Isomorphespolynomial eng

Polynomial algorithm for isomorphic graphs, all cases
Mohamed mimouni
20 street kadissia 60000 Oujda Morocco
mimouni.mohamed@gmail.com
April 12, 2019
Abstract
In this paper, I propose an algorithm capable of solving the problem of isomorphic graphs
in polynomial time. First, I define a pseudo tree that allows us to define for each vertex a
label or a label. Secondly, I apply the pseudo tree for the first graph then I calculate the
labels of each vertex of the first graph, then I do the same for the second graph. Thirdly, I
look for for each graph vertex1 the graph vertices2 which have the same label, if or less a first
graph vertex its label is not in the second graph vertices we deduce that the two pseudo trees
are not isomorphic. In other cases I generate solutions and check them in polynomial time...
This algorithm therefore allows isomorphic graphs to calculate the image of each vertex in
polynomial time.
MOTS-CLES : isomorphic graphs, isomorphism, automorphism, hypergraphs, isomorphs
1

Contents
1 Deﬁnitions of the terms 3
1.1 A graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 A hypergraph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 An isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Building a pseudo tree 3
3 Label 3
4 Algorithme 3
4.1 For the graph G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
4.2 For the graph H . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
5 Mathematical proof 4
5.1 principal theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
5.1.1 proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
5.2 The algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
5.2.1 The veriﬁcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
5.2.2 The complexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
6 Algorithm complexity 4
7 Examples 5
7.1 connected graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
7.1.1 Pseudo tree of head a for graph G . . . . . . . . . . . . . . . . . . . . . . . 5
7.1.2 Label of each graph vertex G . . . . . . . . . . . . . . . . . . . . . . . . . . 5
7.1.3 Pseudo tree of head 2,4,5 and 7 for graph H . . . . . . . . . . . . . . . . . . 5
7.1.4 The labels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
7.1.5 Check the solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
8 pseudo codes 7
8.1 pseudo-arbres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
8.2 The label . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
8.3 Validate a solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.4 comment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
9 Conclusion 8
2

1 Definitions of the terms
1.1 A graph
A graph is a set of points called nodes (sometimes vertices or cells) connected by lines (segments)
or arrows called edges (or links or bows).
1.2 A hypergraph
A hypergraph is a graph whose edges connect one or two or more vertices.
1.3 An isomorphism
An isomorphism f between the graphs G and H is a bijection between the vertices of G and those
of H, such as a pair of vertices u, v of G is an edge of G if and only if (u), (v) is an edge of H.
2 Building a pseudo tree
Let’s say G a simple graph and s a vertex. The pseudo tree header s is done in the following way :
1. The vertex s in level 0.
2. The adjacents of s in level 1.
3. The adjacencies of each vertex of level 1 without repetition, which are not already in the
other levels, in the level 2.
4. Repeat step 3 to build the other levels.
5. End when there are no new adjacent ones.
If in the case (case of not connected graph) where there are still graph vertices G, which are not
in the pseudo-trees, other pseudo-trees must be generated.
3 Label
For each vertex of G a label is formed of four elements :
1. n the level it is at.
2. h the number of adjacent ones who are in the upper level.
3. m the number of adjacents who are in the same level.
4. b the number of adjacent ones who are in the lower level.
4 Algorithme
Let’s say G a graph {g1, g2, g3, , , gn} and H a graph {h1, h2, h3, , , hn}
The two graphs G and H are isomorphic if for the isomorphism function f(gi) = hi, we have the
label of gi equals the label of hi.
label(gi) = label(hi) = n − h − m − b
4.1 For the graph G
Construct one or more pseudo-trees folds so as to have a label for each vertex of the graph G.
4.2 For the graph H
Build pseudo trees and calculate labels for each vertex of H. Search in the graph G vertices that
have the same label. If at least one vertex of H has an unequal label of the labels of the vertices
of H, the header is modified and the labels of the vertices of H. If each vertex of H has a label
equal to graph labels G, we do a validation or verification of the solutions. If the solution is not
validated, the headers are changed. End if a solution is validated or check.
3

5 Mathematical proof
5.1 principal theorem
Let’s say P(a) and Q(a ) two pseudo isomorphic trees.
If f(a) = a we have:
For each vertex s of P(a) : label(s) = label(s)
5.1.1 proof
Isomorphism keeps the distance, therefore:
1. n = d(a, s) = d(a , f(s)) so f(s) is also in level n.
2. for h, there are h vertices of level n-1 adjacent to s such as d(s, th) = d(s , th) and d(a, th) =
d(a , th) = n − 1, according to the definition there is also h vertices in Q(a’)
3. for m, there are m vertices of level n adjacent to s so d(s, tm) = d(s , tm) and d(a, tm) =
d(a, tm) = n, according to the definition there are also m vertices in Q(a’).
4. for b, there are b vertices of level n+1 adjacent to s so d(s, tb) = d(s , tb) and d(a, tb) =
d(a , tb) = n + 1, according to the definition there is also b vertices in Q(a’) .
which shows that label(s) = label(s)
5.2 The algorithm
According to principal theorem we have two cases:
Case 1 label(s) = label(s ) =⇒ f(s) = s .
Case 2 label(s) = label(s ), it is necessary to check f(s)=s’ (We can’t say that f(s) = s )
5.2.1 The verification
So after finding for each vertex of G one or more images of isomorphism f, we must do one last
step: solution verification, this verification and polynomial.
5.2.2 The complexity
All the steps of the algorithm are polynomial, so the algorithm proposed in this paper and Poly-
nomial, so the isomorphism is in P.
6 Algorithm complexity
Each vertex must be in a pseudo-tree, so the complexity is O(n2
).
For the second graph the complexity can reach O(n3
).
The calculation of the labels of each vertex is O(n).
The solution validation is to O(n3
).
So this algorithm is polynomial.
4

