HAL Id: hal-01822100
https://hal.archives-ouvertes.fr/hal-01822100v5
Preprint submitted on 26 Jan 2020
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Algorithme polynomial pour les graphes isomorphes, tous les cas
Mohamed Mimouni
To cite this version:
Mohamed Mimouni. Algorithme polynomial pour les graphes isomorphes, tous les cas. 2020. �hal-
01822100v5�
Polynomial algorithm for isomorphic graphs, all cases
Mohamed mimouni
20 street kadissia 60000 Oujda Morocco mimouni.mohamed@gmail.com
January 26, 2020
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
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 graphG . . . 3
4.2 For the graphH . . . 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 graphG . . . 5
7.1.2 Label of each graph vertexG . . . 5
7.1.3 Pseudo tree of head 2,4,5 and 7 for graphH. . . 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
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 isomorphismf between the graphsGandHis a bijection between the vertices of Gand those of H, such as a pair of verticesu,vof Gis an edge of Gif and only if (u), (v) is an edge of H.
2 Building a pseudo tree
Let’s sayGa simple graph andsa vertex. The pseudo tree headersis done in the following way : 1. The vertexsin level 0.
2. The adjacents of sin level 1.
3. The adjacencies of each vertex of level 1 without repetition, which are not already in the other levels, in thelevel 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 ofGa label is formed of four elements : 1. nthe level it is at.
2. hthe number of adjacent ones who are in the upper level.
3. mthe number of adjacents who are in the same level.
4. bthe number of adjacent ones who are in the lower level.
4 Algorithme
Let’s sayGa graph{g1, g2, g3, , , gn} andHa graph{h1, h2, h3, , , hn}
The two graphsGandH are isomorphic if for the isomorphism functionf(gi) =hi, we have the label ofgi equals the label ofhi.
label(g) =label(h) = − − −
5 Mathematical proof
5.1 principal theorem
Let’s sayP(a) andQ(a′) two pseudo isomorphic trees.
Iff(a) =a′ we have:
For each vertexsof 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 leveln.
2. forh, there arehvertices oflevel n-1adjacent tossuch asd(s, th) =d(s′, t′h) andd(a, th) = d(a′, t′h) =n−1, according to the definition there is alsohvertices inQ(a’)
3. form, there arem vertices of level nadjacent tos so d(s, tm) =d(s′, t′m) and d(a, tm) = d(a, t′m) =n, according to the definition there are alsomvertices inQ(a’).
4. for b, there are bvertices of level n+1 adjacent to sso d(s, tb) = d(s′, t′b) and d(a, t′b) = d(a′, t′b) =n+ 1, according to the definition there is alsobvertices inQ(a’) .
which shows thatlabel(s) =label(s)
5.2 The algorithm
According toprincipal theoremwe have two cases:
Case 1 label(s)6=label(s′) =⇒f(s)6=s′.
Case 2 label(s) =label(s′), it is necessary to check f(s)=s’ (We can’t say thatf(s) =s′)
5.2.1 The verification
So after finding for each vertex of Gone or more images of isomorphismf, we must do one last step: solution verification, this verification andpolynomial.
5.2.2 The complexity
All the steps of the algorithm are polynomial, so the algorithm proposed in this paper and Poly- nomial, sothe isomorphism is in P.
6 Algorithm complexity
Each vertex must be in a pseudo-tree, so the complexity isO(n2).
For the second graph the complexity can reachO(n3).
The calculation of the labels of each vertex isO(n).
The solution validation is toO(n3).
So this algorithm is polynomial.
7 Examples
7.1 connected graphs
Either the two graphsGandH, present by lists of the following adjacents:
GraphG GraphH 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 GraphG 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 graphG, we move on to graphH: Label (a)=0-0-0-0-4 so for graphH, 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 sof(a)6= 5.
7.1.4 The labels
vertex adjacent level top same bottom label images
head 7
vertex adjacent level top same bottom label images
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 havef(2) =hand not d. andf(5) =dand 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 graphsGandH 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
8 pseudo codes
8.1 pseudo-arbres
Here is a pseudo code that generates a pseudo tree.
Algorithm 1:generate a pseudo treeP(a) heada for a graphG Data: A graphGand a vertexa of G
Result: A pseudo treeP(a)
1 initialization; the vertexa in thelevel 0;
2 the adjacentfrom a inlevel 1;
3 the vertices of Gin arraygraph[];
4 delete the vertices of thelevel 0and1arraygraph[];
5 niv←1 ;
6 adjacents←adjacent toa;
7 whilegraph[]is not empty do
8 initialize arraylist[]empty ;
9 niv←niv+ 1 ;
10 forevery elementof 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[]andlist[];
16 if graph[] is emptythen
17 break;
18 end
19 end
20 if lise[] is emptythen
21 break;
22 else
23 list in levelniv;
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 sin a pseudo tree Data: A vertexsin a pseudo tree
Result: slabel 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 leveln;
4 nivb←determine the vertices that are in the level n-1;
5 h←the number of adjacent verticessthat are innivh;
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 forEach f(a)=a’ proposed do
4 adjacent(a)←the adjacent verticesa;
5 adjacent(a′)←the adjacent verticesa’;
6 foreach adjacent to ado
7 x←an adjacent vertex adjacenta;
8 x′ ←the images of xby isomorphism;
9 Intersection←the intersection between the images of xby 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 err6= 0then
16 break;
17 end
18 end
19 if err= 0then
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.
