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Inglês Português

The concept of graph expression was created in 2015 by Glauber Donizeti Gasparotto Junior and aims at the representation of a graph in the form of a mathematical expression.

The concept aims to explore the benefits of a mathematical expression swapping numbers for entities. With this, we can create a new way to carry data and especially create a new kind of research in complex or circular graphs.

It is important to highlight that the concept does not have the goal of being better or worse than other existing ones. The goal is to be just a new way to view a graph and its information.

A graph is composed of 4 basic elements and various information that we will see in this document.

Example 1:

(A + B + C + D)

The elements that make up an expression are:

  • Entity: it is the fundamental element of expression, determines a unit, a vertex in graph theory.
    • Are unique, but may appear "N" times in expression and in different positions.
    • Are represented by a literal, in the above case, the letters: "A", "B", "C" and "D".
  • Sum operator: is the element that adds an entity in another entity.
    • Making an analogy with the theory of graphs, the "+" operator can be seen as an edge.
  • Subtraction operator: is the element that removes an entity to another entity.
  • Parentheses: are used to group the daughters of entities an entity determines.
    • In graph, expression are named: expression Group.

These elements are the same as a mathematical expression, the difference is that in place of numbers we have entities that will be added or removed. In addition, the goal of the result has its differences.

This expression represents the following graph:

A 
----B
----C
----D

The resolution is always from left to right, where the entity left adds or removes the entity from the right and the result of this sum is the entity itself on the left, and so on until you reach the end of the expression.

Simple example (symbolic of the resolution Steps):

  1. (A + B)
  2. Final result of the expression:A

Final graph of the entityA

A 
----B

Complex example (symbolic of the resolution Steps):

  1. (A + B + C + D)
  2. (A + C + D)
  3. (A + D)
  4. Final result of the expression:A

Final graph of the entityA

A 
----B
----C
----D

We saw that each step resolution "entity" disappears and the "left" prevails until the entities remain your right.

It is obvious that each step resolution left entity is changed internally, it adds the right entity.

In a graph, the entities are unique, but they can be in several places at once. For example, there are no two entities with the same name. But the same entity may appear at various points in the graph.

(A + (B + C + A) + C)

Note that the entities "A" and "C" are repeated. They represent the same entity, but in different positions. Each instance contains some information that are unique to that position. We will see this in the topic Information of an occurrence.

The operation of sum uses the "+" operator, as stated, it works as an edge that connects a vertex to any other vertex. In expression of graph, we say that the entity of the "left" add the entity of "right" and without restrictions, for example:

  • The entity of the left can add yourself as many times as it takes:
Expression: A + A + A + A
Graph:
            A 
            ----A
            ----A
            ----A
  • The entity X can add the entity "Y" and "Y" can add the entity "X" as many times as needed.
Expression: X + (Y + X + X) + Y
Graph:
            X 
            ----Y
                ----X
                ----X
            ----Y

The subtraction operation uses the "-" operator. In expression of graph, we say that the entity of the left removes the entity from the right causing the right entity ceases to be your daughter.

For each subtraction operation, only one occurrence will be removed at a time, even if the entity on the left has more than one child of the same entity. For example:

  • The entity of the left removes one of the daughters "B"
Graph 1:
            A 
            ----B
            ----B
            ----B

Expression: A - B

Graph 2:
            A 
            ----B
            ----B

Note that one of the occurrences of the entity "B" was removed from "the entity". Based on the same example, if we wanted to remove all occurrences of the entity "B" would have to do the subtraction operation 3 times, which is equivalent to the amount of times that entity "B" exists within the "entity".

It is still possible to mix the operations of addition and subtraction.

Graph 1:
            A 
            ----B
            ----B
            ----B

Expression: A - B - B - B + (C + Y)

Graph 2:
            A
            ----C    
                ----Y

In this example, remove all occurrences of the entity "B" the entity "A" and add a new daughter ("C") that contains the entity "Y".

The groups of expression is delimited by parentheses: ( to open and ) to close.

The first entity of the Group (after opening parentheses) determines the parent entity of that group, that is, all subsequent entities will be their daughters until they close the parentheses.

Example 1:

(A + B + C)
  • The entity "A" is the parent entity of your expression and the entity group "B" and "C" are their daughters.

Example 2:

(A + B + (C + D))
  • The entity "A" is the parent entity of your expression and the entity group "B" and "C" are their daughters.
  • The "C" entity is the parent entity of your expression and the entity group "D" is your daughter.

The first group of expression is called the root expression group.

It is not mandatory the use of parentheses in the expression. We will see that in the following examples both expression are correct:

(A + B)

Or

A + B

A group of expression can contain other groups of expression within it and the logic is the same for the Sub-group:

(A + B + (C + D))

In this example the entity "A" will be father of "B" and "C" and "C" will be father of entity "D".

We call it "statement" the first moment in which an entity is written, i.e. your first occurrence.

If this entity contains children should declare all your group of expression at the same time, namely, adding his children within the parentheses.

There is no obligation for the Declaration of the Group at the first occurrence, but this helps to simplify the discovery of some information in a faster way.

For example, to find out if the entity "B" contains children in the following expression, you will need to check all of its instances, as it is not possible to say which events your group was declared.

A + B + (C + (B + D)) + B

Now, if we know that groups of expressions were written in the first instances, so we can check only the first occurrence of the entity "B" to see if it contains or not children:

A + (B + D) + (C + B) + B

A group of expression cannot be declared more than once in the expression.

For example:

A + B + (C + D + E) + (I + C)
  • The entity "C" have children "D" and "e"
  • The entity "I" has a daughter to "C" entity, however it is not necessary to redeclare the daughters of entities "C".

Wrong:

A + B + (C + D + E) + (I + (C + D + E))

The parent entity is always the first expression group, she who gives rise to the graph of that group.

