Abstract:
Methods and apparatus teach a digital spectrum of a data file. The digital spectrum is used to map a file&#39;s position in multi-dimensional space. This position relative to another file&#39;s position reveals closest neighbors. Certain of the closest neighbors are grouped together to define a set. Overlapping members in the groups may be further differentiated from one another by partitioning. An optimized partition of set S of N overlapping groups yields a maximum strength for groups and members in that partition. Among other things, the optimized partition includes relative strengths of every individual member in every possible partition and weighting functions applied to the relative strengths and to subgroups of files within the partitions.

Description:
[0001]    This utility application claims priority to U.S. Provisional Application Ser. Nos. 61/236,571 and 61/271,079, filed Aug. 25, 2009, and Jul. 16, 2009, respectively. Their contents are expressly incorporated herein as if set forth herein. 
     
    
     FIELD OF THE INVENTION 
       [0002]    The present invention relates generally to compression/decompression of data. More particularly, it relates to defining a digital spectrum of a data file, such as a compressed file, in order to determine properties that can be compared to other files. In this manner, file similarity, file adjacency and file grouping, to name a few, can be determined. Grouped files with overlapped members are further partitioned to achieve optimization. 
       BACKGROUND OF THE INVENTION 
       [0003]    Recent data suggests that nearly eighty-five percent of all data is found in computing files and growing annually at around sixty percent. One reason for the growth is that regulatory compliance acts, statutes, etc., (e.g., Sarbanes-Oxley, HIPAA, PCI) force companies to keep file data in an accessible state for extended periods of time. However, block level operations in computers are too lowly to apply any meaningful interpretation of this stored data beyond taking snapshots and block de-duplication. While other business intelligence products have been introduced to provide capabilities greater than block-level operations, they have been generally limited to structured database analysis. They are much less meaningful when acting upon data stored in unstructured environments. 
         [0004]    Unfortunately, entities the world over have paid enormous sums of money to create and store their data, but cannot find much of it later in instances where it is haphazardly arranged or arranged less than intuitively. Not only would locating this information bring back value, but being able to observe patterns in it might also prove valuable despite its usefulness being presently unknown. However, entities cannot expend so much time and effort in finding this data that it outweighs its usefulness. Notwithstanding this, there are still other scenarios, such as government compliance, litigation, audits, etc., that dictate certain data/information be found and produced, regardless of its cost in time, money and effort. Thus, a clear need is identified in the art to better find, organize and identify digital data, especially data left in unstructured states. 
         [0005]    In search engine technology, large amounts of unrelated and unstructured digital data can be quickly gathered. However, most engines do little to organize the data other than give a hierarchical presentation. Also, when the engine finds duplicate versions of data, it offers few to no options on eliminating the replication or migrating/relocating redundancies. Thus, a further need in the art exists to overcome the drawbacks of search engines. 
         [0006]    When it comes to large amounts of data, whether structured or not, compression techniques have been devised to preserve storage capacity, reduce bandwidth during transmission, etc. With modern compression algorithms, however, they simply exist to scrunch large blocks of data into smaller blocks according to their advertised compression ratios. As is known, some do it without data loss (lossless) while others do it “lossy.” None do it, unfortunately, with a view toward recognizing similarities in the data itself. 
         [0007]    From biology, it is known that highly similar species have highly similar DNA strings. In the computing context, consider two word processing files relating to stored baseball statistics. In a first file, words might appear for a baseball batter, such as “batting average,” “on base percentage.” and “slugging percentage.” while a second file might have words for a baseball pitcher, such as “strikeouts,” “walks,” and “earned runs.” Conversely, a third file wholly unrelated to baseball, statistics or sports, may have words such as “environmental protection.” “furniture,” or whatever comes to mind. It would be exceptionally useful if, during times of compression, or upon later manipulation by an algorithm if “mapping” could recognize the similarity in subject matter in the first two files, although not exact to one another, and provide options to a user. Appreciating that the “words” in the example files are represented in the computing context as binary bits (1&#39;s or 0&#39;s), which occurs by converting the English alphabet into a series of 1&#39;s and 0&#39;s through application of ASCII encoding techniques, it would be further useful if the compression algorithm could first recognize the similarity in subject matter of the first two files at the level of raw bit data. The reason for this is that not all files have words and instead might represent pictures (e.g., .jpeg) or spread sheets of numbers. 
         [0008]    Appreciating that certain products already exist in the above-identified market space, clarity on the need in the art is as follows. One, present day “keyword matching” is limited to select set of words that have been pulled from a document into an index for matching to the same exact words elsewhere. Two, “Grep” is a modern day technique that searches one or more input files for lines containing an identical match to a specified pattern. Three, “Beyond Compare,” and similar algorithms, are line-by-line comparisons of multiple documents that highlight differences between them. Four, block level data de-duplication has no application in compliance contexts, data relocation, or business intelligence. 
         [0009]    The need in the art, on the other hand, needs to serve advanced notions of identifying new business intelligence, conducting operations on completely unstructured or haphazard data, and organizing it, providing new useful options to users, providing new user views, providing new encryption products, and identifying highly similar data, to name a few. As a byproduct, solving this need will create new opportunities in minimizing transmission bandwidth and storage capacity, among other things. Naturally, any improvements along such lines should contemplate good engineering practices, such as stability, ease of implementation, unobtrusiveness, etc. 
       SUMMARY OF THE INVENTION 
       [0010]    The foregoing and other problems are solved by applying the principles and teachings associated with optimized partitions for grouping and differentiating files of data. Broadly, embodiments of the invention identify self-organizing, emergent patterns in data sets without any keyword, semantic, or metadata processing. Among certain advantages is that: the strongest possible memberships are determined given the input of groups of overlapping members; the smallest number of groups are defined while still allowing members to be in more than one group (not forcing mutually exclusive groups); observable, new, clear, and distinct groups are created from less distinct and less clear “fuzzy” overlapping groups; and a specific algorithm is presented for determining the weight of each member in the fuzzy overlapping groups and the number of fuzzy overlapping groups to determine a strength for every possible new partition of groups created from the members of the fuzzy overlapping groups in order to mathematically (objectively) determine the optimal set of groups (partitioning of the original groups). 
         [0011]    In representative embodiments, methods and apparatus teach a digital spectrum of a compressed file. The digital spectrum is used to map a file&#39;s position in multi-dimensional space. This position relative to another file&#39;s position reveals closest neighbors. Certain of the closest neighbors are grouped together to define a set. Overlapping members in the groups may be further differentiated from one another by partitioning. An optimized partition of set S of N overlapping groups yields a maximum strength for groups and members in that partition. The optimized partition includes relative strengths of every individual member in every possible partition and weighting functions applied to the relative strengths and to subgroups of files within the partitions. 
         [0012]    Using semi formal mathematical notation, the strength T=SP(P) of a partition P is given as: SP(Partition)=Σ SGS(GS i )*WGS(GS i ), for all groups GS i  in the partition and where SGS(GS)=Σ SM(M j )*WM(M j ), for all members M j  in the groups and where SM(Member)=(Number of total groups having that member)/(total numbers of groups); and where WGS( ) is the relative strength of this set of groups compared with the total number of original groups and WM( ) is the relative strength of this member compared with the total number of members in all the groups of this set of groups. Specifically WGS(GS)=this number of groups/number of all original groups and WM(M)=(1.0)/(total number of members in this set of groups). 
         [0013]    In a computing context, files are stored on one or more computing devices. As representatively encoded, they have pluralities of symbols representing an underlying data stream of original bits of data. These symbols define the multi-dimensional space in which file positions are examined relative to one another. Distances between the files in this space and matrix sorts of the distances are still other embodiments that are ultimately used to facilitate file grouping/differentiating, and then the partitioning. 
         [0014]    Executable instructions loaded on one or more computing devices for undertaking the foregoing are also contemplated as are computer program products available as a download or on a computer readable medium. The computer program products are also available for installation on a network appliance or an individual computing device. 
         [0015]    These and other embodiments of the present invention will be set forth in the description which follows, and in part will become apparent to those of ordinary skill in the art by reference to the following description of the invention and referenced drawings or by practice of the invention. The claims, however, indicate the particularities of the invention. 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         [0016]    The accompanying drawings incorporated in and forming a part of the specification, illustrate several aspects of the present invention, and together with the description serve to explain the principles of the invention. In the drawings: 
           [0017]      FIG. 1  is a table in accordance with the present invention showing terminology; 
           [0018]      FIG. 2  a table in accordance with the present invention showing a tuple array and tuple nomenclature; 
           [0019]      FIG. 3  is a table in accordance with the present invention showing the counting of tuples in a data stream; 
           [0020]      FIG. 4  is a table in accordance with the present invention showing the Count from  FIG. 3  in array form; 
           [0021]      FIG. 5  is Pythagorean&#39;s Theorem for use in resolving ties in the counts of highest occurring tuples; 
           [0022]      FIG. 6  is a table in accordance with the present invention showing a representative resolution of a tie in the counts of three highest occurring tuples using Pythagorean&#39;s Theorem; 
           [0023]      FIG. 7  is a table in accordance with the present invention showing an alternative resolution of a tie in the counts of highest occurring tuples; 
           [0024]      FIG. 8  is an initial dictionary in accordance with the present invention for the data stream of  FIG. 9 ; 
           [0025]      FIGS. 8-60  are iterative data streams and tables in accordance with the present invention depicting dictionaries, arrays, tuple counts, encoding, and the like illustrative of multiple passes through the compression algorithm; 
           [0026]      FIG. 61  is a chart in accordance with the present invention showing compression optimization; 
           [0027]      FIG. 62  is a table in accordance with the present invention showing compression statistics; 
           [0028]      FIGS. 63-69  are diagrams and tables in accordance with the present invention relating to storage of a compressed file; 
           [0029]      FIGS. 70-82   b  are data streams, tree diagrams and tables in accordance with the present invention relating to decompression of a compressed file; 
           [0030]      FIG. 83  is a diagram in accordance with the present invention showing a representative computing device for practicing all or some the foregoing; 
           [0031]      FIGS. 84-93  are diagrams in accordance with a “fast approximation” embodiment of the invention that utilizes key information of an earlier compressed file for a file under present consideration having patterns substantially similar to the earlier compressed file; 
           [0032]    FIGS.  94 - 98 A-B are definitions and diagrams in accordance with the present invention showing a “digital spectrum” embodiment of an encoded file, including grouping of files; and 
           [0033]      FIGS. 99A-99B ,  100 A- 100 B and  101 - 105  are diagrams, graphs and flow charts in accordance with the present invention for grouping and differentiating files and determining stopping functions during grouping. 
       
    
    
     DETAILED DESCRIPTION OF THE ILLUSTRATED EMBODIMENTS 
       [0034]    In the following detailed description of the illustrated embodiments, reference is made to the accompanying drawings that form a part hereof, and in which is shown by way of illustration, specific embodiments in which the invention may be practiced. These embodiments are described in sufficient detail to enable those skilled in the art to practice the invention and like numerals represent like details in the various figures. Also, it is to be understood that other embodiments may be utilized and that process, mechanical, electrical, arrangement, software and/or other changes may be made without departing from the scope of the present invention. In accordance with the present invention, methods and apparatus are hereinafter described for optimizing data compression of digital data. 
         [0035]    In a representative embodiment, compression occurs by finding highly occurring patterns in data streams, and replacing them with newly defined symbols that require less space to store than the original patterns. The goal is to eliminate as much redundancy from the digital data as possible. The end result has been shown by the inventor to achieve greater compression ratios on certain tested files than algorithms heretofore known. 
         [0036]    In information theory, it is well understood that collections of data contain significant amounts of redundant information. Some redundancies are easily recognized, while others are difficult to observe. A familiar example of redundancy in the English language is the ordered pair of letters QU. When Q appears in written text, the reader anticipates and expects the letter U to follow, such as in the words queen, quick, acquit, and square. The letter U is mostly redundant information when it follows Q. Replacing a recurring pattern of adjacent characters with a single symbol can reduce the amount of space that it takes to store that information. For example, the ordered pair of letters QU can be replaced with a single memorable symbol when the text is stored. For this example, the small Greek letter alpha (α) is selected as the symbol, but any could be chosen that does not otherwise appear in the text under consideration. The resultant compressed text is one letter shorter for each occurrence of QU that is replaced with the single symbol (α). e.g., “αeen,” “αick,” “acαit,” and “sαare.” Such is also stored with a definition of the symbol alpha (α) in order to enable the original data to be restored. Later, the compressed text can be expanded by replacing the symbol with the original letters QU. There is no information loss. Also, this process can be repeated many times over to achieve further compression. 
       DEFINITIONS 
       [0037]    With reference to  FIG. 1 , a table  10  is used to define terminology used in the below compression method and procedure. 
       DISCUSSION 
       [0038]    Redundancy is the superfluous repetition of information. As demonstrated in the QU example above, adjacent characters in written text often form expected patterns that are easily detected. In contrast, digital data is stored as a series of bits where each bit can have only one of two values: off (represented as a zero (0)) and on (represented as a one (1)). Redundancies in digital data, such as long sequences of zeros or ones, are easily seen with the human eye. However, patterns are not obvious in highly complex digital data. The invention&#39;s methods and procedures identify these redundancies in stored information so that even highly complex data can be compressed. In turn, the techniques can be used to reduce, optimize, or eliminate redundancy by substituting the redundant information with symbols that take less space to store than the original information. When it is used to eliminate redundancy, the method might originally return compressed data that is larger than the original. This can occur because information about the symbols and how the symbols are encoded for storage must also be stored so that the data can be decompressed later. For example, compression of the word “queen” above resulted in the compressed word “αeen.” But a dictionary having the relationship QU=α also needed to be stored with the word “αeen”, which makes a “first pass” through the compression technique increase in size, not decrease. Eventually, however, further “passes” will stop increasing and decrease so rapidly, despite the presence of an ever-growing dictionary size, that compression ratios will be shown to greatly advance the state of the art. By automating the techniques with computer processors and computing software, compression will also occur exceptionally rapidly. In addition, the techniques herein will be shown to losslessly compress the data. 
       The Compression Procedure 
       [0039]    The following compression method iteratively substitutes symbols for highly occurring tuples in a data stream. An example of this process is provided later in the document. 
       Prerequisites 
       [0040]    The compression procedure will be performed on digital data. Each stored bit has a value of binary 0 or binary 1. This series of bits is referred to as the original digital data. 
       Preparing the Data 
       [0041]    The original digital data is examined at the bit level. The series of bits is conceptually converted to a stream of characters, referred to as the data strewn that represents the original data. The symbols 0 and 1 are used to represent the respective raw bit values in the new data stream. These symbols are considered to be atomic because all subsequently defined symbols represent tuples that are based on 0 and 1. 
         [0042]    A dictionary is used to document the alphabet of symbols that are used in the data stream. Initially, the alphabet consists solely of the symbols 0 and 1. 
       Compressing the Data Stream 
       [0043]    The following tasks are performed iteratively on the data stream:
       Identifying all possible tuples that can occur for the set of characters that are in the current data stream.   Determining which of the possible tuples occurs most frequently in the current data stream. In the case of a tie, use the most complex tuple. (Complexity is discussed below.)   Creating a new symbol for the most highly occurring tuple, and add it to the dictionary.   Replacing all occurrences of the most highly occurring tuple with the new symbol.   Encoding the symbols in the data stream by using an encoding scheme, such as a path-weighted Huffman coding scheme.   Calculating the compressed file size.   Determining whether the compression goal has been achieved.   Repeating for as long as necessary to achieve optimal compression. That is, if a stream of data were compressed so completely that it was represented by a single bit, it and its complementary dictionary would be larger than the original representation of the stream of data absent the compression. (For example, in the QU example above, if “α” represented the entire word “queen,” the word “queen” could be reduced to one symbol, e.g., “α.” However, this one symbol and its dictionary (reciting “queen=α” is larger than the original content “queen.”) Thus, optimal compression herein recognizes a point of marginal return whereby the dictionary grows too large relative to the amount of compression being achieved by the technique.
 
Each of these steps is described in more detail below.
       
