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Official PYQ · 2015

GATE CS 2015 Set 2 Q28 — Graphs: Connectivity, Matching, Colouring

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A graph is self-complementary if it is isomorphic to its complement. For all self-complementary graphs on nn vertices, nn is

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Answer: D

Let G=(V,E)G = (V, E) be a self-complementary graph on nn vertices, and let Gˉ=(V,Eˉ)\bar{G} = (V, \bar{E}) be its complement.

By definition, a graph is self-complementary if it is isomorphic to its complement (G≅GˉG \cong \bar{G}). Isomorphic graphs must have the same number of edges, so:

∣E∣=∣Eˉ∣|E| = |\bar{E}|

The total number of edges in a complete graph KnK_n on nn vertices is given by:

∣E∣+∣Eˉ∣=n(n−1)2|E| + |\bar{E}| = \frac{n(n-1)}{2}

Substituting ∣E∣=∣Eˉ∣|E| = |\bar{E}| into the equation:

2∣E∣=n(n−1)2  ⟹  ∣E∣=n(n−1)42|E| = \frac{n(n-1)}{2} \implies |E| = \frac{n(n-1)}{4}

Since the number of edges ∣E∣|E| must be an integer, the term n(n−1)n(n-1) must be divisible by 44.

Because nn and n−1n-1 are consecutive integers, one of them is even and the other is odd. Thus, the factor of 44 must divide either nn or n−1n-1 entirely:

  • If 44 divides nn, then n≡0(mod4)n \equiv 0 \pmod 4.
  • If 44 divides n−1n-1, then n≡1(mod4)n \equiv 1 \pmod 4.

Hence, nn must be congruent to 0(mod4)0 \pmod 4 or 1(mod4)1 \pmod 4.

Examples:

  • For n=4n = 4 (4≡0(mod4)4 \equiv 0 \pmod 4), the path graph P4P_4 is self-complementary.
  • For n=5n = 5 (5≡1(mod4)5 \equiv 1 \pmod 4), the cycle graph C5C_5 is self-complementary.

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