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Albertson–Berman Induced Forest Conjecture

The ratio 15/31 is not claimed to be optimal, and the paper makes no claim that 31 vertices is the smallest possible counterexample. The construction produces separating triangles by design, so it says nothing about the 4-connected case.

Exact FrontierDelta

Prior state unknown→disproved

Scope and record

Occurred: Aug 11, 2026

Delta type: SOURCE CLAIM

Assumptions: VibeMathed verification: site-confirmed. Publication: preprint. AI contribution: ai-co-developed. Editorial metadata: CC BY 4.0; linked content retains its own rights.

Canonical aliases: Albertson–Berman Induced Forest Conjecture · Albertson–Berman

Confidence: Not scored

Registry verification: site confirmed · preprint · resolved

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Attribution

VibeMathed
registry · event recorded by

Heejae Jung
human · human collaborator

GPT-5.6 Sol
model · ai model contributor · OpenAI

Evidence vector

SourceReplayReproducedFormal proofStatement auditExternal checkExpert reviewPeer review

Named evidence

VibeMathed record: Albertson–Berman Induced Forest Conjecture

specialist registry · passed

Reproduced here on 12 August 2026. The refutation is a single finite object, so it is checkable outright rather than on trust. The 31-vertex seed $T$ was rebuilt from the paper's own definitions - the 14-vertex gadget's cyclic neighbour lists, the pentagonal-bipyramid base, the decorated rim edges, the stated labelling and the two completion edges - without running the author's code. That yields a simple 3-connected planar graph on 31 vertices with $87 = 3n-6$ edges, hence a triangulation, with the paper's degree multiset $4^1 5^{17} 6^6 7^7$ . Its maximum induced forest was then computed exactly by two independent algorithms: an ILP with lazy cycle-elimination cuts, and a branch-and-bound minimum feedback vertex set with no LP involved. Both give $a(T) = 15$ , equivalently a minimum feedback vertex set of exactly 16, against the 15.5 the conjecture requires. The two finite inputs to the symbolic argument were separately brute-forced - the terminal profile $(6,6,6,5)$ over all $2^{12}$ internal subsets, and $\beta = 3$ over all $2^7$ subsets of the core - and $M_k$ for $k = 2..5$ confirmed planar on $31k$ vertices with $93k-6$ edges, minimum degree five, every seed induced. Worth noting what the shipped verifier does not do: it certifies the gadget embedding, the profile, $\beta$ and the sphere certificates, but never computes $a(T)$ or $a(M_k)$ , and says so. That computation is the one this site supplied. Not peer-reviewed, not on arXiv, no independent expert review.