STATE MACHINES IN ARTIFICIAL INTELLIGENCE

Giovanni Alestra
STATE MACHINES
IN ARTIFICIAL
INTELLIGENCE
Graph, Hash, Mathematical Inference,
State Analysis
@ Copyright 2010 Giovanni Alestra

to my father
3

Contents
Introduction 7
1.1 Definition and graph representation 8
1.2 Linking data 9
1.3 A real example of graph application 10
1.4 Subgraph 12
1.5 An algoritm to individuate subgraph 13
1.6 Sliding Connections 17
1.7 Filtering connections 18
1.8 Fractal triangle 19
1.9 Keeping a subgraph 19
1.10Query as subgraph 20
1.11Graph programming 22
2.1 Bit definitions 24
2.2 Transcription 24
2.3 Transcription graph 26
2.4 Memory 26
2.5 Transformation 27
2.6 Paths, and circuits 28
2.7 Inverse transcription 30
3.1 Algoritm to calculate the hash of a graph 32
3.2 Counting 34
4.1 Introduction to linked hash 36
4.2 Example of discrimination 39
4.3 Hash links 42
4.4 How to dimensionete the hash space 43
4.5 Identity Transformation 45
4.6 Hash links collision (T,T+1 rule) 45
4.7 Deletion of zero-record 47
4.8 File system 47
4.9 Identity table 48
5.1 Mathematical problem representation 49
5.2 First examples of mathematical deduction 52
5

5.3 Further examples of mathematical deduction 55
5.4 Structured Pieces 57
5.5 Reaching the goal (checkmate) 60
6.1 States of the problem 63
6.2 Invariance 65
6.3 Problem setting 66
6.4 How to demonstrate a rule 67
6.5 Forcing variant 67
6.6 Real states of problem 70
6.7 A mate variant 70
6.8 Objective functions 61
6.9 All in a number 72
6.10Considerations 72
7.1 Reasoning with graph 73
Conclusions 75
6

INTRODUCTION
The aim of this book is to build and implement intelligent state machines for the
information management, The data structure, the memorization and inference
techniques are proposed in order to have and effective implementation in real
contexts, another purpose of this book is to explore states computing and to obtain
notable advantages in application development. We are going to show a coherent
and realizable theory that can improve the knowledge of engineers. The reader can
further contribute to the content of this book in order to bring this project in to
reality.
7

PART I
DATA STRUCTURE
CHAPTER 1 : GRAPH
1.1 Definition and graph representation
A graph is defined as Set ( Vertices, edges ) where the edges connect the vertices. A
graph is represented matematically in the form of matrix VxV in witch each entry
is either zero and one. If a connection exists between vertexs i and vertex j a one is
present at position (i,j) in the matrix. If connection between thes vertices does not
exist we write a zero at position (i,j). If a connection exists between i and i, at
8

position (i,i) we have a one, and this is possible for all vertices. It is easy to
individuate this kind of connection from-to the same vertex looking for the values
on the diagonal. An example of graph is :
In this example graph, bi-directional edges are represented like an edge without
arrow, in the graph, but it's evident that we have two different values in the
matrix..
1.2 Linking data
In ooperations research the values in the matix are tipically associated with the
flow from a vertex A to a vertex B. In this book the approach to the assocition of
data in edges and vertices is different The reason is simple, we want to have the
possibilty to associate an indeterminate number of information to verteces and
edges.
We can connect more than one information to verteces and edges and this data is
9

stored in two different tables. The first table is about the information connected to
verteces, where we have the vertex identifier and the list of information associated
to the vertex. In the second table we have the edges, each of witch has an identifier
formed by the union of origin vertex identifier and the destination vertex
identifier . For every edge there is a list of information associated with the edge.
1.3 A real example of graph application
Suppose we have an application for travelling. For example a travel agency
operating in Italy, with many travel destinations. For example:
10

We can model this situation through a graph, in which we have the state, the
regions, the provinces and the tourist places. In this graph are also present airline
compani train and buses operating between these places. We can model all
problems wtih the aid of a graph and create a tool application for maintenance of
the travel to tourist places. This application assist the user with addinf, modify or
deleting routes and tourist places. We can paint the following graph to represent
this problem. :
We interpose a node between two places, that indicates how we can travel (by
airbus, train or bus) and we have another edge and node in witch indicating
running time / travel time in order to have a complete inference about time of
departure and travel time.
11

Throught such a graph we can query the system with the following question :
– How can I travel from Trapani at 7 a.m. to Milan at 8 p.m. ?
The query has as response something like this :
You can go:
from Trapani to Palermo at 9:00 o'clock by Train ,
from Palermo to Pisa at 12:00 o'clock by Airbus,
from Pisa to Milano at 5.30 p.m. by Train arriving at 7.30 p.m.
...
The inferntial engine can be builded using algoritm that finds subgraphs. This
concept is shown below. By simply adding vertices and edges we can model any
kind of situation.
1.4 Subgraph
A subgraph is a graph with this property : Set ( V, E ) Subgraph = Sub Set ( V, E )
graph. A subgraph is a selection of graph vertices with a selection of edges related
to selected vertices. If we have data linked to vertices and edges the information on
the subgraph must be a selection of the information of the related vertices and
edges.
12

also if the concept is simple, findind a subgraph is a complex problem. Hereafier
we show an algoritm that helps finding a subgraph in graph. Identifying a subgraph
is a fundamental operation in graph theory.
1.5 An algoritm to individute a subgraph
Suppose we have a graph and a subgraph :
The algorithm to find a subgraph in a graph operates like this. At the first step we
select a vertex in the subgraph. The selection can be random or intelligent. A
random selection can select for example the vertices with information B, an
intelligent selection is the node A. In this example we operate in intelligent way. If
we have selected the vertices with information A we must find all vertices with the
same information in the graph. If we cannot find vertices with the same
information we cannot create any hypothesis of association between the vertex and
the algorithm terminates with fault. In our example we have an only vertex with
information A, so we have the hypothesis:
H:[10=1]
At the second step we select for each hypothesis the connection with vertices that
are not in the hypothesis. We call connection an edge with the destination vertex.
Now we enumerate the connection by information of edge and information of
13

