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Statement
Let be a finite nonempty bipartite graph with a fixed coloring and let . For every and integer , there is an integer such that every -free graph on vertices with maximum degree at most an integer satisfies
Here uses the admissibility thresholds from good paths. The same works for every and all these lengths simultaneously. It depends only on .
Proof
Let be the constants for in heavy-path assembly. Choose an integer , so , and take
We prove the stronger integer inequality
for every .
First suppose . Every length- endpoint fiber has at most members, by the walk bound. Since and ,
The last inequality follows directly from the positive threshold recurrence for . Thus no heavy fiber exists and (3) holds. This includes and graphs with no vertices.
Otherwise put
The assumption gives . The division inequalities give
If , apply heavy common neighborhoods with , ratio parameter , and target constant . Its hypotheses hold by (2) and (4). It supplies the common-heavy configuration and a path family of size greater than . Write by Euclidean division. Because , one has and . Since , heavy-path assembly then produces a copy of , contradicting its exclusion.
Therefore . Multiplying by and using proves (3). Finally implies (1). All constants were chosen before the host graph or the length , so the quantifiers are uniform as stated.
Source and scope
Complete reconstruction of AdmissibleFiberDegree and
HubAdmissiblePruning.scaled and pruning, pinned Lean lines 9786–9926.
The exposition, p. 5, describes this
as discarding only a controlled fraction of long paths; the calculation
above supplies a common threshold for every shorter length used there.
No assumption that heavy paths themselves are mutually disjoint enters
the count.
Used by. Proposition 4.1.
Bears on. #571.