Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-27
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A maximal pure blockade with large total a-mass must already have at least ϵ−2 blocks

Statement

Let ϵ∈(0,12), let a≥1, and let G be a graph with the property that every induced subgraph F of G with ∣F∣≥ϵ2a∣G∣ contains a complete or anticomplete (k,∣F∣/ka)-blockade for some k∈[2,ϵ−1].

Suppose q is maximal subject to the existence of a pure blockade (A1,…,Aq) in G whose pattern graph is P4-free, such that ∣Ai∣≥ϵ3a∣G∣ for every i and

∑i=1q∣Ai∣1/a≥∣G∣1/a.

Then q≥ϵ−2.

Facts & Assumptions

Given: The hypotheses of the statement and a maximal blockade (A1,…,Aq).

Proof

technique · direct
1.1assume-contragiven

Suppose for contradiction that q<ϵ−2. Reorder the blocks so that ∣A1∣=max⁡i∣Ai∣. Then q∣A1∣1/a≥∑i=1q∣Ai∣1/a≥∣G∣1/a, so ∣A1∣≥∣G∣/qa≥ϵ2a∣G∣. By the hypothesis on G, the induced subgraph G[A1] contains a complete or anticomplete (k,∣A1∣/ka)-blockade (B1,…,Bk) for some k∈[2,ϵ−1].

2.1step 1.1given

Replace the block A1 by B1,…,Bk, and keep the other blocks A2,…,Aq. Because (A1,…,Aq) was pure, every outside block is either complete or anticomplete to A1, hence to each Bj⊆A1. The new blockade is still pure, its pattern graph is obtained by substituting a complete or edgeless graph for the vertex corresponding to A1, and so it is still P4-free.

3.1step 2.1algebra

Every new block satisfies ∣Bj∣≥∣A1∣/ka≥ϵa∣A1∣≥ϵ3a∣G∣, while ∑j=1k∣Bj∣1/a≥k(∣A1∣ka)1/a=∣A1∣1/a. So the new blockade still satisfies the lower bound on every block and on the total a-mass, but it has q−1+k>q blocks. This contradicts the maximality of q.

4.1discharge-contradictionstep 3.1∎

Therefore q≥ϵ−2.

Depends on

Used by

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Sources