# Self-Complementary Graph

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Let G be a graph. Then G = (V, E), where V and E are the vertex set and edge set, respectively, of G.

The complement of G, which we will refer to as "G bar," is the graph (V, E bar), where V is the vertex set of G bar (i.e., the vertex set of G bar is identical to the vertex set of G) and E bar is the edge set of G bar. The edge set E bar is defined as follows:

For distinct vertices v1, v2, there is an edge that connects v1 and v2 in G bar if and only if there is no edge that connects v1 and v2 in G.

Definition: A graph G is self-complementary if G is isomorphic to G bar.

Do the following:

(a) Show that the graph G = ({a, b, c, d}, {ab, bc, cd}) is self-complementary.

(b) Find a self-complementary graph with five vertices.

(c) Prove that if a self-complementary graph has n vertices, then either n is congruent to 0 (mod 4) or n is congruent to 1 (mod 4).

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##### Solution Summary

This solution provides step-by-step proofs of parts (a) and (c) are given, and an example of a self-complementary graph with five vertices (including a detailed justification) is provided in an attached Word document.

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(a) Proof that the graph G = ({a, b, c, d}, {ab, bc, cd}) is self-complementary:

By the definition of the complement of G, we have G bar = ({a, b, c, d}, {ac, ad, bd}).

To show that G is self-complementary, define a mapping f(V) -> V as follows:

f(a) = b; f(b) = d; f(c) = a; f(d) = c

Proof that f is a graph isomorphism:

f is a bijection (one-to-one, onto mapping from V to V), because every element of V is mapped (via f) to exactly one element of V, and every element of V is the image (via f) of exactly one element of V.

For every pair v1, v2 of distinct vertices in V, there is an edge connecting v1 and v2 in G if and only if there is an edge connecting f(v1) and f(v2) in G bar:

There is an edge connecting a and b in G, and an edge connecting f(a) = b and f(b) = d in G bar.

There is an edge connecting b and c in G, and an edge connecting f(b) = d and f(c) = a in G bar.

There is an edge connecting c and d in G, and an edge connecting f(c) = a and ...

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- AB, Hood College
- PhD, The Catholic University of America
- PhD, The University of Maryland at College Park

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