as follows:





  
    
      Pr
      ⁢
      
        {
        
          
            X
            nk
          
          |
          j
        
        }
      
    
    =
    
      Pr
      ⁢
      
        
          {
          
            
              X
              nk
            
            |
            j
          
          }
        
        /
        
          
            ∑
            k
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  X
                  nk
                
                |
                j
              
              }
            
          
        
      
    
  
  ,




the Pr {Xnk|j} on the right side are the values before the normalization;
(138) After satisfying the normalization, Pr{Xnk|j} becomes the conditional probability Pnk;nj in the standard implicit representation mode;
(139) Connect the {X,B} type direct cause variables of Xn, through or not through logic gates with Xn according to the implicit representation mode, the DUCG conditioned on E is transformed as the IDUCG.

39. The medium according to claim 36, wherein, further including the following steps:
(144) outspread the evidence events Eh, included in E, which determine the states of the {B,X} type variables, and the events Hkj in concern, and in the process of outspread, break the logic cycles;
(145) based on the outspreaded logic expressions of Eh and Hkj, further outspread





  
    
      ∏
      h
    
    ⁢
    
      
        E
        h
      
      ⁢
      
        
      
      ⁢
      and
      ⁢
      
        
      
      ⁢
      
        H
        kj
      
      ⁢
      
        
          ∏
          h
        
        ⁢
        
          E
          h
        
      
    
  
  ;



(146) calculate the state probability and the rank probability of the concerned event Hkj conditioned on E according to the following equations:
The state probability:





  
    
      h
      kj
      s
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            E
          
          }
        
      
      
        Pr
        ⁢
        
          {
          E
          }
        
      
    
  
  ;



The rank probability:





  
    h
    kj
    r
  
  =
  
    
      
        h
        kj
        s
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          h
          kj
          s
        
      
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            E
          
          }
        
      
      
        
          ∑
          
            
              H
              kj
            
            ∈
            S
          
        
        ⁢
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              E
            
            }
          
        
      
    
  



Where S is the set of all the events in concern.

40. The medium according to claim 39, wherein, the said step (145) including:
(151) Express the evidence set





  
    ∏
    h
  
  ⁢
  
    E
    h
  




indicating the states or the {B,X} type variables as E′E′′, in which




  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      E
      i
      ′
    
  




is the evidence set composed of the evidence events indicating the abnormal states of variables, and




  
    E
    ′′
  
  =
  
    
      ∏
      
        i
        ′
      
    
    ⁢
    
      E
      
        i
        ′
      
      ′′
    
  




is the evidence set composed of the evidence events indicating the normal states of variables;
(152) Outspread





  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      E
      i
      ′
    
  




and determine the possible solution set S conditioned on E, where every possible solution Hkj is an event in concern for the problem to be solved;
And further the said step (146) including:
(153) Calculate two types of the state probability and rank probability of Hkj conditioned on E:
The state probability with incomplete information:





  
    
      h
      kj
      
        s
        ′
      
    
    =
    
      
        Pr
        ⁢
        
          {
          
            
              H
              kj
            
            ⁢
            
              E
              ′
            
          
          }
        
      
      
        Pr
        ⁢
        
          {
          
            E
            ′
          
          }
        
      
    
  
  ;



The state probability with complete information:





  
    
      h
      kj
      s
    
    =
    
      
        
          h
          kj
          
            s
            ′
          
        
        ⁢
        
          
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                |
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
          
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                |
                
                  E
                  ′
                
              
              }
            
          
        
      
      =
      
        
          
            h
            kj
            
              s
              ′
            
          
          ⁢
          Pr
          ⁢
          
            {
            
              
                E
                ′′
              
              ❘
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
              
            
            }
          
        
        
          
            ∑
            j
          
          ⁢
          
            
              h
              kj
              
                s
                ′
              
            
            ⁢
            Pr
            ⁢
            
              {
              
                
                  E
                  ′′
                
                |
                
                  
                    H
                    kj
                  
                  ⁢
                  
                    E
                    ′
                  
                
              