For example:

(A + B + (C + D))

  • In this example, we have two entities father: "A" and "C".
  • The element "+" is used as a symbol for an entity (son) in another entity (parent).

The first entity of the expression is the root entity . An expression can only contain a root entity.

A + B + (C + A)
  • The entity "A" is the root of the expression entity and will be the top of the graph.

An entity that does not have expression groups in your level is called the end-entity. This does not mean that the entity does not have children, see:

End entity without children:

(A + B + C + (D + E))
  • The entities "B", "C" and "e" are end entities.

End entity with children:

(A + (B + C) + (D + B))
  • The entity "C" is final and does not contain children
  • The last instance of entity B, expression of the entity group "D", but it also contains children.

All entity contains a path that must be traversed to get to your position. To represent this way we can use the following notation:

A.B.C.D

This notation indicates the location of the entity "D" inside the expression below:

A + (B + (C + D))
  • The entity "D" is the daughter of entity "C"
  • The entity "C" entity's daughter "B"
  • The entity "B" is the daughter of "the entity"

The notation uses the "." character between the parent entity and the child entity. The entity of the left will be the father and the son right.

Other examples:

Expression:

(A + A + (B + C) + (D + B))

Paths of A entity:

  • 1 Occurrence:A
  • 2 Occurrence:A.A

In the second instance we have a cyclic relationship, so the notation is interrupted when this happens, otherwise we'd have an infinite path.

Paths of B entity:

  • 1 Occurrence:A.B
  • 2 Occurrence:A.D.B

When an entity is father of herself, or a descendant entity is the father of some entity ascendant, that determines that there is a cyclic path between the entities. In this case, the expression must only repeat the name of the entity, that is enough to discover that there is a cyclical situation.

Note that the graph contains two cyclic paths:

A + A + B + (C + A)
  • A ( A + A ): where the entity "A" is her own father.
  • A hint ( C + A ): Where "C" is the father of a rising entity in case the entity "A".

An entity may appear several times within an expression for each occurrence and we have a set of information that will be seen in this topic.

This information is very important and we will see examples of this in the topic Research.

An expression has two types of levels: General level and Level in the expression.

The General level is the level in relation to the graphic hierarchy. The level starts at "1" and "+1" increments until you reach the last level.

For example:

A (Level: 1)
----B (Level: 2)
    ----C (Level: 3)
    ----D (Level: 3)
        ----B (Level: 4)
----E (Level: 2)
    ----A (Level: 3)

The level in the expression determines in what level the entity is related to the expression. The level starts at "1" and "+1" increments until you reach the last level.

For example:

                        A + B + C + ( D + E + ( F + G ) )
Level in expression:    1   1   1     2   2     3   3    
Level:                  1   2   2     2   3     3   4   

Note that the level of expression is very similar to General level. The only difference is the value of the parent entity, the General level this number is always less than the General level of their children and at the level of the expression they are equal.

An expression has two types of indexes: Index on expression and the Index level.

The index of the expression determines what position the entity is about expression. The index starts at "0" and "+1" increments until you reach the last entity of the expression.

For example:

A + B + C + ( D + E + ( F  + G ) ) 
0   1   2     3   4     5    6

The index of the level determines what position the entity is in relation to your level. The index starts at "0" and "+1" increments until you reach the last peer entity.

For example:

                A + B + C + ( D + E + ( F + G + Y ) )
Level:          1   2   2     2   3     3   4   4
Level Index:    0   0   1     2   0     1   0   1

Graph:

A (Level Index: 0)
----B (Level Index: 0)
----C (Level Index: 1)
----D (Level Index: 2)
    ----E (Level Index: 0)
    ----F (Level Index: 1)
        ----G (Level Index: 0)
        ----Y (Level Index: 1)
  • The entity "A" is the root of the expression and your "index" level is zero. Note that the root entity, she will have no other entities in your level.
  • The entity "B" is the first of the second level and have zero position. She is the daughter of the entity "A".
  • The entity "C" is the second of the second level and have the position 1. She is the daughter of the entity "A".
  • The entity "D" is the third of the second level and have the position 2. She is the daughter of the entity "A".
  • The entity "and" is the first of the third level, and will have the position 0. She is the daughter of entity "D".
  • The entity "F" is the second of the third level, and will have the position 1. She is the daughter of entity "D".
  • The entity "G" is the first of the fourth level and have the 0 position. She is the daughter of entity "F".
  • The entity "Y" is the second of the fourth level and will have the position 1. She is the daughter of entity "F".

All entity, with the exception of the last expression, is aware of the next entity in the expression.

In the example below, we have a map of knowledge to all entities to the right of the current entity:

A + B + C + ( D + E + ( F + G ) )
B   C   D     E   F     G

In the example, the entity "A" has knowledge of the entity. Note that the entity "(B)" daughter of "the", but that doesn't influence, because the idea is to know the next entity of the expression and not your level.

All entity, with the exception of the first of the expression (the root entity) has knowledge of the previous entity in the expression. In the example below, we have a map of knowledge of all entities on the left of the current entity:

A + B + C + ( D + E + ( F + G ) ) 
    A   B     C   D     E   F

The normalizations are designed to improve the visualization of expressions.

The standardization of type 1 aims to wipe away expression groups that belong to the same parent entity and that are in different places in the expression.

For example:

A + (B + Y) + (D + (B + C))
     ^              ^

Note that in the expression above, the "B" has two groups of expression in different places. In practice, this has no problem, but will be visually better if we apply the standardization by eliminating one of the principal groups "B", see:

A + (B + Y + C) + (D + B)

It must be said that no changes in expression must modify the your final graph. It is noticeable that in the example that did not occur, just organizations were reorganized.

In the next example, we will see an expression that can lead to confusion at the time of normalization:

A + (B + Y) + (D + (B + Y))
     ^              ^

In this example, it is natural to think that one of the principal groups "B" can be eliminated because they are equal, but that thinking is wrong. If we eliminate one of the groups, we will be modifying the final graph and this is not the point.