destination vertex. We have these connections :
A → B with multiplicity 3 in graph and subgraph
A → D with multiplicity 1 in graph and subgraph
A → E with multiplicity 1 in graph and subgraph
If a multiplicity of subgraph is greater of a multiplicity of a graph the algorithm
terminates with fault. If this does happen, with must associate the combinations of
vertices of graph and subgraph. In this example we have:
A → B
[2=11; 3=12; 4=13]
[2=11; 3=13; 4=12]
[2=12; 3=11; 4=13]
[2=12; 3=13; 4=11]
[2=13; 3=11; 4=12]
[2=13; 3=12; 4=11]
A → D
[6=15]
A → E
[7=16]
The final hypothesis are:
[2=11; 3=12; 4=13; 6=15;7=16]
[2=11; 3=13; 4=12; 6=15;7=16]
[2=12; 3=11; 4=13; 6=15;7=16]
[2=12; 3=13; 4=11; 6=15;7=16]
[2=13; 3=11; 4=12; 6=15;7=16]
[2=13; 3=12; 4=11; 6=15;7=16]
14

At the next step we repeat the previous step looking for those vertices not present
in the hypothesis. In this case we have only one connection:
B → C
For the first hypothesis we have:
[2=11; 3=12; 4=13; 6=15;7=16] [5=14]
for the second hypothesis we have :
[2=11; 3=13; 4=12; 6=15;7=16] [5=14]
the other hypothesis has a fault:
[2=12; 3=11; 4=13; 6=15;7=16] [ 5=/ ]
in fact if we associate the vertex 3 to vertex 11 there are no desired connections.
Finally we have:
[2=11; 3=12; 4=13; 6=15;7=16] [5=14]
[2=11; 3=13; 4=12; 6=15;7=16] [5=14]
[2=12; 3=11; 4=13; 6=15;7=16] [5=/]
[2=12; 3=13; 4=11; 6=15;7=16] [5=/]
[2=13; 3=11; 4=12; 6=15;7=16] [5=/]
[2=13; 3=12; 4=11; 6=15;7=16] [5=/]
At this point we have no vertex without hypothesis and we can perform the
matching step within the matrix where the identifier has changed according to the
hypothesis
H 1 : [2=11; 3=12; 4=13; 6=15;7=16;5=14].
15

Matrix: Graph-Subgraph
Since the matrices are the same we have found a subgraph.
H 2 : [2=11; 3=13; 4=12; 6=15;7=16;5=14].
Matrix: Graph-Subgraph
The matching is negative which means that we do not have a subgraph for this
hypothesis.
16

1.7 Sliding connections
A sliding connection is a connection of one or more edges that connect two
verteces with defined information. For example:
The sliding connection indicates a connection of one or multiple edges that connect
the vertex with information A to the vertex with information B. Suppose we have
the following graph:
The resulting connection is composed by two connections: A → X and X → B
We have two different kinds of sliding connections: kind 1 and kind 2. The kind of
connection is indicated in the graph by a number above the connection, like this:
With the connection of kind 1 we are able to find only the first occurrence of the
destination vertex, while with the connection of kind 2 we are able to find all the
occurrence of destination vertex. For example:
17

.
The connection of kind 1 finds only the first occurrence of connection between A
and B, the connection of kind 2 find the first and the second occurrence of
connection between A and B.
1.9 Filtering connections
Different informatuib can be associtated to an edge. If information is associated to
a sliding connection it occurs that only compatible connection can be selectedith
sliding happen that only compatible conncetion can be selected. We call this
concept “filtering connections”. This example shows the concept:
The concept is simple, filtering connection means to select only edges having the
18

information of the sliding connection. This is really useful, indeed since we can use
a simple sliding connection without information if we want to select all the edges,
and adding information if we want a selection with major semantics.
1.8 Fractal triangle
The fractal triangle is often shown as a game of intelligence and used to
determinate the I.Q. of humans. By the use of sliding connection we can solve the
problem to determinate how many triangles are present in a fractal triangle. The
problem is to find subgraph with sliding connection of kind 2. The number of
subgraphs that we find is the number of triangles in a fractal triangle.
The image shows the subgraph definition and the fractal triangle.
Finding subgraphs we find the number of triangles in the fractal triangle.
1.9 Keeping a subgraph
When we solve a sliding connection we can keep all vertices that are in the path
from origin vertex to the destination vertex. This operation is called “Keeping a
Subgraph”. Keeping Subgraph is really useful and adds notable power to our
19

system. For example:
The resulting subgraph is:
We show by an example the power of keeping subgraph. We recall the first
example in this book.
1.10 Query as a subgraph
In the first example we showed how to model a travel problem by using a graph.
Now we show queries mat be represented as subgraphs with sliding connections h
sliding and the query respose may be obtained from a graph..
20

Suppose we want to go from Trapani to Pisa. The query graph is this subgraph:
The response of the query is the keeping graph:
To have a string as response with the calculation of the correct travel time we need
a graph programming, a programming with function to visit the graph and extract
information about the problem. In order to make the reader able to program using
graph functions, it is necessary to discuss a bit graph programming.
21

