Problem detail · source-aware

WOWII Conjecture 72: Two induced trees pin down tree($ G $)

candidateconfidence 50%

VibeMathed reports this item as candidate. VibeMath preserves that report as a source assertion and has not independently authored a plain-language mathematical explanation.

Precise statement

For a connected graph $ G $, let $ t= $ tree($ G $) (order of a largest induced tree), $ A= $ average eccentricity, and $ L= $ maximum independence number of a neighbourhood. Then $ \lceil (A+L)/3 \rceil \le t. $ (The evenly-divided reading of the conjecture holds; a stronger reading that divides only $ L $ by three is false.)

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 model was given a short prompt (copied from a successful earlier run on a different WOWII conjecture) and asked to find a solution or counterexample to an open conjecture of its choosing. It selected WOWII Conjecture 72, produced the two-lemma argument (diametral path induces a tree on $ D+1 $ vertices; maximum independent neighbourhood induces a star on $ L+1 $ vertices), derived the bound $ \lceil(A+L)/3\rceil\le t $, and explicitly distinguished the evenly-divided reading (true) from the stronger reading that divides only $ L $ by three (false). The human then posted the diagram and commentary.

Provider: OpenAI · Prompt public: unknown · Independence: unknown

Verification boundary

unreviewed

Re-derived in full by this site on 17 August 2026 - the argument is elementary and correct, and short enough to state: a shortest path between two vertices at maximum distance D is induced, so it induces a path (a tree) on D+1 vertices, giving tree(G) >= D+1 >= A+1 since average eccentricity is at most D; a maximum independent set in a neighbourhood plus its centre induces a star on L+1 vertices, giving tree(G) >= L+1; hence A + L <= 2*tree(G) - 2, and ceil((A+L)/3) <= tree(G) follows by integrality. The stronger reading (dividing only L by three) fails on the claimed counterexample family. What keeps this Candidate is not the mathematics but the statement: Conjecture 72's canonical wording is not publicly pinned (no formal statement exists in the Formal Conjectures repository), so which reading DeLaViña intended is unconfirmed, and the X post plus transcript is the only artifact. Site-confirmed records this site's independent re-derivation of the proved reading.

Correctness: unknown · statement fidelity: unaudited · peer review: none

Timeline

  1. X post by @4thRT (with attached diagram and ChatGPT conversation link)

    The evenly-divided reading of WOWII Conjecture 72 holds: $ \lceil(A + L)/3\rceil \le t, $ where $ t = $ tree($ G $) (order of a largest induced tree), $ A = $ average eccentricity and $ L = $ maximum neighbourhood independence number. A stronger reading that divides only $ L $ by three is false. The argument rests on two elementary observations (a diametral path is chordless and therefore induces a tree on $ D+1 $ vertices; a maximum independent set in a neighbourhood induces a star on $ L+1 $ vertices). The original conjecture’s precise wording is not yet pinned in a public formal repository, so statement fidelity remains to be audited.

Known method families

argument (source-reported)

Source-reported tools: argument.

Independent: unknown · difference confidence: 0

What remains uncertain

VibeMath has not independently audited the mathematical statement, proof, or novelty claim.

  • The source status is candidate and must not be represented as solved.
  • 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.