ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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The self-complementary five-cycle satisfies $hom (C_{5}) = 2$

Example

The five-cycle is self-complementary and satisfies $hom (C_{5}) = 2$ .

Facts & Assumptions

Given: The graph $C_{5}$ on vertices $0, 1, 2, 3, 4$ .

[L1]

The homogeneous number is $\max {ω, α}$ (Homogeneous vertex sets and the homogeneous number $hom (G) = \max {ω (G), α (G)}$ ).

[L2]

In $C_{5}$ , precisely the consecutive pairs modulo $5$ are edges (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

[L3]

A graph isomorphism is a bijection preserving adjacency in both directions, and the complement contains precisely the missing pairs (Graph isomorphisms, automorphisms and graph complements).

Verification

technique · direct

1.1

The pair ${0, 1}$ is an edge and ${0, 2}$ is a nonedge, so $C_{5}$ has both a two-vertex clique and a two-vertex stable set.

1.2

Any three vertices on the cycle contain a consecutive pair, hence an edge; their complement has two omitted vertices, so among the three cyclic gaps one has length at least two, giving a nonconsecutive pair and hence a nonedge. Thus no three vertices are homogeneous.

L2algebra

1.3

The map $i \mapsto 2 i (m o d 5)$ sends consecutive differences $\pm 1$ to differences $\pm 2$ , exactly the nonedges of $C_{5}$ , so it is an isomorphism $C_{5} ≅ \overline{C_{5}}$ .

L2L3algebra

2.1

Steps 1.1 and 1.2 give $ω (C_{5}) = α (C_{5}) = 2$ , hence $hom (C_{5}) = 2$ by [L1]; step 1.3 gives self-complementarity.

step 1.1step 1.2step 1.3L1∎

Depends on

Used by

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Sources

M. Chudnovsky, The Erdos-Hajnal Conjecture: A Survey, sec. 2 (standard reference, not scraped)

The self-complementary five-cycle satisfies hom⁡(C5)=2hom(C5​)=2