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Statement
Let be a finite connected bipartite graph with a fixed two-coloring, let be an integer, and let . If , then its two-hub replacement satisfies
Here each old edge is replaced by a path of length , the hubs meet the two old color classes, and there is no hub-hub edge. This is a single-forbidden-graph upper bound, with the constant depending on and the old upper bound. It does not assert a matching lower bound for an arbitrary .
Proof
Make the old upper bound uniform for by increasing its constant to ; the finitely many smaller positive inputs permit this. Set
Choose an integer with and put . All these choices precede the host graph. For almost-regular hosts, use
The uniform pruning lemma with supplies a threshold parameter , independent of the host, so that for every in an -free graph of maximum degree at most .
Let such a host have vertices and an integer with
If , then . Otherwise put and . The light-path count applies and gives
where is its fixed threshold-dependent loss. The choice of gives precisely the displayed error coefficient. Since , subtraction yields
Both and are positive, and , so
Choose . This covers both degree ranges and proves for every required almost-regular host. In particular it applies to all bipartite such hosts. The bipartite form of regularization then gives for every -vertex -free graph. Avoidance is hereditary under all subgraph choices made there. At the edge count is zero. Taking the maximum over hosts proves the stated asymptotic bound.
Source and scope
Exposition, Proposition 4.1,
pp. 4–5. Its compressed path-pruning argument is completed by the linked
same-source lemmas. The final formal deductions are
HubPathDegree.almost_regular, BipartiteRegularization.bound, and
HubPathBounds.edge_bound and upper_isBigO, pinned Lean lines
9966–10175. The source invokes a more general PowerErrorAbsorption
interface with its error parameter set to zero; the division and positive
power extraction above prove exactly that needed instance directly.
No logarithmic term is hidden in the notation.
Used by. Proposition 4.2.
Bears on. #571.