Wrong:

A + (B + Y) + (D + B)

Correct:

A + (B + Y + Y) + (D + B)

The standardization of type 2 aims to organize, where possible, end entities at the beginning of your expression group to aid in the visualization of expression.

A + (B + (C + D) + E) + F + G
                   ^    ^   ^

After the normalization would look like this:

A + F + G + (B + E + (C + D))
    ^   ^        ^    
  • Note that the entities "F" and "G" were for the beginning of your group.
  • The entity "and" also was reorganized to the beginning of your group.

The standardization of type 3 aims to declare as soon as possible all the groups of expressions. This theme was also discussed in the topic Entity declarations.

Example:

A + B + (C + G + (B + F)) + (G + F)
    ^             ^    
             ^               ^

Note that the entities "B" and "G" are used before their groups are declared and after standardisation we have:

A + (B + F) + (C + (G + F) + B) + G
  • After normalization, the groups of "B" and "G" were declared at the first moment they were used.
  • The entity "B" within the group "C", and "G" it's lonely at the end of the speech, turned to End entity and because of this, we can apply Standardisation-2 type to improve viewing, see:
A + G + (B + F) + (C + B + (G + F))
  • Note that now the entity "G" that was at the end of the expression was moved to the beginning. Therefore, we must apply again to Standardization-3 type:
A + (G + F) + (B + F) + (C + B + G)

With this we conclude the normalization and obtain an expression more readable.

The goal of denormalization is to generate a new expression where the groups of expressions are declared every time your parent entity is used.

After the denormalization will be impossible to go back to the original expression, this process.

Consider the following expression:

A + (B + D) + (E + B)
  • Note that the "B" has two parents: "A" and "e"
  • After the denormalization will have the following expression:
A + (B + D) + (E + (B + D))
                    ^
  • After the denormalization the entity "B" had your expression group redeclared when was used again as the son of entity "D".

As stated, it is impossible to go back to the original expression, because we can't distinguish which groups were expressions of the original expression. Therefore, we cannot say that an original expression is equal to your expression denormalized.

See an example of how they are different:

Original:       A + (B + D) + (E + B)
Final Graph:
                A
                ---B
                ------D
                ---E
                ------B

If we take the expression denormalized and extract the your graph, we will have a graph differs from the original graph:

Original:                       A + (B + D) + (E + (B + D))
After normalization of type 1:  A + (B + D + D) + (E + B)
Final Graph:
                                A
                                ---B
                                ------D
                                ------D
                                ---E
                                ------B

Therefore, we cannot consider that a denormalized expression is used as an original expression, this changes the final graph. Besides, she breaks rule topic Repetitions of expression group.

Research on expression of graph can be divided into two types: superficial Research and deep Research.

The next few topics we will address the difference between these types of searches, but first, you must understand what a array of information. This is a common issue among both types of searches.

We can represent an expression of graph in a vertical array with all information of an expression.

With the vision in the form of array we got a better view of the graph and understand better how does research on complex graphs using the concept of expression of graph.

Let's see an example:

Expression:

Expression:     A + B + ( C + Y ) + ( D + E + ( F + ( G + B + C ) + Y ) + Z )
Level:          1   2     2   3       2   3     3     4   5   5     4     3
Level Index:    0   0     1   0       2   0     1     0   0   1     1     2

Hierarchy:

A (Level Index: 0)
----B (Level Index: 0)
----C (Level Index: 1)
    ----Y (Level Index: 0)
----D (Level Index: 2)
    ----E (Level Index: 0)
    ----F (Level Index: 1)
        ----G (Level Index: 0)
            ----B (Level Index: 0)
            ----C (Level Index: 1)
        ----Y (Level Index: 1)
    ----Z (Level Index: 2)

Array of information:

Index   | Entity | Level | Level Index
#00     | A      | 1     | 0
#01     | B      | 2     | 0
#02     | C      | 2     | 1
#03     | Y      | 3     | 0
#04     | D      | 2     | 2
#05     | E      | 3     | 0
#06     | F      | 3     | 1
#07     | G      | 4     | 0
#08     | B      | 5     | 0
#09     | C      | 5     | 1
#10     | Y      | 4     | 1
#11     | Z      | 3     | 2

Note that the expression changed from horizontal to vertical orientation and all entities were stacked on each other and respecting the same order as they had in the expression.

Indeed, this is an important rule: Never change the order of the rows, that completely changes the graph.

The elements of sum and parentheses were removed, they are not needed, because only with the indexes and information levels, it is possible to identify all the groups of expressions.

And is based on this array of information and knowledge of neighbouring entities, i.e. those that are positioned on your left or your right, regardless of the level, we can create searches and navigations.

The deep search is intended to return the largest possible amount of results and that she considers all paths that an entity traverses in a graph.

In order to create a deep research, we need to use a denormalized expression. This is necessary, because only the denormalized expression contains all paths that an entity has in the graph since the original version of the expression does not repeat the expression groups (and nor should it).

Let's follow the same example used in the topic Array of information, but now, the expression was denormalized:

Expression:

Original:       A + B + ( C + Y ) + ( D + E + ( F + ( G + B + C ) + Y ) + Z )
                          ^                                   ^
Denormalized:   A + B + ( C + Y ) + ( D + E + ( F + ( G + B + ( C + Y ) ) + Y ) + Z )
                                                                    ^
Level:          1   2     2   3       2   3     3     4   5     5   6       4     3 
Level Index:    0   0     1   0       2   0     1     0   0     1   0       1     2

Hierarchy:

A (Level Index: 0)
----B (Level Index: 0)
----C (Level Index: 1) 
    ----Y (Level Index: 0)
----D (Level Index: 2)
    ----E (Level Index: 0)
    ----F (Level Index: 1)
        ----G (Level Index: 0)
            ----B (Level Index: 0)
            ----C (Level Index: 1)
                ----Y (Level Index: 0) *
        ----Y (Level Index: 1)
    ----Z (Level Index: 2)
  • Was applied to denormalization and the entity "C" had your group of expression within the entity "redeclared G".
  • After the denormalization a new path was created for the entity "Y":
    • Before:
      • First occurrence:A.C.Y
      • Second occurrence:A.D.F.G.Y
    • After:
      • First occurrence:A.C.Y
      • Second occurrence:A.D.F.G.C.Y
      • Third occurrence:A.D.F.G.Y

Denormalized Array:

See how was the denormalized expression in the form of array:

Index   | Entity | Level | Level Index
#00     | A      | 1     | 0 
#01     | B      | 2     | 0 
#02     | C      | 2     | 1 
#03     | Y      | 3     | 0 
#04     | D      | 2     | 2 
#05     | E      | 3     | 0 
#06     | F      | 3     | 1 
#07     | G      | 4     | 0 
#08     | B      | 5     | 0 
#09     | C      | 5     | 1 
#10     | Y *    | 6     | 0
#11     | Y      | 4     | 1 
#12     | Z      | 3     | 2 
  • A new line was created with respect to original version: the line "#10" contains the new path.

On superficial Research do not consider the paths that have already been declared (or driven), i.e. does not get applied to the denormalization to create these new paths. This greatly reduces the time of the survey, but in some cases will not have the same accuracy of deep Research.

For example, if we want to return all occurrences of the entity "Y", we would have the following difference between the types of searches:

Example expression:

A + B + ( C + Y ) + ( D + E + ( F + ( G + B + C ) + Y ) + Z )

Deep search:

First, apply the denormalization:

A + B + ( C + Y ) + ( D + E + ( F + ( G + B + ( C + Y ) ) + Y ) + Z )
  • First occurrence:A.C.Y
  • Second occurrence: A.D.F.G.C.Y -> new path
  • Third occurrence:A.D.F.G.Y

Surface search:

Uses the original expression:

  • First occurrence:A.C.Y
  • Second occurrence:A.D.F.G.Y

In this type of research we have no entity as a reference and the search will be made throughout the array.

As there are endless research option within a graph, we will discuss just a few examples of unreferenced research.

To find the root entity of the expression, we need to return the entity that has the General index equal to "0".

Attention: This search shows no differences between the two types of research: Research deep and superficial Research.

Based on the following expression, we can affirm that the entity "A" is the root entity .

        A + B + C
Index:  0   1   2

To find all parent entities of the graph, we must apply the following procedure:

  1. Retrieve the previous entities of all entities whose level index is equal to "0".
  2. For each line found, return to your previous entity that will always be a parent entity.

Attention: This research presents differences in types: deep Search and superficial Research. However, the deep research can return duplicate entities in cases of groups of expressions that were declared more than once. You will need to remove the duplication.

Deep search

We'll use in this example the denormalized array of topic about Deep search.

  1. First, we find all rows with the index of the level equal to zero:
  • #00 (A)
  • #01 (B)
  • #03 (Y)
  • #05 (E)
  • #07 (G)
  • #08 (B)
  • #10 (Y)
  1. For each line found, return to your previous entity that will be a parent entity:
  • NULL-> #00 (A) : contains no previous entity, therefore does not return anything.
  • #00 (A)-> #01 (B) : Returns the entity "A" as your previous
  • #02 (C)-> #03 (Y) : Returns the entity "C" as your previous
  • #04 (D)-> #05 (E) : Returns the entity "D" as your previous
  • #06 (F)-> #07 (G) : Returns the entity "F" as your previous
  • #07 (G)-> #08 (B) : Returns the entity "G" as your previous
  • #09 (C)-> #10 (Y) : Returns the entity "C" as your previous

With that, after we remove the repetitions (in this case, the "C" entity that appears in the lines "#2" and "#09"), we obtain as a result "entities", "C", "D", "F" and "G" as being the only entities with children in the expression.

Superficial research

The logic is the same as the deep search, however we will not have duplication because on superficial research there are no groups of repeated expressions.

The reference research assumes that the entity or one of its occurrences already found. On this basis, we can take actions such as: checks, navigations or research on their ascendants and descendants.

As there are endless research option using an entity, we will discuss just a few examples of research with references.

To find out if an entity is the first of your group (first within the parentheses), check if your General level is lower than the General level of the next entity, if it is, it is the first of your group of expression.

Attention: This search shows no differences between the two types of research: Research deep and superficial Research.

        A + B + ( C + Y ) + (D + C)
                  ^
Level:  1   2     2   3      2   3
Index:  0   1     2   3      4   5

In the example above, the "C" #02 "," index has the General level equal to "2" and your next entity "Y" has the same general level "3", and that is why she is the first inside your parentheses.

Note:

Do not confuse this technique as being the solution to verify that an entity contains children. We will see this in the topic Finding all the descendants of an entity.

To find out if an entity is the last of your group (last inside the parentheses), check whether your General level is greater than the General level of the next entity, if it is, it is the last of your group.

Attention: This search shows no differences between the two types of research: Research deep and superficial Research.

        A + B + ( C + Y ) + (D + C) + U
                      ^
Level:  1   2     2   3      2   3    2
Index:  0   1     2   3      4   5    6

In the example above, the "Y" of the index "#03", has the same general level "3" and your next entity "D" has the General level equal to "2", and that is why she is the last inside your parentheses.

  • The entity "U" of the index "#06" does not have a next entity, so she is the last of your expression group, even though he is omitted because we're in the root expression group.

To return the previous entity of a given entity, we must subtract the your General index minus one (" -1").

Attention: This search shows no differences in types: deep Search and surface Research

We'll use in this example the denormalized array of topic about Deep search.