 
       Identifying all Possible Tuples 
       [0052]    From  FIG. 1 , a “tuple” is an ordered pair of adjoining characters in a data stream. To identify all possible tuples in a given data stream, the characters in the current alphabet are systematically combined to form ordered pairs of symbols. The left symbol in the pair is referred to as the “first” character, while the right symbol is referred to as the “last” character. In a larger context, the tuples represent the “patterns” examined in a data stream that will yield further advantage in the art. 
         [0053]    In the following example and with any data stream of digital data that can be compressed according to the techniques herein, two symbols (0 and 1) occur in the alphabet and are possibly the only symbols in the entire data stream. By examining them as “tuples,” the combination of the 0 and 1 as ordered pairs of adjoining characters reveals only four possible outcomes, i.e., a tuple represented by “00,” a tuple represented by “01,” a tuple represented by “10,” and a tuple represented by “11.” 
         [0054]    With reference to  FIG. 2 , these four possibilities are seen in table  12 . In detail, the table shows the tuple array for characters 0 and 1. In the cell for column 0 and row 0, the tuple is the ordered pair of 0 followed by 0. The shorthand notation of the tuple in the first cell is “0&gt;0”. In the cell for column 0 and row 1, the tuple is 0 followed by 1, or “0&gt;1”. In the cell for column 1 and row 0, the tuple is “1&gt;0”. In the cell for column 1 and row 1, the tuple is “1&gt;1”. 
       Determining the Most Highly Occurring Tuple 
       [0055]    With  FIG. 2  in mind, it is determined which tuple in a bit stream is the most highly occurring. To do this, simple counting occurs. It reveals how many times each of the possible tuples actually occurs. Each pair of adjoining characters is compared to the possible tuples and the count is recorded for the matched tuple. 
         [0056]    The process begins by examining the adjacent characters in position one and two of the data stream. Together, the pair of characters forms a tuple. Advance by one character in the stream and examine the characters in positions two and three. By incrementing through the data stream one character at a time, every combination of two adjacent characters in the data stream is examined and tallied against one of the tuples. 
         [0057]    Sequences of repeated symbols create a special case that must be considered when tallying tuples. That is, when a symbol is repeated three or more times, skilled artisans might identify instances of a tuple that cannot exist because the symbols in the tuple belong to other instances of the same tuple. The number of actual tuples in this case is the number of times the symbol repeats divided by two. 
         [0058]    For example, consider the data stream  14  in table  16  ( FIG. 3 ) having 10 characters shown as “0110000101.” Upon examining the first two characters 01, a tuple is recognized in the form 0 followed by 1 (0&gt;1). Then, increment forward one character and consider the second and third characters 11, which forms the tuple of 1 followed by 1 (1&gt;1). As progression occurs through the data stream, 9 possible tuple combinations are found: 0&gt;1, 1&gt;1, 1&gt;0, 0&gt;0, 0&gt;0, 0&gt;0, 0&gt;1, 1&gt;0, and 0&gt;1 (element  15 ,  FIG. 3 ). In the sequence of four sequential zeros (at the fourth through seventh character positions in the data stream “0110000101”), three instances of a 0 followed by a 0 (or 0&gt;0) are identified as possible tuples. It is observed that the second instance of the 0&gt;0 tuple (element  17 ,  FIG. 3 ) cannot be formed because the symbols are used in the 0&gt;0 tuple before and after it, by prescribed rule. Thus, there are only two possible instances in the COUNT  18 ,  FIG. 3 , of the 0&gt;0 tuple, not 3. In turn, the most highly occurring tuple counted in this data stream is 0&gt;1, which occurs 3 times (element  19 ,  FIG. 3 ). Similarly, tuple 1&gt;1 occurs once (element  20 ,  FIG. 3 ), while tuple 1&gt;0 occurs twice (element  21 ,  FIG. 3 ). 
         [0059]    After the entire data stream has been examined, the final counts for each tuple are compared to determine which tuple occurs most frequently. In tabular form, the 0 followed by a 1 (tuple 0&gt;1) occurs the most and is referenced at element  19  in table  22 ,  FIG. 4 . 
         [0060]    In the situation of a tie between two or more tuples, skilled artisans must choose between one of the tuples. For this, experimentation has revealed that choosing the tuple that contains the most complex characters usually results in the most efficient compression. If all tuples are equally complex, skilled artisans can choose any one of the tied tuples and define it as the most highly occurring. 
         [0061]    The complexity of a tuple is determined by imagining that the symbols form the sides of a right triangle, and the complexity is a measure of the length of the hypotenuse of that triangle. Of course, the hypotenuse is related to the sum of the squares of the sides, as defined by the Pythagorean Theorem,  FIG. 5 . 
         [0062]    The tuple with the longest hypotenuse is considered the most complex tuple, and is the winner in the situation of a tie between the highest numbers of occurring tuples. The reason for this is that less-complex tuples in the situation of a tie are most likely to be resolved in subsequent passes in the decreasing order of their hypotenuse length. Should a tie in hypotenuse length occur, or a tie in complexity, evidence appears to suggest it does not make a difference which tuple is chosen as the most highly occurring. 
         [0063]    For example, suppose that tuples 3&gt;7, 4&gt;4 and 1&gt;5 each occur 356 times when counted (in a same pass). To determine the complexity of each tuple, use the tuple symbols as the two sides of a right triangle and calculate the hypotenuse,  FIG. 6 . In the instance of 3&gt;7, the side of the hypotenuse is the square root of (three squared (9) plus seven squared (49)), or the square root of 58, or 7.6. In the instance of 4&gt;4, the side of the hypotenuse is the square root of (four squared (16) plus four squared (16), of the square root of 32, or 5.7. Similar, 1&gt;5 calculates as a hypotenuse of 5.1 as seen in table  23  in the Figure. Since the tuple with the largest hypotenuse is the most complex. 3&gt;7&#39;s hypotenuse of 7.6 is considered more complex than either of the tuples 4&gt;4 or 1&gt;5. 
         [0064]    Skilled artisans can also use the tuple array to visualize the hypotenuse by drawing lines in the columns and rows from the array origin to the tuple entry in the array, as shown in table  24  in  FIG. 7 . As seen, the longest hypotenuse is labeled  25 , so the 3&gt;7 tuple wins the tie, and is designated as the most highly occurring tuple. Hereafter, a new symbol is created to replace the highest occurring tuple (whether occurring the most outright by count or by tie resolution), as seen below. However, based on the complexity rule, it is highly likely that the next passes will replace tuple 4&gt;4 and then tuple 1&gt;5. 
       Creating a Symbol for the Most Highly Occurring Tuple 
       [0065]    As before, a symbol stands for the two adjacent characters that form the tuple and skilled artisans select any new symbol they want provided it is not possibly found in the data stream elsewhere. Also, since the symbol and its definition are added to the alphabet, e.g., if “α=QU,” a dictionary grows by one new symbol in each pass through the data, as will be seen. A good example of a new symbol for use in the invention is a numerical character, sequentially selected, because numbers provide an unlimited source of unique symbols. In addition, reaching an optimized compression goal might take thousands (or even tens of thousands) of passes through the data stream and redundant symbols must be avoided relative to previous passes and future passes. 
         [0000]    Replacing the Tuple with the New Symbol 
         [0066]    Upon examining the data stream to find all occurrences of the highest occurring tuple, skilled artisans simply substitute the newly defined or newly created symbol for each occurrence of that tuple. Intuitively, substituting a single symbol for two characters compresses the data stream by one character for each occurrence of the tuple that is replaced. 
       Encoding the Alphabet 
       [0067]    To accomplish this, counting occurs for how many times that each of the symbols in the current alphabet occurs in the data stream. They then use the symbol count to apply an encoding scheme, such as a path-weighted Huffman coding scheme, to the alphabet. Huffman trees should be within the purview of the artisan&#39;s skill set. 
         [0068]    The encoding assigns bits to each symbol in the current alphabet that actually appears in the data stream. That is, symbols with a count of zero occurrences are not encoded in the tree. Also, symbols might go “extinct” in the data stream as they are entirely consumed by yet more complex symbols, as will be seen. As a result, the Huffman code tree is rebuilt every time a new symbol is added to the dictionary. This means that the Huffman code for a given symbol can change with every pass. The encoded length of the data stream usually decreases with each pass. 
       Calculating the Compressed File Size 
       [0069]    The compressed file size is the total amount of space that it takes to store the Huffman-encoded data stream plus the information about the compression, such as information about the file, the dictionary, and the Huffman encoding tree. The compression information must be saved along with other information so that the encoded data can be decompressed later. 
         [0070]    To accomplish this, artisans count the number of times that each symbol appears in the data stream. They also count the number of bits in the symbol&#39;s Huffman code to find its bit length. They then multiply the bit length by the symbol count to calculate the total bits needed to store all occurrences of the symbol. This is then repeated for each symbol. Thereafter, the total bit counts for all symbols are added to determine how many bits are needed to store only the compressed data. To determine the compressed file size, add the total bit count for the data to the number of bits required for the related compression information (the dictionary and the symbol-encoding information). 
         [0000]    Determining Whether the Compression Goal has been Achieved 
         [0071]    Substituting a tuple with a single symbol reduces the total number of characters in a data stream by one for each instance of a tuple that is replaced by a symbol. That is, for each instance, two existing characters are replaced with one new character. In a given pass, each instance of the tuple is replaced by a new symbol. There are three observed results:
       The length of the data stream (as measured by how many characters make up the text) decreases by half the number of tuples replaced.   The number of symbols in the alphabet increases by one.   The number of nodes in the Huffman tree increases by two.       
 
         [0075]    By repeating the compression procedure a sufficient number of times, any series of characters can eventually be reduced to a single character. That “super-symbol” character conveys the entire meaning of the original text. However, the information about the symbols and encoding that is used to reach that final symbol is needed to restore the original data later. As the number of total characters in the text decreases with each repetition of the procedure, the number of symbols increases by one. With each new symbol, the size of the dictionary and the size of the Huffman tree increase, while the size of the data decreases relative to the number of instances of the tuple it replaces. It is possible that the information about the symbol takes more space to store than the original data it replaces. In order for the compressed file size to become smaller than the original data stream size, the text size must decrease faster than the size increases for the dictionary and the Huffman encoding information. 
         [0076]    The question at hand is then, what is the optimal number of substitutions (new symbols) to make, and how should those substitutions be determined? 
         [0077]    For each pass through the data stream, the encoded length of the text decreases, while the size of the dictionary and the Huffman tree increases. It has been observed that the compressed file size will reach a minimal value, and then increase. The increase occurs at some point because so few tuple replacements are done that the decrease in text size no longer outweighs the increase in size of the dictionary and Huffman tree. 
         [0078]    The size of the compressed file does not decrease smoothly or steadily downward. As the compression process proceeds, the size might plateau or temporarily increase. In order to determine the true (global) minimum, it is necessary to continue some number of iterations past the each new (local) minimum point. This true minimal value represents the optimal compression for the data stream using this method. 
         [0079]    Through experimentation, three conditions have been found that can be used to decide when to terminate the compression procedure: asymptotic reduction, observed low, and single character. Each method is described below. Other terminating conditions might be determined through further experimentation. 
       Asymptotic Reduction 
       [0080]    An asymptotic reduction is a concession to processing efficiency, rather than a completion of the procedure. When compressing larger files (100 kilobytes (KB) or greater), after several thousand passes, each additional pass produces only a very small additional compression. The compressed size is still trending downward, but at such a slow rate that additional compute time is not warranted. 
         [0081]    Based on experimental results, the process is terminated if at least 1000 passes have been done, and less than 1% of additional data stream compression has occurred in the last 1000 passes. The previously noted minimum is therefore used as the optimum compressed file. 
       Observed Low 
       [0082]    A reasonable number of passes have been performed on the data and in the last reasonable number of passes a new minimum encoded file size has not been detected. It appears that further passes only result in a larger encoded file size. 
         [0083]    Based on experimental results, the process is terminated if at least 1000 passes have been done, and in the last 10% of the passes, a new low has not been established. The previously noted minimum is then used as the optimum compressed file. 
       Single Character 
       [0084]    The data stream has been reduced to exactly one character. This case occurs if the file is made up of data that can easily reduce to a single symbol, such a file filled with a repeating pattern. In cases like this, compression methods other than this one might result in smaller compressed file sizes. 
       How the Procedure Optimizes Compression 
       [0085]    The representative embodiment of the invention uses Huffman trees to encode the data stream that has been progressively shortened by tuple replacement, and balanced against the growth of the resultant Huffman tree and dictionary representation. 
         [0086]    The average length of a Huffman encoded symbol depends upon two factors:
       How many symbols must be represented in the Huffman tree   The distribution of the frequency of symbol use       
 
