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TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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Complete or anticomplete blockade hypotheses force an ϵ-restricted induced subgraph

Statement

Let ϵ(0,12) and a1. Let G be a graph such that for every induced subgraph F of G with Fϵ2aG, there exists k[2,ϵ1] and a complete or anticomplete (k,F/ka)-blockade in F. Then G has an ϵ-restricted induced subgraph with at least ϵ3aG vertices.

Facts & Assumptions

Given: The hypotheses of the statement.

Proof

technique · direct
1.1

Let q be maximal subject to the existence of a pure blockade (A1,,Aq) whose pattern graph is P4-free, every block has size at least ϵ3aG, and i=1qAi1/aG1/a. By A maximal pure blockade with large total a-mass must already have at least ϵ2 blocks, one has qϵ2.

givenchoose
2.1

By Pure blockades with P4-free patterns contain complete or anticomplete subblockades of square-root length, this blockade has a complete or anticomplete subblockade indexed by a set I[q] with Iqϵ1. For each iI, choose SiAi with Si=ϵ3aG, and put S:=iISi. Then S=Iϵ3aGϵ3aG.

step 1.1choose
3.1

If the chosen subblockade is anticomplete, then every vertex of Si has neighbors in S only inside Si, so its degree in G[S] is at most Si1S/IϵS. Hence S is ϵ-sparse, and therefore ϵ-restricted.

step 2.1given
3.2

If the chosen subblockade is complete, then in the complement G[S] every vertex of Si has neighbors only inside Si, so the same estimate shows that G[S] is ϵ-sparse. Therefore G[S] is ϵ-dense, and again ϵ-restricted.

step 2.1given
4.1

In either case G has an ϵ-restricted induced subgraph on at least ϵ3aG vertices, namely G[S].

step 3.1step 3.2

Depends on

Used by

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Dependency tree · two levels

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