  1. To obtain the previous entity entity "Y" the line "#03", we've got your general index ("3"), and subtract " -1". With the result ("2"), found in the array, the entity that is in that position, in this case, one goes back to "C" entity.
Index   | Entity | Level | Level Index
#02     | C      | 2     | 1 
#03     | Y      | 3     | 0 
  • If the result is less than zero, is because we're in the root entity and there is no previous authority.

To return the next entity of a given entity, we must add the your General index plus one ("+1").

Attention: This search shows no differences in types: deep Search and superficial Research.

We'll use in this example the denormalized array of topic about Deep search.

  1. To get the next entity entity "Y" the line "#03", we've got your general index ("3") and add "+1". With the result ("4"), found in the array, the entity that is in that position, in this case, one goes back to entity "D".
Index   | Entity | Level | Level Index
#03     | Y      | 3     | 0 
#04     | D      | 2     | 2 
  • If the result is greater than the maximum quantity of items in the array is the last expression entity and there is no next entity.

To find all occurrences of an entity, we have to traverse the entire array from index "0" to last position in the array.

Attention: This research presents differences in types: deep Search and superficial Research. However, the deep search may return a larger quantity of occurrences. This is because, in this type of research, the groups of expressions are declared every time the parent entity is used.

Therefore, it is recommended to use the deep search if your need is to obtain the largest possible number of paths.

Deep search

We'll use in this example the denormalized array of topic about Deep search.

  1. If we want to get all occurrences of "Y" entity within the graph, we would find the lines:
  • #03 (Y)
  • #10 (Y): This occurrence is derived from denormalization.
  • #11 (Y)

Superficial research

The logic is the same as the deep search, however we will not have any occurrences resulting from redeclarações of expression.

We'll use in this example the original array of topic about Array of information.

  1. If we want to get all occurrences of "Y" entity within the graph, we would find the lines:
  • #03 (Y)
  • #10 (Y)
  • Note that was found an occurrence unless on deep research.

If we want to find the descendants of an entity, if the General level is lower than the General level of the next entity, if it is, that entity is a descendant of the current entity. This is the same technique used in the topic Verifying that an entity is the first expression Group (first within the parenthesis).

Should we continue navigating forward until the next entity has the General level equal or lower the General level of the current entity or if the expression does not have more entities.

Attention: This search can be done using the two types of research: Research deep and superficial Research. However, there are different approaches for each of them. In addition, we should have a special treatment for entities containing an ascendant of the entity itself, namely a cyclic path.

Entity with cyclic path:

We must take some care to find the descendants of entities with cyclic paths. This is because the groups of expressions cannot be declared again.

For example, how can we find the descendants of the "entity" that is at index "#05"?

        A + B + (C + Y) + (D + A + C)
                               ^
Level:  1   2    2   3     2   3   3
Index:  0   1    2   3     4   5   6
  • The "entity" that is at index "#05" was not declared again to avoid a cyclic path.
  • Note that the entity "A" contains descendants (is the root entity), but it's impossible to know that analyzing only the occurrence of the index "#05".

The answer would be:

  • Find all occurrences of the entity "A".
  • Among the occurrences found, we must find and use the first containing descendants and ignore the others.
    • 1 Occurrence:
      • #00: The entity "A" has the General level equal to "1".
      • #01: The entity "B" is the next entity after "A" and your General level is "2", is descended.
      • Ready! We found the occurrence which has the Declaration of the entity's expression group "A".
    • 2 Occurrence:
      • #05: You do not need to check the second occurrence of the entity "," because I've found the your statement.
  • Return the descendants of "the entity" the index "#00":
    • #00: The entity "A" has the General level equal to "1".
    • #01: The entity "B" is the next entity after "A" and your General level is "2", is descended.
    • #02: The entity "C" is the next entity after "B" and your General level is "2", is descended.
    • #03: The entity "Y" is the next entity after "C" and your General level is "3", is descended.
    • #04: The entity "D" is the next entity after "Y" and your General level is "2", is descended.
    • #05: The entity "A" is the next entity after "D" and your General level is "3", is descended.
    • #06: The entity "C" is the next entity after "A" and your General level is "3", is descended.
    • Just the expression
    • The following were found: A, B, C, Y, D, A, C .
  • Remove occurrences that are duplicated:C
  • Return the result:A, B, C, Y, D, A

Deep search

If an entity does not have a cyclic path, we can simply continue the search for descendants of the current occurrence, since it is guaranteed that your group of expression was redeclared.

Superficial research

On superficial research, some care is needed. Note that in the expression below in a similar scenario to the scenario of entities with cyclic paths.

For example, how can we return the descendants of the "C" entity "#02 index"?

        A + B + C + (D + A + (C + Y)) + Z
                ^              
Level:  1   2   2    2   3    3   4     2
Index:  0   1   2    3   4    5   6     7
  • The "C" entity that is at index "#02" was not declared again, because we are using the superficial research.
  • This expression is not normalized, the entity "C" should have been declared as soon as possible, but this did not occur.
  • The "C" entity contains descendants. His expression is declared in the index "#05".

In this case we have two options:

Option 1:

Using the same logic that was explained to entities with cyclic paths. With that evaluates all occurrences of the entity "C" until we find the occurrence that declares the Group of your expression.

  • Would be found occurrence of the index "#05" and "#02 index" would be discarded.
  • Now that we found the correct occurrence, we should return the descendants:
    • #05: The entity "C" has the General level of "3".
    • #06:The entity "Y" is the next entity after "C" and your General level is "4", is descended.
    • #07: The entity "Z" is the next entity after "Y" and your General level is "2", she's not a descendant.
    • The expression did not finish, but was interrupted after the negative result of the index "#07".
    • The following entity was found: Y .
  • Remove occurrences that are duplicated, in this case, we didn't have any.
  • Return the result:Y

Option 2:

The second option can present a better performance if the expression is already normalized, if this is granted, we don't have to perform the first step.