         [0089]    The average encoded symbol length grows in a somewhat stepwise fashion as more symbols are added to the dictionary. Because the Huffman tree is a binary tree, increases naturally occur as the number of symbols passes each level of the power of 2 (2, 4, 8, 16, 32, 64, etc.). At these points, the average number of bits needed to represent any given symbol normally increases by 1 bit, even though the number of characters that need to be encoded decreases. Subsequent compression passes usually overcome this temporary jump in encoded data stream length. 
         [0090]    The second factor that affects the efficiency of Huffman coding is the distribution of the frequency of symbol use. If one symbol is used significantly more than any other, it can be assigned a shorter encoding representation, which results in a shorter encoded length overall, and results in maximum compression. The more frequently a symbol occurs, the shorter the encoded stream that replaces it. The less frequently a symbol occurs, the longer the encoded stream that replaces it. 
         [0091]    If all symbols occur at approximately equal frequencies, the number of symbols has the greater effect than does the size of the encoded data stream. Supporting evidence is that maximum compression occurs when minimum redundancy occurs, that is, when the data appears random. This state of randomness occurs when every symbol occurs at the same frequency as any other symbol, and there is no discernable ordering to the symbols. 
         [0092]    The method and procedure described in this document attempt to create a state of randomness in the data stream. By replacing highly occurring tuples with new symbols, eventually the frequency of all symbols present in the data stream becomes roughly equal. Similarly, the frequency of all tuples is also approximately equal. These two criteria (equal occurrence of every symbol and equal occurrence of ordered symbol groupings) is the definition of random data. Random data means no redundancy. No redundancy means maximum compression. 
         [0093]    This method and procedure derives optimal compression from a combination of the two factors. It reduces the number of characters in the data stream by creating new symbols to replace highly occurring tuples. The frequency distribution of symbol occurrence in the data stream tends to equalize as oft occurring symbols are eliminated during tuple replacement. This has the effect of flattening the Huffman tree, minimizing average path lengths, and therefore, minimizing encoded data stream length. The number of newly created symbols is held to a minimum by measuring the increase in dictionary size against the decrease in encoded data stream size. 
       Example of Compression 
       [0094]    To demonstrate the compression procedure, a small data file contains the following simple ASCII characters: 
         [0095]    aaaaaaaaaaaaaaaaaaaaaaaaaaabaaabaaaaaaaababbbbbb 
         [0096]    Each character is stored as a sequence of eight bits that correlates to the ASCII code assigned to the character. The bit values for each character are: 
         [0097]    a=01100001 
         [0098]    b=01100010 
         [0099]    The digital data that represents the file is the original data that we use for our compression procedure. Later, we want to decompress the compressed file to get back to the original data without data loss. 
       Preparing the Data Stream 
       [0100]    The digital data that represents the file is a series of bits, where each bit has a value of 0 or 1. We want to abstract the view of the bits by conceptually replacing them with symbols to form a sequential stream of characters, referred to as a data stream. 
         [0101]    For our sample digital data, we create two new symbols called 0 and 1 to represent the raw bit values of 0 and 1, respectively. These two symbols form our initial alphabet, so we place them in the dictionary  26 ,  FIG. 8 . 
         [0102]    The data stream  30  in  FIG. 9  represents the original series of bits in the stored file, e.g., the first eight bits  32  are “01100001” and correspond to the first letter “a” in the data file. Similarly, the very last eight bits  34  are “01100010” and correspond to the final letter “b” in the data file, and each of the 1&#39;s and 0&#39;s come from the ASCII code above. 
         [0103]    Also, the characters in data stream  30  are separated with a space for user readability, but the space is not considered, just the characters. The space would not occur in computer memory either. 
       Compressing the Data Stream 
       [0104]    The data stream  30  of  FIG. 9  is now ready for compression. The procedure will be repeated until the compression goal is achieved. For this example, the compression goal is to minimize the amount of space that it takes to store the digital data. 
       Initial Pass 
       [0105]    For the initial pass, the original data stream and alphabet that were created in “Preparing the Data Stream” are obtained. 
       Identifying all Possible Tuples 
       [0106]    An easy way to identify all possible combinations of the characters in our current alphabet (at this time having 0 and 1) is to create a tuple array (table  35 ,  FIG. 10 ). Those symbols are placed or fitted as a column and row, and the cells are filled in with the tuple that combines those symbols. The columns and rows are constructed alphabetically from left to right and top to bottom, respectively, according to the order that the symbols appear in our dictionary. For this demonstration, we will consider the symbol in a column to be the first character in the tuple, and the symbol in a row to be the last character in the tuple. To simplify the presentation of tuples in each cell, we will use the earlier-described notation of “first&gt;last” to indicate the order of appearance in the pair of characters, and to make it easier to visually distinguish the symbols in the pair. The tuples shown in each cell now represent the patterns we want to look for in the data stream. 
         [0107]    For example, the table  35  shows the tuple array for characters 0 and 1. In the cell for column 0 and row 0, the tuple is the ordered pair of 0 followed by 0. The shorthand notation of the tuple in the first cell is “0&gt;0”. In the cell for column 0 and row 1, the tuple is 0 followed by 1, or “0&gt;1”. In the cell for column 1 and row 0, the tuple is “1&gt;0”. In the cell for column 1 and row 1, the tuple is “1&gt;1” (As skilled artisans will appreciate, most initial dictionaries and original tuple arrays will be identical to these. The reason is that computing data streams will all begin with a stream of 1&#39;s and 0&#39;s having two symbols only.) 
       Determining the Highly Occurring Tuple 
       [0108]    After completion of the tuple array, we are ready to look for the tuples in the data stream  30 ,  FIG. 9 . We start at the beginning of the data stream with the first two characters “01” labeled element  37 . We compare this pair of characters to our known tuples, keeping in mind that order matters. We match the pair to a tuple, and add one count for that instance. We move forward by one character, and look at the pair of characters 38 in positions two and three in the data stream, or “11.” We compare and match this pair to one of the tuples, and add one count for that instance. We continue tallying occurrences of the tuples in this manner until we reach the end of the data stream. In this instance, the final tuple is “10” labeled  39 . By incrementing through the data stream one character at a time, we have considered every combination of two adjacent characters in the data stream, and tallied each instance against one of the tuples. We also consider the rule for sequences of repeated symbols, described above, to determine the actual number of instances for the tuple that is defined by pairs of that symbol. 
         [0109]    For example, the first two characters in our sample data stream are 0 followed by 1. This matches the tuple 0&gt;1 so we count that as one instance of the tuple. We step forward one character. The characters in positions two and three are 1 followed by 1, which matches the tuple 1&gt;1. We count it as one instance of the 1&gt;1 tuple. We consider the sequences of three or more zeros in the data stream (e.g., 01100001 . . . ) to determine the actual number of tuples for the 0&gt;0 tuple. We repeat this process to the end of the data set with the count results in table  40 ,  FIG. 11 . 
         [0110]    Now that we have gathered statistics for how many times each tuple appears in the data stream  30 , we compare the total counts for each tuple to determine which pattern is the most highly occurring. The tuple that occurs most frequently is a tie between a 1 followed by 0 (1&gt;0), which occurs 96 times, and a 0 followed by 1 (0&gt;1), which also occurs 96 times. As discussed above, skilled artisans then choose the most complex tuple and do so according to Pythagorean&#39;s Theorem. The sum of the squares for each tuple is the same, which is 1 (1+0) and 1 (0+1). Because they have the same complexity, it does not matter which one is chosen as the highest occurring. In this example, we will choose tuple 1&gt;0. 
         [0111]    We also count the number of instances of each of the symbols in the current alphabet as seen in table  41 ,  FIG. 12 . The total symbol count in the data stream is 384 total symbols that represent 384 bits in the original data. Also, the symbol 0 appears 240 times in original data stream  30 ,  FIG. 9 , while the symbol 1 only appears 144 times. 
       Pass 1 
       [0112]    In this next pass, we replace the most highly occurring tuple from the previous pass with a new symbol, and then we determine whether we have achieved our compression goal. 
       Creating a Symbol for the Highly Occurring Tuple 
       [0113]    We replace the most highly occurring tuple from the previous pass with a new symbol and add it to the alphabet. Continuing the example, we add a new symbol 2 to the dictionary and define it with the tuple defined as 1 followed by 0 (1&gt;0). It is added to the dictionary  26 ′ as seen in  FIG. 13 . (Of course, original symbol 0 is still defined as a 0, while original symbol 1 is still defined as a 1. Neither of these represent a first symbol followed a last symbol which is why dashes appear in the dictionary  26 ′ under “Last” for each of them.) 
         [0000]    Replacing the Triple with the New Symbol 
         [0114]    In the original data stream  30 , every instance of the tuple 1&gt;0 is now replaced with the new, single symbol. In our example data stream  30 ,  FIG. 9 , the 96 instances of the tuple 1&gt;0 have been replaced with the new symbol “2” to create the output data stream  30 ′.  FIG. 14 , that we will use for this pass. As skilled artisans will observe, replacing ninety-six double instances of symbols with a single, new symbol shrinks or compresses the data stream  30 ′ in comparison to the original data stream  30 ,  FIG. 8 . 
       Encoding the Alphabet 
       [0115]    After we compress the data stream by using the new symbol, we use a path-weighted Huffman coding scheme to assign bits to each symbol in the current alphabet. 
         [0116]    To do this, we again count the number of instances of each of the symbols in the current alphabet (now having “0,” “1” and “2.”) The total symbol count in the data stream is 288 symbols as seen in table  41 ′,  FIG. 15 . We also have one end-of-file (EOF) symbol at the end of the data stream (not shown). 
         [0117]    Next, we use the counts to build a Huffman binary code tree. 1) List the symbols from highest count to lowest count. 2) Combine the counts for the two least frequently occurring symbols in the dictionary. This creates a node that has the value of the sum of the two counts. 3) Continue combining the two lowest counts in this manner until there is only one symbol remaining. This generates a Huffman binary code tree. 
         [0118]    Finally, label the code tree paths with zeros (0s) and ones (1s). The Huffman coding scheme assigns shorter code words to the more frequent symbols, which helps reduce the size length of the encoded data. The Huffman code for a symbol is defined as the string of values associated with each path transition from the root to the symbol terminal node. 
         [0119]    With reference to  FIG. 16 , the tree  50  demonstrates the process of building the Huffman tree and code for the symbols in the current alphabet. We also create a code for the end of file marker that we placed at the end of the data stream when we counted the tuples. In more detail, the root contemplates 289 total symbols, i.e., the 288 symbols for the alphabet “0,” “1” and “2” plus one EOF symbol. At the leaves, the “0” is shown with its counts 144, the “1” with its count of 48, the “2” with its count of 96 and the EOF with its count of 1. Between the leaves and root, the branches define the count in a manner skilled artisans should readily understand. 
         [0120]    In this compression procedure, we will re-build a Huffman code tree every time we add a symbol to the current dictionary. This means that the Huffman code for a given symbol can change with every compression pass. 
       Calculating the Compressed File Size 
       [0121]    From the Huffman tree, we use its code to evaluate the amount of space needed to store the compressed data as seen in table  52 ,  FIG. 17 . First, we count the number of bits in the Huffman code for each symbol to find its bit length  53 . Next, we multiply a symbol&#39;s bit length by its count  54  to calculate the total bits  55  used to store the occurrences of that symbol. We add the total bits  56  needed for all symbols to determine how many bits are needed to store only the compressed data. As seen, the current data stream  30 ′.  FIG. 14  requires 483 bits to store only the information. 
         [0122]    To know whether we achieved optimal compression, we must consider the total amount of space that it takes to store the compressed data plus the information about the compression that we need to store in order to decompress the data later. We also must store information about the file, the dictionary, and the Huffman tree. The table  57  in  FIG. 18  shows the total compression overhead as being 25 bits, which brings the compressed size of the data stream to 508 bits, or 483 bits plus 25 bits. 
         [0000]    Determining Whether the Compression Goal has been Achieved 
         [0123]    Finally, we compare the original number of bits ( 384 ,  FIG. 12 ) to the current number of bits ( 508 ) that are needed for this compression pass. We find that it takes 1.32 times as many bits to store the compressed data as it took to store the original data, table  58 ,  FIG. 19 . This is not compression at all, but expansion. 
         [0124]    In early passes, however, we expect to see that the substitution requires more space than the original data because of the effect of carrying a dictionary, adding symbols, and building a tree. On the other hand, skilled artisans should observe an eventual reduction in the amount of space needed as the compression process continues. Namely, as the size of the data set decreases by the symbol replacement method, the size grows for the symbol dictionary and the Huffman tree information that we need for decompressing the data. 
       Pass 2 
       [0125]    In this pass, we replace the most highly occurring tuple from the previous pass (pass 1) with still another new symbol, and then we determine whether we have achieved our compression goal. 
       Identifying all Possible Tuples 
       [0126]    As a result of the new symbol, the tuple array is expanded by adding the symbol that was created in the previous pass. Continuing our example, we add 2 as a first symbol and last symbol, and enter the tuples in the new cells of table  35 ′  FIG. 20 . 
       Determining the Highly Occurring Tuple 
       [0127]    As before, the tuple array identifies the tuples that we look for and tally in our revised alphabet. As seen in table  40 ′,  FIG. 21 , the Total Symbol Count=288. The tuple that occurs most frequently when counting the data stream  30 ′,  FIG. 14 , is the character 2 followed by the character 0 (2&gt;0). It occurs 56 times as seen circled in table  40 ′,  FIG. 21 . 
       Creating a Symbol for the Highly Occurring Tuple 
       [0128]    We define still another new symbol “3” to represent the most highly occurring tuple 2&gt;0, and add it to the dictionary  26 ″,  FIG. 22  for the alphabet that was developed in the previous passes. 
         [0000]    Replacing the Tuple with the New Symbol 
         [0129]    In the data stream  30 ′.  FIG. 14 , we replace every instance of the most highly occurring tuple with the new single symbol. We replace the 56 instances of the 2&gt;0 tuple with the symbol 3 and the resultant data stream  30 ′″ is seen in  FIG. 23 . 
       Encoding the Alphabet 
       [0130]    As demonstrated above, we count the number of symbols in the data stream, and use the count to build a Huffman tree and code for the current alphabet. The total symbol count has been reduced from 288 to 234 (e.g., 88+48+40+58, but not including the EOF marker) as seen in table  41 ″,  FIG. 24 . 
       Calculating the Compressed File Size 
       [0131]    We need to evaluate whether our substitution reduces the amount of space that it takes to store the data. As described above, we calculate the total bits needed ( 507 ) as in table  52 ′,  FIG. 25 . 
         [0132]    In table  57 ′,  FIG. 26 , the compression overhead is calculated as 38 bits. 
         [0000]    Determining Whether the Compression Goal has been Achieved 
         [0133]    Finally, we compare the original number of bits ( 384 ) to the current number of bits (545=507+38) that are needed for this compression pass. We find that it takes 141% or 1.41 times as many bits to store the compressed data as it took to store the original data. Compression is still not achieved and the amount of data in this technique is growing larger rather than smaller in comparison to the previous pass requiring 132%. 
       Pass 3 
       [0134]    In this pass, we replace the most highly occurring tuple from the previous pass with a new symbol, and then we determine whether we have achieved our compression goal. 
       Identifying all Possible Tuples 
       [0135]    We expand the tuple array  35 ″,  FIG. 28  by adding the symbol that was created in the previous pass. We add the symbol “3” as a first symbol and last symbol, and enter the tuples in the new cells. 
       Determining the Highly Occurring Topic 
       [0136]    The tuple array identifies the tuples that we look for and tally in our revised alphabet. In table  40 ″,  FIG. 29 , the Total Symbol Count is 232, and the tuple that occurs most frequently is the character 1 followed by character 3 (1&gt;3). It occurs 48 times, which ties with the tuple of character 3 followed by character 0. We determine that the tuple 1&gt;3 is the most complex tuple because it has a hypotenuse length  25 ′ of 3.16 (SQRT(1 2 +3 2 )), and tuple 3&gt;0 has a hypotenuse of 3 (SQRT(0 2 +3 2 )). 
       Creating a Symbol for the Highly Occurring Triple 
       [0137]    We define a new symbol 4 to represent the most highly occurring tuple 1&gt;3, and add it to the dictionary  26 ′″,  FIG. 30 , for the alphabet that was developed in the previous passes. 
         [0000]    Replacing the Tuple with the New Symbol 
         [0138]    In the data stream, we replace every instance of the most highly occurring tuple from the earlier data stream with the new single symbol. We replace the 48 instances of the 1&gt;3 tuple with the symbol 4 and new data stream  30 - 4  is obtained,  FIG. 31 . 
       Encoding the Alphabet 
       [0139]    We count the number of symbols in the data stream, and use the count to build a Huffman tree and code for the current alphabet as seen in table  41 ′″,  FIG. 32 . There is no Huffman code assigned to the symbol 1 because there are no instances of this symbol in the compressed data in this pass. (This can be seen in the data stream  30 - 4 ,  FIG. 31 .) The total symbol count has been reduced from 232 to 184 (e.g., 88+0+40+8+48, but not including the EOF marker). 
       Calculating the Compressed File Size 
       [0140]    We need to evaluate whether our substitution reduces the amount of space that it takes to store the data. As seen in table  52 ″,  FIG. 33 , the total bits are equal to 340. 
         [0141]    In table  57 ″,  FIG. 34 , the compression overhead in bits is 42. 
         [0000]    Determining Whether the Compression Goal has been Achieved 
         [0142]    Finally, we compare the original number of bits ( 384 ) to the current number of bits ( 382 ) that are needed for this compression pass. We find that it takes 0.99 times as many bits to store the compressed data as it took to store the original data. Compression is achieved. 
       Pass 4 
       [0143]    In this pass, we replace the most highly occurring tuple from the previous pass with a new symbol, and then we determine whether we have achieved our compression goal. 
       Identifying all Possible Tuples 
       [0144]    We expand the tuple array  35 ′″,  FIG. 36 , by adding the symbol that was created in the previous pass. We add the symbol 4 as a first symbol and last symbol, and enter the tuples in the new cells. 
       Determining the Highly Occurring Tuple 
       [0145]    The tuple array identifies the tuples that we look for and tally in our revised alphabet. In table  40 ′″,  FIG. 37 , the Total Symbol Count=184 and the tuple that occurs most frequently is the character 4 followed by character 0 (4&gt;0). It occurs 48 times. 
       Creating a Symbol for the Highly Occurring Tuple 
       [0146]    We define a new symbol 5 to represent the 4&gt;0 tuple, and add it to the dictionary  26 - 4 ,  FIG. 38 , for the alphabet that was developed in the previous passes. 
         [0000]    Replacing the Tuple with the New Symbol 
         [0147]    In the data stream, we replace every instance of the most highly occurring tuple with the new single symbol. We replace the 48 instances of the 40 tuple in data stream  30 - 4 ,  FIG. 31 , with the symbol 5 as seen in data stream  30 - 5 ,  FIG. 39 . 
       Encoding the Alphabet 
       [0148]    As demonstrated above, we count the number of symbols in the data stream, and use the count to build a Huffman tree and code for the current alphabet. There is no Huffman code assigned to the symbol 1 and the symbol 4 because there are no instances of these symbols in the compressed data in this pass. The total symbol count has been reduced from 184 to 136 (e.g., 40+0+40+8+0+48, but not including the EOF marker) as seen in table  41 - 4 ,  FIG. 40 . 
       Calculating the Compressed File Size 
       [0149]    We need to evaluate whether our substitution reduces the amount of space that it takes to store the data. As seen in table  52 ′″,  FIG. 41 , the total number of bits is 283. 
         [0150]    As seen in table  57 ′″,  FIG. 42 , the compression overhead in bits is 48. 
         [0000]    Determining Whether the Compression Goal has been Achieved 
         [0151]    Finally, we compare the original number of bits ( 384 ) to the current number of bits ( 331 ) that are needed for this compression pass as seen in table  58 ′″,  FIG. 43 . In turn, we find that it takes 0.86 times as many bits to store the compressed data as it took to store the original data. 
       Pass 5 
       [0152]    In this pass, we replace the most highly occurring tuple from the previous pass with a new symbol, and then we determine whether we have achieved our compression goal. 
       Identifying all Possible Tuples 
       [0153]    We expand the tuple array by adding the symbol that was created in the previous pass. We add the symbol 5 as a first symbol and last symbol, and enter the tuples in the new cells as seen in table  35 - 4 ,  FIG. 44 . 
       Determining the Highly Occurring Tuple 
       [0154]    The tuple array identifies the tuples that we look for and tally in our revised alphabet as seen in table  40 - 4 ,  FIG. 45 . (Total Symbol Count=136) The tuple that occurs most frequently is the symbol 2 followed by symbol 5 (2&gt;5), which has a hypotenuse of 5.4. It occurs 39 times. This tuple ties with the tuple 0&gt;2 (hypotenuse is 2) and 5&gt;0 (hypotenuse is 5). The tuple 2&gt;5 is the most complex based on the hypotenuse length  25 ″ described above. 
       Creating a Symbol for the Highly Occurring Tuple 
       [0155]    We define a new symbol 6 to represent the most highly occurring tuple 2&gt;5, and add it to the dictionary for the alphabet that was developed in the previous passes as seen in table  26 - 5 ,  FIG. 46 . 
         [0000]    Replacing the Tuple with the New Symbol 
         [0156]    In the data stream, we replace every instance of the most highly occurring tuple with the new single symbol. We replace the 39 instances of the 2&gt;5 tuple in data stream  30 - 5 ,  FIG. 39 , with the symbol 6 as seen in data stream  30 - 6 ,  FIG. 47 . 
       Encoding the Alphabet 
       [0157]    As demonstrated above, we count the number of symbols in the data stream, and use the count to build a Huffman tree and code for the current alphabet as seen in table  41 - 5 ,  FIG. 48 . There is no Huffman code assigned to the symbol 1 and the symbol 4 because there are no instances of these symbols in the compressed data in this pass. The total symbol count has been reduced from 136 to 97 (e.g., 40+1+8+9+39, but not including the EOF marker) as seen in table  52 - 4 ,  FIG. 49 . 
       Calculating the Compressed File Size 
       [0158]    We need to evaluate whether our substitution reduces the amount of space that it takes to store the data. As seen in table  52 - 4 ,  FIG. 49 , the total number of bits is 187. 
         [0159]    As seen in table  57 - 4 ,  FIG. 50 , the compression overhead in bits is 59. 
         [0000]    Determining Whether the Compression Goal has been Achieved 
         [0160]    Finally, we compare the original number of bits ( 384 ) to the current number of bits ( 246 , or 187+59) that are needed for this compression pass as seen in table  58 - 4 ,  FIG. 51 . We find that it takes 0.64 times as many bits to store the compressed data as it took to store the original data. 
       Pass 6 
       [0161]    In this pass, we replace the most highly occurring tuple from the previous pass with a new symbol, and then we determine whether we have achieved our compression goal. 
       Identifying all Possible Tuples 
       [0162]    We expand the tuple array  35 - 5  by adding the symbol that was created in the previous pass as seen in  FIG. 52 . We add the symbol 6 as a first symbol and last symbol, and enter the tuples in the new cells. 
       Determining the Highly Occurring Tuple 
       [0163]    The tuple array identifies the tuples that we look for and tally in our revised alphabet. (Total Symbol Count=97) The tuple that occurs most frequently is the symbol 0 followed by symbol 6 (0&gt;6). It occurs 39 times as seen in table  40 - 5 ,  FIG. 53 . 
       Creating a Symbol for the Highly Occurring Tuple 
       [0164]    We define a new symbol 7 to represent the 0&gt;6 tuple, and add it to the dictionary for the alphabet that was developed in the previous passes as seen in table  26 - 6 ,  FIG. 54 . 
         [0000]    Replacing the Tuple with the New Symbol 
         [0165]    In the data stream, we replace every instance of the most highly occurring tuple with the new single symbol. We replace the 39 instances of the 0&gt;6 tuple in data stream  30 - 6 ,  FIG. 47 , with the symbol 7 as seen in data stream  30 - 7 ,  FIG. 55 . 
       Encoding the Alphabet 
       [0166]    As demonstrated above, we count the number of symbols in the data stream, and use the count to build a Huffman tree and code for the current alphabet as seen in table  41 - 6 ,  FIG. 56 . There is no Huffman code assigned to the symbol 1, symbol 4 and symbol 6 because there are no instances of these symbols in the compressed data in this pass. The total symbol count has been reduced from 97 to 58 (e.g., 1+0+1+8+0+9+0+39, but not including the EOF marker). 
         [0167]    Because all the symbols 1, 4, and 6 have been removed from the data stream, there is no reason to express them in the encoding scheme of the Huffman tree  50 ′,  FIG. 57 . However, the extinct symbols will be needed in the decode table. A complex symbol may decode to two less complex symbols. For example, a symbol 7 decodes to 0&gt;6. 
         [0168]    We need to evaluate whether our substitution reduces the amount of space that it takes to store the data. As seen in table  52 - 5 ,  FIG. 58 , the total number of bits is 95. 
         [0169]    As seen in table  57 - 5 ,  FIG. 59 , the compression overhead in bits is 71. 
         [0000]    Determining Whether the Compression Goal has been Achieved 
         [0170]    Finally, we compare the original number of bits ( 384 ) to the current number of bits ( 166 , or 95+71) that are needed for this compression pass as seen in table  58 - 5 ,  FIG. 60 . We find that it takes 0.43 times as many bits to store the compressed data as it took to store the original data. 
       Subsequent Passes 
       [0171]    Skilled artisans will also notice that overhead has been growing in size while the total number of bits is still decreasing. We repeat the procedure to determine if this is the optimum compressed file size. We compare the compression size for each subsequent pass to the first occurring lowest compressed file size. The chart  60 ,  FIG. 61 , demonstrates how the compressed file size grows, decreases, and then begins to grow as the encoding information and dictionary sizes grow. We can continue the compression of the foregoing techniques until the text file compresses to a single symbol after 27 passes. 
       Interesting Symbol Statistics 
       [0172]    With reference to table  61 ,  FIG. 62 , interesting statistics about the symbols for this compression are observable. For instance, the top 8 symbols represent 384 bits (e.g., 312+45+24+2+1) and 99.9% (e.g., 81.2+11.7+6.2+0.5+0.3%) of the file. 
       Storing the Compressed File 
       [0173]    The information needed to decompress a file is usually written at the front of a compressed file, as well as to a separate dictionary only file. The compressed file contains information about the file, a coded representation of the Huffman tree that was used to compress the data, the dictionary of symbols that was created during the compression process, and the compressed data. The goal is to store the information and data in as few bits as possible. 
         [0174]    This section describes a method and procedure for storing information in the compressed file. 
       File Type 
       [0175]    The first four bits in the file are reserved for the version number of the file format, called the file type. This field allows flexibility for future versions of the software that might be used to write the encoded data to the storage media. The file type indicates which version of the software was used when we saved the file in order to allow the file to be decompressed later. 
         [0176]    Four bits allows for up to 16 versions of the software. That is, binary numbers from 0000 to 1111 represent version numbers from 0 to 15. Currently, this field contains binary 0000. 
       Maximum Symbol Width 
       [0177]    The second four bits in the file are reserved for the maximum symbol width. This is the number of bits that it takes to store in binary form the largest symbol value. The actual value stored is four less than the number of bits required to store the largest symbol value in the compressed data. When we read the value, we add four to the stored number to get the actual maximum symbol width. This technique allows symbol values up to 20 bits. In practical terms, the value 2̂20 (2 raised to the 20 th  power) means that about 1 million symbols can be used for encoding. 
         [0178]    For example, if symbols 0-2000 might appear in the compressed file, the largest symbol ID (2000) would fit in a field containing 11 bits. Hence, a decimal 7 (binary 0111) would be stored in this field. 
         [0179]    In the compression example, the maximum symbol width is the end-of-file symbol 8, which takes four bits in binary (1000). We subtract four, and store a value of 0000. When we decompress the data, we add four to zero to find the maximum symbol width of four bits. The symbol width is used to read the Huffman tree that immediately follows in the coded data stream. 
       Coded Huffman Tree 
       [0180]    We must store the path information for each symbol that appears in the Huffman tree and its value. To do this, we convert the symbol&#39;s digital value to binary. Each symbol will be stored in the same number of bits, as determined by the symbol with the largest digital value and stored as the just read “symbol width”. 
         [0181]    In the example, the largest symbol in the dictionary in the Huffman encoded tree is the end-of-file symbol 8. The binary form of 8 is 1000, which takes 4 bits. We will store each of the symbol values in 4 bits. 
         [0182]    To store a path, we will walk the Huffman tree in a method known as a pre-fix order recursive parse, where we visit each node of the tree in a known order. For each node in the tree one bit is stored. The value of the bit indicates if the node has children (1) or if it is a leaf with no children (0). If it is a leaf, we also store the symbol value. We start at the root and follow the left branch down first. We visit each node only once. When we return to the root, we follow the right branch down, and repeat the process for the right branch. 
         [0183]    In the following example, the Huffman encoded tree is redrawn as  50 - 2  to illustrate the prefix-order parse, where nodes with children are labeled as 1, and leaf nodes are labeled as 0 as seen in  FIG. 63 . 
         [0184]    The discovered paths and symbols are stored in the binary form in the order in which they are discovered in this method of parsing. Write the following bit string to the file, where the bits displayed in bold/underline represent the path, and the value of the 0 node are displayed without bold/underline. The spaces are added for readability; they are not written to media. 
         [0000]    110 0101 110 0000 10 1000 0 0010 0 0011 0 0111 
       Encode Array for the Dictionary 
       [0185]    The dictionary information is stored as sequential first/last definitions, starting with the two symbols that define the symbol 2. We can observe the following characteristics of the dictionary:
       The symbols 0 and 1 are the atomic (non-divisible) symbols common to every compressed file, so they do not need to be written to media.   Because we know the symbols in the dictionary are sequential beginning with 2, we store only the symbol definition and not the symbol itself.   A symbol is defined by the tuple it replaces. The left and right symbols in the tuple are naturally symbols that precede the symbol they define in the dictionary.   We can store the left/right symbols of the tuple in binary form.   We can predict the maximum number of bits that it takes to store numbers in binary form. The number of bits used to store binary numbers increases by one bit with each additional power of two as seen, for example, in table  62 ,  FIG. 64 :       
 
