VibeMathed reports this item as resolved. VibeMath preserves that report as a source assertion and has not independently authored a plain-language mathematical explanation.
Precise statement
Albertson and Berman conjectured that for every simple planar graph $G$ on $n$ vertices, the largest vertex set inducing a forest has size at least $n/2$. The standing lower bound since the same year has been Borodin's $2n/5$, from his acyclic five-colour theorem. False: there is an explicit $31$-vertex simple $3$-connected maximal planar graph $T$ whose largest induced forest has exactly $15$ vertices, and an infinite family $M_k$ on $31k$ vertices with induced-forest number exactly $15k$, giving the ratio $15/31 < 1/2$ even for triangulations of minimum degree five.
The source statement is reproduced for indexing with attribution. Mathematical correctness requires domain-expert or mechanical review. VibeMath has not independently audited statement
fidelity, correctness, priority, or novelty.
What AI did
GPT-5.6 Sol
The paper's "Acknowledgments and AI disclosure" section states that the two-terminal gadget "was discovered, and substantial parts of the proof strategy were developed, through interaction with OpenAI GPT-5.6 Sol", while the author "selected the research problem, directed the computational search and subsequent proof development, and checked the resulting mathematical arguments and computational certificates". GPT-5.6 Sol also assisted in preparing the manuscript and the verification code.
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.
A 15/31 Counterexample Family to the Albertson-Berman 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.
Known method families
construction (source-reported)
Source-reported tools: construction.
Independent: unknown · difference confidence: 0
What remains uncertain
VibeMath has not independently audited the mathematical statement, proof, or novelty claim.
VibeMath has not independently verified the mathematical claim.
AI-attempt independence and training-data exposure are unknown unless explicitly documented.
VibeMath has not independently audited the mathematical statement, proof, or novelty claim.