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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck 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 a≥1. Let G be a graph such that for every induced subgraph F of G with ∣F∣≥ϵ2a∣G∣, there exists k∈[2,ϵ−1] and a complete or anticomplete (k,∣F∣/ka)-blockade in F. Then G has an ϵ-restricted induced subgraph with at least ϵ3a∣G∣ vertices.

Facts & Assumptions

Given: The hypotheses of the statement.

Proof

technique · direct
1.1givenchoose

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 ϵ3a∣G∣, and ∑i=1q∣Ai∣1/a≥∣G∣1/a. By A maximal pure blockade with large total a-mass must already have at least ϵ−2 blocks, one has q≥ϵ−2.

2.1step 1.1choose

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 ∣I∣≥q≥ϵ−1. For each i∈I, choose Si⊆Ai with ∣Si∣=⌈ϵ3a∣G∣⌉, and put S:=⋃i∈ISi. Then ∣S∣=∣I∣⌈ϵ3a∣G∣⌉≥ϵ3a∣G∣.

3.1step 2.1given

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 ∣Si∣−1≤∣S∣/∣I∣≤ϵ∣S∣. Hence S is ϵ-sparse, and therefore ϵ-restricted.

3.2step 2.1given

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.

4.1step 3.1step 3.2∎

In either case G has an ϵ-restricted induced subgraph on at least ϵ3a∣G∣ vertices, namely G[S].

Depends on

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