  • Apply Standardization-3 type to ensure that all entities are being declared the first use. This step is not necessary if the expression is already normalized.
        A + B + (C + Y) + (D + A + C) + Z
                 ^              
Level:  1   2    2   3     2   3   3    2
Index:  0   1    2   3     4   5   6    7
  • Find the first occurrence of the entity "C". After normalization, we will find the occurrence you're in the index "#02".
  • Retrieve the descendants of the first occurrence of the entity "C" index "#02".
    • #02: The entity "C" has the General level of "2".
    • #03: The entity "Y" is the next entity after "C" and your General level is "3", is descended.
    • #04: The entity "D" is the next entity after "Y" and your General level is "2", she's not a descendant.
    • The expression did not finish, but was interrupted after the negative result of the index "#04".
    • The following entity was found: Y .
  • Remove occurrences that are duplicated, in this case, we didn't have any.
  • Return the result:Y

Finally, it is possible to say that we don't need to assign a special treatment for entities with cyclic paths if we were using a superficial research. We have seen that the solution is the same in both situations.

This theme was also addressed, superficially, in the topic Entity declarations.

To start this topic you need to understand the topic Finding all the descendants of an entity.

The logic is exactly the same as the descendants search, the only difference is the General level: [General current entity-level] + 1

We'll use in this example the denormalized array of topic about Deep search.

Based on this array, if we want to find all the daughters of entity "D" the line "#04":

  • The entity "D" has the General level of "2".
  • The entity "and" is the next entity after "D" and your General level is 3, is the daughter of "D".
  • The entity "F" is the next entity after "and" and your General level is also 3, is the daughter of "D".
  • The next entities after "F" are: "G", "B", "C", "Y" and "Y", all levels greater than 3, then will be ignored.
  • The entity "Z" is the next entity after "Y" and your General level is also 3, is the daughter of "D".

The term has come to an end and at the end we will have the result:E, F, Z

To find the upside of an entity, we must check if the previous entity has your General level is lower than the General level of the desired entity. If you have, this entity is an ascendant.

                A + B
Level:          1   2
                ^   *
Parent of B:    A

If the entity is the same level of the entity you want to, you should ignore it and continue navigating backwards until you find the first entity with the General level lower than the General level of the desired entity.

                A + B + J
Level:          1   2   2
                ^       *
Parent of J:    A

After finding the first ancestor, one should continue navigating back, but the General level to be considered now is the first ancestor and not more of the desired entity. This process should continue until you reach the root entity.

                A + B + (J + Y)
Level:          1   2    2   3
                ^        ^   *
Parents of Y:   J, A

Attention: This research has differences in types: deep Search and superficial Research. However, the deep search may return a larger quantity of occurrences. This is why this type of search expressions groups are declared every time the parent entity is used.

For example, if we want to catch the ascendants of the entity "C" considering all its instances:

First occurrence:

  • The entity "C" of "#02" has the General level of "2".
  • #01: The entity "B" has the General level of "2". It's not a relative in the ascending line.
  • #00: The entity "A" has the General level equal to "1" (lowest) is therefore the first ascending (parent entity). Now the level to be considered will be the "level 1" and no longer "level 2".

The term came to an end and we will have the following ancestors:A

Second occurrence:

  • The entity "C" of "#09" has the General level of "5".
  • #08: The entity "B" has the General level of "5" is not a.
  • #07: The entity "G" has the General level of "4" (is less than) is therefore the first ascending (parent entity). Now the level to be considered will be the level "4" and no longer "level 5".
  • #06: The entity "F" has the General level of "3". She has the General level less than the entity "G" is therefore an ascendant. Now the level to be considered will be the "level 3" and no longer "level 4".
  • #05: The entity "and" has the General level of "3". It's not an ascendant.
  • #04: The entity "D" has the General level of "2". She is an ascendant. Now the level to be considered will be the "level 2" and "3".
  • #03: The entity "Y" has the General level of "3". It's not an ascendant.
  • #02: The entity "C" has the General level of "2". It's not an ascendant.
  • #01: The entity "B" has the General level of "2". It's not an ascendant.
  • #00: The entity "A" has the General level equal to "1". She is an ascendant. Now the level to be considered will be the "level 1" and no longer "level 2".

The term has come to an end and we will have the following ancestors:G, F, D, A

Following the logic of the topic Finding all the ascendants of an entity, to find just the father of the "Y" entity, we would need to limit the overall level of the ancestors: [General current entity-level]-1; or the first entity with the General level less than the desired entity.

Attention: This research presents differences in types: deep Search and superficial Research. However, the deep search may return a larger quantity of occurrences. This is because, in this type of research, the groups of expressions are declared every time the parent entity is used.

As there are 3 occurrences of the entity "Y", we will have a parent entity for occurrence:

First occurrence:

  • The entity "Y" the line "#3" has the General level of "3".
  • #02: The "C" entity is the entity before "Y" and has the General level of "2", therefore, she is the parent of the entity "Y".

Second occurrence:

  • The entity "Y" the line "#10" has the General level equal to "6".
  • #09: The "C" entity is the entity before "Y" and has the General level of "5", so she is the parent of the entity "Y".

Third occurrence:

  • The entity "Y" the line "#11" has the General level of "4".
  • #10: The entity "Y" has the General level equal to "6". It's not an ascendant.
  • #09: The entity "C" has the General level of "5". It's not an ascendant.
  • #08: The entity "B" has the General level of "5". It's not an ascendant.
  • #07: The entity "G" has the General level of "4". It's not an ascendant.
  • #06: The "F" entity is the entity entity's previous "G" and has the General level of "3", so she is the parent of the entity "Y".

This topic will demonstrate in practice some examples of implementations of some of the concepts we study.

Use the C# programming language due to your ability to overload mathematical operators.

In this example we will demonstrate how to create a graph using Graph expression. We will do this in the most simple and objective as possible.

Will be used a circular entity, i.e., an entity that relates to herself.

