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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)cXY, 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
1.1

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)88 cross-edges. Therefore each such pair is weakly 1/16-sparse by [L2].

L1L2givenalgebra
1.2

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.

givenL2
2.1

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

step 1.1step 1.2
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. B0B1 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,YB2 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
1.1

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.

L1L2given
1.2

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 B0B1. Therefore every earlier block in (B0,B1,X,Y) is anticomplete to every later block.

givenL2
2.1

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.

step 1.1step 1.2
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=210.

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)4G and returns a (cy)2-restricted induced subgraph of size at least (cy)4F. The iterative lemma then yields an x-restricted induced subgraph of size at least x8G.

Facts & Assumptions

Given: The numerical choices c3=c4=4, c=14, and x=210, 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 xb1b2G (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
1.1

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.

L1givenalgebra
2.1

The graph G itself supplies the starting c2-restricted subgraph required by [L2], because G(c2)b2G. 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 Fλb2G=λ4G to a λb1-restricted subgraph of size at least λb3F=λ4F. Thus both hypotheses of [L2] hold with starting constant c2, and it gives an x-restricted induced subgraph of size at least xb1b2G=x8G.

step 1.1L1L2givenalgebra
3.1

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

step 2.1L3

Sources