7 Examples
7.1 connected graphs
Either the two graphs G and H, present by lists of the following adjacents:
Graph G Graph H
a:b,d,e,h 1:4,6,7,8,9
b:a,c,e,g,i 2:6,7,8,9
c:b,f,g,h,i 3:5,6
d:a,e,f,i 4:1,5,6,9
e:a,b,d,h,i 5:3,4,7,9
f:c,d 6:1,2,3,4,8
g:b,c,h 7:1,2,5,9
h:a,c,e,g 8:1,2,6
i:b,c,d,e 9:1,2,4,5,7
7.1.1 Pseudo tree of head a for graph G
Graph G levels
a level 0
b,d,e,h level 1
f,i,c,g level 2
7.1.2 Label of each graph vertex G
vertex adjacent level top even bottom label
a b,d,e,h 0 empty a b,d,e,h 0-0-0-4
b a,c,e,g,i 1 a b,d,e,h f,i,c,g 1-1-1-3
d a,e,f,i 1 a b,d,e,h f,i,c,g 1-1-1-2
e a,b,d,h,i 1 a b,d,e,h f,i,c,g 1-1-3-1
h a,c,e,g 1 a b,d,e,h f,i,c,g 1-1-1-2
f c,d 2 b,d,e,h f,i,c,g empty 2-1-1-0
i b,c,d,e 2 b,d,e,h f,i,c,g empty 2-3-1-0
c b,f,g,h,i 2 b,d,e,h f,i,c,g empty 2-2-3-0
g b,c,h 2 b,d,e,h f,i,c,g empty 2-2-1-0
explanations After ﬁnishing for graph G, we move on to graph H : Label (a)=0-0-0-0-4 so for
graph H, we test f(a)=2, f(a)=4, f(a)=5 and f(a)=7.
7.1.3 Pseudo tree of head 2,4,5 and 7 for graph H
2 4 5 7 level 0 a
6,7,8,9 1,5,6,9 3,4,7,9 1,2,5,9 level 1 b,d,e,h
3,1,4,5 3,8,2,7 6,1,2 6,8,3,4 level 2 f,i,c,g
8 level 3
We are forward to vertex 5 four levels so f(a) = 5.
7.1.4 The labels
head 2
vertex adjacent level top same bottom label images
2 6,7,8,9 0 empty 2 6,7,8,9 0-0-0-4 a
6 1,2,3,4,8 1 2 6,7,8,9 3,1,4,5 1-1-1-2 d,h
7 1,2,5,9 1 2 6,7,8,9 3,1,4,5 1-1-1-2 d,h
8 1,2,6 1 2 6,7,8,9 3,1,4,5 1-1-1-1 empty
At the vertex 8, the label (1-1-1-1-1) does not exist in graph labels G, so f(a) = 2.
head 4
4 1,5,6,9 0 empty 4 1,5,6,9 0-0-0-4 a
1 4,6,7,8,9 1 4 1,5,6,9 3,8,2,7 1-1-2-2 empty
For vertex 1 the label (1-1-2-2) does not exist in the graph labels G, so f(a) = 4.
5