         [0191]    Because the symbol represents a tuple made up of lower-level symbols, we will increase the bit width at the next higher symbol value; that is, at 3, 5, 9, and 17, instead of at 2, 4, 8, and 16. 
         [0192]    We use this information to minimize the amount of space needed to store the dictionary. We store the binary values for the tuple in the order of first and last, and use only the number of bits needed for the values. 
         [0193]    Three dictionary instances have special meanings. The 0 and 1 symbols represent the atomic symbols of data binary 0 binary 1, respectively. The last structure in the array represents the end-of-file (EOF) symbol, which does not have any component pieces. The EOF symbol is always assigned a value that is one number higher than the last symbol found in the data stream. 
         [0194]    Continuing our compression example, the table  63 ,  FIG. 65 , shows how the dictionary is stored. 
         [0195]    Write the following bit string to the file. The spaces are added for readability; they are not written to media. 
         [0000]    10 1000 0111 100000 010101 000110 
       Encoded Data 
       [0196]    To store the encoded data, we replace the symbol with its matching Huffman code and write the bits to the media. At the end of the encoded bit string, we write the EOF symbol. In our example, the final compressed symbol string is seen again as  30 - 7 ,  FIG. 66 , including the EOF. 
         [0197]    The Huffman code for the optimal compression is shown in table  67 ,  FIG. 67 . 
         [0198]    As we step through the data stream, we replace the symbol with the Huffman coded bits as seen at string  68 ,  FIG. 68 . For example, we replace symbol 0 with the bits  0100  from table  67 , replace symbol 5 with 00 from table  67 , replace instances of symbol 7 with 1, and so on. We write the following string to the media, and write the end of file code at the end. The bits are separated by spaces for readability; the spaces are not written to media. 
         [0199]    The compressed bit string for the data, without spaces is: 
         [0000]    010000111111111111111111111111111011001110110011111111011001011000110001 10001100011000101101010 
       Overview of the Stored File 
       [0200]    As summarized in the diagram  69 ,  FIG. 69 , the information stored in the compressed file is the file type, symbol width, Huffman tree, dictionary, encoded data, and EOF symbol. After the EOF symbol, a variable amount of pad bits are added to align the data with the final byte in storage. 
         [0201]    In the example, the bits  70  of  FIG. 70  are written to media. Spaces are shown between the major fields for readability; the spaces are not written to media. The “x” represents the pad bits. In  FIG. 69 , the bits  70  are seen filled into diagram  69   b  corresponding to the compressed file format. 
       Decompressing the Compressed File 
       [0202]    The process of decompression unpacks the data front the beginning of the file  69 ,  FIG. 69 , to the end of the stream. 
       File Type 
       [0203]    Read the first four bits of the file to determine the file format version. 
       Maximum Symbol Width 
       [0204]    Read the next four bits in the file, and then add four to the value to determine the maximum symbol width. This value is needed to read the Huffman tree information. 
       Huffman Tree 
       [0205]    Reconstruct the Huffman tree. Each 1 bit represents a node with two children. Each 0 bit represents a leaf node, and it is immediately followed by the symbol value. Read the number of bits for the symbol using the maximum symbol width. 
         [0206]    In the example, the stored string for Huffman is: 
         [0207]    11001011100000101000000100001100111 
         [0208]    With reference to  FIG. 71 , diagram  71  illustrates how to unpack and construct the Huffman tree using the pre-fix order method. 
       Dictionary 
       [0209]    To reconstruct the dictionary from file  69 , read the values for the pairs of tuples and populate the table. The values of 0 and 1 are known, so they are automatically included. The bits are read in groups based on the number of bits per symbol at that level as seen in table  72 ,  FIG. 72 . 
         [0210]    In our example, the following bits were stored in the file: 
         [0000]    1010000111101000010101000110 
         [0211]    We read the numbers in pairs, according to the bits per symbol, where the pairs represent the numbers that define symbols in the dictionary: 
         [0000]    
       
         
               
               
               
             
           
               
                   
                   
               
               
                   
                 Bits 
                 Symbol 
               
               
                   
                   
               
             
             
               
                   
                 1 0 
                 2 
               
               
                   
                 10 00 
                 3 
               
               
                   
                 01 11 
                 4 
               
               
                   
                 100 000 
                 5 
               
               
                   
                 010 101 
                 6 
               
               
                   
                 000 110 
                 7 
               
               
                   
                   
               
             
          
         
       
     
         [0212]    We convert each binary number to a decimal number: 
         [0000]    
       
         
               
               
               
             
           
               
                   
                   
               
               
                   
                 Decimal Value 
                 Symbol 
               
               
                   
                   
               
             
             
               
                   
                 1 0 
                 2 
               
               
                   
                 2 0 
                 3 
               
               
                   
                 1 3 
                 4 
               
               
                   
                 4 0 
                 5 
               
               
                   
                 2 5 
                 6 
               
               
                   
                 0 6 
                 7 
               
               
                   
                   
               
             
          
         
       
     
         [0213]    We identify the decimal values as the tuple definitions for the symbols: 
         [0000]    
       
         
               
               
               
             
           
               
                   
                   
               
               
                   
                 Symbol 
                 Tuple 
               
               
                   
                   
               
             
             
               
                   
                 2 
                 1 &gt; 0 
               
               
                   
                 3 
                 2 &gt; 0 
               
               
                   
                 4 
                 1 &gt; 3 
               
               
                   
                 5 
                 4 &gt; 0 
               
               
                   
                 6 
                 2 &gt; 5 
               
               
                   
                 7 
                 0 &gt; 6 
               
               
                   
                   
               
             
          
         
       
     
         [0214]    We populate the dictionary with these definitions as seen in table  73 ,  FIG. 73 . 
       Construct the Decode Tree 
       [0215]    We use the tuples that are defined in the re-constructed dictionary to build the Huffman decode tree. Let&#39;s decode the example dictionary to demonstrate the process. The diagram  74  in  FIG. 74  shows how we build the decode tree to determine the original bits represented by each of the symbols in the dictionary. The step-by-step reconstruction of the original bits is as follows: 
         [0216]    Start with symbols 0 and 1. These are the atomic elements, so there is no related tuple. The symbol 0 is a left branch from the root. The symbol 1 is a right branch. (Left and right are relative to the node as you are facing the diagram—that is, on your left and on your right.) The atomic elements are each represented by a single bit, so the binary path and the original path are the same. Record the original bits  0  and  1  in the decode table. 
         [0217]    Symbol 2 is defined as the tuple 1&gt;0 (symbol 1 followed by symbol 0). In the decode tree, go to the node for symbol 1, then add a path that represents symbol 0. That is, add a left branch at node  1 . The terminating node is the symbol 2. Traverse the path from the root to the leaf to read the branch paths of left (L) and right (R). Replace each left branch with a 0 and each right path with a 1 to view the binary forum of the path as LR, or binary 10. 
         [0218]    Symbol 3 is defined as the tuple 2&gt;0. In the decode tree, go to the node for symbol 2, then add a path that represents symbol 0. That is, add a left branch at node  2 . The terminating node is the symbol 3. Traverse the path from the root to the leaf to read the branch path of RLL. Replace each left branch with a 0 and each right path with a 1 to view the binary form of the path as 100. 
         [0219]    Symbol 4 is defined as the tuple 1&gt;3. In the decode tree, go to the node for symbol 1, then add a path that represents symbol 3. From the root to the node for symbol 3, the path is RLL. At symbol 1, add the RLL path. The terminating node is symbol 4. Traverse the path from the root to the leaf to read the path of RRLL, which translates to the binary format of 1100. 
         [0220]    Symbol 5 is defined as the tuple 4&gt;0. In the decode tree, go to the node for symbol 4, then add a path that represents symbol 0. At symbol 4, add the L path. The terminating node is symbol 5. Traverse the path from the root to the leaf to read the path of RRLLL, which translates to the binary format of 11000. 
         [0221]    Symbol 6 is defined as the tuple 2&gt;5. In the decode tree, go to the node for symbol 2, then add a path that represents symbol 5. From the root to the node for symbol 5, the path is RRLLL. The terminating node is symbol 6. Traverse the path from the root to the leaf to read the path of RLRRLLL, which translates to the binary format of 1011000. 
         [0222]    Symbol 7 is defined as the tuple 0&gt;6. In the decode tree, go to the node for symbol 0, then add a path that represents symbol 6. From the root to the node for symbol 6, the path is RLRRLLL. The terminating node is symbol 7. Traverse the path from the root to the leaf to read the path of LRLRRLLL, which translates to the binary format of 01011000. 
       Decompress the Data 
       [0223]    To decompress the data, we need the reconstructed Huffman tree and the decode table that maps the symbols to their original bits as seen at  75 ,  FIG. 75 . We read the bits in the data file one bit at a time, following the branching path in the Huffman tree from the root to a node that represents a symbol. 
         [0000]    The compressed file data bits are:
 
0100001111111111111111111111111101100111011001111111101100101000110001 10001100011000101101010
 
         [0224]    For example, the first four bits of encoded data 0100 takes us to symbol 0 in the Huffman tree, as illustrated in the diagram  76 ,  FIG. 76 . We look up 0 in the decode tree and table to find the original bits. In this case, the original bits are also 0. We replace 0100 with the single bit  0 . 
         [0225]    In the diagram  77  in  FIG. 77 , we follow the next two bits  00  to find symbol 5 in the Huffman tree. We look up 5 in the decode tree and table to find that symbol 5 represents original bits of 11000. We replace 00 with 11000. 
         [0226]    In the diagram  78 ,  FIG. 78 , we follow the next bit  1  to find symbol 7 in the Huffman tree. We look up 7 in the decode tree and table to find that symbol 7 represents the original bits  01011000 . We replace the single bit  1  with 01011000. We repeat this for each 1 in the series of 1s that follow. 
         [0227]    The next symbol we discover is with bits  011 . We follow these bits in the Huffman tree in diagram  79 ,  FIG. 79 . We look up symbol 3 in the decode tree and table to find that it represents original bits  100 , so we replace 011 with bits  100 . 
         [0228]    We continue the decoding and replacement process to discover the symbol 2 near the end of the stream with bits  01011 , as illustrated in diagram  80 ,  FIG. 80 . We look up symbol 2 in the decode tree and table to find that it represents original bits  10 , so we replace 01011 with bits  10 . 
         [0229]    The final unique sequence of bits that we discover is the end-of-file sequence of 01010, as illustrated in diagram  81 ,  FIG. 81 . The EOF tells us that we are done unpacking. 
         [0230]    Altogether, the unpacking of compressed bits recovers the original bits of the original data stream in the order of diagram  82  spread across two  FIGS. 82   a  and  82   b.    
         [0231]    With reference to  FIG. 83 , a representative computing system environment  100  includes a computing device  120 . Representatively, the device is a general or special purpose computer, a phone, a PDA, a server, a laptop, etc., having a hardware platform  128 . The hardware platform includes physical I/O and platform devices, memory (M), processor (P), such as a CPU(s), USB or other interfaces (X), drivers (D), etc. In turn, the hardware platform hosts one or more virtual machines in the form of domains  130 - 1  (domain  0 , or management domain),  130 - 2  (domain U 1 ), . . .  130 - n  (domain Un), each having its own guest operating system (O.S.) (e.g., Linux, Windows, Netware, Unix, etc.), applications  140 - 1 ,  140 - 2 , . . .  140 - n , file systems, etc. The workloads of each virtual machine also consume data stored on one or more disks  121 . 
         [0232]    An intervening Xen or other hypervisor layer  150 , also known as a “virtual machine monitor,” or virtualization manager, serves as a virtual interface to the hardware and virtualizes the hardware. It is also the lowest and most privileged layer and performs scheduling control between the virtual machines as they task the resources of the hardware platform, e.g., memory, processor, storage, network (N) (by way of network interface cards, for example), etc. The hypervisor also manages conflicts, among other things, caused by operating system access to privileged machine instructions. The hypervisor can also be type 1 (native) or type 2 (hosted). According to various partitions, the operating systems, applications, application data, boot data, or other data, executable instructions, etc., of the machines are virtually stored on the resources of the hardware platform. Alternatively, the computing system environment is not a virtual environment at all, but a more traditional environment lacking a hypervisor, and partitioned virtual domains. Also, the environment could include dedicated services or those hosted on other devices. 
         [0233]    In any embodiment, the representative computing device  120  is arranged to communicate  180  with one or more other computing devices or networks. In this regard, the devices may use wired, wireless or combined connections to other devices/networks and may be direct or indirect connections. If direct, they typify connections within physical or network proximity (e.g., intranet). If indirect, they typify connections such as those found with the internet, satellites, radio transmissions, or the like. The connections may also be local area networks (LAN), wide area networks (WAN), metro area networks (MAN), etc., that are presented by way of example and not limitation. The topology is also any of a variety, such as ring, star, bridged, cascaded, meshed, or other known or hereinafter invented arrangement. 
         [0234]    In still other embodiments, skilled artisans will appreciate that enterprises can implement some or all of the foregoing with humans, such as system administrators, computing devices, executable code, or combinations thereof. In turn, methods and apparatus of the invention further contemplate computer executable instructions, e.g., code or software, as part of computer program products on readable media, e.g., disks for insertion in a drive of a computing device  120 , or available as downloads or direct use from an upstream computing device. When described in the context of such computer program products, it is denoted that items thereof, such as modules, routines, programs, objects, components, data structures, etc., perform particular tasks or implement particular abstract data types within various structures of the computing system which cause a certain function or group of function, and such are well known in the art. 
         [0235]    While the foregoing produces a well-compressed output file, e.g.,  FIG. 69 , skilled artisans should appreciate that the algorithm requires relatively considerable processing time to determine a Huffman tree, e.g., element  50 , and a dictionary, e.g., element  26 , of optimal symbols for use in encoding and compressing an original file. Also, the time spent to determine the key information of the file is significantly longer than the time spent to encode and compress the file with the key. The following embodiment, therefore, describes a technique to use a file&#39;s compression byproducts to compress other data files that contain substantially similar patterns. The effectiveness of the resultant compression depends on how similar a related file&#39;s patterns are to the original file&#39;s patterns. As will be seen, using previously created, but related key, decreases the processing time to a small fraction of the time needed for the full process above, but at the expense of a slightly less effective compression. The process can be said to achieve a “fast approximation” to optimal compression for the related files. 
         [0236]    The definitions from  FIG. 1  still apply. 
         [0237]    Broadly, the “fast approximation” hereafter 1) greatly reduces the processing time needed to compress a file using the techniques above, and 2) creates and uses a decode tree to identify the most complex possible pattern from an input bit stream that matches previously defined patterns. Similar to earlier embodiments, this encoding method requires repetitive computation that can be automated by computer software. The following discusses the logical processes involved. 
       Compression Procedure Using a Fast Approximation to Optimal Compression 
       [0238]    Instead of using the iterative process of discovery of the optimal set of symbols, above, the following uses the symbols that were previously created for another file that contains patterns significantly similar to those of the file under consideration. In a high-level flow, the process involves the following tasks:
       1. Select a file that was previously compressed using the procedure(s) in  FIGS. 2-82   b . The file should contain data patterns that are significantly similar to the current file under consideration for compression.   2. From the previously compressed file, read its key information and unpack its Huffman tree and symbol dictionary by using the procedure described above, e.g.,  FIGS. 63-82   b.      3. Create a decode tree for the current file by using the symbol dictionary from the original file.   4. Identify and count the number of occurrences of patterns in the current file that match the previously defined patterns.   5. Create a Huffman encoding tree for the symbols that occur in the current file plus an end-of-file (EOF) symbol.   6. Store the information using the Huffman tree for the current file plus the file type, symbol width, and dictionary from the original file.
 
Each of the tasks is described in more detail below. An example is provided thereafter.
       
 
       Selecting a Previously Compressed File 
       [0245]    The objective of the fast approximation method is to take advantage of the key information in an optimally compressed file that was created by using the techniques above. In its uncompressed form of original data, the compressed file should contain data patterns that are significantly similar to the patterns in the current file under consideration for compression. The effectiveness of the resultant compression depends on how similar a related tile&#39;s patterns are to the original file&#39;s patterns. The way a skilled artisan recognizes a similar file is that similar bit patterns are found in the originally compressed and new file yet to be compressed. It can be theorized a priori that files are likely similar if they have similar formatting (e.g., text, audio, image, powerpoint, spreadsheet, etc), topic content, tools used to create the files, file type, etc. Conclusive evidence of similar bit patterns is that similar compression ratios will occur on both files (i.e. original file compresses to 35% of original size, while target file also compresses to about 35% of original size). It should be noted that similar file sizes are not a requisite for similar patterns being present in both files. 
         [0246]    With reference to  FIG. 84 , the key information  200  of a file includes the file type, symbol width, Huffman tree, and dictionary from an earlier file, e.g., file  69 ,  FIG. 69 . 
       Reading and Unpacking the Key Information 
       [0247]    From the key information  200 , read and unpack the File Type, Maximum Symbol Width, Huffman Tree, and Dictionary fields. 
       Creating a Decode Tree for the Current File 
       [0248]    Create a pattern decode tree using the symbol dictionary retrieved from the key information. Each symbol represents a bit pattern from the original data stream. We determine what those bits are by building a decode tree, and then parsing the tree to read the bit patterns for each symbol. 
         [0249]    We use the tuples that are defined in the re-constructed dictionary to build the decode tree. The pattern decode tree is formed as a tree that begins at the root and branches downward. A terminal node represents a symbol ID value. A transition node is a placeholder for a bit that leads to terminal nodes. 
       Identifying and Counting Pattern Occurrences 
       [0250]    Read the bit stream of the current file one bit at a time. As the data stream is parsed from left to right, the paths in the decode tree are traversed to detect patterns in the data that match symbols in the original dictionary. 
         [0251]    Starting from the root of the pattern decode tree, use the value of each input bit to determine the descent path thru the pattern decode tree. A “0” indicates a path down and to the left, while a “1” indicates a path down and to the right. Continue descending through the decode tree until there is no more descent path available. This can occur because a branch left is indicated with no left branch available, or a branch right is indicated with no right branch available. 
         [0252]    When the end of the descent path is reached, one of the following occurs:
       If the descent path ends in a terminal node, count the symbol ID found there.   If the descent path ends in a transition node, retrace the descent path toward the root, until a terminal node is encountered. This terminal node represents the most complex pattern that could be identified in the input bit stream. For each level of the tree ascended, replace the bit that the path represents back into the bit stream because those bits form the beginning of the next pattern to be discovered. Count the symbol ID found in the terminal node.       
 