1.11 Graph programming
To implement graph programming functions we need a software library with the
following methods :
1. Editing Graph
2. Finding Subgraph
3. Keeping Subgraph
4. Surfing Subgraph
Editing a graph is a graphic tool; Finding a subgraph and keeping a subgraph are
two algorithms. Surfing subgraph is a system using cursor as pointer to point a
vertex can surf throw the edges and keep information about vertices and edges.
The complete program solves the travel example like this:
MAIN FUNCTION
1. Print “Start Form:”; Input StartPlace;
2. Print “At:”; Input StartTime
3. Print “To:”; Input ToPlace
4. Graph Query = new Graph (StartPlace, 1);
5. Query.AddConnection (1, SECONDKIND, “” , 2 );
6. Query.AddDataToNode ( 2, EndPlace );
7. Graph[] KeeptedGraph = FindAndKeepSubgraph ( MainGraph, Query );
8. For Each KeeptedGraph KG { SolvePath ( StartPlace, StartTime, KG,
ToPlace ); }
9. End
SOLVEPATH FUNCTION : String SolvePath ( String StartPlace, Time StartTime,
Graph KG, String To Place )
10. if (StartPlace==ToPlace) Return “.”;
11. Cursor P = KG.FindVertex ( StartPlace );
12. String By; Time UsefulTime
22

13. Return “From “ + StartPlace + “ by “ + GetInformation ( Cursor.moveTo
(NEXT) , StartTime, By, UsefulTime + “ At “ + UsefulTime + “ To “ +
(ArriveAt=Cursor.moveTo (NEXT).GetData()) + “\n” + SolvePath
( ArriveAt, UsefulTime, KG, ToPlace );
14. End
GETINFORMATION FUNCTION : void GetInformation ( Cursor C, Time
StartTime, String & By, Time & UsefulTime )
15. String[] Inf = C.GetData();
16. int FirstRun = null;
17. for ( int i = 0; i < inf.lentgh(); i++ ) { if ( Inf[i] == “Train” ) By= “Train”
else if ( Inf[i] == “Airbus” ) By = “Airbus”; else if ( FirstRun == null }
FirstRun = Inf[i]; else if ( FirstRun > Inf[i] ) FirstRun=Inf[i]; if
( StartTime < Inf[i] and (Inf[i] < UsefulTime or UsefultTime == null ) )
Usefultime = Inf[i]; }
18. if ( UsefulTime = null ) UsefulTime = FirstRun; //Next day
19. Return
20. End
So our example is completely solved
Start From: Trapani At: 8:00 To: Pisa
– From Trapani by Airbus at 8 a.m. To Rome, From Rome by Airbus at 12
a.m To Milan From Milan by Train at 3,30 p.m. To Pisa
23

CHAPTER 2 : SEMANTIC GRAPH
2.1 Bit definitions
A bit definition is associated to information of a single vertex or a single edge. If
we have a vertex with defined information we can add a semantic information. For
example, we can associate a unique semantic information to different vertices
:
The semantic information added means that the vertex is a delimitator.
Conventionally we indicate this concept with the simbol -|-. Bit definitions are very
fast to solve and can be applied to vertices and edges. They are really powerful in
the transcription process..
2.2 Transcription
A transctipyion is a continuous process of finding subgraphs to whom vertices and
edges are added, since we have a complete representation characterized by
impossibility to find subgraphs not transcripted before.
24

y transcription, semantic informatuin is added automatically to a graph and it is
very powerful in creating graphs with regularity that are fundamental in our
system.. Suppose we have defined a set of graphs with respective processes of
adding vertices and edges as following picture:
Based on these definitions the following graph:
is transcripted with the next one graph:
25

The first graph, the graph without transcription can be created automatically from
assertions. For example the assertion :
– Set Colours ( Red, Green, Blue )
The graph transcripted has a regular structure and a major semantics of the simple
assertion graph. Now we show how a defined graph or a transcription of a graph
can be associated to a simple expression or to a simple subgraph.
2.3 Transcription graph
A graph composed of more graphs is called “Transcription graph”. The
transcription graph definition is a list of graphs that we find in the transcription
process. By the transcription graph definition we can get all the structure of a
transcripted graph and so is more useful as we show the next paragraph.
2.4 Memory
When we find a subgraph we can link this subgraph with a graph in memory. This
is a process of associative memory and is very powerful in analysing the meaning
of an assertion. For example we can associate to the simple graph Set Colours the
entire transcription of the Set Colours shown before:
26

In this way, when we have an assertion or a representation of reality, we are able to
recall a great number of concets in order to have a good inference or an upgrade of
the knowledge of our system.
2.5 Transformation
A transformation is a process associated tipically to the recognition of a subgraph
or to a command called trigger in a session of graph programming. A
transformation modifies a graph by adding or deleting vertices and edges. The
concept of transformation is very easy to understand by an example. An example of
transformation can be the deletion of the last element of a SET.
27

Associated with assertion :
It is evident that the process of transformation is a process of graph programming
that must be implemented, but more complex processes can be obtained by the reuse
of transformation functions. For example the delection of all elements of a set
can be considered as the iteration of process of deleting the last element since the
set is empty.
2.6 Paths and circuits
If we delete a vertex of a graph we obtain a minor of a graph but in the process of
deleting vertices and edges we can compromise the structure of the graph.
28

This problem can be solved by the definition of paths and circuits. Paths,and
circuits are graphs having tipically sliding connections and their definition can
solve the problem of maintenance of the coherence of the graph. We show the
concept by an example. Suppose we want to delete the second element of a set. We
obtain a graph like the previous : If we define two subgraphs with sliding
connection like this :
We obtained the graph
By the fusion of two consecutive delimitators we have:
29