[DebuggerDisplay("{Name}")]
public class Entity : List<Entity>
{
    public string Name { get; private set; }
    public Entity(string identity) => this.Name = identity;

    public static Entity operator +(Entity a, Entity b)
    {
        a.Add(b);
        return a;
    }

    public static Entity operator -(Entity a, Entity b)
    {
        a.Remove(b);
        return a;
    }
}
  • The class inherits from a generic list of the class itself, our intention is to create a cyclic instance.
  • The class requires a name as input parameter, is the name of the entity
  • The operators "+" and "-" have been overwritten, now it can be used within an expression.
    • When there's a sum ("+"), the entity of the right will be added in the list of the entity from the left, and the left will be returned as a result. This is the basis of the concept of expression of graph.
    • When there is a minus sign ("-"), the entity of the right will be removed from the list of the entity from the left, and the left will be returned as a result.

To use is simple, just use as if it were a mathematical expression:

class Program
{
    static void Main(string[] args)
    {
        var A = new Entity("A");
        var B = new Entity("B");
        var C = new Entity("C");
        var D = new Entity("D");
        var E = new Entity("E");
        var F = new Entity("F");
        var Y = new Entity("Y");
        var H = new Entity("H");

        // expression1
        A = A + B + (C + (D + E + F)) + (Y + H);

        // expression2
        D = D - E;
    }
}

After running the first expression, we have the following graph:

A
----B
----C
    ----D
        ----E
        ----F
----Y
    ----H

After the implementation of the second term, we see that the "D" no longer has the entity "and" as a daughter, she was subtracted/removed:

A
----B
----C
    ----D
        ----F
----Y
    ----H

Note that the expression is exactly the same as all the expressions that we saw during this study. This shows that you can enjoy circular entities of this concept without using large blocks of code.

For more complex entities, would not be as simple as the use of operators. There would be the need to create mechanisms for reflection and string for the creation and processing of the expression.

In this example we will show how to convert an array of information back to expression of graph.

It is important to note that this code is simple and specific to our example. Although it can be useful for several purposes due to your ability to identify the correct start and end times of an iteration of an entity.

[DebuggerDisplay("{Entity.Name}")]
public class EntityItem
{
    private readonly Expression expression;

    public EntityItem(Expression expression)
    {
        this.expression = expression;
    }

    public int Index { get; set; }
    public int IndexAtLevel { get; set; }
    public int Level { get; set; }
    public int LevelAtExpression { get; set; }
    public Entity Entity { get; set; }

    public EntityItem Previous { get => expression.ElementAtOrDefault(Index - 1); }
    public EntityItem Next { get => expression.ElementAtOrDefault(Index + 1); }
    public EntityItem Parent
    {
        get
        {
            var previous = this.Previous;
            while(previous != null)
            {
                if (previous.Level < this.Level)
                    return previous;
                previous = previous.Previous;
            }
            return null;
        }
    }
}
public class Expression : List<EntityItem>
{
    public string ToExpressionAsString()
    {
        var parenthesisToClose = new Stack<EntityItem>();
        var output = "";
        foreach (var item in this)
        {
            var next = item.Next;
            var isFirstInParenthesis = next != null && item.Level < next.Level;
            var isLastInParenthesis = next == null || item.Level > next.Level;
            var isNotRoot = item.Index > 0;

            if (isNotRoot) output += " + ";

            if (isFirstInParenthesis)
            {
                output += "(";
                parenthesisToClose.Push(item);
            }

            output += item.Entity.Name.ToString();

            if (isLastInParenthesis)
            {
                int countToClose;

                if (next == null)
                    countToClose = parenthesisToClose.Count;
                else
                    countToClose = item.Level - next.Level;

                for (var i = countToClose; i > 0; i--)
                {
                    parenthesisToClose.Pop();
                    output += ")";
                }
            }
        }

        return output;
    }
}

class Program 
{
    static void Main(string[] args)
    {
        var A = new Entity("A");
        var B = new Entity("B");
        var C = new Entity("C");
        var Y = new Entity("Y");
        var D = new Entity("D");
        var E = new Entity("E");
        var F = new Entity("F");
        var G = new Entity("G");
        var Z = new Entity("Z");

        var expression = new Expression();
        expression.Add(new EntityItem(expression) { Entity = A, Index = 0, IndexAtLevel = 0, Level = 1 });
        expression.Add(new EntityItem(expression) { Entity = B, Index = 1, IndexAtLevel = 0, Level = 2 });
        expression.Add(new EntityItem(expression) { Entity = C, Index = 2, IndexAtLevel = 1, Level = 2 });
        expression.Add(new EntityItem(expression) { Entity = Y, Index = 3, IndexAtLevel = 0, Level = 3 });
        expression.Add(new EntityItem(expression) { Entity = D, Index = 4, IndexAtLevel = 2, Level = 2 });
        expression.Add(new EntityItem(expression) { Entity = E, Index = 5, IndexAtLevel = 0, Level = 3 });
        expression.Add(new EntityItem(expression) { Entity = F, Index = 6, IndexAtLevel = 1, Level = 3 });
        expression.Add(new EntityItem(expression) { Entity = G, Index = 7, IndexAtLevel = 0, Level = 4 });
        expression.Add(new EntityItem(expression) { Entity = B, Index = 8, IndexAtLevel = 0, Level = 5 });
        expression.Add(new EntityItem(expression) { Entity = C, Index = 9, IndexAtLevel = 1, Level = 5 });
        expression.Add(new EntityItem(expression) { Entity = Y, Index = 10, IndexAtLevel = 1, Level = 4 });
        expression.Add(new EntityItem(expression) { Entity = Z, Index = 11, IndexAtLevel = 2, Level = 3 });
        var expressionString = expression.ToExpressionAsString();
    }
}

The Main method we have to call our function. Note that we are creating the array of information manually. This array must represent the following expression:

(A + B + (C + Y) + (D + E + (F + (G + B + C) + Y) + Z))

The function ToExpressionAsString will be responsible for making all the iteration and arrive at our goal: return a string containing our expression.