         [0255]    Return to the root of the decode tree and continue with the next bit in the data stream to find the next symbol. 
         [0256]    Repeat this process until all of the bits in the stream have been matched to patterns in the decode tree. When done, there exists a list of all of the symbols that occur in the bit stream and the frequency of occurrence for each symbol. 
       Creating a Huffman Tree and Code for the Current File 
       [0257]    Use the frequency information to create a Huffman encoding tree for the symbols that occur in the current file. Include the end-of-file (EOF) symbol when constructing the tree and determining the code. 
       Storing the Compressed File 
       [0258]    Use the Huffman tree for the current file to encode its data. The information needed to decompress the file is written at the front of the compressed file, as well as to a separate dictionary only file. The compressed file contains:
       The file type and maximum symbol width information from the original file&#39;s key   A coded representation of the Huffman tree that was created for the current file and used to compress its data,   The dictionary of symbols from the original file&#39;s key,   The Huffman-encoded data, and   The Huffman-encoded EOF symbol.       
 
       Example of “Fast Approximation” 
       [0264]    This example uses the key information  200  from a previously created but related compressed file to approximate the symbols needed to compress a different file. 
       Reading and Unpacking the Key Information 
       [0265]    With reference to table  202 ,  FIG. 85 , a representative dictionary of symbols (0-8) was unpacked from the key information  200  for a previously compressed file. The symbols 0 and 1 are atomic, according to definition ( FIG. 1 ) in that they represent bits  0  and  1 , respectively. The reading and unpacking this dictionary from the key information is given above. 
         [0000]    Construct the Decode Tree from the Dictionary 
         [0266]    With reference to  FIG. 86 , a diagram  204  demonstrates the process of building the decode tree for each of the symbols in the dictionary ( FIG. 85 ) and determining the original bits represented by each of the symbols in the dictionary. In the decode tree, there are also terminal nodes, e.g.,  205 , and transition nodes, e.g.,  206 . A terminal node represents a symbol value. A transition node does not represent a symbol, but represents additional bits in the path to the next symbol. The step-by-step reconstruction of the original bits is described below. 
         [0267]    Start with symbols 0 and 1. These are the atomic elements, by definition, so there is no related tuple as in the dictionary of  FIG. 85 . The symbol 0 branches left and down from the root. The symbol 1 branches right and down from the root. (Left and right are relative to the node as you are facing the diagram—that is, on your left and on your right.) The atomic elements are each represented by a single bit, so the binary path and the original path are the same. You record the “original bits”  0  and  1  in the decode table  210 , as well as its “branch path.” 
         [0268]    Symbol 2 is defined from the dictionary as the tuple 1&gt;0 (symbol 1 followed by symbol 0). In the decode tree  212 , go to the node for symbol 1 (which is transition node  205  followed by a right path R and ending in a terminal node  206 , or arrow  214 ), then add a path that represents symbol 0 (which is transition node  205  followed by a left path L and ending in a terminal node  206 , or path  216 ). That is, you add a left branch at node  1 . The terminating node  220  is the symbol 2. Traverse the path from the root to the leaf to read the branch paths of right (R) and left (L). Replace each left branch with a 0 and each right path with a 1 to view the binary form of the path as RL, or binary 10 as in decode table  210 . 
         [0269]    Symbol 3 is defined as the tuple 2&gt;0. In its decode tree  230 , it is the same as the decode tree for symbol 2, which is decode tree  212 , followed by the “0.” Particularly, in tree  230 , go to the node for symbol 2, then add a path that represents symbol 0. That is, you add a left branch (e.g., arrow  216 ) at node  2 . The terminating node is the symbol 3. Traverse the path from the root to the leaf to read the branch path of RLL. Replace each left branch with a 0 and each right path with a 1 to view the binary format of 100 as in the decode table. 
         [0270]    Similarly, the other symbols are defined with decode trees building on the decode trees for other symbols. In particular, they are as follows: 
         [0271]    Symbol 4 from the dictionary is defined as the tuple 1&gt;3. In its decode tree, go to the node for symbol 1, then add a path that represents symbol 3. From the root to the node for symbol 3, the path is RLL. At symbol 1, add the RLL path. The terminating node is symbol 4. Traverse the path from the root to the leaf to read the path of RRLL, which translates to the binary format of 1100 as in the decode table. 
         [0272]    Symbol 5 is defined as the tuple 4&gt;0. In its decode tree, go to the node for symbol 4, then add a path that represents symbol 0. At symbol 4, add the L path. The terminating node is symbol 5. Traverse the path from the root to the leaf to read the path of RRLLL, which translates to the binary format of 11000. 
         [0273]    Symbol 6 is defined as the tuple 5&gt;3. In its decode tree, go to the node for symbol 5, then add a path that represents symbol 3. The terminating node is symbol 6. Traverse the path from the root to the leaf to read the path of RRLLLRLL, which translates to the binary format of 11000100. 
         [0274]    Symbol 7 is defined from the dictionary as the tuple 5&gt;0. In its decode tree, go to the node for symbol 5, then add a path that represents symbol 0. From the root to the node for symbol 5, the path is RRLLL. Add a left branch. The terminating node is symbol 7. Traverse the path from the root to the leaf to read the path of RRLLLL, which translates to the binary format of 110000. 
         [0275]    Finally, symbol 8 is defined in the dictionary as the tuple 7&gt;2. In its decode tree, go to the node for symbol 7, then add a path that represents symbol 2. From the root to the node for symbol 7, the path is RRLLLL. Add a RL path for symbol 2. The terminating node is symbol 8. Traverse the path from the root to the leaf to read the path of RRLLLLRL, which translates to the binary format of 11000010. 
         [0276]    The final decode tree for all symbols put together in a single tree is element  240 ,  FIG. 87 , and the decode table  210  is populated with all original bit and branch path information. 
       Identifying and Counting Pattern Occurrences 
       [0277]    For this example, the sample or “current file” to be compressed is similar to the one earlier compressed who&#39;s key information  200 .  FIG. 84 , was earlier extracted. It contains the following representative “bit stream” (reproduced in  FIG. 88 , with spaces for readability): 
         [0000]    011000010110001001100001011000100110000101100001011000100110000101100010 011000010110000101100010011000010110001001100001011000100110000101100010 011000100110001001100010011000100110000101100001011000100110000101100010 0110000101100010 
         [0278]    We step through the stream one bit at a time to match patterns in the stream to the known symbols from the dictionary  200 ,  FIG. 85 . To determine the next pattern in the bit stream, we look for the longest sequence of bits that match a known symbol. To discover symbols in the new data bit stream, read a single bit at a time from the input bit stream. Representatively, the very first bit,  250   FIG. 88 , of the bit stream is a “0.” With reference to the Decode Tree,  240  in  FIG. 87 , start at the top-most (the root) node of the tree. The “0” input bit indicates a down and left “Branch Path” from the root node. The next bit from the source bit stream at position  251  in  FIG. 88 , is a “1,” indicating a down and right path. The Decode Tree does not have a defined path down and right from the current node. However the current node is a terminal node, with a symbol ID of 0. Write a symbol 0 to a temporary file, and increment the counter corresponding to symbol ID 0. Return to the root node of the Decode Tree, and begin looking for the next symbol. The “1” bit that was not previously usable in the decode (e.g.,  251  in  FIG. 88 ) indicates a down and right. The next bit “1” ( 252  in  FIG. 88 ) indicates a down and right. Similarly, subsequent bits “000010” indicate further descents in the decode tree with paths directions of LLLLRL, resulting in path  254  from the root. The next bit “1” (position  255 ,  FIG. 88 ) denotes a further down and right path, winch does not exist in the decode tree  240 , as we are presently at a terminal node. The symbol ID for this terminal node is 8. Write a symbol 8 to the temporary file, and increment the counter corresponding to symbol ID 8. 
         [0279]    Return to the root node of the Decode Tree, and begin looking for the next symbol again starting with the last unused input stream hit, e.g., the bit “ 1 ” at position  255 ,  FIG. 88 . Subsequent bits in the source bit stream. “11000100,” lead down through the Decode Tree to a terminal node for symbol 6. The next bit, “ 1 ”, at position  261 ,  FIG. 88 , does not represent a possible down and right traversal path. Thus, write a symbol 6 to the temporary file, and increment the counter corresponding to symbol ID 6. Again, starting back at the root of the tree, perform similar decodes and book keeping to denote discovery of symbols 86886868868686866666886868. Starting again at the root of the Decode Tree, parse the paths represented by input bits “1100010” beginning at position  262 . There are no more bits available in the input stream. However, the current position in the Decode Tree, position  268 , does not identify a known symbol. Thus, retrace the Decode Tree path upward toward the root. On each upward level node transition, replace a bit at the front of the input bit stream with a bit that represents that path transition; e.g. up and right is a “0” up and left is a “1”. Continue the upward parse until reaching a valid symbol ID node, in this case the node  267  for symbol ID 5. In the process, two bits (e.g., positions  263  and  264 ,  FIG. 88 ) will have been pushed back onto the input stream, a “0”, and then a “1.” As before, write a symbol 5 to a temporary file, and increment the counter corresponding to symbol ID 5. Starting back at the root of the tree, bits are pulled from the input stream and parsed downward, in this case the “1” and then the “0” at positions  263  and  264 . As we are now out of input bits, after position  264 , examine the current node for a valid symbol ID, which in this case does exist at node  269 , a symbol ID of 2. Write a symbol 2 to the temporary files, increment the corresponding counter. All input bits have now been decoded to previously defined symbols. The entire contents of the temporary file are symbols: 
         [0000]    “0868688686886868686666688686852.” 
         [0280]    From here, the frequency of occurrence of each of the symbols in the new bit stream is counted. For example, the symbols “0” and “2” are each found occurring once at the beginning and end of the new bit stream. Similarly, the symbol “5” is counted once just before the symbol “2.” Each of the symbols “6” and “8” are counted fourteen times in the middle of the new bit stream for a total of thirty-one symbols. Its result is shown in table  275 ,  FIG. 89 . Also, one count for the end of file (EOF) symbol is added that is needed to mark the end of the encoded data when we store the compressed data. 
       Creating a Huffman Tree and Code for the Current File 
       [0281]    From the symbol “counts” in  FIG. 89 , a Huffman binary code tree  280  is built for the current file, as seen in  FIG. 90 . There is no Huffman code assigned to the symbol 1, symbol 3, symbol 4, and symbol 7 because there are no instances of these symbols in the new bit stream. However, the extinct symbols will be needed in the decode table for the tree. The reason for this is that a complex symbol may decode to two less complex symbols. For example, it is known that a symbol 8 decodes to tuple 7&gt;2, e.g.,  FIG. 85 . 
         [0282]    To construct the tree  280 , list first the symbols from highest count to lowest count. In this example, the symbol “8” and symbol “6” tied with a count of fourteen and are each listed highest on the tree. On the other hand, the least counted symbols were each of symbol “0,” “2,” “5,” and the EOF. Combine the counts for the two least frequently occurring symbols in the dictionary. This creates a node that has the value of the sum of the two counts. In this example, the EOF and 0 are combined into a single node  281  as are the symbols 2 and 5 at node  283 . Together, all four of these symbols combine into a node  285 . Continue combining the two lowest counts in this manner until there is only one symbol remaining. This generates a Huffman binary code tree. 
         [0283]    Label the code tree paths with zeros (0s) and ones (1s). To encode a symbol, parse from the root to the symbol. Each left and down path represents a 0 in the Huffman code. Each right and down path represents a 1 in the Huffman code. The Huffman coding scheme assigns shorter code words to the more frequent symbols, which helps reduce the size length of the encoded data. The Huffman code for a symbol is defined as the string of values associated with each path transition from the root to the symbol terminal node. 
         [0284]    With reference to  FIG. 91 , table  290  shows the final Huffman code for the current file, as based on the tree. For example, the symbol “8” appears with the Huffman code 0. From the tree, and knowing the rule that “0” is a left and down path, the “8” should appear from the root at down and left, as it does. Similarly, the symbol “5” should appear at “1011” or right and down, left and down, right and down, and right and down, as it does. Similarly, the other symbols are found. There is no code for symbols 1, 3, 4, and 7, however, because they do not appear in the current file. 
       Storing the Compressed File 
       [0285]    The diagram in  FIG. 92  illustrates how we now replace the symbols with their Huffman code value when the file is stored, such as in file format element  69 ,  FIG. 69 . As is seen, the diagram  295  shows the original bit stream that is coded to symbols or a new bit stream, then coded to Huffman codes. For example, the “0” bit at position  250  in the original hit stream coded to a symbol “0” as described in  FIG. 88 . By replacing the symbol 0 with its Huffman code (1001) from table  290 ,  FIG. 91 , the Huffman encoded bits are seen, as: 
         [0000]    1001 0 11 0 11 0 0 11 0 11 0 0 11 0 11 0 0 11 11 11 11 11 0 0 11 0 11 0 1011 1010 1000 
         [0286]    Spaces are shown between the coded bits for readability; the spaces are not written to media. Also, the code for the EOF symbol (1000) is placed at the end of the encoded data and shown in underline. 
         [0287]    With reference to  FIG. 93 , the foregoing information is stored in the compressed file  69 ′ for the current file. As skilled artisans will notice, it includes both original or re-used information and new information, thereby resulting in a “fast approximation.” In detail, it includes the file type from the original key information ( 200 ), the symbol width from the original key information ( 200 ), the new Huffman coding recently created for the new file, the dictionary from the key information ( 200 ) of the original file, the data that is encoded by using the new Huffman tree, and the new EOF symbol. After the EOF symbol, a variable amount of pad bits are added to align the data with the final byte in storage. 
         [0288]    In still another alternate embodiment, the following describes technology to identify a file by its contents. It is defined, in one sense, as providing a tile&#39;s “digital spectrum.” The spectrum, in turn, is used to define a file&#39;s position in an N-dimensional universe. This universe provides a basis by which a file&#39;s position determines similarity, adjacency, differentiation and grouping relative to other files. Ultimately, similar files can originate many new compression features, such as the “fast approximations” described above. The terminology defined in  FIG. 1  remains valid as does the earlier-presented information for compression and/or fast approximations using similar files. It is supplemented with the definitions in  FIG. 94 . Also, the following considers an alternate use of the earlier described symbols to define a digital variance in a file. For simplicity in this embodiment, a data stream under consideration is sometimes referred to as a “file.” 
         [0289]    The set of values that digitally identities the file, referred to as the file&#39;s digital spectrum, consists of several pieces of information found in two scalar values and two vectors. 
         [0000]    The scalar values are:
       The number of symbols in the symbol dictionary (the dictionary being previously determined above.)   The number of symbols also represents the number of dimensions in the N-dimensional universe, and thus, the number of coordinates in the vectors.   The length of the source file in bits.   This is the total number of bits in the symbolized data stream after replacing each symbol with the original bits that the symbol represents.
 
The vectors are:
   An ordered vector of frequency counts, where each count represents the number of times a particular symbol is detected in the symbolized data stream.
           F x =(F 0x , F 1x , F 2x , F 3x , . . . , F Nx ),   
           where F represents the symbol frequency vector. 0 to N are the symbols in a file&#39;s symbol dictionary, and x represents the source file of interest.   An ordered vector of bit lengths, where each hit length represents the number of bits that are represented by a particular symbol.
           B x =(B 0x , B 1x , B 2x , B 3x , . . . , B Nx ),   where B represents the bit-length vector, 0 to N are the symbols in a file&#39;s symbol dictionary, and x represents the source file of interest.   
               
 
         [0300]    The symbol frequency vector can be thought of as a series of coordinates in an N-dimensional universe where N is the number of symbols defined in the alphabet of the dictionary, and the counts represent the distance from the origin along the related coordinate axis. The vector describes the file&#39;s informational position in the N-dimension universe. The meaning of each dimension is defined by the meaning of its respective symbol. 
         [0301]    The origin of N-dimensional space is an ordered vector with a value of 0 for each coordinate: 
         [0302]    F O =(0, 0, 0, 0, 0, 0, 0, 0, . . . , 0). 
         [0303]    The magnitude of the frequency vector is calculated relative to the origin. An azimuth in each dimension can also be determined using ordinary trigonometry, which may be used at a later time. By using Pythagorean geometry, the distance from the origin to any point F x  in the N-dimensional space can be calculated, i.e.: 
         [0000]        D   ox =square root((( F   0x   −F   0o )̂2)+(( F   1x   −F   1o )̂2)+(( F   2x   −F   2o )̂2)+(( F   3x   −F   3o )̂2)+ . . . +(( F   Nx   −F   No )̂2))
 
         [0304]    Substituting the 0 at each coordinate for the values at the origin, the simplified equation is: 
         [0000]        D   ox =square root(( F   0x )̂2)+( F   1x )̂2)+( F   2x )̂2)+( F   3x )̂2)+ . . . +( F   Nx )̂2))
 
         [0305]    As an example, imagine that a file has 10 possible symbols and the frequency vector for the file is: 
       F x =(3, 5, 6, 1, 0, 7, 19, 3, 6, 22). 
       [0306]    Since this vector also describes the file&#39;s informational position in this 10-dimension universe, its distance from the origin can be calculated using the geometry outlined. Namely, 
         [0000]        D   ox =square root(((3−0)̂2)+((5−0)̂2)+((6−0)̂2)+((6−0)̂2)+((1−0)̂2)+((0−0)̂2)+((7−0)̂2)+((19−0)̂2)+((3−0)̂2)+((6−0)̂2)+((22−0)̂2))
 
         [0000]      Dox=31.78. 
       Determining a Characteristic Digital Spectrum 
       [0307]    To create a digital spectrum for a file under current consideration, we begin with the key information  200 ,  FIG. 84 , which resulted from an original file of interest. The digital spectrum determined for this original file is referred to as the characteristic digital spectrum. A digital spectrum for a related file of interest, on the other hand, is determined by its key information from another file. Its digital spectrum is referred to as a related digital spectrum. 
         [0308]    The key information actually selected for the characteristic digital spectrum is considered to be a “well-suited key.” A “well-suited key” is a key best derived from original data that is substantially similar to the current data in a current file or source file to be examined. The key might even be the actual compression key for the source file under consideration. However, to eventually use the digital spectrum information for the purpose of file comparisons and grouping, it is necessary to use a key that is not optimal for any specific file, but that can be used to define the N-dimensional symbol universe in which all the files of interest are positioned and compared. The more closely a key matches a majority of the files to be examined, the more meaningful it is during subsequent comparisons. 
         [0309]    The well-suited key can be used to derive the digital spectrum information for the characteristic file that we use to define the N-dimensional universe in which we will analyze the digital spectra of other files. From above, the following information is known about the characteristic digital spectrum of the file:
       The number of symbols (N) in the symbol dictionary   The length of the source file in bits   An ordered vector of symbol frequency counts
           F i =(F 0i , F 1i , F 2i , F 3i , . . . , F Ni ),   where F represents the symbol frequency, 0 to N are the symbols in the characteristic file&#39;s symbol dictionary, and i represents the characteristic file of interest.   
           An ordered vector of bit lengths
           B i =(B 0i , B 1i , B 2i , B 3i , . . . , B Ni ),   where B represents the bit-length vector, 0 to N are the symbols in the characteristic file&#39;s symbol dictionary and i represents the characteristic file of interest.   
               