The circuits are graphs with sliding connections having circles. The concept is the
same, we can define paths, and circuits to have the maintenance of the structure of
the graph after deleting of vertices and edges.
2.6 Inverse transcription
The inverse trascription is the process to extract a string from a graph. If the graph
has a regular structure the inverse transcription can be a composed function. How
to define an inverse transcription is a problem of graph programming. Now we are
going to explain the concept. Suppose we have the graph:
We want that the output of the inverse transcription is:
– The Set Colours has the elements : red , green , blue
we can easily define a function for the inverse transcription like this:
1. Graph ITSubgraphs = FindSubgraph ( MainGraph, TranscriptionOfSet )
2. If ( ITSubgraphs == null ) Return
3. For Each ITSubgraph SET { InverseTranscriptionOfSet ( SET );
FUNCTION InverseTranscriptionOfSet ( Graph SET )
4. Cursor C = SET.FindVertex ( “Set” );
30

5. Print “The Set “; String ( C.moveTo ( “NEXT” ) ); Print “ has elements : “
6. Graph FirstElement = FindSubgraph ( FirstSetElement );
7. Cursor E = FirstSetElement.FindVertex ( “Sete”);
8. PrintElementsOfSet ( E )
FUNCTION PrintElementsOfSet ( Cursor E )
9. Print String ( E.moveTo ( “Begin” ) );
10. E = E.moveTo ( next ); if ( E == null ) return else { Print “,”;
PrintElementsOfSet ( E );
11. End
The library function String prints the String between two delimitators.
31

CHAPTER 3 : HASH OF A GRAPH
3.1 Algorithm to calculate the hash of a graph
It is possible to calculate the hash of a graph with a simple algorithm. We indicate
in this book the hash of a representation with a letter between two pipes like this:
Hash = | R |. If we use a function based on trucament we have this property :
| x 1 + x 2 + ... + x n | = | | x 1 | + | x 2 | + | ... | + | x n | |
This expression can be read as the hash of sum that is equal to the hash of sum of
the hash and introduce the commutative property of the sum to the sum of the hash.
Based on this function if we have a graph like this, we can calculate the hash on
this way :
In the first step of the algorithm we change the rows and columns and associate the
vertex idenfifier to the values, from major to minor or viceversa, of the hash of the
information linked to the vertex. We can see that there are values of hash that are
equals. In this case we must calculate all combinatory association between
32

idenfier and values. Finally we apply the hash of the hash sum calculate the final
hash value of the graph. Now we show an example of application.
In the first step we associate the identifier to the value of the hash of information
from the minor to major.
H : 0403 : O → C
H : 3789 : O → B
H : 8532 : O → A
H : 9111 : O → D
we can have a first association of vertices like this:
[1=2] , [ 2=1; 3=3; 4=5] , [5=6; 6=7], [7=4]
The matrix changes on this way :
Now we must calculate all combinatory associative combination as:
[1=2] , [ 2=3; 3=1; 4=5] , [ 5=6, 6=7] , [ 7=4 ]
[1=2] , [ 2=3; 3=4; 4=3] , [ 5=6, 6=7] , [ 7=4 ]
...
33

[1=2] , [ 4=1; 3=3; 2=5] , [ 5=7; 6=5 ] , [ 7=4 ]
We show the last combinatory association as follows. Every association as an hash
that is component of the final hash of the sum of the hash.
For the last combinatory association :
[1=2] , [ 4=1; 3=3; 2=5] , [ 5=7; 6=6 ] ,[ 7=4 ]
we have this matrix :
The final value of the hash is : | |4333| + |...| + …. + |...| + |2121| |.
3.2 Counting
Pratically, without the counting process we are not able to apply the algorithm
shown before because the combinations of combinatory process, in case we have a
number of information replied around twenty, are too much to computate. With the
counting process that consist in the association of a numeric information to the
vertex having equals information, the problem can be solved. Suppose we have the
Set graph, we can add the counting information so :
34

The counting is a graph programming process very easy to implement but having
fundamental function.
35

PART II
SPACE OF STATES
CHAPTER 4 : LINKED HASH
4.1 Introduction to linked hash
If we have a graph or a representation we know how to calculate correctly the hash,
and this representation is related to a logical context characterized by logic
transformation; we can store the evolution of the states of the logical representation
just by storing the hash value of all states and linking these values toghether.
36

Suppose, for example, we have a chessboard and we want to memorize the opening
moves for chess game. We can operate in this way :
We see that the hash value is composed of :
| | Transformation | + | Transformed Representation | |
We can delete the reppresentation and leave only the hash linked. Suppose we have
a structure like this, in which we have multiple values of the same hash related to a
different rappresentation.
37

How can we determinate if a representation is related to one of the multiple hash
values ?
An example of collision can be this :
To determinate the hash relative to actual representation we can take into account
the following logical steps. If the hash correctly corresponds the first value to |234|,
(the value on the top), then a trasformation that produces |987| must exist. If such
trasformation exists the transformed representation releated to |987| may be further
transformed to obtain |864|. It is possibile to visit the entire structure with certain
pair to :
Hs is the Hash space tipically 2 k
m
Hs m /∏ T i
i=1
, m is the number consecutive link we
analize and T i the number of possible trasformation of each step.
The error rate is equal to :
38

m
Hs m /∏ i=1
Ti −1
We can set the error according to our task, for example 10 −10
4.2 Example of discrimination
Suppose we have a situation like this :
Suppose we have a chessboard in the initial configuration.. In order to discriminate
the hash |234| we operate in this way : We calculate all possible transformation of
the representation, and apply the rule we learn above. If the representation is really
what we search, there must exist a transformation that produce |987|. The appliable
transformation produces hash |987|, |877| and again |987|. I this case we can
observe a collision. Applying the same rule we are able to discriminate in this
case :
.
39

How we can see, there exist two transformation having the desired value. For the
discrimination of the two, we calculate the transformation for each one :
Possible Transformation of : d2-d4