  • The class Expression represents a graph expression as a whole. This class inherits from a EntityItem type list. That approach this class of the concept: a set of entity instances with your information.
  • The method ToExpressionAsString returns a string that will be our expression.
  • The list (containing all instances of entities) will be covered completely. The zero position until the end of the list. Each iteration can contain various levels of expression.
  • The variable parenthesisToClose stores a list of all the brackets that were opened and need to be closed. The list has to be in the format: last in, first out.
  • For each iteration:

With these code snippets we saw how simple it is to iterate in a graph expression and understand their moments. In addition to open paths to more complete implementations such as: research on expression of graph.

In the previous example we saw how to generate a graph from an array of manual and information that was represented by the class Expression .

In this example, let's discuss an implementation that creates this array.

public class Expression : List<EntityItem>
{
    public bool Deep { get; }

    public Expression(Entity root, bool deep = true)
    {
        Deep = deep;

        if (root != null)
            Build(root);
    }

    private void Build(Entity parent, int level = 1)
    {
        // only when is root entity
        if (Count == 0)
        {
            var rootItem = new EntityItem(this)
            {
                Entity = parent,
                Index = 0,
                IndexAtLevel = 0,
                LevelAtExpression = level,
                Level = level
            };

            Add(rootItem);
        }

        var indexLevel = 0;
        var parentItem = this.Last();

        level++;
        foreach (var child in parent.Children)
        {
            var previous = this.Last();
            var childItem = new EntityItem(this)
            {
                Entity = child,
                Index = Count,
                IndexAtLevel = indexLevel++,
                Level = level,
            };

            Add(childItem);

            // if:   IS 'deep' and the entity already declareted in expression, don't build the children of item.
            // else: if current entity exists in ancestors (to INFINITE LOOP), don't build the children of item.
            var continueBuild = true;
            if (Deep)
                continueBuild = !HasAncestorEqualsTo(childItem);
            else
                continueBuild = !IsEntityDeclared(childItem);

            if (continueBuild && child.Children.Count() > 0)
            {
                childItem.LevelAtExpression = parentItem.LevelAtExpression + 1;
                Build(child, level);
            }
            else
            {
                childItem.LevelAtExpression = parentItem.LevelAtExpression;
            }
        }
    }

    private bool HasAncestorEqualsTo(EntityItem entityItem)
    {
        var ancestor = entityItem.Parent;
        while (ancestor != null)
        {
            if (entityItem.Entity == ancestor.Entity)
                return true;

            ancestor = ancestor.Parent;
        }

        return false;
    }

    private bool IsEntityDeclared(EntityItem entityItem)
    {
        return this.Any(e => e != entityItem && e.Entity == entityItem.Entity);
    }

    public string ToMatrixAsString()
    {
        var s = "";
        s += "Index    | Entity  | Level    | Level Index     | LevelAtExpression \r\n";

        foreach (var i in this)
        {
            s += $"{i.Index.ToString("00")}       ";
            s += $"| {i.Entity.Name}       ";
            s += $"| {i.Level.ToString("00")}       ";
            s += $"| {i.IndexAtLevel.ToString("00")}              ";
            s += $"| {i.LevelAtExpression.ToString("00")} \r\n";
        }
        return s;
    }
}

class Program 
{
    static void Main(string[] args)
    {
        var A = new Entity("A");
        var B = new Entity("B");
        var C = new Entity("C");
        var Y = new Entity("Y");
        var D = new Entity("D");
        var E = new Entity("E");
        var F = new Entity("F");
        var G = new Entity("G");
        var Z = new Entity("Z");
        A = A + (B + (C + A) + A) + (D + D + E + (F + (G + A + C) + Y) + Z) + G;
        var expression = new Expression(A, false);
        var matrix = expression.ToMatrixAsString();
    }
}

The method ToMatrixAsString will be used to check the result of our example. And after processing of the graph of entity "A", we have the following array of information:

Index    | Entity  | Level    | Level Index     | LevelAtExpression 
00       | A       | 01       | 00              | 01 
01       | B       | 02       | 00              | 02 
02       | C       | 03       | 00              | 03 
03       | A       | 04       | 00              | 03 
04       | A       | 03       | 01              | 02 
05       | D       | 02       | 01              | 02 
06       | D       | 03       | 00              | 02 
07       | E       | 03       | 01              | 02 
08       | F       | 03       | 02              | 03 
09       | G       | 04       | 00              | 04 
10       | A       | 05       | 00              | 04 
11       | C       | 05       | 01              | 04 
12       | Y       | 04       | 01              | 03 
13       | Z       | 03       | 03              | 02 
14       | G       | 02       | 02              | 01
  • The class gets into your constructor, the root entity. From this instance, let's navigate in your graph.
  • The Deep parameter determines if scan will be deep or not and what was explained in the topicDeep search
  • The first if inside the function Build checks to see if it is the root entity, if it is, we must create the first item. At this point, the information is fixed (for being the root entity), will be the initial values.
  • In the second part of the function, we start reading the children of the entity parent .
  • Will be incremented +1 in General level as deepens in the entity. This value is passed by parameter, because it transcends the entire graph.
  • Will be incremented +1 in the index level. This value is closed only in foreach scope, that is, only for the children of the entity.
  • For each iteration, it is checked if the property Deep is true , if it is, we should keep the navigation even if current entity has already been covered in some point of the expression. However, the only situation that limits the continuation is if the current entity has relationships with herself in one of their ancestors. If so, stops the continuation.
  • If the property Deep is false , so we should just check if the entity has already been covered at some point of the term, if it was, then continue.
  • Property LevelAtExpression (expression level) is populated with the expression level of the parent entity. Added "+1" when the entity has children and doesn't add anything when I haven't.

With that, we found the three main examples of the concept and that can be the basis for more complex implementations such as research in graph expression.


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