 
       Determining a Related Digital Spectrum 
       [0318]    Using the key information and digital spectrum of the characteristic file, execute the process described in the fast approximation embodiment for a current, related file of interest, but with the following changes:
       1. Create a symbol frequency vector that contains one coordinate position for the set of symbols described in the characteristic file&#39;s symbol dictionary.
           F j =(F 0j , F 1j , F 2j , F 3j , . . . , F Nj ),   where F represents the symbol frequency, 0 to N are the symbols in the characteristic file&#39;s symbol dictionary, and j represents the related file of interest.   
           Initially, the count for each symbol is zero (0).   2. Parse the data stream of the related file of interest for symbols. As the file is parsed, conduct the following:
           a. Tally the instance of each discovered symbol in its corresponding coordinate position in the symbol frequency vector. That is, increment the respective counter for a symbol each time it is detected in the source file.   b. Do not Huffman encode or write the detected symbol.   c. Continue parsing until the end of the file is reached.   
           3. At the completion of the source file parsing, write a digital spectrum output file that contains the following:
           a. The number of symbols (N) in the symbol dictionary   b. The length of the source file in bits   c. The symbol frequency vector developed in the previous steps.
               F j =(F 0v , F 1j , F 2j , F 3j , . . . , F Nj ),   where F represents the frequency vector, 0 to N are the symbols in the characteristic file&#39;s symbol dictionary, and the j represents the file of interest.   
               d. The bit length vector
               B j =(B 0j , B 1j , B 2j , B 3j , . . . , B Nj ),   where B represents the bit-length vector, 0 to N are the symbols in the characteristic file&#39;s symbol dictionary, and j represents the file of interest.   
               
               
 
       Advantages of Digital Spectrum Analysis 
       [0336]    The digital spectrum of a file can be used to catalog a file&#39;s position in an N-dimensional space. This position in space, or digital spectrum, can be used to compute “distances” between file positions, and hence similarity, e.g., the closer the distance, the closer the similarity. The notion of a digital spectrum may eventually lead to the notion of a self-cataloging capability of digital files, or other. 
       Begin: Example Defining a File&#39;s Digital Spectrum 
       [0337]    To demonstrate the foregoing embodiment, the digital spectrum will be determined for a small data file that contains the following simple ASCII characters: 
         [0000]      aaaaaaaaaaaaaaaaaaaaaaaaaaabaaabaaaaaaaababbbbbb  (eqn. 100)
 
         [0338]    Each character is stored as a sequence of eight bits that correlates to the ASCII code assigned to the character. The bit values for each character are: 
         [0000]      a=01100001  (eqn. 101)
 
         [0000]      b=01100010  (eqn. 102)
 
         [0339]    By substituting the bits of equations 101 and 102 for the “a” and “b” characters in equation 100, a data stream  30  results as seen in  FIG. 9 . (Again, the characters are separated in the Figure with spaces for readability, but the spaces are not considered, just the characters.) 
         [0340]    After performing an optimal compression of the data by using the process defined above in early embodiments, the symbols remaining in the data stream  30 - 7  are seen in  FIG. 55 . Alternatively, they are shown here as: 
         [0000]      0 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 3 5 7 7 7 3 5 7 7 7 7 7 7 7 7 3 5 7 3 5 3 5 3 5 3 5 3 5 2  (eqn. 103)
 
         [0341]    With reference to  FIG. 95 , table  300  identifies the symbol definitions from equation 103 and the bits they represent. The symbol definition  302  identifies the alphabet of symbols determined from the data during the compression process. The symbols 0 and 1 are atomic symbols and represent original bits  0  and  1 , by definition. The subsequent symbols, i.e., 2-7, are defined by tuples, or ordered pairs of symbols, that are represented in the data, e.g., symbol 4 corresponds to a “1” followed by a “3,” or 1&gt;3. In turn, each symbol represents a series or sequence of bits  304  in the data stream of equation 103 (the source file), e.g., symbol 4 corresponds to original bits  1100 . 
         [0342]    With reference to table  310 ,  FIG. 96 , the number of occurrences of each symbol is counted in the data stream (equation 103) and the number of bits represented by each symbol is counted. For example, the symbol “7” in equation 103 appears thirty nine (39) times. In that its original bits  304 , correspond to “01011000,” it has eight (8) original bits appearing in the data stream for every instance of a “symbol 7” appearing. For a grand total of numbers of bits, the symbol count  312  is multiplied by the bit length  314  to arrive at a bit count  316 . In this instance, thirty nine (39) is multiplied by eight (8) to achieve a bit count of three-hundred twelve (312) for the symbol 7. A grand total of the number of bit counts  316  for every symbol  320  gives a length of the source file  325  in numbers of bits. In this instance, the source file length (in bits) is three-hundred eighty-four (384). 
         [0343]    In turn, the scalar values to be used in the tile&#39;s digital spectrum are:
       Source File Length in bits=384   Number of Symbols=8 total (or symbols 0 through 7, column 320,  FIG. 96 )       
 
         [0346]    The vectors to be used in the file&#39;s digital spectrum are:
       Frequency spectrum, F x , represented by the ordered vector of counts for each symbol, from column 312,  FIG. 96 :
           F x =(1, 0, 1, 8, 0, 9, 0, 39)   
           Bit length spectrum, Bx, is represented by the ordered vector of counts for the original bits in the file that are represented by each symbol, from column 314,  FIG. 96 :
           B x =(1, 1, 2, 3, 4, 5, 7, 8)   
               
 
         [0351]    The digital spectrum information can be used to calculate various useful characteristics regarding the file from which it was derived, as well as its relationship to other spectra, and the files from which the other spectra were derived. As an example, the frequency spectrum F(x) shown above, may be thought to describe a file&#39;s informational position in an 8-dimension universe, where the meaning of each dimension is defined by the meaning of its respective symbols. 
         [0352]    Since the origin of the 8-dimensional space is an ordered vector with a value of 0 at each symbol position, e.g., F(0)=(0, 0, 0, 0, 0, 0, 0, 0), the informational position in 8-dimensional space can be defined as an azimuth and distance from the origin. The magnitude of the position vector is calculated using Pythagorean geometry. Dist(x,0)=sqrt (((F(x,0)−F(00)̂2)+ . . . (F(x,7)−F(0,7)̂2)). Simplified, this magnitude becomes Dist(x,0)=sqrt((F(x,0)̂2+F(x,2)̂2+F(x,3)̂2 . . . F(x,7)̂2)). Using the values above in F x , the magnitude of the Dist(x,0)=40.84, or D x0 =square root (((1)̂2)+((0)̂2)+((1)̂2)+((8)̂2)+((0)̂2)+((9)̂2)+((0)̂2)+((39)̂2))=square root (1+0+1+64+0+81+0+1521)=40.84. Azimuth of the vector can be computed using basic trigonometry. 
         [0353]    Using information found in the digital spectra of a group of files, an analysis can be done to determine similarity, or not, of two or more subject files. Information from the digital spectrum is used to create an information statistic for a file. Statistics found to be pertinent in doing this analysis include at least: 
         [0354]    S 1 ) Frequency of occurrence of each possible symbol (FREQ) 
         [0355]    S 2 ) Normalized frequency of occurrence of each possible symbol (NORM FREQ) 
         [0356]    S 3 ) Informational content of occurrence of each symbol (INFO) 
         [0357]    S 4 ) Normalized information content of occurrence of each symbol (NORM INFO) 
         [0358]    For ease of reference, statistic S 1  can be called FREQ for frequency, statistic S 2  can be called NORM FREQ for normalized frequency, statistic S 1  can be called INFO for informational content, and statistic S 4  can be called NORM INFO for normalized informational content. A further discussion is given below for each of these statistical values. 
         [0359]    As a first example, a digital spectra of three files, F 1 , F 2 , and F 3  is given with respect to a common set of “N” symbols, e.g., symbols 1, symbol 2 and symbol 3. Each file is processed looking for the number of times each symbol is found in the file. The frequency of each symbol as it is found in each file is recorded along with a total number of symbols in each file. For this example, their respective spectra are: 
         [0000]    
       
         
               
               
               
               
               
               
             
               
               
               
               
               
               
             
           
               
                   
               
               
                 File 
                 Description 
                 Total 
                 Symbol 1 
                 Symbol 2 
                 Symbol 3 
               
               
                   
               
             
             
               
                   
               
             
          
           
               
                 File 1 
                 Number of Symbols 
                 3 
                   
                   
                   
               
               
                   
                 Sum of all Symbol 
                 9 
                   
                   
                   
               
               
                   
                 Occurrences 
                   
                   
                   
                   
               
               
                   
                 Symbol frequencies 
                   
                 2 
                 4 
                 3 
               
               
                   
                 Symbol bits sized 
                   
                 7 
                 6 
                 10 
               
               
                 File 2 
                 Number of Symbols 
                 3 
                   
                   
                   
               
               
                   
                 Sum of all Symbol 
                 8 
                   
                   
                   
               
               
                   
                 Occurrences 
                   
                   
                   
                   
               
               
                   
                 Symbol frequencies 
                   
                 4 
                 2 
                 2 
               
               
                   
                 Symbol bits sized 
                   
                 7 
                 6 
                 10 
               
               
                 File 3 
                 Number of Symbols 
                 3 
                   
                   
                   
               
               
                   
                 Sum of all Symbol 
                 27 
                   
                   
                   
               
               
                   
                 Occurrences 
                   
                   
                   
                   
               
               
                   
                 Symbol frequencies 
                   
                 8 
                 11 
                 8 
               
               
                   
                 Symbol bits sized 
                   
                 7 
                 6 
                 10 
               
               
                   
               
             
          
         
       
     
         [0360]    Using a relevant pattern-derived statistic (possibly including S 1 , S 2 , S 3 , or S 4  above), a vector of values is calculated for the N symbol definitions that may occur in each file. A position in N-dimensional space is determined using this vector, where the distance along each axis in N-space is determined by the statistic describing its corresponding symbol. 
         [0361]    Specifically in this example, we will use statistic S 1  (FREQ) and we have three (3) common symbols that we are using to compare these files and so a 3-dimensional space is determined. Each file is then defined as a position in this 3-dimensional space using a vector of magnitude 3 for each file. The first value in each vector is the frequency of symbol 1 in that file, the second value is the frequency of symbol 2, and the third value is the frequency of symbol 3. 
         [0362]    With reference to  FIG. 97 , these three example files are plotted. The frequency vectors are F 1 =(2, 4, 3), F 2 =(4, 2, 2), and F 3 =(8, 11, 8). The relative position in 3-space (N=3) for each of these files is readily seen. 
         [0363]    A matrix is created with the statistic chosen to represent each file. A matrix using the symbol frequency as the statistic looks like the following: 
         [0000]    
       
         
               
               
               
               
               
             
               
               
               
               
               
             
           
               
                   
                   
               
               
                   
                 File ID 
                 Sym1 
                 Sym2 
                 Sym3 
               
               
                   
                   
               
             
             
               
                   
               
             
          
           
               
                   
                 F1 
                 2 
                 4 
                 3 
               
               
                   
                 F2 
                 4 
                 2 
                 2 
               
               
                   
                 F3 
                 8 
                 11 
                 8 
               
               
                   
                   
               
             
          
         
       
     
         [0364]    Using Pythagorean arithmetic, the distance (D) between the positions of any two files (Fx, Fy) is calculated as 
         [0000]    
       
         
           
             
               
                 
                   
                     D 
                      
                     
                       ( 
                       
                         Fx 
                         , 
                         Fy 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       
                         
                           ( 
                           
                             
                               Fx 
                               1 
                             
                             - 
                             
                               Fy 
                               1 
                             
                           
                           ) 
                         
                         2 
                       
                       + 
                       
                         
                           ( 
                           
                             
                               Fx 
                               2 
                             
                             - 
                             
                               Fy 
                               2 
                             
                           
                           ) 
                         
                         2 
                       
                       + 
                       
                         
                           ( 
                           
                             
                               Fx 
                               n 
                             
                             - 
                             
                               Fy 
                               n 
                             
                           
                           ) 
                         
                         2 
                       
                     
                   
                 
               
               
                 
                   ( 
                   1 
                   ) 
                 
               
             
           
         
       
     
         [0000]    In the example above, the distance between the position of F 1  and F 2  is 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         
                           ( 
                           
                             2 
                             - 
                             4 
                           
                           ) 
                         
                         2 
                       
                       + 
                       
                         
                           ( 
                           
                             4 
                             - 
                             2 
                           
                           ) 
                         
                         2 
                       
                       + 
                       
                         
                           ( 
                           
                             3 
                             - 
                             2 
                           
                           ) 
                         
                         2 
                       
                     
                   
                   = 
                   
                     
                       
                         ( 
                         
                           4 
                           + 
                           4 
                           + 
                           1 
                         
                         ) 
                       
                     
                     = 
                     
                       
                         9 
                       
                       = 
                       3.00 
                     
                   
                 
               
               
                 
                   ( 
                   2 
                   ) 
                 
               
             
           
         
       
     
         [0365]    Similarly, the distance between F 1  and F 3  is found by 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         
                           ( 
                           
                             2 
                             - 
                             8 
                           
                           ) 
                         
                         2 
                       
                       + 
                       
                         
                           ( 
                           
                             4 
                             - 
                             11 
                           
                           ) 
                         
                         2 
                       
                       + 
                       
                         
                           ( 
                           
                             3 
                             - 
                             2 
                           
                           ) 
                         
                         2 
                       
                     
                   
                   = 
                   
                     
                       
                         
                           ( 
                           
                             36 
                             + 
                             49 
                             + 
                             25 
                           
                           ) 
                         
                       
                       + 
                       
                         110 
                       
                     
                     = 
                     10.49 
                   
                 
               
               
                 
                   ( 
                   3 
                   ) 
                 
               
             
           
         
       
     
         [0366]    A matrix of distances between all possible files is built. In the above example this matrix would look like this: 
         [0000]    
       
         
               
               
               
               
             
               
               
               
               
               
             
           
               
                   
                   
               
               
                   
                 F1 
                 F2 
                 F3 
               
               
                   
                   
               
             
             
               
                   
               
             
          
           
               
                   
                 F1 
                 0.00 
                 3.00 
                 10.49 
               
               
                   
                 F2 
                 3.00 
                 0.00 
                 11.53 
               
               
                   
                 F3 
                 10.49 
                 11.53 
                 0.00 
               
               
                   
                   
               
               
                   
                 Distance between files 
               
             
          
         
       
     
         [0367]    It can be seen graphically in  FIG. 97 , that the position of File  1  is closer to File  2  than it is to File  3 . It can also be seen in  FIG. 97  that File  2  is closer to File  1  than it is to File  3 . File  3  is closest to File  1 ; File  2  is slightly further away. 
         [0368]    Each row of the matrix is then sorted, such that the lowest distance value is on the left, and the highest value is on the right. During the sort process, care is taken to keep the File ID associated with each value. The intent is to determine an ordered distance list with each file as a reference. The above matrix would sort to this: 
         [0000]    
       
         
               
               
               
             
               
               
               
               
             
           
               
                   
               
               
                 File 
                 Distance 
               
               
                   
               
             
             
               
                   
               
             
          
           
               
                 F1 
                 F1 (0.00) 
                 F2 (3.00) 
                 F3 (10.49) 
               
               
                 F2 
                 F2 (0.00) 
                 F1 (3.00) 
                 F3 (11.53) 
               
               
                 F3 
                 F3 (0.00) 
                 F1 (10.49) 
                 F2 (11.53) 
               
               
                   
               
               
                 Sorted Distance between files 
               
             
          
         
       
     
         [0369]    Using this sorted matrix, the same conclusions that were previously reached by visual examination can now be determined mathematically. Exclude column 1, wherein it is obvious that the closest file to a given file is itself (or a distance value of 0.00). Column 2 now shows that the closest neighbor to F 1  is F 2 , the closest neighbor to F 2  is F 1 , and the closest neighbor the F 3  is F 1 . 
         [0370]    Of course, this concept can be expanded to hundreds, thousands, or millions or more of files and hundreds, thousands, or millions or more of symbols. While the matrices and vectors are larger and might take more time to process, the math and basic algorithms are the same. For example, consider a situation in which there exist 10,000 files and 2,000 symbols. 
         [0371]    Each file would have a vector of length  2000 . The statistic chosen to represent the value of each symbol definition with respect to each file is calculated and placed in the vector representing that file. An information position in 2000-space (N-2000) is determined by using the value in each vector position to represent the penetration along the axis of each of the 2000 dimensions. This procedure is done for each file in the analysis. With the statistic value matrix created, the distances between each file position are calculated using the above distance formula. A matrix that has 10,000 by 10,000 cells is created, for the 10,000 files under examination. The content of each cell is the calculated distance between the two files identified by the row and column of the matrix. The initial distance matrix would be 10,000×10,000 with the diagonal values all being 0. The sorted matrix would also be 10,000 by 10,000 with the first column being all zeros. 
         [0372]    In a smaller example, say ten files, the foregoing can be much more easily demonstrated using actual tables represented as text tables in this document. An initial matrix containing the distance information of ten files might look like this. 
         [0000]    
       
         
               
             
               
               
               
               
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
               
             
           
               
                   
               
               
                 Distance Matrix 
               
             
          
           
               
                 Files 
                 F1 
                 F2 
                 F3 
                 F4 
                 F5 
                 F6 
                 F7 
                 F8 
                 F9 
                 F10 
               
               
                   
               
             
          
           
               
                 F1 
                 0.0 
                 17.4 
                 3.5 
                 86.4 
                 6.7 
                 99.4 
                 27.6 
                 8.9 
                 55.1 
                 19.3 
               
               
                 F2 
                 17.4 
                 0.0 
                 8.6 
                 19.0 
                 45.6 
                 83.2 
                 19.9 
                 4.5 
                 49.2 
                 97.3 
               
               
                 F3 
                 3.5 
                 8.6 
                 0.0 
                 33.7 
                 83.6 
                 88.6 
                 42.6 
                 19.6 
                 38.2 
                 89.0 
               
               
                 F4 
                 86.4 
                 19.0 
                 33.7 
                 0.0 
                 36.1 
                 33.6 
                 83.9 
                 36.2 
                 48.1 
                 55.8 
               
               
                 F5 
                 6.7 
                 45.6 
                 83.6 
                 36.1 
                 0.0 
                 38.0 
                 36.9 
                 89.3 
                 83.4 
                 28.9 
               
               
                 F6 
                 99.4 
                 83.2 
                 88.6 
                 33.6 
                 38.0 
                 0.0 
                 38.4 
                 11.7 
                 18.4 
                 22.0 
               
               
                 F7 
                 27.6 
                 19.9 
                 42.6 
                 83.9 
                 36.9 
                 38.4 
                 0.0 
                 22.6 
                 63.3 
                 35.7 
               
               
                 F8 
                 8.9 
                 4.5 
                 19.6 
                 36.2 
                 89.3 
                 11.7 
                 22.6 
                 0.0 
                 8.1 
                 15.3 
               
               
                 F9 
                 55.1 
                 49.2 
                 38.2 
                 48.1 
                 83.4 
                 18.4 
                 63.3 
                 8.1 
                 0.0 
                 60.2 
               
               
                 F10 
                 19.3 
                 97.3 
                 89.0 
                 55.8 
                 28.9 
                 22.0 
                 35.7 
                 15.3 
                 60.2 
                 0.0 
               
               
                   
               
             
          
         
       
     
         [0373]    The distances in each row are then sorted such that an ordered list of distances, relative to each file, is obtained. The file identity relation associated with each distance is preserved during the sort. The resulting matrix now looks like this: 
         [0000]    
       
         
               
             
               
               
               
               
               
               
               
               
               
               
               
             
           
               
                   
               
               
                 Sorted Distance Matrix 
               
             
          
           
               
                   
                 1 
                 2 
                 3 
                 4 
                 5 
                 6 
                 7 
                 8 
                 9 
                 10 
               
               
                   
               
               
                 F1 
                 F1(0.0) 
                 F3(3.5) 
                 F5(6.7) 
                 F8(8.9) 
                 F2(17.4) 
                 F10(19.3) 
                 F7(27.6) 
                 F9(55.1) 
                 F4(86.4) 
                 F6(99.4) 
               