41
The existence of the desired transformation is evident. Now we can see that the
disired information for the other representation does not exist.
Possible Transformation of : c2-c4
We have shown how the discrimination process works. And we have to
demonstrate the formulas of centain and error that we remember are :
and
m
Hs m /∏ T i
i=1
m
Hs m /∏ i=1
Ti −1
After the analysis of ten links we have a certain around : 20 10 and the error rate
is around 1/20 10
.
41

4.3 Hash links
We can realize hash links and create a simple record for every reppresentation we
want to store in our system composed by the hash of the representation and the
hash links. We can realize hash links by the use of four prime numbers two, three,
five and seven.
Suppose we have a structure like this :
For each level of the tree we have a repetition of the sequence : two-six-three-six.
To delete the link you need operate as follow:
42

We have a connection between two representation if there exists one and only one
prime number common in the hash link and the hashes of the link sre different If
we want to delete a connection we can insert another prime number like five or
seven. The number of bits necessary for the implementation are four.
4.4 How to dimesionate the hash space
The record space is the number of records (hash) we can store in memory. It is
related to the memory and to the number of bits necessary to store a record. The
hash space must be dimensioned using this formula :
Hs≫T max∗Rs
43

T max is the maximum number of transformations applicsble to the
representation. We can call dim the ratio :
so that Hs=dim∗Rs .
dim=Hs/Rs
When we operate the hash of link the certain and the error rate change. The
certain,in the worst case, is equal to :
m
Hs/N r m /∏ i=1
2∗T i
N r is the number of records stored in a memory. If the hash space has been
dimensionated correctly, then we have:
and error rate is equal to :
If N r=R s we have:
R s / N r m ∗dim m /∏ 2∗T i i=1
R s / N r m ∗dim m /∏ 2∗T i i=1
−1
m
dim m /∏ i=1
m
m
2∗T i .
.
44

4.5 Identity Transformation
The identity transformation associate another hash value based to a transformation
that modifies the representation in order to obtain a different hash value. We
indicate the identity transformation as :
Identity=∣T R∣
In this way an inderminate number of identity hash values can be associated to a
transformation doing more transformation of tranformation. For example :
Identity=∣T T R∣
4.6 Hash links collision (T, T+1 rule)
When we store in memory a record composed by the hash of representatio and the
hash of links it could happen that the record is stored have hash representation
equals and different hash links. In this case we must not store the record in a
memory and we must apply the T, T+1 rule. This rule says that we can store the
record using an identity transformation for the hash of the representation in order
to not have collisions. So we must define an algorithm for the correct storiang. I
want to explain the algorithm with an example.
Suppose we have an empty record space.. We can store the first record as follows.
00100110010-0001
At this point we must store another record that is composed by the identity
transformation hash value, by the zero in the hash of links.
10101001010-0000.
This last record is called zero-record. The rule says that if we have a
representation R 1 its transformation R 2 is related to R 1 if and
45

only if the Identity transformation of R 2 ,is stored as a zero-record.
Suppose we have another record we want to store and we have an hash links
collision :
00100110010-0001
10101001010-0000.
00100110010-0101
in this case we cannot store the record but we store its Identity transformation and
we add a zero record with the identity transformation of the identity
transformation.
00100110010-0001
10101001010-0000.
01110110010-0101
11010101010-0000
Suppose we have the hash relative to the first record have stored in a memory.
Applying the rule we can determinate its relation. If we have the hash relative to
the second record stored in memory, the rule has negative result for the first record.
In fact the identity transformation of the second record is not a zero-record, but it is
the record we are searching. So we can consider it as hash related to the
representation if its identity transformation is a zero-record.. So we can reassume
that the T,T+1 rule, say that in order to stare a record we must to find a
transformation or an indentity trasformation with this storing property :
T R=Unique ,T 1 R=Unique
T+1(R) is the zero-record. When we want to discriminate the relation, we must
apply the same rule :
∃T R ,∃T 1 R as zero−record
46

4.7 Deletion of zero-record
When we have stored all the records in memory we can delete the zero records.
The T,T+1 rule changes and become:
∃T R not zero−record , not ∃T 1 R
Not all zero records can be deleted, Suppose we have a record:
000101001010-0000
and we cannot store the record :
000101001010-0010
so we store the records based on identity transformation and T,T+1 rule:
011110101101-0010
10001010100-0000
So the forst zero-record is not deletable, while the last zero-record is deletable. We
can mark the deletable of a record using a bit.
4.8 File System
The file system is structred as follows. The file name is the most significative bit of
the hash of the representation.
In the file we have the remaining bits of the hash of representation with the four
bits of the hash of links. The physical or logical position of the record represents
the identifier:
File Name : 1011010101111010001010111
47

01011111-00000011
11101010-00000110
11100001-00001010
11101010-00001010
The values are not sorted because the phical or logical position is the identifier, but
we can have multiple values of the same hash relative to different representation.
We know how the system works. If we want to use a database engine and a
database table to store the records, then the identifier can be a primary key.
4.6 Identity Table
We know that the hash value is composed by the sum of two hash values :
| | Transformation | + | Transformed Representation | |
So two equal representations can be a different hash because they have different
transformations. In this situation we use a table called “Identity Table”, that
indicates two hashes have the same representation. The Identity Table is so
composed :
The identity table and his processes about are fundamental for finding a solution
for loop problems in recursive process.
48

