VibeMathedMath problems solved with AI

Levit–Mandrescu Unimodality Conjecture

A graph on nn vertices is very well-covered if every maximal independent set has size n/2n/2. Levit and Mandrescu conjectured that the independence polynomial i(G,x)i(G,x) of every very well-covered graph is unimodal, i.e. its coefficient sequence is nondecreasing and then nonincreasing.

Result
Disproved
Status
Resolved
AI contribution
AI-discovered
Method
Construction
Field
Graph Theory, Independence Polynomials
Posed by
Vadim E. Levit, Eugen Mandrescu
Year posed
2006
Years open
20y
Solved
2026-07-22
Model
GPT-5.6 Sol, Claude Fable 5
Vendor
Collaborators
Lucas B.
Verification
Unreviewed
Publication
Announced
Significance
10 / 100
Disclosed cost
Wikipedia
No dedicated article

What the AI did

The models produced an explicit counterexample: the whiskering of E588(39K1185K12)E_{588} \vee (39K_{11} \sqcup 85K_{12}), a very well-covered graph on 4,074 vertices, whose independence polynomial (1+x)1449(1+2x)588+(1+x)1913(1+12x)39(1+13x)85(1+x)2037(1+x)^{1449}(1+2x)^{588} + (1+x)^{1913}(1+12x)^{39}(1+13x)^{85} - (1+x)^{2037} has a strict local valley at a1095a_{1095}. Per the announcement, the search took a few hours once the question was posed.

Verification

Announced on LinkedIn by Lucas B. (Head of AI Research, Jump Trading); no preprint yet, and the reviewers credited are internal to the team rather than independent. The stated polynomial was recomputed with exact integer arithmetic: a1094>a1095<a1096a_{1094} > a_{1095} < a_{1096} holds, so the polynomial given is genuinely not unimodal, and its low-order coefficients (a0=1a_0 = 1, a1=4074a_1 = 4074) are consistent with a graph on 4,074 vertices. What remains unchecked is that the whiskered graph's independence polynomial equals the polynomial stated.

Source

Submitted by Rasmus Lindahl on

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