              }
            
          
        
      
    
  
  ;



The rank probability with incomplete information:





  
    
      h
      kj
      
        r
        ′
      
    
    =
    
      
        
          h
          kj
          
            s
            ′
          
        
        
          
            ∑
            
              
                H
                kj
              
              ∈
              S
            
          
          ⁢
          
            h
            kj
            
              s
              ′
            
          
        
      
      =
      
        
          Pr
          ⁢
          
            {
            
              
                H
                kj
              
              ⁢
              
                E
                ′
              
            
            }
          
        
        
          
            ∑
            
              
                H
                kj
              
              ∈
              S
            
          
          ⁢
          
            Pr
            ⁢
            
              {
              
                
                  H
                  kj
                
                ⁢
                
                  E
                  ′
                
              
              }
            
          
        
      
    
  
  ;



The rank probability with complete information:





  
    h
    kj
    r
  
  =
  
    
      
        h
        kj
        
          r
          ′
        
      
      ⁢
      Pr
      ⁢
      
        {
        
          
            E
            ′′
          
          ❘
          
            
              H
              kj
            
            ⁢
            
              E
              ′
            
          
        
        }
      
    
    
      
        ∑
        
          
            H
            kj
          
          ∈
          S
        
      
      ⁢
      
        
      
      ⁢
      
        
          h
          kj
          
            r
            ′
          
        
        ⁢
        Pr
        ⁢
        
          {
          
            
              E
              ′′
            
            ❘
            
              
                H
                kj
              
              ⁢
              
                E
                ′
              
            
          
          }
        
      
    
  



In which, if HkjE′ is null, Pr{E′′|HkjE′}≡0.

41. The medium according to claim 38, wherein, further including the following steps:
(140) use the BN method to calculate the state probability distribution of the variables in concern conditioned on E.

42. The medium according to claim 40, wherein, further including the following steps to outspread E, E′, HkjE or HkiE′, and to outspread the evidence Eh, indicating the states of the {B,X} type variables and the X type variables included in Hkj, and breaks the logic cycles during the outspread:
(171) When Eh indicates that Xn is in its state k, then Eh=Xnk; if Eh is the virtual consequence variable of Xn,





  
    
      E
      h
    
    =
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            E
            ;
            nk
          
        
        ⁢
        
          X
          nk
        
      
    
  
  ;




when Eh indicates that Bi is in its state j, then Eh=Bij; if Eh is the virtual consequence variable of Bi,




  
    
      E
      h
    
    =
    
      
        ∑
        j
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            E
            ;
            ij
          
        
        ⁢
        
          B
          ij
        
      
    
  
  ;



(172) Outspread Xnk according to





  
    
      X
      nk
    
    =
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        
          F
          
            nk
            ;
            i
          
        
        ⁢
        
          V
          i
        
      
    
  
  ,




where Vi are the direct cause variables of Xn, i∈SEXn, V∈{X,B,G,D};
(173) When Vi is a logic gate, the input variables of Vi are outspreaded according to the truth value table of this logic gate; if the input variables are logic gates again, outspread these input variables in the same way;
(174) Consider every non-F type variable in the logic expression outspreaded from (172) and (173): {circle around (1)} If it is such an X type variable that has not appeared in the causality chain, repeat the logic outspread process described in (172) and (173); {circle around (2)} If it is a {B,D} type variable or such an X type variable that has appeared in the causality chain, no further outspread is needed;
(175) In the said step (174), {circle around (2)} the X type variable that has appeared in the causality chain is called the repeated variable; in the dynamical case, the repeated variable is the same variable but is in the near earlier moment; the probability distribution of this variable is known according to the computation or the observed evidence in the earlier moment; in the static case, the repeated variable as cause is treated as null, i.e. {circle around (1)} if the repeated variable as cause is connected to the consequence variable by only an F type variable without any logic gate, this F type variable is eliminated, meanwhile the relationship corresponding to this F type variable is eliminated from rn; {circle around (2)} f the repeated variable as cause is connected with the consequence variable by being an input variable of a logic gate in which the repeated variable is logically combined with other input variables, this repeated variable is eliminated from the input variables of the logic gate.