PART III
LOGICAL INFERENCE
CHAPTER 5 : Mathematic demostrations
5.1 Mathematical problem representation
Through a methematical problem representation we can solve a problem using the
mathematical logics taht is int the mathematical formulas. We are going to analize
chess inferences.
49

mathematical representation that consists in formulas about possibile moves of
pieces, and concept about the winning or losing. After we have the mathematical
representation is evident that we can apply all mathematic theories, proofs and
algebra.
We analize a chess final, the final of rook composed by the interaction of two
pieces, king and rook of a colour, and the only king of the opponent. The formulas
express that the possibile moves are :
Possible Move of King:
X k t1=X k t∏ 1
b
a
−1∗X j ∗T :1 X k t18
Y k t1=Y k t∏ −1∗Y j∗T :1Y k t18
1
with a∈[1,2]
with b∈[1,2]
with 1 X jY j2with X j∈[0,1]∧Y j∈[0,1] with T ∈[0,1]
a indicate the direction of moving to the left or to the right
b indicate the direction of up or down movement
X j indicate if we want to move on the selected direction to the left or to the
right
Y j indicates if we want to move on the selected direction up or down
1 X j Y j 2 means that we must move in one of the selected directions
T indicates if we can move or if the opponent can move
50

Possible Move of Rook:
X r t1= X r t X d ∗a∗1−T :1 X r t18
Y r t1=Y r t Y d ∗1−a∗1−T :1Y r t18
with a∈[0,1]
with−7 X d −1∨1 X d 7
with−7Y d −1∨1Y d 7
with T ∈[0,1]
a indicates the perpendicular or parallel direction
X d is a discrete number that indicates how many positions we want to move
Y d has the same function of X d
T indicate if we can move or if the opponent can move
The concept about winning or losing is the possible capture of the pieces. We have:
The King captures the Rook
X k t1=X rt ,Y k t1=Y rt
The Rook captures the King
X rt1=X k t ,Y rt1=Y k t
51

This is the representation of the problem. After the representation we are able to
deduce matematically rules for the game.
5.2 First examples of mathematical deduction
With the first example, we introduce a fundamental concept called : “Condition
sine qua non”, that is a condition taht can be used in the reasoning process. We
want to deduce when a rook can capture viceversa. For humans the concept is very
simple and obvious, but the computer needs to do a mathematical computation to
understand when it is possible.
In the first example we can find the answer of when the rook can capture the king
by solving this formula :
X r t ∗T X r t1∗1−T = X k t∗1−T X k t1∗T
and
Y r t ∗T Y r t1∗1−T =Y k t∗1−T Y k t1∗T
that for T =0 becomes :
X r t =X k t1∗T
Y r t =Y k t1∗T
and for T =1 becomes:
X r t1= X k t
Y r t1=Y k t
we must analize all the combinations of possible variable state in order to find the
conditio sine qua non.
Case T =1
52

X r t X d ∗a= X k t
Y r t Y d ∗1−a=Y k t
Case a=0
X r t =X k t This is a condition IF
Y r t Y d =Y k t
Case a=1
X r t X d = X k t
Y r t =Y k t This is a condition IF
The conditions we have found are conditio sine qua non.
We have the first answer : it is possible to capture the rook if T =1 and
X r t =X k t or Y r t =Y k t .
It is clear that for T =1 it is not possible to capture the king.
A second example is the possible capture of the rook by the king.
a
X k t ∏ 1
b
Y k t ∏ 1
−1∗X j ∗T = X r t
−1∗Y j ∗T =Y r t
we analize only the case that T =0
Case a=1 b=1 X j =1 Y j =0
53

X k t −1= X r t X k t= X r t 1
Y k t =Y r t
Case a=1b=1 X j =0Y j =1
X r t =X k t
Y r t −1=Y k t Y r t=Y k t 1
Case a=1 b=1 X j =1 Y j =1
X k t −1= X r t X k t= X r t 1
Y r t −1=Y k t Y r t=Y k t 1
Case a=2 b=1 X j =1Y j =0
X k t 1= X r t X k t= X r t −1
Y k t =Y r t
Case a=2 b=1 X j =0Y j =1
X r t =X k t
Y r t −1=Y k t Y r t=Y k t 1 equals to previous case
….
By the complete analysis of the combination we find eight positions in which the
rook can be captured. Now we can deduce that the capture is possible if :
∣X r t− X k t ∣1∧∣Y r t−Y k t∣1
After that we have the inverse concept, that is when it is not possible to capture the
roook..
∣X r t− X k t ∣1∨∣Y r t−Y k t∣1
54

X r t − X k t Y r t −Y k t
-1 0
-1 -1
0 -1
1 -1
1 0
1 1
0 1
-1 1
5.3 Further examples of matematical
deduction
How to move to defence the rook from the attack of the opponent's king ?
We know that a rook can be captured if :
∣X r t− X k t ∣1∧∣Y r t−Y k t∣1
So we can say that we choose correctly :
X d ∧Y d :∣X r t− X k t ∣1∨∣Y r t−Y k t∣1
Another way to solve this problem is to calculate a unique formula derived from
the system of two formulas :
55

X r t1= X r t X d ∗a :∣X r t1− X k t∣1
Y r t1=Y r t Y d ∗1−a:∣Y r t1−Y k t∣1
Case a=0 :
X r t1= X r t X d
X r t11 X k t X r t1=2 X k t X a with X a ≥0
2 X k t X a =X r t X d
X d =2 X k t −X r t X a
X r t1=2 X k t X r t− X r t X a
X r t1=2 X k t X a
Case a=1 :
Y r t1=Y r t Y d
Y r t11Y k t Y r t1=2Y k tY a with Y a ≥0
2Y k tY a =Y r tY d
Y d =2Y k t −Y r tY a
Y r t1=2Y k t Y r t−Y r tY a
Y r t1=2Y k t Y a
The absolute value impose to analize −X d ,−Y d we will find :
X r t1=−2X k t− X a
Y r t1=−2Y k t−Y a
This demonstration shows how we must move the rook in order not to have a
sacrifice. The picture shows the concept :
56

5.4 Structured Pieces
A rook can have a position near the opponent's king, if and only if the capture of
the rook implicates the capure of the king. I want to leave this demonstration to the
reader and write a table for the three cases represented :
X rt=−1 X k1t∧Y rt−Y k t=−1
in this case
57

