Chromatic bounds from locally small bipartization costs

Can a graph have infinite chromatic number while every finite n-vertex subgraph can be made bipartite by deleting only f(n) edges, for an arbitrarily slowly divergent profile f? The target asks whether some divergent profile instead forces finite chromatic number.

Why it matters and what counts as progress

Why it matters: The question asks how closely finite subgraphs may resemble bipartite graphs before the entire graph must have bounded chromatic number.

What counts as progress: Construct an infinite-chromatic example for a given profile or prove that a specific divergent profile forces a finite, potentially three-color, bound.

How the problem developed

1982 · Open question [1][2]
Erdős, Hajnal, and Szemerédi Asked whether graphs of arbitrarily high chromatic number can have every sufficiently small induced subgraph made bipartite by deleting a slowly growing number of edges.
August 2026 · Reported counterexample [3]
Tomasz Adamczewski Posted a disproof of Erdős problem 74: for a suitable diverging local deletion bound, graphs satisfying the local condition have bounded chromatic number; the repository credits GPT-6 Astra with the proof.

AI contribution

GPT-6 Astra resolves Erdős Problem 74 on local bipartiteness

There is a divergent integer-valued function f such that if every finite n-vertex subgraph of a graph becomes bipartite after at most f(n) edge deletions, then the graph has finite chromatic number. The formal proof gives three-colourability.

AI involvementAI-autonomous

ImportanceNotable

Technical verificationFormal (Lean 4 with Mathlib and Comparator)

Human understandingPartial