43. The medium according to claim 42, wherein, the said step (175) {circle around (2)} including the following steps to eliminate an input variable of a logic gate is involved: suppose the variable to be eliminated from the logic gate is Vi, then,
(181) When the logic gate is a virtual logic gate, eliminate the direct cause variable Vi in the corresponding implicit mode first, reconstruct the conditional probability table and then transform the new implicit mode case to a new virtual logic gate and a new virtual functional variable; correspondingly, the new virtual functional variable may be introduced;
(182) When the logic gate is not a virtual logic gate, make the logic gate as the most simplified logic gate first; based on the most simplified logic gate, calculate the logic expression in every input row in the truth value table by treating any state of Vi as null, eliminate the input row along with the corresponding logic gate state when this row is calculated as null; the functional or conditional functional events with this logic gate state as their input events are also eliminated;
(183) If all the input variables of a non-virtual logic gate are eliminated, or all the input rows of the truth value table are eliminated, this logic gate becomes null;
(184) Repeat the above steps to treat the case when more than one input variables are eliminated.

44. The medium according to claim 42, wherein further including the following steps to outspread E, E′ HkjE or HkjE′:
(191) to simplify DUCG and to outspread the X type variables for breaking logic cycles, it may change the input variables and the truth value table of the logic gate in EDUCG; after the change, make the expression in the truth value table of the logic gate as the exclusive expression; then, the logic gate is outspreaded according to the exclusive expressions of the input rows in the truth value table;
(192) The result of the AND operation of different initiating events is null “0”;
(193) If the logical outspread to the default state Xnη of Xn is necessary, while the direct cause variables of Xnη are not represented, outspread Xnη according to





  
    
      X
      
        n
        ⁢
        
          
        
        ⁢
        η
      
    
    =
    
      
        ∏
        
          k
          ≠
          η
        
      
      ⁢
      
        
      
      ⁢
      
        
          X
          _
        
        nk
      
    
  
  ;



(194) If Xnk, k≠η, does not have input or the input is null, Xnk=0;
(195) When the condition Cn;i of the conditional functional variable Fn;i becomes invalid during the outspread, Fn;i is eliminated.

45. The medium according to claim 40, wherein, the said step (152) further including the following steps to find the possible solution set S:
(201) Outspread





  
    E
    ′
  
  =
  
    
      ∏
      i
    
    ⁢
    
      
    
    ⁢
    
      E
      i
      ′
    
  




so as to main me sum-of-product type logic expression composed of only the {B,D,F} type events, where “product” indicates the logic AND, “sum” indicates the logic OR, and a group of events ANDed together is an “item”;
(202) After Eliminating the {F,D} type events and other inevitable events in all items, further simplify the outspreaded expression by logically absorbing or combining the physically same items;
(203) After finishing the above steps, every item in the final outspreaded expression is composed of only the B type events and every item is a possible solution event;
all these items compose the possible solution set S conditioned on E, in which the item with same B type variables is denoted as Hk, and the item with same B type variables but in different states is denoted as Hkj. Hkj is a possible solution.

46. The medium according to claim 40, wherein, further including the following steps to extend the method to include the dynamical case involving more than one time point, that is, transform the case that the process system dynamically changes according to time as the static cases at sequential discrete time points, and perform the computation for each time point; then, combine all the static computation results at different time points together so as to correspond the dynamical change of the process system:
(211) Classify the time as discrete time points t1, t2, . . . tn; for each time point ti, collect the static evidence E(ti) at that time point; find all the possible solutions Hkj conditioned on E(ti), these possible solutions compose the static possible solution set S(ti) at time ti; wherein: treat E(ti) as E, {circle around (1)} Construct the E(ti) conditional original DUCG; {circle around (2)} Simplify the E(ti) conditional original DUCG; {circle around (3)} transform the simplified DUCG as EDUCG; {circle around (4)} Outspread





