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✓ 3 results · all verified · 1 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Generalized Niceness and Reduction Outcomes -- Examples

1 · Prerequisites

2 · Summary

These examples check the batch-15 mechanisms on explicit finite data. The first shows how a weakly sparse four-block configuration can be thinned to equal subblocks with directional sparsity. The second shows the maximal-blockade extension step behind the pure-pair extraction lemma. The third records the numerical exponent choice used in the final iterative restricted-sparsification argument.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04Open item page →

Thinning a four-block weakly sparse blockade to directional sparse subblocks

Example

Let G have four pairwise disjoint blocks A1={a1,…,a8},A2={b1,…,b8},A3={c1,…,c8},A4={d1,…,d8}.

and suppose the only cross-edges between noncomplete pairs are

a1b1,a2c1,a3d1,b2c2,b3d2,c3d3.

Then (A1,A2,A3,A4) is a blockade of width 8, every noncomplete pair is weakly 1/16-sparse, and the subblocks

D1={a7,a8},D2={b7,b8},D3={c7,c8},D4={d7,d8}

are pairwise complete or pairwise anticomplete. In particular every formerly weakly sparse pair becomes directionally 0-sparse after thinning.

Facts & Assumptions

Given: The graph G and the four blocks A1,A2,A3,A4 described in the example.

[L1]

A blockade is an ordered sequence of pairwise disjoint nonempty vertex sets, and its width is the minimum block size (Blockades, their length, their width, and their support).

[L2]

A weakly c-sparse pair (X,Y) satisfies eG(X,Y)≤c∣X∣∣Y∣, while directional sparsity bounds the neighbours of each single vertex into the opposite set (Sparsity of one vertex set to another, and weak sparsity of a pair).

Verification

technique · direct adjacency count
1.1L1L2givenalgebra

The four blocks are pairwise disjoint and nonempty, each has size 8, so (A1,A2,A3,A4) is a blockade of width 8 by [L1]. Every noncomplete pair listed in the example has exactly one cross-edge, hence at most 1=(1/16)⋅8⋅8 cross-edges. Therefore each such pair is weakly 1/16-sparse by [L2].

1.2givenL2

None of the vertices a7,a8,b7,b8,c7,c8,d7,d8 appears in any of the six displayed cross-edges. Hence every noncomplete pair among D1,D2,D3,D4 has no cross-edge at all, so each vertex in one chosen subblock has 0 neighbours in the other. By [L2], those pairs are directionally 0-sparse.

2.1step 1.1step 1.2∎

Therefore the thinning (A1,A2,A3,A4)⇝(D1,D2,D3,D4) exhibits exactly the weak-to-directional sparsity conversion claimed in the example.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-04Open item page →

A large almost-pure pair extends an anticomplete blockade

Example

Let

B0={u1,u2},B1={v1,v2},B2={w1,w2,w3,w4,w5,w6},

and assume:

  1. B0 is anticomplete to B1;
  2. B0∪B1 is anticomplete to B2;
  3. inside B2 there are disjoint subsets X={w1,w2},Y={w3,w4,w5,w6}, with Y anticomplete to X.

Then (B0,B1,B2) is an anticomplete blockade, and replacing the last block B2 by the pair (X,Y) produces the longer anticomplete blockade (B0,B1,X,Y).

Facts & Assumptions

Given: The blocks B0,B1,B2 and the subsets X,Y⊆B2 with the adjacency relations stated in the example.

[L1]

A blockade is an ordered sequence of pairwise disjoint nonempty vertex sets (Blockades, their length, their width, and their support).

[L2]

A pair of disjoint vertex sets is anticomplete exactly when there are no edges between them (Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs).

Verification

technique · direct
1.1L1L2given

The sets B0,B1,B2 are pairwise disjoint and nonempty, so (B0,B1,B2) is a blockade by [L1]. Hypotheses 1 and 2 say that each earlier block is anticomplete to every later block, so [L2] makes it an anticomplete blockade.

1.2givenL2

The sets X and Y are disjoint nonempty subsets of B2, and hypothesis 3 says that Y is anticomplete to X. Hypothesis 2 also implies that both X and Y are anticomplete to B0∪B1. Therefore every earlier block in (B0,B1,X,Y) is anticomplete to every later block.

2.1step 1.1step 1.2∎

By steps 1.1 and 1.2, replacing the last block B2 by the anticomplete pair (X,Y) extends the original anticomplete blockade by one step.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-04Open item page →

A numeric run of the Lemma 3.3 exponent choice

Example

Suppose the three-outcome constants for a generalized nice, leaf-reducible, wonderful family satisfy

c3=c4=4,c=14,x=2−10.

Then the exponent choices in the final restricted-sparsification step become

b1=c42=2,b2=4c3c4=4,b3=c3=4,

so

b1b2=8=b2+b3.

Let G be a c2-restricted F‾-free graph satisfying the two global failure hypotheses of the helper lemma. If cy∈[x,c2], then the helper asks for a cy-restricted induced subgraph F of size at least (cy)4∣G∣ and returns a (cy)2-restricted induced subgraph of size at least (cy)4∣F∣. The iterative lemma then yields an x-restricted induced subgraph of size at least x8∣G∣.

Facts & Assumptions

Given: The numerical choices c3=c4=4, c=14, and x=2−10, and a graph G satisfying the conditional hypotheses in the Example.

[L1]

The helper claim uses the substitutions b1=c4/2, b2=4c3/c4, and b3=c3 (A large cy-restricted subgraph in the three-outcome theorem forces a smaller-scale restricted subgraph).

[L2]

The iterative restricted-sparsification lemma concludes with an x-restricted induced subgraph of size at least xb1b2∣G∣ (Iterated restricted sparsification reaches the target scale).

[L3]

The final generalized-niceness lemma is obtained by exactly this choice of b1,b2,b3 (Constant-scale restricted generalized niceness yields an x-scale restricted subgraph, a polynomial clique or stable set, or a blockade).

Verification

technique · direct arithmetic
1.1L1givenalgebra

Substituting the given values into [L1] gives b1=2, b2=4, and b3=4. Therefore b1b2=8=b2+b3, so the numerical inequality required by the iterative lemma holds exactly.

2.1step 1.1L1L2givenalgebra

The graph G itself supplies the starting c2-restricted subgraph required by [L2], because ∣G∣≥(c2)b2∣G∣. For every λ∈[x,c2], write λ=cy. The helper [L1], under the two global failure hypotheses in the Given data, sends each λ-restricted F with ∣F∣≥λb2∣G∣=λ4∣G∣ to a λb1-restricted subgraph of size at least λb3∣F∣=λ4∣F∣. Thus both hypotheses of [L2] hold with starting constant c2, and it gives an x-restricted induced subgraph of size at least xb1b2∣G∣=x8∣G∣.

3.1step 2.1L3∎

This is exactly the numerical exponent pattern used again in [L3].

Sources