# Algorithm for Option Number Selection in Stochastic Paired Comparison Models
## 3 Comparison of Models Allowing Different Options in Choices from a Theoretical Aspect
### The Case of \(s=2\) and \(s+2=4\)
#### c.4.3.  
Let us choose a node from which a'much better' edge starts (\(i_{1}\)). It goes to the object \(i_{2}\). C.2.1. implies that there is a path along the directed edges from \(i_{2}\) to \(i_{1}\) in \(GC^{(A^{(2)})}\). We denote it by saying that \((i_{2},i_{3},\ldots,i_{1})\) and \(i_{j}\) are different. These edges are also included as'slightly better' or'much better' edges in \(GC^{(A^{(4)})}\). Therefore, the cycle \((i_{1},i_{2},\ldots,i_{1})\) forms a cycle along the'slightly better' and'much better' edges in \(GC^{(A^{(4)})}\) and it contains at least one'much better' edge.  
We proved that all three conditions, C.4.1., C.4.2., and C.4.3., hold; consequently, data matrix \(A^{(4)}\) is evaluable.  
(Case B) This case may occur. A simple example can be seen in Figure 2. The justification is as follows:  
Data are evaluable in the four-option model as  
C.4.1. is obviously satisfied; \(A^{(4)}_{1,2,4}=1\) and \(A^{(4)}_{2,3,2}=1\).  
C.4.2. In the case of every partition, the subsets are connected either by'much better' edges, both forward and backward, or a'slightly better' edge. More exactly, if \(S=\{1\}\), \(\overline{S}=\{2,3\}\), or \(S=\{1,3\}\), \(\overline{S}=\{2\}\), then \(A^{(4)}_{1,2,4}=1\), \(A^{(4)}_{2,1,4}=A^{(4)}_{1,2,1}=1\). If \(S=\{1,2\}\) and \(\overline{S}=\{3\}\), then \(A^{(4)}_{3,2,3}=A^{(4)}_{2,3,1}=1\). When \(S\) and \(\overline{S}\) are changed, the situation remains the same.  
C.4.3. The required cycle is (1,2,1).  
Data cannot be evaluated using the two-option model as C.2.1. does not hold. See \(S=\{1,2\}\) and \(\overline{S}=\{3\}\); moreover, take into the consideration that \(A^{(2)}_{1,2,1}=A^{(2)}_{1,2,2}=2\), \(A^{(2)}_{2,3,1}=A^{(2)}_{3,2,2}=1\), \(A^{(2)}_{2,3,2}=A^{(2)}_{3,2,1}=0\), \(A^{(2)}_{1,3,1}=A^{(2)}_{3,1,2}=A^{(2)}_{1,3,2}=A^{(2)}_{3,1,1}=0\).  
In this case, the situation is clear; there is no choice.  
(Case C) The data given by \(A^{(4)}\) and its transformed version \(A^{(2)}\) can be evaluated in the four- and two-option model, respectively; see Figure 3. This is a very common case. This simple example is a modification of the data presented in Figure 2. The comparison results in Figure 2 are supplemented by a'slightly better' edge from object 2 to object 3; see Figure 3. The data are obviously evaluable in the four-option model. Moreover, C.2.1. is also satisfied; hence, data can be evaluated in the two-option model too.  
In this case, we can decide which model will be used for the data evaluation. In Section 4, we investigate these cases' practical aspects using computer simulations.  
(Case D) Data cannot be evaluated in either the two-option or the four-option model.  
Let \(A^{(4)}_{1,2,1}=A^{(4)}_{1,2,2}=A^{(4)}_{1,2,3}=A^{(4)}_{1,2,4}=1\), \(A^{(4)}_{2,3,1}=A^{(4)}_{2,3,2}=A^{(4)}_{2,3,3}=0\), \(A^{(4)}_{2,3,4}=1\), \(A^{(4)}_{1,3,k}=0\), and \(k=1,2,3,4\). The graph of the results belonging to the data matrix \(A^{(4)}\) can be seen in Figure 4.  
Figure 3: Comparison results which can be evaluated in both models (objects to be compared are denoted by 1, 2, and 3).  
Figure 2: Comparison results which can be evaluated using the four-option model but not in the two-option model (the objects to be compared are denoted by 1, 2, and 3).  
The reason why the data are not evaluable is clear: object 3 is 'worse' than objects 2 and 1, but we do not know by how much they differ. Formally, conditions C.2.1. and C.4.2. do not hold: they take the partition \(S=\{1,2\}\) and \(\overline{S}=\{3\}\) in both cases.