X k2 t Y k2t
X k2 t= X rt−1 Y k2t=Y rt−1
X k2 t= X rt−1 Y k2t=Y rt
X k2 t= X rt−1 Y k2t=Y rt1
It is easy to demonstrate that in this case reassumed in the table, the opponent king
can move only to a=2, b=2 because if a=1∧ X j=1∨b=1∧Y j=1
we have the capture of the king. I want to demonstrate this inference because it is
fundamental in the process of finding the forcing variant I explain in the next
chapter :
X k1t1=X k1t ∏ 1
Y k1t1=Y k1t∏ 1
a 1
b 1
−1∗X j1
−1∗Y j1
a=1, X j1=1 impossible, and b=1,Y j1=1 impossible. Now we can
consider to solve the problem the position we have found. We must demonstrate
that there exists a possible move of the rook that implies the capture of the rook in
the direction we analyze. Remember that the capture of the rook occurs when :
X rt1=X k1t X r t Xd∗a= X k1t withY rt=Y k1t
or
Y rt1=Y k1tY rtYd∗1−a=Y k1t with X rt=X k1t
58

if
X rt=−1 X k1t∧Y rt−Y k1t=−1
we have
−1X k1t Xd∗a= X k1t with Y rt=Y k1t
solved for X d=1 ,a=1
and
−1Y k1tYd∗1−a=Y k1twith X rt= X k1t
solved for Yd =1 , a=0
We must analize the case : a=1, X j1=1 , in this case :
X rt= X k1t ,Y rt=Y k t−1Y d=1 , a=0 capture.
case b=1,Y j1=1
Y rt=Y k1t , X rt=X k1t−1 Xd =1,a=1 capture
case a=1, X j1=1,b=1,Y j1=1 implicates the capture of the opponent's
king. We have to demonstrate that for T =0 moves the king, and it exists a
position in witch:
X k1t1=X k1t 1
or / and
Y k1t1=Y k1t1
59
59

5.5 Reaching goal (checkmate)
Before showing how we can reach the checkmate in the rook's final , we must
demonstrate when a king cannot move in the opponent's direction. The condition
is:
with: a=1 and X j =1
X k2 t ∏ 1
b2
Y k2 t ∏ 1
a2
a
X k t ∏ 1
b
Y k t ∏ 1
−1∗X j ∗T
−1∗Y j ∗T
X k t −1∗T capture
−1∗X j2 ∗T = X k1 t−1∗1−T
−1∗Y j2 ∗T =Y k2 t∏ 1
b1
−1∗Y j1 1−T
60

The capture is possible when :
we begin analizying the first condition :
X k2 t1∏ 1
we analize for T =0
is solved for :
a2
X k2 t1= X k1 t1
Y k2 t1=Y k t1
−1∗ X j2 ∗T = X k1 t1−1∗1−T
X k2 t =X k1 t−1
b1
Y k2 t =Y k2 t∏ 1
X k2 t =X k1 t−1
Y k2 t =Y k2 t
−1∗Y j1
Indeed, to determine this last formula we analize all possible cases of combination
of the variable b 1, b 2, Y j1 , Y j2 . We can compile a table and verify the only
case possible to capture for T =0 and T =1 in all cases is
Y k2 t =Y k2 t
61

1 Y j1 Y k2 =Y k1 Y k2 =Y k1 −1 Y k2 =Y k1 1
1 0 OK OK OK
1 1 OK OK --
2 1 OK -- OK
2 0 OK OK OK
We can inference that :
Y k2 t =Y k2 t
So the goal of the game can be calculated by this reasoning :
If X k1 =8 a=2, X j1 =1 impossible
if X k2= X k1−1∧Y k2=Y k1 a=1, x j1=1 impossible
if Y r =Y k1 ± 2 X a b=1 ,Y j1 =1 impossible and b=2, y j1 =1
impossible.
This means that there does not exist a ,b , X j1 , Y j1 without capture. The
position is checkmate. The system can calculate automatically the goal of the
game.
62

PART IV
STATE ANALYSIS
CHAPTER 6 : THE PROBLEM SOLVER
6.1 States of the problem
The number of states of the problem is related to all possible configurations of the
pieces. A configuration can be expressed by the use of distance variables. In our
problem we have six significative distances like the following picture shows.
Therefore a state of the problem can be represented by a tuple :
d 1, d 2, d 3,d 4, d 5, d 6
63

All possible states are : 1310848. The next picture shows the significative
distances:
An evalutation of the number of states can be done considering that :
0≤d1≤7
0≤d2≤7
−7≤d3≤7
−7≤d4≤7
−7≤d5≤7
−7≤d6≤7
the number of the states without considering the real possible configurations can be
calculated as follow : 8 2 15 4 =3240000 . In order to obtain the only
consistent configuration of pieces we must add the following conditions:
d1∣d3∣≤7
d2∣d4∣≤7
d1∣d5∣≤7
d2∣d6∣≤7
not d3=0∧d4=0
64

not d5=0∧d6=0
not d5=d6∧d7=d8
In this way we can obtain the number of states we have mentioned above.
6.2 Invariance
Rotationg the chessboard we obtain another one that is logically related to the first
one and we can apply all knowledge about of problem to one or the other
chessboard. This concept is called invariance.
Now we choose the instance of the chessboard where d 1 ∧d 2 are inferior. We
can see that the correct move in this example is Y rt1=Y rt2 .
65