head 7
7 1,2,5,9 0 empty 7 1,2,5,9 0-0-0-4 a
1 4,6,7,8,9 1 7 1,2,5,9 6,8,3,4 1-1-1-3 b
2 6,7,8,9 1 7 1,2,5,9 6,8,3,4 1-1-1-2 d,h
5 3,4,7,9 1 7 1,2,5,9 6,8,3,4 1-1-1-2 d,h
9 1,2,4,5,7 1 7 1,2,5,9 6,8,3,4 1-1-3-1 e
6 1,2,3,4,8 2 1,2,5,9 6,8,3,4 empty 2-2-3-0 c
8 1,2,6 2 1,2,5,9 6,8,3,4 empty 2-2-1-0 g
3 5,6 2 1,2,5,9 6,8,3,4 empty 2-1-1-0 f
4 1,5,6,9 2 1,2,5,9 6,8,3,4 empty 2-3-1-0 i
7.1.5 Check the solutions
vertex adjacent images images adjacent vertex adjacent state
7 1,2,5,9 a (b),(d,h),(d,h),(e) a b,d,e,h ok
1 4,6,7,8,9 b (i),(c),(a),(g),(e) b a,c,e,g,i ok
2 6,7,8,9 d (c),(a),(g),(e) d a,e,f,i no
2 6,7,8,9 h (c),(a),(g),(e) h a,c,e,g ok
5 3,4,7,9 d (f),(i),(a),(e) d a,e,f,i ok
5 3,4,7,9 h (f),(i),(a),(e) h a,c,e,g no
9 1,2,4,5,7 e (b),(d,h),(i),(d,h),(a) e a,b,d,h,i ok
6 1,2,3,4,8 c (b),(d,h),(f),(i),(g) c b,f,g,h,i ok
8 1,2,6 g (b),(d,h),(c) g b,c,h ok
3 5,6 f (d,h),(c) f c,d ok
4 1,5,6,9 i (b), (d,h),(c),(e) i b,c,d,e ok
explanations In this table we have f(2) = h and not d. and f(5) = d and not h.
The check to give 2 errors, we must then do another check
vertex adjacent images images adjacent vertex adjacent state
7 1,2,5,9 a (b),(,h),(d),(e) a b,d,e„h ok
1 4,6,7,8,9 b (i),(c),(a),(g),(e) b a,c,e,g,i ok
2 6,7,8,9 h (c),(a),(g),(e) h a,c,e,g ok
5 3,4,7,9 d (f),(i),(a),(e) d a,e,f,i ok
9 1,2,4,5,7 e (b),(,h),(i),(d),(a) e a,b,d„h,i ok
6 1,2,3,4,8 c (b),(,h),(f),(i),(g) c b,f,g„h,i ok
8 1,2,6 g (b),(,h),(c) g b,c„h ok
3 5,6 f (d),(c) f c,d ok
4 1,5,6,9 i (b),(d),(c),(e) i b,c,d,e ok
explanations Now with no errors so both graphs G and H are isomorphic and here is the ﬁnal
solution:
vertex Image
a 7
b 1
c 6
d 5
e 9
f 3
g 8
h 2
i 4
6

8 pseudo codes
8.1 pseudo-arbres
Here is a pseudo code that generates a pseudo tree.
Algorithm 1: generate a pseudo tree P(a) head a for a graph G
Data: A graph G and a vertex a of G
Result: A pseudo tree P(a)
1 initialization; the vertex a in the level 0;
2 the adjacent from a in level 1;
3 the vertices of G in array graph[];
4 delete the vertices of the level 0 and 1 array graph[];
5 niv ← 1 ;
6 adjacents ← adjacent to a ;
7 while graph[] is not empty do
8 initialize array list[] empty ;
9 niv ← niv + 1 ;
10 for every element of adjacent do
11 adjacents1 ← the adjacents of element ;
12 adjacents1 ← the intersections between adjacent1 and graph[] ;
13 liste[] ← list[]+adjacents1 ;
14 liste[] ← liste[] order and without duplication ;
15 graph[] ← the diﬀerence between graph[] and list[] ;
16 if graph[] is empty then
17 break;
18 end
19 end
20 if lise[] is empty then
21 break;
22 else
23 list in level niv ;
24 end
25 end
For unconnected graphs, it is necessary to generate other pseudo-trees for the other remaining
vertices.
8.2 The label
Here’s a pseudo code that calculates the label of a vertex in a pseudo tree.
Algorithm 2: Calculate the label of a vertex s in a pseudo tree
Data: A vertex s in a pseudo tree
Result: s label for a pseudo tree
1 n ← determine the level of s;
2 nivh ← determine the vertices that are in the level n+1;
3 nivm ← determine the vertices that are in the level n;
4 nivb ← determine the vertices that are in the level n-1;
5 h ← the number of adjacent vertices s that are in nivh;
6 m ← the number of adjacent vertices s that are in nivm;
7 b ← the number of adjacent vertices s that are in nivb;
8 label(s) ← "n-h-m-b";
7

8.3 Validate a solution
Here’s a pseudo code to check out a proposed solution.
Algorithm 3: Check a solution
Data: A proposed solution
Result: validated or erroneous solution
1 initialization;
2 err ← 0;
3 for Each f(a)=a’ proposed do
4 adjacent(a) ← the adjacent vertices a;
5 adjacent(a ) ← the adjacent vertices a’;
6 for each adjacent to a do
7 x ← an adjacent vertex adjacent a;
8 x ← the images of x by isomorphism;
9 Intersection ← the intersection between the images of x by isomorphism and the
adjacent ones of a’;
10 if intersection is empty then
11 err ← err+1;
12 break;
13 end
14 end
15 if err = 0 then
16 break;
17 end
18 end
19 if err = 0 then
20 "The solution is validated";
21 else
22 "The solution is not validé";
23 end
8.4 comment
For the not connected graphs it is necessary of course to study the isomorph between the connected
components.
9 Conclusion
In this paper I presented a version of a polynomial algorithm, which gives us the images of each
vertices by isomorphic bijection, or to say that the two graphs are not isomorphic.
8

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