  
    
      E
      ⁡
      
        (
        
          t
          i
        
        )
      
    
    =
    
      
        ∏
        k
      
      ⁢
      
        
      
      ⁢
      
        
          E
          k
        
        ⁡
        
          (
          
            t
            i
          
          )
        
      
    
  
  ,




then obtain me possible solution set Si at time ti;
(212) Calculate





  
    
      S
      ⁡
      
        (
        
          t
          n
        
        )
      
    
    =
    
      
        ∏
        
          i
          =
          1
        
        n
      
      ⁢
      
        
      
      ⁢
      
        S
        i
      
    
  
  ,




S(tn) is called the dynamical possible solution at time tn;
(213) Eliminate the other possible solutions included in EDUCG but not included in S(tn), further simplify the EDUCG;
(214) Based on the above simplified EDUCG, calculate the static state probabilities with incomplete and complete information hkjs′(ti) and hkjs(ti) respectively, the static rank probabilities with incomplete and complete information hkjr′(ti) and hkjr(ti) respectively, of Hkj in S(tn), as well as the unconditional probability hkj(t0)=Pr{Hkj};
(215) Calculate the dynamical state and rank probabilities with incomplete and complete information of Hkj included in S(tn) as follows:
{circle around (1)} The dynamical state probabilities with incomplete and complete information:




  
    
      
        h
        kj
        
          s
          ′
        
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              kj
              
                s
                ′
              
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          
            ∑
            j
          
          ⁢
          
            
              ∏
              
                i
                =
                1
              
              n
            
            ⁢
            
              
                
                  h
                  kj
                  
                    s
                    ′
                  
                
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              /
              
                
                  (
                  
                    
                      h
                      kj
                    
                    ⁡
                    
                      (
                      
                        t
                        0
                      
                      )
                    
                  
                  )
                
                
                  n
                  -
                  1
                
              
            
          
        
        ⁢
        
          
        
      
    
  




  
    
      
        h
        kj
        s
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              kj
              s
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          
            ∑
            j
          
          ⁢
          
            
              ∏
              
                i
                =
                1
              
              n
            
            ⁢
            
              
            
            ⁢
            
              
                
                  h
                  kj
                  s
                
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              /
              
                
                  (
                  
                    
                      h
                      kj
                    
                    ⁡
                    
                      (
                      
                        t
                        0
                      
                      )
                    
                  
                  )
                
                
                  n
                  -
                  1
                
              
            
          
        
        ⁢
        
          
        
      
    
  


In which, when hkj(t0)=0, hkjs′(ti)/(hkj(t0))n-1=0 and hkjs(ti)/(hkj(t0))n-1=0;
{circle around (2)} The dynamical rank probabilities with incomplete and complete information:




  
    
      
        h
        kj
        
          r
          ′
        
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              kj
              
                r
                ′
              
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          
            ∑
            
              
                H
                kj
              
              ∈
              
                S
                ⁡
                
                  (
                  
                    t
                    n
                  
                  )
                
              
            
          
          ⁢
          
            
              ∏
              
                i
                =
                1
              
              n
            
            ⁢
            
              
            
            ⁢
            
              
                
                  h
                  kj
                  
                    r
                    ′
                  
                
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              /
              
                
                  (
                  
                    
                      h
                      kj
                    
                    ⁡
                    
                      (
                      
                        t
                        0
                      
                      )
                    
                  
                  )
                
                
                  n
                  -
                  1
                
              
            
          
        
        ⁢
        
          
        
      
    
  




  
    
      
        h
        kj
        r
      
      ⁡
      
        (
        t
        )
      
    
    =
    
      
        
          ∏
          
            i
            =
            1
          
          n
        
        ⁢
        
          
        
        ⁢
        
          
            
              h
              kj
              r
            
            ⁡
            
              (
              
                t
                i
              
              )
            
          
          /
          
            
              (
              
                
                  h
                  kj
                
                ⁡
                
                  (
                  
                    t
                    0
                  
                  )
                
              
              )
            