The invariance reduces the number of possible states so that the distances
d3 ,d4 become positive 0≤d3≤7 , 0≤d4≤7 .
In this way have a reduction of the number of possible states.
6.3 Problem Setting
We can represent the problem as a different sequence of possible moves of white
and black pieces. The move of the white pieces is indicated with W i , while the
move of black pieces is indicated with B i . Thus, the interaction of three pieces
two of the same colour and one of a different one can be set as follow :
W i=W 0i ,W 1i:W 0iW 1i=1∗T with w 0∈[0,1] , w 1∈[0,1]
B i=B 0i : B 0i=1∗1−T with B 0i∈[0,1]
T =m mod 2
W 0a , b , X j ,Y j indicate all possible moves of white king
W 1a , Xd ,Yd indicate all possible moves of white rook
B 0a ,b , X j ,Y j indicate all possible moves of black king
we can create two different sequences of movies, one of white moves and one of
black moves :
SW i=[W 0,W 1,... ,W m]
SB i=[B 0, B 1,... , B m]
the index indicates the move, considering T =m mod 2 the sequence has this
structure:
66

SW i=[1,0 ,0,0 ,0,1 ,0,0 ,1,0 ,... ,0,0 ,0,1]
SB i=[1 ,0 ,1 ,0 ,1 ,... ,0 ,1]
a set of possible sequences of moves of white or black pieces are indicated like
this:
SW =[SW 0, SW 1,... , SW p]
SB=[SB 0, SB 1,... , SB p]
6.4 How to demonstrate a rule
In our system we try to demonstrate some rules that can be used in reasoning by
the inference machine. A process demonstrating the rules rules is this :
∃m , SW :∀ SB i∈SB ∃ SW i∈SW goal
In other words we can affirm the existence of a number of moves “m” and a set of
possible moves of white pieces. This implicate that to every move of black piece
corresponds one or more sequences of white moves
6.5 Forcing variant
The forcing variant is the sequence of possible moves that implicate d 1=0 .
The invariance implicates that d 1=0 or d 2=0 , so the forcing variant has
as goal the condition d 1=0∨d 2=0 .Now we must demonstrate the existence
of a number of moves necessary to achieve this goal.
∃m , SW :∀ SB i∈SB ∃ SW i∈Sw d 1=0∨d 2=0
67

The problem solver finds a several number of possible sequences of moves like
this:
68

…
We can write the formulas we have found based on the distance by this sobstitution
:
X k1=8−d1 ,Y k2=8−d 2
X k2=8−d1−d3 ,Y k2=8−d2−d4
X r=8−d1−d5 ,Y r=8−d2−d6
by this substitution we can inference that the forcing variant is not the function of
variables d 1 , d 2 . For example in the case we have show in paragraph about
structured pieces we have :
X rt=−1 X k1t∧Y rt−Y k t=−1
becomes
8−d1−d5=−18−d1∧8−d2−d6−8−d2=−1
d5=1∧d6=1
69

And the same occurs for the rest of formulas. That implicates the indipendence of
the goal from variables d 1 , d 2.
6.6 Real states of the problem
At this point we can say that the states of the problem are related only to the
condiction: d 1=0∨d 2=0 and the significative distances are only
d 3, d 4,d 5,d 6 . Thus the number of states of the problem is : 10108. The
optimal solution for the rook's final in the chess game are around 10 4
states.
6.7 A mate variant
The mate variant has the condition d 1=0 , and states related to the distances
d 3, d 4,d 5,d 6 . The picture shows the sequence of correnct movies :
70

Therefore we know how the system works to find the solution of the problem. The
number of states is 10108.
6.8 Objective functions
We can analize a problem and insert the distance for the problem solution in every
state. Choosing a state means to select, from the current state,
71

the one having the inferior distance from the solution of the problem. If we have
the same states for more problems we can insert a lot of objective functions in the
states. So a state becomes :
| | Obj. Fun. 1 : Dist. 1 | + | Obj. Fun. 2 : Dist. 2 | … + | Transform. | + | Rapp. | |
6.9 All in a number
We can memorize all the object functions with relative distances, the
transformation and the representation in a number. If we want to move from a state
to another state we must to do the hypothesis related to the distance of all objective
function and not only of the function in which we are interested. In our example we
can operate like this :When we have a representation in order to find the state
stored in memory we must to do an hypothesis about the possible transformation
that produces the representation and the hypothesis of object function distances.
We do not need to use identity table for this application.
6.10 Considerations
The current memory size is terabytes. How many abilities we can store in memory
of terabytes using the methods showed in this book? For a memory of only one
terabyte I evaluete that we can store around 10000000 of abilities of medium
complexity. I think this is the moment to implement intelligent state machines.
72

PART V
HUMAN KNOWLEDGE
CHAPTER 7 : KNOW REPRESENTATION
7.1 Reasoning with graph
When we do not have a mathematic representation of a problem or it it is not
sufficient we can create a system using only graphs or an hybrid system using
graphs and mathematical representations.
73

We know that it is possible to do the hash of a graph so that a graph can be used to
represent a state. If it is possible to have transformation functions of a graph we
can link the hash of graph with the the transformed graph in order to have a state
machine that solve a problem. These methods use massively graphs and subgraphs
with sliding connections. The nature of the problem is determinable by the kind of
subgraph and it's combination found in the graph of the problem. Methods like
transcription are more useful, and we can apply all graph theories. The problem
solver by using graph is simple and it is related to abstaction of graph in witch
trannsformations are applicable.
74

CONCLUSIONS
At the beginnig of the present work we explained how a data structure like a graph
can be used for the representation of real world. If we can apply a hash operation
on a representation in order to store his evolution through different states then we
can build a stete machine. Applying mathematical logics to inference rules and
goals assists in solving mathematical representation problems. The number of
states of a problem can be reduce by identifying subgoals, rules and inference
methods. If a mathematical representation does not exist and we need knowledge
concerning the problem we can use graphs as literature to teach.
In an only server using memory of several terabytes we can store a great number of
problems solutions. Which i's the most important success of this theory. How to
build the inferential engine requires an entire book for the game game and a lot of
books for problems of different nature.
75