1 Number Theory and Graph Theory Chapter 6 Basic concepts and definitions of graph theory By A. Satyanarayana Reddy Department of Mathematics Shiv Nadar University Uttar Pradesh, India E-mail: [email protected] 2 Module-5: Graph isomorphism and automorphism group of a graph Objectives • Isomorphic graphs • Self complementary graphs • Automorphism group of a graph and asymmetric graphs The graph isomorphism problem can be easily stated: check to see if two graphs that look differently are actually the same. • The problem of determining whether two given graphs are isomorphic is an old problem and it has many applications, e.g., in chemistry, listing all molecules in a certain class. All attempts to find a collection of easily computable graph parameters that are sufficient to distinguish any pair of non-isomorphic graphs have failed. • This problem occupies a rare position in the world of complexity theory. This problem is clearly in NP but is not known to be in P and it is not known to be NP-complete. Many sub-disciplines of mathematics, such as topology theory and group theory, can be brought to bear on the problem, and yet only for special classes of graphs polynomial-time algorithms been discovered. Definition 1. Two graphs X1 = (V1;E1) and X2 = (V2;E2) are said to be isomorphic (to each other) if there is a function f : V(X1) ! V(X2) such that • f is one-one, • f is onto, and • for each pair of vertices u;v 2 V(X1), fu;vg 2 E(X1) if and only if f f (u); f (v)g 2 E(X2). 3 The following pairs of graphs are isomorphic. c1 c1 c2 c2 3 4 c d c5 c5 c c 1 2 a b 3 3 C4 K2;2 c4 c4 1. We can often show that two simple graphs are not isomorphic by showing that they do not share a property called invariant under isomorphism. For example, If X1 and X2 are isomorphic, then both have the same • number of vertices, • number of edges, • degree sequences, etc. 2. The relation isomorphism between graphs is an equivalence relation. Given a graph X, the set of graphs isomorphic to X is called isomorphism class of X. The isomorphism classes partition the set of graphs with vertex set V. The size of isomorphism class containing X is n! jGj [by orbit-stabilizer theorem], where G is the automorphism group of the graph X, which we define shortly. Example 2. Is the graph K3;3 isomorphic to a prism graph (see the graph given below) on 6 vertices. Solution:It is easy to see that both the graphs have the same number of vertices and the same 4 number of edges. Also, both have the same degree sequence as both are 3-regular. But, they are not isomorphic as the girth of K3;3 is 4, whereas the girth of the prism graph is 3 (or K3;3 is bipartite but the prism graph is not bipartite). It is interesting to show that every 3-regular graph on six vertices is isomorphic to one of these graphs. Example 3. Find all non-isomorphic graphs on four vertices. Solution:There are 11 graphs with four vertices which are not isomorphic. In other words any graph with four vertices is isomorphic to one of the following 11 graphs. Exercises 4. A connected graph is isomorphic to its line graph if and only if it is a cycle graph. Definition 5. A simple graph X is called self complementary if X and Xc are isomorphic. The following are the first few examples of self complementary graphs. 5 5 3 4 3 4 3 4 1 1 2 1 2 1 2 K1 P4 C5 A-graph Lemma 6. If a simple graph on n vertices is self complementary, then show that 4 divides n(n − 1). Proof. Let X be a self complementary graph on n vertices. Then X is isomorphic to its complement. c c So, the number of edges in X and X are equal, say k. Further X [ X = Kn, the complete graph n(n−1) with vertices. Hence, 2k = 2 . Or equivalently n(n − 1) = 4k: 5 • The converse of the above result is not true. The graph C4 satisfies 4 divides 12 = n(n − 1) but is not self-complementary. • Let X be a self complementary graph on n vertices. Then 4 divides n(n−1). But the numbers n and n − 1 are coprime and hence either 4 divides n or 4 divides n − 1. That is, either n is of the form 4k or 4k + 1: Exercises 7. 1. For what values of n the cycle graph Cn is self complementary? 2. Can a simple graph with 7 vertices be isomorphic to its complement? 3. Show that the line graph of K3;3 is self complementary. 4. Show that Payley graphs are self complementary. Procedure to produce some self-complementary graphs Step-1 Let P4 be a path graph with vertices v1v2v3v4. Let Y be any self-complementary graph. Step-2 Join vertices v2 and v3 of degree 2 of P4 to all the vertices of Y. Then it is easy to see that the resulting graph will have jV(Y)j + 4 vertices and is self-complementary. For example, we know that the cycle graph C5 is self complementary. Hence, using the above procedure, the following graph is a self complementary graph on 9 vertices. 6 The problem that two graphs are isomorphic or not can be converted to finding an automorphism group of a graph. 0.1 Automorphism of a graph Definition 8. Automorphism of a graph: An isomorphism from a graph X = (V;E) to itself is called an automorphism of X. • An automorphism is therefore a permutation of the vertices of X that maps edges to edges and non edges to non edges. • G = Aut(X) = the set of all automorphisms of X. It can be easily verified that G forms a group and is called the automorphism group of X. • A graph is said to be asymmetric if its automorphism group consists of only the identity. The following graph is the smallest asymmetric graph. • Verify that Aut(Kn) = Sn = Aut(K¯n) = Aut(K1;n) and Aut(Cn)=D2n (the dihedral group con- sisting of 2n elements). • For any x 2 V(X) and g 2 Aut(X), note that deg(x) = deg(g(x)). Thus, the automorphism group of a graph permutes the vertices of equal degree among themselves. • If x;y 2 V(X) and g 2 Aut(X) then d(x;y) = d(g(x);g(y)), where d(x;y) is the length of the shortest path between x and y. • It can be easily verified that Aut(X) = Aut(Xc). That is, the automorphism group of a graph is equal to that of its complement graph. 7 • If X is a directed graph then the automorphism preserves the direction of the edges as well. Example 9. We now give an example of a regular asymmetric graph. l k i g d j f h e a b c Solution:Consider the graph given above. We now prove that every automorphism fixes all the vertices. To do that, we provide special properties of each vertex. • There are exactly two triangles ade and i jk. • g is the only vertex having no neighbor on a triangle. • l is the only vertex with neighbors on both the triangles. • f is the unique neighbor of both g and l. Thus, if we fix these three vertices, namely f ;g and l, then it is easy to show that other vertices will also get fixed. Consider two algorithmic questions. Graph isomorphism Instance: Graphs X and Y. Question: Is X =∼ Y? Note that it is sufficient to consider both graphs to be connected. As automorphism group of graph is equal to its complement. Further for every graph Z either Z or its complement is connected. 8 Automorphism Group Instance: A graph X Output : Finding automorphisms of X. First problem is a decision problem and the second requires output. Both problems are related. First problem can be reduced to the second. Let Z be a graph which is a disjoint union of X and Y: Then Z has two connected components X and Y. If we were able to compute automorphism group of the graph Z and if that contains a permutation which interchanges two components of Z then X and Y are isomorphic..