            
              n
              -
              1
            
          
        
      
      
        
          
            ∑
            
              
                H
                kj
              
              ∈
              
                S
                ⁡
                
                  (
                  
                    t
                    n
                  
                  )
                
              
            
          
          ⁢
          
            
              ∏
              
                i
                =
                1
              
              n
            
            ⁢
            
              
            
            ⁢
            
              
                
                  h
                  kj
                  r
                
                ⁡
                
                  (
                  
                    t
                    i
                  
                  )
                
              
              /
              
                
                  (
                  
                    
                      h
                      kj
                    
                    ⁡
                    
                      (
                      
                        t
                        0
                      
                      )
                    
                  
                  )
                
                
                  n
                  -
                  1
                
              
            
          
        
        ⁢
        
          
        
      
    
  


In which, when hkj(t0)=0, hkjr′(ti)/(hkj(t0))n-10 and hkjr(ti)/(hkj (t0))n-1=0.

47. A non-transitory computer-readable medium containing executable instructions that, when executed by a machine, cause the machine to implement a method for constructing an intelligent system for processing the uncertain causality information, the method includes: representing the causalities among the things in the explicit representation mode, specifically including the following steps:
(1) Establish a representation system about the various cause variables Vi and consequence variables Xn in concern with the problem to be solved, wherein: {circle around (1)} Let V represent two type variables B and X, i.e. V∈{B,X}, in which B is the basic variable that is only the cause variable and X is the consequence variable that can be also the cause variable of the other consequence variables; {circle around (2)} No matter the states of the variable Vi or Xn are discrete or not, represent them all as the discrete or fuzzy discrete states, so as to be dealt with by using the same manner, that is, represent the different states of Vi and Xn as Vij and Xnk respectively, where i and n index variables while j and k index the discrete or fuzzy discrete states of the variables; {circle around (3)} When Vi or Xn is continuous, the membership of an arbitrary value ei of Vi or en of Xn, belonging to Vij or Xnk respectively, is mij(ei) or mnk(en) respectively, and they satisfy





  
    
      
        ∑
        j
      
      ⁢
      
        
      
      ⁢
      
        
          m
          ij
        
        ⁡
        
          (
          
            e
            i
          
          )
        
      
    
    =
    
      
        1
        ⁢
        
          
        
        ⁢
        and
        ⁢
        
          
        
        ⁢
        
          
            ∑
            j
          
          ⁢
          
            
          
          ⁢
          
            
              m
              ij
            
            ⁡
            
              (
              
                e
                i
              
              )
            
          
        
      
      =
      1
    
  
  ;




{circle around (4)} Vij and Xnk are treated as events, i.e., Vij represents the event that Vi is in its state j and Xnk represents the event that Xn, is in its state k; meanwhile, if j≠j and k≠k′, Vij is exclusive with Vij′ and Xnk is exclusive with Xnk′; {circle around (5)} If i≠i′, Bij and Bij′ are independent events, and their occurrence probabilities bij satisfies




  
    
      
        ∑
        j
      
      ⁢
      
        
      
      ⁢
      
        b
        ij
      
    
    ≤
    1
  
  ;



(2) For the consequence variable Xn, determine its direct cause variables Vi, i∈SEXn, SEXn is the index set of the {B,X} type direct variables of Xn in the explicit representation mode;
(3) The conditional probability table (CPT) is used to represent the causality between the consequence variable Xn and its direct cause variables Vi, I∈SIXn, wherein:
{circle around (1)} When no cause variable will be eliminated, CPT is composed of only the conditional probabilities pnk;ij, where pnk;ij≡Pr{Xnk|j} and j indexes the state combination of the cause variables Vi, i∈SIXn, {circle around (2)} When part or even all cause variables may be eliminated, CPT is composed of three parameters: pnk;ij, qnk;ij and dn;j, satisfying pnk;ij=qnk;ij/dn;j, so that CPT can be reconstructed when some of its cause variables are eliminated, where qnk;ij and qn;nj are the sample number and occurrence number of Xnk respectively, conditioned on the state combination indexed by j of the cause variables.

