Abstract

The isoperimetric number of a graph , denoted by , was introduced by Mohar (1987). A graph and a subset of its vertices are given, and let denote the edge boundary of , the set of edges which connects vertices in to vertices not in . The isoperimetric number of is defined as . In this paper, some results about the isoperimetric number of graphs obtained by graph operations are given.

1. Introduction

Given a graph and a subset of its vertices, let denote the edge boundary of that is, the set of edges which connects vertices in with vertices not in . The isoperimetric number is defined as Clearly, can be defined in a more symmetric form by where the minimum runs over all partitions of into nonempty subsets and , and are the edges between and .

As examples of isoperimetric numbers, we consider the following.(i)The isoperimetric number of the complete graph with vertices is .(ii)The isoperimetric number of the cycle with vertices is .(iii)The isoperimetric number of the path with vertices is .(iv)The isoperimetric number of the complete bipartite graph with vertices is It can be briefly shown as .

The isoperimetric number is also closely related to the notion of bisection width, , of a graph . This is the minimum number of edges that must be removed from in order to split into two equal-sized (within one if the number of vertices in is odd) subsets: where . If known, one can use the isoperimetric number of a graph to establish a lower bound for its bisection width using the fact that

See [1].

The importance of lies in various interesting interpretations of this number by Mohar as follows [2].(a)From (2), it is evident that, in trying to determine , we have to find a small edge-cut separating as large a subset (assume ) as possible from the remaining larger part . So, it is evident that can serve as measure of connectivity of graphs. It seems that there might be possible applications in problems concerning connected networks,and the ways to “destroy” them are by removing a large portion of the network by cutting only a few edges.(b)The problem of the partitioning into two equally sized subsets (to within one element), in such a way that the number of the edges in the cut is minimal, is known as the bisectionwidth problem. It is important in VLSI design and some other practical applications. Clearly, it is related to isoperimetric number.

Theorem 1 (see [2]). Some of the theorems that Mohar stated are given below.(a) if and only if is disconnected.(b)If is k-edge-connected then .(c)If is the minimal degree of vertices in then .(d)If is an edge of and then .(e)If is the maximum vertex degree in then . If G has a cycle with almost half the vertices of G then .

If a set with reaches the minimum we call it an isoperimetric set. For denoted by , the subgraph of is induced on [2].

Proposition 2 (see [2]). If is a connected graph then it has an isoperimetric set such that and are connected subgraphs of .

In the next section, we prove a upper bound for isoperimetric number of lexicographic product of graphs.

2. Lexicographic Product

The lexicographic product of two graphs and has its vertex set with adjacent to if either adjacent to in or and are adjacent to in . Note that unlike the union, join, and Cartesian product, this operation is not commutative.

Theorem 3. Let be a graph with vertices, and let edges and be a graph with vertices. Then,

Proof. Let and be the isoperimetric set of and edge boundary of . We know that includes copies of . If then . Hence, So,
Similarly, let with and . We know that includes copies of . If then we have . Therefore, . So, . The function takes its minimum value at . Since then . We have The proof is completed by (8) and (9).

We have according to the upper bounds of , and hence we get .

Corollary 4. Let be a graph with vertices, and let edges and be a graph with vertices. If is even and and then

Proof. If is even and then . Hence, we have . Since then . So, we have . Therefore, we have

Corollary 5. Let be a graph with vertices, and let edges and be a graph with vertices. If is odd and then

Corollary 6. Let be a tree with vertices, and let be a graph with vertices. If is even then

Proof. For and , we know that . Then, . Since and then we have . Therefore, we have

Corollary 7. Let be a tree with vertices, and let be a graph with vertices. If is odd and then .

Corollary 8. Let be a graph with vertices, edges that is not a tree, and let be a path graph with vertices. If is even and then

Proof. If is even and then . Thus, . Therefore, . So,

Corollary 9. Let be a graph with vertices, edges that is not a tree, and let be a path graph with vertices. If is odd and then

Corollary 10. Let be a graph with vertices, edges that is not a tree, and let be a cycle graph with vertices. If is even and then

Proof. If is even and then So,

Corollary 11. Let be a graph with vertices, edges that is not a tree, and let be a cycle graph with vertices. If is odd and then

Corollary 12. Let be a graph with vertices, edges that is not a tree and let be a complete graph with vertices.(a)If is even and then (b)If is odd and then

Proof. (a) If is even and then we have The proof for even is very similar to the proof for odd.

Theorem 13. Let be a path graph with vertices, and let edges and be a graph with vertices. Then,

Proof. Let and with . For let . Hence, is the disjoint union of . Let and . To prove this theorem, we have three cases.
Case  1. Let be an even integer. To prove this case, we have three subcases.
Subcase 1.1. If where then . Therefore, . The function has its minimum value at , and we have
Subcase 1.2. If and and where then . Therefore, . The function has its minimum value at . If then . Thus,
Subcase 1.3. If or or and where then Thus, . The function has its minimum value at , and we have
By (26), (27), and (29), if is even then .
Case  2. Let be an odd, and let be an even integer. To prove this case, we have four subcases.
Subcase 2.1. If where then . Therefore, . The function has its minimum value at , and we have
Subcase 2.2. If and and where then . Therefore, . The function has its minimum value at . If then . Thus,
Subcase 2.3. If and and where then . Therefore, . The function has its minimum value at . Thus,
Subcase 2.4. If or or and where then Thus, . The function takes minimum value at , and we have
By (30), (31), (32), and (34), we have that if is odd and is even and then , and if is odd and is even and then .
Case  3. The proofs of the case in which and are odd are similar to that of Case 2.

The isoperimetric number of is given in the following corollary.

Corollary 14. Let and be positive integers. Then,

The isoperimetric number of is given in the following corollary.

Corollary 15. Let and be positive integers. Then,

The isoperimetric number of is given in the following corollary.

Corollary 16. Let and be positive integers. Then,

The isoperimetric number of is given in the following corollary.

Corollary 17. Let and be positive integers. Then,

The isoperimetric number of is given in the following corollary.

Corollary 18. Let and be positive integers. Then,

The isoperimetric number of is given in the following corollary.

Corollary 19. Let and be positive integers. Then,