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Property (*) and leaf reducibility yield a long x-sparse or complete blockade, or a better outcome

Statement

Suppose that F has property () and that F is leaf-reducible. Then there exist constants c1,c2,c3>0 and c4,c54 such that, with c:=24c5, for every 0<xyc and every cy3-restricted F-free graph G, at least one of the following holds:

  1. G has an x-sparse or complete blockade of length at least y1 and width at least yc4+2G;
  2. G has a 2y4-restricted induced subgraph with at least yc4+2G vertices;
  3. G has a clique or stable set of size at least (x10G)c1;
  4. G has a complete or anticomplete (k,G/kc2+7/c3)-blockade for some real kyc3;
  5. G has a pure (,G/9)-blockade for some real [y1,x2].

Facts & Assumptions

Given: A finite family F with property () and leaf-reducible, parameters 0<xyc, and a cy3-restricted F-free graph G.

[L1]

The five-outcome lemma provides constants c1,c2,c3>0 and c4,c54 for y3-restricted graphs (Property (*) and leaf reducibility yield five comb outcomes in a restricted graph).

[L2]

If every induced subgraph F of G with FcG has disjoint sets X,Y with Xyc4F, Y(1c5y)F, and Y x-sparse or complete to X, then G has an x-sparse or complete blockade of length at least y1 and width at least yc4+2G (Large sparse-pair hypotheses yield an x-sparse or complete blockade).

[L3]

If a graph is cy3-restricted and F is an induced subgraph with FcG, then F is y3-restricted (c-sparse, c-dense and c-restricted vertex sets).

Proof

Proof technique: either every large induced subgraph satisfies the large pair hypothesis of [L2], or choose a counterexample F and apply the five-outcome lemma inside it.

1.1

Let c1,c2,c3>0 and c4,c54 be the constants from [L1], and put c:=24c5.

L1choose
2.1

If Gy(c4+2), then yc4+2G1. Any single vertex spans an induced subgraph that is 0-restricted, hence 2y4-restricted, so outcome 2 holds. Therefore we may assume Gy(c4+2).

step 1.1givenalgebracases
2.2

[assume-case universal-pair] Suppose that every induced subgraph F of G with FcG has disjoint sets X,Y with Xyc4F, Y(1c5y)F, and Y x-sparse or complete to X. Then [L2] gives outcome 1.

step 1.1L2
2.3

[assume-case obstruction] Assume instead that there is an induced subgraph F of G with FcG for which no such pair X,Y exists. By [L3], the graph F is y3-restricted, so [L1] applies to F. Because the first outcome of [L1] fails for this specific F, one of the remaining four outcomes of [L1] holds inside F.

step 1.1L1L3
3.1

If [L1] gives a 2y4-restricted induced subgraph of F with at least yc4F vertices, then that subgraph has at least yc4cG=yc4+1Gyc4+2G vertices because yc<1. Hence outcome 2 holds in G.

step 2.3L1algebra
3.2

If [L1] gives a clique or stable set of size at least (x9F)c1, then (x9F)c1(x9cG)c1(x9xG)c1=(x10G)c1, because xyc. So outcome 3 holds.

step 2.3L1algebra
3.3

If [L1] gives a complete or anticomplete (k,F/kc2+6/c3)-blockade with kyc3, then Fkc2+6/c3cGkc2+6/c3yGkc2+6/c3Gkc2+7/c3, because kyc3 implies k1/c3y1. Hence outcome 4 holds.

step 2.3L1algebra
3.4

If [L1] gives a pure (,F/8)-blockade with [y1,x2], then F8cG8yG8G9, again because y1. Thus outcome 5 holds.

step 2.3L1algebra
4.1

The exhaustive alternatives 2.1 and 2.2, together with steps 3.1-3.4, show that one of the five stated outcomes always holds.

step 2.2step 2.3step 3.1step 3.2step 3.3step 3.4cases-exhaustive

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