48. The medium according to claim 47, wherein the said step (3) including the following steps:
(231) In the implicit representation mode, the cause variables Vi, i∈SIXn, can be separated as several groups, every group uses the implicit representation mode to represent the uncertain causality to Xn;
(232) Give the relationship rXn between every group of direct cause variables to the consequence variable Xn;
(233) If some cause variables in the group are eliminated for any reason, the CPT of this group can be reconstructed as follows: Suppose the variable to be eliminated is Vi, before the elimination, there are several subgroups of the state combinations of the input variables indexed by j′; in subgroup j′, the states of all the variables are same except the states of Vi; denote the index set of the state combination j in subgroup j′ as Sij′, then





  
    
      q
      
        nk
        ;
        
          j
          ′
        
      
    
    =
    
      
        ∑
        
          j
          ∈
          
            S
            
              ij
              ′
            
          
        
      
      ⁢
      
        
      
      ⁢
      
        q
        
          nk
          ;
          j
        
      
    
  
  ,
  
    
      d
      
        nj
        ′
      
    
    =
    
      
        ∑
        
          j
          ⁢
          
            I
            ^
          
          ⁢
          
            S
            
              ij
              ′
            
          
        
      
      ⁢
      
        
      
      ⁢
      
        d
        nj
      
    
  
  ,
  
    
      p
      
        nk
        ;
        
          j
          ′
        
      
    
    =
    
      
        q
        
          nk
          ;
          
            j
            ′
          
        
      
      /
      
        d
        
          n
          ;
          
            j
            ′
          
        
      
    
  



In which j′ is the new index of the remnant state combinations after the elimination of Vi;
(234) Repeat (233) to deal with the case in which more than one cause variable is eliminated.

49. The medium according to claim 47, wherein the method further includes:
representing the causalities among the things in the explicit representation mode, specifically including the following steps:
(4) The functional variable Fn;i is used to represent the causality between Vi, i∈SEXn, and Xn. Vi is the input or cause variable of Fn;i and Xn is the output or consequence variable of Fn;i wherein: {circle around (1)} The causality uncertainty between Vi and Xn is represented by the occurrence probability fnk;ij of the specific value Fnk;ij of Fn;i. Fnk;jj is a random event representing the uncertain functional mechanism of Vij causing Xnk. fnk;ij is the probability contribution of Vij to Xnk; {circle around (2)} fnk;ij=(rn;i/rn)ank;ij, where rn;i is called the relationship between Vi and Xn, rn is the normalization factor and





  
    
      r
      n
    
    =
    
      
        ∑
        i
      
      ⁢
      
        
      
      ⁢
      
        r
        
          n
          ;
          i
        
      
    
  
  ,




ank;ij is the probability of the event that Vij causes Xnk regardless of any other cause variables and ank;ij and rn can be the function of time; {circle around (3)} ank satisfies




  
    
      
        ∑
        k
      
      ⁢
      
        
      
      ⁢
      
        a
        
          nk
          ;
          ij
        
      
    
    ≤
    1
  
  ;
  
    
      4
      ⁢
      
        ◯
        ⁢
        Pr
      
      ⁢
      
        {
        
          X
          nk
        
        }
      
    
    =
    
      
        ∑
        
          i
          ,
          j
        
      
      ⁢
      
        
      
      ⁢
      
        
          f
          
            nk
            ;
            ij
          
        
        ⁢
        Pr
        ⁢
        
          
            {
            
              V
              ij
            
            }
          
          .

50. The medium according to claim 49, wherein the functional variable Fn;i in the explicit representation mode can be the conditional functional variable, the conditional functional variable is used to represent the functional relation between the cause variable Vi and the consequence variable Xn conditioned on Cn;i, wherein:
(1) Cn;i has only two states: true or false, and its state can be found according to the observed information or the computation results;
(2) When Cn;i is true, the conditional functional variable becomes the functional variable;
(3) When Cn;i is false, the conditional functional variable is eliminated.