1 Number Theory and

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: 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 ∈ V(X1), {u,v} ∈ E(X1) if and only if { f (u), f (v)} ∈ 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! |G| [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 if and only if it is a 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 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 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 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 |V(Y)| + 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 ∈ V(X) and g ∈ 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 ∈ V(X) and g ∈ 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 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.