               
                 F2 
                 F2(0.0) 
                 F8(4.5) 
                 F3(8.6) 
                 F1(17.4) 
                 F4(19.0) 
                 F7(19.9) 
                 F5(45.6) 
                 F9(49.2) 
                 F6(83.2) 
                 F10(97.3) 
               
               
                 F3 
                 F3(0.0) 
                 F1(3.5) 
                 F2(8.6) 
                 F8(19.6) 
                 F4(33.7) 
                 F9(38.2) 
                 F7(42.6) 
                 F5(83.6) 
                 F6(88.6) 
                 F10(89.0) 
               
               
                 F4 
                 F4(0.0) 
                 F2(19.0) 
                 F6(33.6) 
                 F3(33.7) 
                 F5(36.1) 
                 F8(36.2) 
                 F9(48.1) 
                 F10(55.8) 
                 F1(86.4) 
                 F7(83.9) 
               
               
                 F5 
                 F5(0.0) 
                 F1(6.7) 
                 F10(28.9) 
                 F4(36.1 
                 F7(36.9) 
                 F6(38.0) 
                 F2(45.6) 
                 F9(83.4) 
                 F3(83.6) 
                 F8(89.3) 
               
               
                 F6 
                 F6(0.0) 
                 F8(11.7) 
                 F9(18.4) 
                 F10(22.0) 
                 F4(33.6) 
                 F5(38.0) 
                 F7(38.4 
                 F2(83.2) 
                 F3(88.6 
                 F1(99.4) 
               
               
                 F7 
                 F7(0.0) 
                 F2(19.9) 
                 F8(22.6) 
                 F1(27.6) 
                 F5(36.9) 
                 F10(35.7) 
                 F6(38.4) 
                 F3(42.6) 
                 F9(63.3 
                 F4(83.9) 
               
               
                 F8 
                 F8(0.0) 
                 F2(4.5) 
                 F9(8.1) 
                 F1(8.9) 
                 F6(11.7) 
                 F10(15.3 
                 F3(19.6 
                 F7(22.6) 
                 F4(36.2) 
                 F5(89.3) 
               
               
                 F9 
                 F9(0.0 
                 F8(8.1) 
                 F6(18.4) 
                 F3(38.2) 
                 F4(48.1) 
                 F2(49.2) 
                 F1(55.1) 
                 F10(60.2) 
                 F7(63.3) 
                 F5(83.4) 
               
               
                 F10 
                 F10(0.0 
                 F8(15.3 
                 F1(19.3) 
                 F6(22.0) 
                 F5(28.9) 
                 F7(35.7) 
                 F4(55.8) 
                 F9(60.2) 
                 F3(89.0) 
                 F2(97.3) 
               
               
                   
               
             
          
         
       
     
         [0374]    Using the information in columns 1 and 2 a relationship graph can be created of closest neighbor files. From the above matrix, skilled artisans will note the following: 
         [0375]    F 1 &#39;s nearest neighbor is F 3 . Create a group, G 1 , assign these two files to that group. 
         [0376]    F 2 &#39;s nearest neighbor is F 8 . Create a group, G 2 , assign these two files to that group. 
         [0377]    F 3  has already been assigned, its nearest neighbor is F 1 , and they belong to group G 1 . 
         [0378]    F 4 &#39;s nearest neighbor is F 2 , which already belongs to G 2 . Assign F 4  to G 2  as well. 
         [0379]    F 5 &#39;s nearest neighbor is F 1 , which already belongs to G 1 . Assign F 5  to G 1  as well. 
         [0380]    F 6 &#39;s nearest neighbor is F 8 , which already belongs to G 2 . Assign F 6  to G 2  as well. 
         [0381]    F 7 &#39;s nearest neighbor is F 2 , which already belongs to G 2 . Assign F 7  to G 2  also. 
         [0382]    F 8 &#39;s has already been assigned, It&#39;s nearest neighbor is F 2 , and they belong to G 2 . 
         [0383]    F 9 &#39;s nearest neighbor is F 8 , which already belongs to G 2 . Assign F 9  to G 2  also. 
         [0384]    F 10 &#39;s nearest neighbor is F 8  which already belongs to G 2 . Assign F 10  to G 2  also. 
         [0385]    The above “nearest neighbor” logic leads to the conclusion that two groups (G 1  and G 2 ) of files exist. Group G 1  contains F 1 , F 3 , F 5 , while Group G 2  contains F 2 , F 4 , F 6 , F 7 , F 8 , F 9 , and F 10 . 
         [0386]    An algorithm for determining groups based on adjacent neighbors is given in  FIG. 98A . For each file in the scope of analysis  900 , a closest neighbor is determined,  910 . From the example, this includes using the distance values that have been sorted in columns 1 and 2. If a closest neighbor already belongs to a group at  920 , the file joins that group at  930 . Else, if the closest neighbor belongs to no group at  940 , a new group is created at  950  and both files are added to the new group at  960 . From the example, F 1 &#39;s nearest neighbor is F 3  and no groups exist at  940 . Thus, a new group G 1  is created at  950  and both F 1  and F 3  are assigned, or added. Similarly, F 2 &#39;s nearest neighbor is F 8 , but only group G 1  exists. Thus, a new group G 2  is created at  950  for files F 2  and F 8  at  960 . Later, it is learned that F 4 &#39;s nearest neighbor is F 2 , which already belongs to G 2  at step  920 . Thus, at  930  file F 4  joins group G 2 . Once all files have been analyzed, the groups are finalized and group processing ceases at  970 . 
         [0387]    With reference to  FIG. 98B , a graph of the relationships can be made, although doing so in 2D space is often impossible. In groups G 1  and G 2  above, a representation of a 2-D graph that meets the neighbor criteria might look like reference numeral  980 . Using this grouping method and procedure, it can be deduced that a group of files are pattern-related and are more closely similar to each other, than to files which find membership in another group. Thus, files F 1 , F 3  and F 5  are more closely similar than those in group G 2 . 
         [0388]    In order to find relationships or groupings of files in alternate embodiments of the invention, the following proposes an embodiment of relevancy groups for arbitrary sets of data files having rightly sized and rightly numbered files. In turn, underlying data of the files is partitioned into meaningful groups without any prior classification schemes or metadata analysis. There is no defining a number of groups before grouping begins and criteria for ascertaining functions for stopping grouping is revealed. 
         [0389]    With reference to  FIGS. 99A and 99B , a sample of one-hundred files (F 1 , F 2  . . . F 100 ) is prospectively taken, and distance values are obtained between them. In  FIG. 99A , the actual distance values are placed into a matrix  450 . As between a file and itself, such as between file F 1  and F 1 , or file F 100  and F 100 , a distance value of 0.0 appears in the matrix cells. As between other files, distance values range upward to 70.0. This figure is similar in matrix form to the earlier table labeled “distance matrix.” In  FIG. 99B , a matrix  452  of sorted distance values is arranged in columns to show a first closest file, then a second closest file, and so on until a last or 100 th  closest file is revealed for each of the one-hundred files (rows). This view is similar to the table example labeled “sorted distance matrix.” 
         [0390]    Generically, skilled artisans create these matrixes as array “distances” that are N×N in an (N) multi-dimensional space. Each Row i  in the table represents file F i , each Col k  in the table represents file F k , and cell at Row i  and Col k  in represents the distance from F i  to F k . The values in the matrix are generically floating point values, depending on the algorithm used to generate the vectors in the N-dimensional symbol universe. Sometimes they range from 0.0 to 1.0 inclusive, or sometimes they range from 1.0 to some MAX value. Either embodiment is acceptable for use with the following grouping algorithm. 
         [0391]    The “sorted” array is also N×N where each row is sorted from the closest file (itself) to its furthest neighbor. In this case, each Row i  in the table still represents F i , but each Col k  now represents the K th  closest file and thus each cell at Row i  and Col k  has a value J that represents F i . This means that the value J is in the cell at Row i  and Col k  then F j  is the K th  closest file to F i . 
         [0392]    Hereafter, the minimum and maximum distances between any two files in the “distances” matrix are calculated. In the ongoing example of  FIGS. 99A and 99B , 0.0 and 70.0 are the minimums and maximums. 
         [0393]    Next, a stepping mechanism is created that moves a tolerance value in rounds between the min and max values in some regular fashion. In one embodiment, this could be a linear progress with (step+1) iterations using a formula like (min+((max−min)/round)*(round−1)) for each round in step+1 rounds. Alternatively, this could be a log or exponential type function that moves much more slowly for “close” values and much more quickly for large values. The latter type of progress is much more useful in N-dimensional symbol universes where some nodes are very far away from a cluster of nodes that are close together. Whatever the stepping function, the tolerance value grows. 
         [0394]    A tolerance is calculated (increasing with each round) and a group of members (e.g., files) is determined for each row in the “Distances” matrix  450 . Only those files that are closer than or equal to the tolerance value are included in that relevancy group for that file for that round. For example, consider Row 1 for file F 1  in the Distances matrix  450  which has ten steps of movement through the minimum and maximum values (from 0.0 to 70.0) with the tolerance value growing in linear fashion starting at 0.0 and moving to 70.0 by a tolerance of 7.0 at each step for 11 (steps+1) rounds. The result is seen in  FIG. 100A . Also, example relevancy groupings include only itself in round 1, since the tolerance value is 0.0 (labeled  470 ). In round 4, on the other hand, four files (F 1 , F 2 , F 99  and F 100 ) are grouped  472  since each has a distance value closer or equal to a tolerance value of 21.0. Namely, file F 1  has a distance value of 0.0, while each of files F 2 , F 99  and F 100  have distance values of 10.0, 20.0 and 5.0, respectively, as seen in  FIG. 99A . Similarly, Row 3, file F 3 , from  FIG. 99A  has groupings given as element  475  in  FIG. 100B . 
         [0395]    Other rows would follow similar patterns. For any row, skilled artisans should observe that a relevancy group grows in members as the tolerances (grow larger as it approaches the MAX value. At each round, any groups that are subsets of any other groups are deleted since they add no more information to the set of relevancy groups. Also, at round 1 there are one-hundred separate groups with each file being in its own dedicated group (or one member per one group). This is expected since the tolerance is 0.0 and the group does not include any other files other than the file itself since no other files are “closer” than the file itself. Unfortunately, this set of one-hundred groups with one file is not interesting from a “relevancy groups” point of view since the number of relevancy groups is the same as the number of individual files. The inventors refer to this situation as “isolation” and it represents one end of the spectrum defining relevancy groups, e.g., one file per group and the number of groups equals the original number of files, in this case 100. 
         [0396]    In the middle of grouping, at Round x  there are some number of groups such that the number of groups is more than one, but less than the total possible numbers of groups, or, 1&lt;=NumberGroups(Round x )&lt;=100. 
         [0397]    In the last round, round 11 or round (steps+1), there is only one group remaining and it includes all files as members. This is due to the fact that the tolerance has now reached the maximum value of any distance in the Distances matrix. So, all files are within that closeness to any other file and each group for each row now includes all files such that every group for every row is a subset of each other and they all contain all files. Thus, at round 1, there were 100 groups, each with one member. At the last round, round 11, there is only 1 group with 100 files in the group. The inventors call the latter “chaos” or “total integration” and it is at the other end of the spectrum defining relevancy groups. It is opposite the spectrum where “isolation” occurs. Also, this single group of 100 files is also not interesting from a “relevancy groups” point of view since all files are integrated into a chaotic mess of only one relevancy group. 
         [0398]    As a result, the inventors have valuably identified a situation as to when the tolerance should stop growing and at which round the grouping should stop so that the resultant relevancy group is rightly sized with the right number of groups. The inventors seek to avoid isolation (or limited grouping) or total integration (or chaos) since neither of these extremes provides optimization. To answer the question of when a correct number of groups occur, the inventors have experimented with various types of actual files and data sets. To date, these include: 100 fairly random text, PDF, MS Office, and Open Office documents; 250 fairly random ASCII short stories; 65 JPG files (with various levels of compression and color); 75 audio files; and various patent files. 
         [0399]    Hereafter, the embodiments of the invention identify at least three situations in which a stopping function is optimized for stopping the adding of members to a relevancy group. With reference to  FIG. 101 , a first step  500  begins grouping files and a second  502  applies a stopping function appropriate for the situation at hand. The stopping function and its features have been learned through experimentation with the actual files and data sets described. 
         [0400]    With reference to graph  510 ,  FIG. 102 , consider the number of groups at each round, such as in  FIGS. 100A and 100B . Upon plotting, skilled artisans should determine when the number of groups changes significantly. Although not exact, this is the point where the number of groups becomes “too integrated” and the groups themselves become less meaningful. 
         [0401]    All of these plots  512 ,  514 ,  516 ,  518 , and  520  were obtained from real data sets (101 files with different symbol counting algorithms) and they show the number of groups at each round for 100 rounds. From the beginning, the plots have 101 groups, while at the end they have but one group. In between, the plots reveal different information. In any situation, however, the generic response is to determine the numbers of groups per round, step  530 ,  FIG. 103 , determine the curve type of the plotted information, step  532 , and apply a stopping function per curve type, step  534 . 
         [0402]    In the example of the plotted curve  512 ,  FIG. 102 . “Sample 1” shows a smooth “Reverse S” shaped curve. This embodiment contemplates a stopping point for curves shaped like “Sample 1” at the round where the number of groups decreases by more than S %, where S % is configurable with any value between 5% and 50% depending on the data set but utilizing a default value of 10% in the absence of any other configuration data. This is an inflection point that shows larger numbers of files starting to be grouped together and so the groups are becoming more chaotic after this point. For “Sample 1.” the stopping point is at round 9. Samples 3, 4 and 5, (plots  516 ,  518 ,  520 ) all are similar and they show a gradual, but “bumpy” decrease until there is a sharp increase. This reflects a “local minimum” defined by a jump in the number of groups by 10% or more. This is a good stopping point since further rounds start to add many new overlapping groups. For these data sets, and the curves that look like these, the stopping point would be at around round 24 or 25. Sample 2 (plot  514 ), on the other hand, shows a steady decrease in the number of groups (which also means there is a steady increase in the number of files in each group which is leading to more and more chaos). For curve “Sample 2” there is no clear significant increase until round 55. However, if grouping stopped there, the groups would be too integrated and less meaningful. In this situation, experimental results have called for a stopping point at one quarter of the total number of rounds (which in this case would be at round 25 or 101/4). 
         [0403]    In a second approach to applying a stopping, function, stopping can occur at a time when there is no overlapping,  FIG. 104 . For example, consider the following “transitive closure” on membership that yields truly mutually exclusive groups. 
         [0404]    Instead of having these overlapping groups, 
         [0000]    {1, 2, 3}
 
{1, 2, 3, 4}
 
{4, 5}
 
{6, 7, 8}
 
{8, 9}
 
         [0405]    a re-defining of the groups reveals 
         [0000]    {1, 2, 3, 4, 5}
 
{6, 7, 8, 9}
 
         [0406]    In this case, the number of groups reduces much more quickly than in case 1 above and so it is impractical to use a metric like a 10% decrease in a steep “Reverse S” curve or a 10% increase in “local minimum” curves. The stopping point for this type of grouping algorithm, then, would be based on a much larger percent. It is recommended that stopping occur at the round where only ⅓ (33%) of the original number of groups is obtained. These groups are often more broad and not tightly clustered groups, but the inventors have found that they are good groups for making a first level segmentation decision. More fine-grained group divisions could then be done amongst each 1st level group, e.g., taking another third of the first third, and so on. 
         [0407]    In a third approach to stopping, a function examines the size of each group and “dismisses a group” when it becomes fairly meaningless. Representatively,  FIG. 105  shows this as steps  540  and  542 . In one embodiment, the dismissal occurs by the following example. A control shows groups whose size is anywhere from 2 members to some percent (say 25%) of the total number of members in the original data set. In the ongoing example of the 100 files data-set, numbers of groups having from 2 to 25 members are examined. Once the membership gets so large, the group becomes “noise” and is not meaningful any more. The max size of the groups that are not dismissed can range anywhere from 20% to 50% based on the data set. This embodiment calls for a variable percent that can be manipulated for a “soft” stopping point. Once there are no more groups that fit the criteria for not being dismissed, adding groups is stopped. 
         [0408]    Among certain advantages of the foregoing stopping approaches is that a right number and right sized relevancy group for an arbitrary set of data files is readily determined. These relevancy groups are then used to partition the data into meaningful groups without any prior classification or metadata analysis. This also occurs according to the following: without any pre-defined or pre-declared conditions about the number or size of the groups; by organizing relevance groups into human understandable relevance relationships without any human intervention; and by optimally combining related files into relevance groups up to a point but not passing the point where the relevance groups become too broad to be useful. 
         [0409]    In still other embodiments, pluralities of files grouped together into relevancy groups may have overlapping members which causes some ambiguity. In these cases, these overlapping relevancy groups may be further differentiated from one another into more clearly defined relevancy groups. For example, consider the following set S of four (4) groups where there are overlapping members in the groups: 
         [0000]    S={A, B, C, D} where 
       A={1, 2, 3} 
     B={2, 3} 
     C={1, 2,} 
     D={1, 4} 
       [0410]    In this example, member  1  is in groups A, C, and D; member  2  is in groups A, B, and C; member  3  is in groups A and B; and member  4  is only in group D. The question arises around the best partition of S into new relevancy groups that show some aggregation into the strongest possible groups with the least amount of overlap within those groups. The problem in doing that is evident in this example. Specifically, member  4  seems to be related to member  1  because of D. However, member  1  also seems to be related to member  2  because of groups A and C. Of note is that the relationship of member  1  to member  4  seems weaker than the relationship of member  1  to member  2  because the declaration that member  1  is related to member  4  occurs only once (in group D) while the declaration that member  1  is related to member  2  occurs twice (in groups A and C). The question that the inventors were faced with in experimental data similar to this example was how to improve the relevancy groups while considering the notion that some members might be more strongly related to each other than to other members. Specifically in this example, member  1  might have a stronger relevancy bonding with member  2  than it does with member  4 . In addition to and simultaneously with that information, the inventors were also faced with the strong or weak relevancy bonding relationships of member  1  with member  3 , member  3  with member  2 , member  3  with member  4 , and so on. It became clear that all of the information declared in the set S must be considered and used to define a partition of S into new relevancy groups that shows the strongest possible relevancy grouping strength with the least amount of overlap. In this example, if members  1  and  2  and  4  were placed in the same relevancy group would that be stronger or weaker (given only the data defined in set S) than if member  4  were not placed in the same relevancy group as members  1  and  2 . Even more to the point, would a partition of S into [{1, 2, 3}, {1, 4}] be stronger than a partition of S into [{1, 2, 3, 4}]. 
         [0411]    This invention teaches a new method for using only information in the set S of overlapping groups to define a new partition of that set into the strongest possible relevancy groups with the least amount of overlap among members of those groups. This is achieved using only the relevancy information contained in that original set—using no other a priori external information or any other pre-determined or semantic information about that set. This method can thus be used across any data set regardless of semantic content or domain of interest specific knowledge. 
         [0412]    First we define the new method in general terms and then we will move back to our specific earlier example to show the application of the new method on that set. The new method finds an optimized partition of any set S of N overlapping groups that yields the maximum strength for relevancy groups and members in that partition. Using semi-formal notation: the best partition P of S has strength T where T=SP(P) and T is greater than all other T′ where T′=SP(P′) and P′ is some other partition of S. Alternatively stated, there should be a partition P of S into groups such that Order(P)=J; 1&lt;=J&lt;=N; and SP(P) is the maximum of all other SP(P′) for all other partitions P′ of S. SP(P) is defined as function to determine the “strength of a partition.” Thus, for all partitions P 1 , P 2 , . . . , P p  of S, and where SP(P) is defined as “strength of partition P” the optimal partition P i  is the one where SP(P i ) yields the largest value. This invention teaches a method where SP(Pi) yields a value 0.0 to 1.0 (where 1.0 is the strongest). SP(P) is defined using a summation of all weighted SGS(GS) for each set of groups in the partition. SGS(GS) is defined as the “strength of a set of groups” and GS is that set of groups. SGS(GS) is defined using a summation of all weighted SM(M) for all members in the set of groups. SM(M) is defined as the “strength of a member” and M is that member. Said another way, SP(P) is the strongest when it defines a partition P that has the strongest set of groups where each group claims membership of the strongest set of members. 
         [0413]    The weighting factors affect the strength of a value by considering the context of that value. As such, we define all of our computational elements as: 
         [0000]    
       
         
               
               
               
             
           
               
                   
               
               
                 NOTATION 
                 NAME 
                 DESCRIPTION 
               
               
                   
               
             
             
               
                 S 
                 Set of Groups 
                 The original set of possibly overlapping groups 
               
               
                 P 
                 Partition or Group of GS 
                 A set of sets of groups 
               
               
                 GS 
                 Group of G 
                 A set of groups 
               
               
                 G 
                 Group of M 
                 A group which is a set of members 
               
               
                 M 
                 Member 
                 A member of the original set S 
               
               
                 SP(P) 
                 Strength of a Partition 
                 The numerical value of the strength of a partition. 
               
               
                   
                   
                 Specifically, this value is the sum of SGS(GS) * WGS(GS) 
               
               
                   
                   
                 for each GS in the partition. 
               
               
                 SGS(GS) 
                 Strength of a Group of G 
                 The numerical value of the strength of a group of G. 
               
               
                   
                   
                 Specifically, this value is the sum of SM(M) * WM(M)) for 
               
               
                   
                   
                 each M in this group of G. 
               
               
                 SM(M) 
                 Strength of a Member 
                 The numerical value of the strength of a member given a set 
               
               
                   
                   
                 of groups containing that member. Specifically this value is 
               
               
                   
                   
                 “number of groups in which this member appears”/“number 
               
               
                   
                   
                 of groups in this group of G” 
               
               
                 WGS(GS) 
                 Weight of a Group of G 
                 The numerical value of the weight that this group of G has 
               
               
                   
                   
                 compared to other groups of G in the partition. Specifically, 
               
               
                   
                   
                 this value is “number of groups in this group of G”/“number 
               
               
                   
                   
                 of groups in the partition” 
               
               
                 WM(M) 
                 Weight of a Member 
                 The numerical value of the weight that this member has 
               
               
                   
                   
                 given a group of G. Specifically. this value is “1.0”/ 
               
               
                   
                   
                 “number of total members in this group of G” 
               
               
                   
               
             
          
         
       
     
         [0414]    Using semi formal mathematical notation, the strength T=SP(P) of a partition P is given as: 
         [0000]    SP(Partition)=Σ SGS(GS i )*WGS(GS i ), for all groups GS i  in the partition and where SGS(GS)=Σ SM(M j )*WM(M j ), for all members M j  in the groups and where SM(Member)=(Number of total groups having that member)/(total numbers of groups); and where WGS(GS) is the relative strength of this set of groups compared with the total number of original groups and WM(M) is the relative strength of this member compared with the total number of members in all the groups of this set of groups. Specifically WGS(GS)=this number of groups/number of all original groups and WM(M)=(1.0)/(total number of members in this set of groups). 
         [0415]    The following is a pseudo-code representation of the method 
         [0000]    
       
         
               
             
               
               
             
               
               
             
               
               
             
               
               
             
               
               
             
               
             
               
               
             
               
               
             
               
               
             
               
             
               
               
             
               
               
             
               
               
             
               
             
           
               
                   
               
             
             
               
                 Main (S) { 
               
             
          
           
               
                   
                 MaxGroups = null; 
               
               
                   
                 MaxStrength = 0.0; 
               
               
                   
                 Groups[ ] partition = null; 
               
               
                   
                 For each partition of S { 
               
             
          
           
               
                   
                 partition_strength = Strength(partition) 
               
               
                   
                 if (partition_strength &gt; MaxStrength) { 
               
             
          
           
               
                   
                 MaxGroups = partition; 
               
               
                   
                 MaxStrength = partition_strength; 
               
             
          
           
               
                   
                 } 
               
             
          
           
               
                   
                 } 
               
               
                   
                 MaxGroups is the strongest partition of S; 
               
             
          
           
               
                 } 
               
               
                 Strength(Groups[ ] partition) { 
               
             
          
           
               
                   
                 retval = 0.0; 
               
               
                   
                 numSets = total number of original groups; 
               
               
                   
                 For each groups in partition { 
               
             
          
           
               
                   
                 groups_strength = Strength(groups) 
               
               
                   
                 groups_weight = groups.Size( ) / numSets; 
               
               
                   
                 groups_val = groups_strength * groups_weight; 
               
               
                   
                 retval += groups_val; 
               
             
          
           
               
                   
                 } 
               
               
                   
                 return retval; 
               
             
          
           
               
                 } 
               
               
                 Strength(Groups groups) { 
               
             
          
           
               
                   
                 numGroups = groups.Size( ); 
               
               
                   
                 numMembers = total number of all members in all of the groups; 
               
               
                   
                 retval = 0.0; 
               
               
                   
                 for each member across all of this set of groups { 
               
             
          
           
               
                   
                 member_count = Count of how many groups this member is in; 
               
               
                   
                 member_strength = member_count / numGroups; 
               
               
                   
                 member_weight = 1 / numMembers; 
               
               
                   
                 member_val = member_strength * member_weight; 
               
               
                   
                 retval += member_val; 
               
             
          
           
               
                   
                 } 
               
               
                   
                 return retval; 
               
             
          
           
               
                 } 
               
               
                   
               
             
          
         
       
     
         [0416]    Applying the foregoing, we revisit our example Set S, where S={A, B, C, D} and where 
       A={1, 2, 3} 
     B={2, 3} 
     C={1, 2} 
     D={1, 4} 
       [0417]    and the desired outcome is the best (strongest) partitioning of S. First we consider what are all possible partitions of set S. To do this we apply well know algorithms in the area of permutations and combinations yielding all possible partitions of S. In this example, the possible partitions of S are: 
       1: {A, B, C, D} 
     2: {A}, {B, C, D} 
     3: {A}, {B}, {C, D} 
     4: {A}, {C}, {B, D} 
     6: {A}, {D}, {B, C} 
     7: {A}, {B}, {C}, {D} 
     8: {B}, {A, C, D} 
     9: {B}, {A}, {C, D} 
     10: {B}, {C}, {A, D} 
     11: {B}, {D}, {A, C} 
     12: {B}, {A}, {C}, {D} 
     13: {C}, {A, B, D} 
     14: {C}, {A}, {B, D} 
     15: {C}, {B}, {A, D} 
     16: {C}, {D}, {A, B} 
     17: {C}, {A}, {B}, {D} 
     18: {D}, {A, B, C} 
     19: {D}, {A}, {B, C} 
     20: {D}, {B}, {A, C} 
     21: {D}, {C}, {A, B} 
     22: {D}, {A}, {B}, {C} 
     23: {A, B}, {C, D} 
     24: {A, B}, {C}, {D} 
     25: {A, C}, {B, D} 
     26: {A, C}, {B}, {D} 
     27: {A, D}, {B, C} 
     28: {A, D}, {B}, {C} 
     29: {B, A}, {C, D} 
     30: {B, A}, {C}, {D} 
     31: {B, C}, {A, D} 
     32: {B, C}, {A}, {D} 
     33: {B, D}, {A, C} 
     34: {B, D}, {A}, {C} 
       [0418]    . . . , and so on. 
         [0419]    Of the foregoing, skilled artisans should observe that many of these partitions are identical (e.g., 7=12=17=22 and 16=24). In turn, only one partition out of an equivalent set of identical partitions need be processed since the strength of identical partitions will be identical. 
         [0420]    In applying the weighting, we will consider just two partitions and compare strength for each and then compare the two strengths to determine which of the two partitions is the strongest. Let the first partition be P 1  which only has one set of groups: 
       Groups 1 
       [0421]    A: {1, 2, 3} 
         [0422]    B: {2, 3} 
         [0423]    C: {1, 2} 
         [0424]    D: {1, 4} 
         [0425]    This partition can be represented using our shorter notation above as Partition 1: {A, B, C, D}. 
         [0426]    Let the second partition be P 2  which has two sets of groups: 
       Groups 1 
       [0427]    A: {1, 2, 3} 
         [0428]    B: {2, 3} 
         [0429]    C: {1, 2,} 
       Groups 2 
       [0430]    D: {1, 4} 
         [0431]    which, in our shorter notation above, is Partition 18: {A, B, C}, {D}. 
         [0432]    In examining the strength of the first partition, P 1 , there are four (4) total groups. There are three (3) instances of file  1  in the four total groups, three (3) instances of file  2  in the four total groups, two (2) instances of file  3  in the four total groups, and one (1) single instance of file  4  in the four total groups. Also, there are four (4) total members in the four total groups. Namely, files  1 ,  2 ,  3 , and  4  are the members spread over the many groups. 
         [0433]    Applying these facts to the foregoing equations, WM(1)=WM(2)=WM(3)=WM(4)=1.0/4. Since member or file  1  appears in three (3) groups, the strength of file  1 , SM(1) becomes three out of four, or SM(1)=3/4. Similarly, SM(2)=3/4, SM(3)=2/4, and SM(4)=1/4. Summing these together, as weighted by WM, we get: (3/4*1/4)+(3/4*1/4)+(2/4*1/4)+(1/4*1/4)=0.56=SGS(Groups 1). WGS(Groups 1)=4.0/4 since there are 4 groups in Groups 1 and there are 4 original groups. The ultimate strength of the partition for P 1  is SP(P 1 )=0.56*4/4=0.56. 
         [0434]    Conversely, consider the strength of the second partition P 2 . For the first major subgroup (Groups 1) there are three total members, so WM(1)=WM(2)=WM(3)=1.0/3. For the second major subgroup (Groups 2) there are two total members, so WM(1)=SM(4)=1.0/2. The strength of the individual files in Groups 1 is SM(1)=2/3, SM(2)=3/3, and SM(3)=2/3, while the strength of the individual files in Groups 2 is SM(1)=1/1 and SM(4)=1/1. Summing these together per their individual subgroups, we get (2/3*1/3)+(3/3*1/3)+(2/3*1/3)=0.78 and (1/1*1/2)+(1/1*1/2)=1.00. Bringing the two major groups together, and weighting them for the entire partition (e.g., WGS(Groups 1) is three groups out of four total groups in the entire partition (or ¾), and WGS(Groups 2) is one group out of four total groups in the entire partition (or ¼)), we get the strength of the partition P 2  as: SP(P 2 )=(0.78*3/4)+(1.0*1/4)=0.83. Since 0.83 for partition P 2  is closer to 1.0 than 0.56 is for partition P 1 , partition P 2  is the strongest partition for set S. The same approach for all possible partitions would occur to reveal the strongest possible partition for the set. 
         [0435]    The following is a more concise summary of the computations involved in this example of S and two partitions of P 1  and P 2  and determining which partition is the strongest: 
         [0436]    P 1 :
       Groups 1={A, B, C, D}
           WM(1)=1/4; WM(2)=1/4; WM(3)=1/4; WM(4)=1/4;   SM(1)=3/4; SM(2)=3/4; SM(3)=2/4; SM(4)=1/4   SGS(Groups 1)=(3/4*1/4)+(3/4*1/4)+(2/4*1/4)+(1/4*1/4)=0.56   WGS(Groups 1)=1/1   
           SP(P 1 )=0.56*1/1=0.56       
 
         [0443]    P 2 :
       Groups 1={A, B, C}
           WM(1)=1/3; WM(2)=1/3; WM(3)=1/3;   SM(1)=2/3; SM(2)=3/3; SM(3)=2/3   SGS(Groups 1)=(2/3*1/3)+(3/3*1/3)+(2/3*1/3)=0.78   WGS(Groups 1)=3/4   
           Groups 2={D}
           WM(1)=1/2; WM(4)=1/2   SM(1)=1/1; SM(4)=1/1   SGS(Groups 2)=(1/1*1/2)+(1/1*1/2)=1.0   WGS(Groups 2)=1/4   
           SP(P 1 )=(0.78*3/4)+(1.0*1/4)=0.83       
 
         [0455]    Among certain advantages, skilled artisans should now appreciate that the foregoing provides: 1) an optimal partitioning of “fuzzy” overlapping relevance groups into clearly distinguishable, minimally overlapping relevance groups; 2) identifiable bridging information that causes the overlap which is helpful in determining the semantic information that can be associated with the new clearly distinguishable groups; 3) fuzzy logic weighting factors that beget crisp and distinct membership information; and 4) new partitions that generate new insights, observations, and cognitions that were not perceivable in the original overlapping groups due to the obfuscation inherent in the fuzziness. 
       Example 
       [0456]    The inventors ran many samples using the JAVA code appended in the code Exhibit. Examples t1-t8 below are a few of the results. In all, the inputs (.in) are groups of files that resulted in the partitioning of two optimal subgroups (.out). However, the algorithm works for breaking the possible output partitions into any number N groups where there are enough members to support N groups. 
         [0000]    - - -
 
t1.in
 
- - -
 
1, 2, 3
 
1, 2
 
2, 3
 
1, 4
 
- - -
 
t1.out
 
- - -
 
1, 4
 
1, 2, 3
 
- - -
 
- - -
 
t2.in
 
- - -
 
20, 29, 30, 35
 
20, 21, 24, 25, 33, 34
 
20, 31, 32, 33, 35
 
20, 25, 29, 30, 31, 33, 34
 
21, 24, 25, 34
 
29, 30, 31, 35
 
20, 25, 21, 33, 34
 
31, 32
 
19, 21, 24, 25
 
19, 27, 28, 34
 
19, 21, 27, 28
 
- - -
 
t2.out
 
- - -
 
20, 25, 29, 30, 31, 32, 33, 34, 35
 
19, 20, 21, 24, 25, 27, 28, 33, 34
 
- - -
 
- - -
 
t3.in
 
- - -
 
1, 2, 3
 
2, 3
 
1, 2
 
4, 5, 6
 
4, 5
 
5, 6
 
- - -
 
t3.out
 
- - -
 
1, 2, 3
 
4, 5, 6
 
- - -
 
- - -
 
t5.in
 
- - -
 
1, 2
 
3, 4
 
- - -
 
t5.out
 
- - -
 
1, 2
 
3, 4
 
- - -
 
- - -
 
t6.in
 
- - -
 
13, 16, 17, 18, 19
 
13, 15, 17, 18, 19
 
15, 16, 17, 18, 19
 
13, 15, 16, 18, 19
 
13, 14, 16, 18, 19
 
- - -
 
t6.out
 
- - -
 
13, 14, 16, 18, 19
 
13, 15, 16, 17, 18, 19
 
- - -
 
- - -
 
t7.in
 
- - -
 
48, 54, 59, 66, 89
 
22, 37, 63, 66, 88
 
37, 63, 66, 79, 88
 
54, 59, 70, 87, 89
 
43, 45, 53, 68, 89
 
44, 56, 59, 86, 89
 
54, 59, 68, 70, 89
 
48, 54, 59, 70, 89
 
20, 54, 59, 70, 89
 
54, 61, 68, 70, 89
 
- - -
 
t7.out
 
- - -
 
22, 37, 43, 44, 45, 53, 56, 59, 63, 66, 68, 79, 86, 88, 89
 
20, 48, 54, 59, 61, 66, 68, 70, 87, 89
 
- - -
 
- - -
 
t8.in
 
- - -
 
1, 4, 5, 6, 35
 
1, 4, 5, 35, 36
 
1, 2, 4, 5, 80
 
1, 5, 25, 34, 36
 
25, 34, 35, 36, 60
 
- - -
 
t8.out
 
- - -
 
1, 2, 4, 5, 6, 35, 80
 
1, 4, 5, 25, 34, 35, 36, 60
 
- - -
 
         [0457]    Upon inspection, skilled artisans should see that examples t7 and t8 show a separation into the two largest groups with non-overlapping nodes with the smallest bridge group. This allows users to see at what point the fuzzy overlapping groups overlap the most and at what point the overlap the least. 
         [0458]    For instance, example t8 reveals: 
         [0000]    Largest Non Overlapping Group 1: 2, 6, 80 (e.g., the group 1, 2, 4, 5, 6, 35, 80 absent the bridge group);
 
Largest Non Overlapping Group 2: 25, 34, 36, 60 (e.g., the group 1, 4, 5, 25, 34, 35, 36, 60 absent the bridge group); and
 
       Smallest Bridge Group: 1, 4, 5, 35. 
       [0459]    From venn diagramming, the smallest bridge group in both Group 1 (1, 2, 4, 5, 6, 35, 80) and Group 2 (1, 4, 5, 25, 34, 35, 36, 60) is the “Smallest Bridge Group: 1, 4, 5, 35.” However, if the bridge group is made a group separate from either groups 1 and 2, the “bridge” can be more readily identified. Said in more mathematical language: 
         [0000]    Let G 1 ′ be all members in G 1  and not in G 2 ;
 
Let G 2 ′ be all members in G 2  and not in G 1 ; and
 
Let B be all members in G 1  and in G 2 .
 
         [0460]    In turn, G 1 ′ and G 2 ′ are no longer overlapping at all and B shows why G 1  and G 2  were overlapping in the first place. Again, this can be expanded to 3, 4, . . . , N groups. The inventors have also proved this with jpeg photographs and other files as seen in their provisional application U.S. Ser. No. 61/271,079, filed Jul. 16, 2009, incorporated herein by reference in its entirety. 
         [0461]    Still further advantages of the partitioning techniques include: defining the strongest possible membership assignments possible given the input, overlapping groups; defining the smallest number of groups while still allowing members to be in more than one group (not forcing mutually exclusive groups)—the natural and social worlds we live in have naturally occurring fuzzy overlapping groups and even though many of those are optimized they stay a bit fuzzy (consider the problem of where to classify and place items in a hardware or grocery store, the ever changing 6 Kingdom or 3 Domain Scientific Classification taxonomies, social and cultural groups that we humans belong to, etc); making observable, new, clear, and distinct groups from less distinct and less clear fuzzy overlapping groups; and a specific algorithm for determining the weight of each member in the fuzzy overlapping groups and the number of fuzzy overlapping groups to determine a strength for every possible new partition of groups created from the members of the fuzzy overlapping groups in order to mathematically (objectively) determine the optimal set of groups (partitioning of the original groups). 
         [0462]    The foregoing has been described in terms of specific embodiments, but one of ordinary skill in the art will recognize that additional embodiments are possible without departing from its teachings. This detailed description, therefore, and particularly the specific details of the exemplary embodiments disclosed, is given primarily for clarity of understanding, and no unnecessary limitations are to be implied, for modifications will become evident to those skilled in the art upon reading this disclosure and may be made without departing from the spirit or scope of the invention. Relatively apparent modifications, of course, include combining the various features of one or more figures with the features of one or more of the other figures.