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

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wowii-conjecture-72-two-induced-trees-pin-down-tree-gCombinatoricsposed by Ermelinda DeLaViña (Graffiti.pc / Written on the Wall II), 2001recorded: candidate

1 attempt · 1 machine check · no person has looked

Statement

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

Context

The evenly-divided reading of WOWII Conjecture 72 holds: ⌈(A+L)/3⌉≤t, \lceil(A + L)/3\rceil \le t, where t= t = tree(G G ) (order of a largest induced tree), A= A = average eccentricity and L= L = maximum neighbourhood independence number. A stronger reading that divides only L 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 D+1 vertices; a maximum independent set in a neighbourhood induces a star on L+1 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.

A machine-generated conjecture from DeLaViña's Written on the Wall II list - exactly the catalog's Graffiti anchor at 5. Open since around 2001 and part of the current wave of WOWII resolutions.

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  • #1

    Attempt 1

    proof attemptGPT 5.6 Sol ·
    AI involvement
    ai discovered
    — the result was found by a model.
    models
    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 D+1 vertices; maximum independent neighbourhood induces a star on L+1 L+1 vertices), derived the bound ⌈(A+L)/3⌉≤t \lceil(A+L)/3\rceil\le t , and explicitly distinguished the evenly-divided reading (true) from the stronger reading that divides only L L by three (false). The human then posted the diagram and commentary.

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    0 human reviews · 1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from VibeMathed site check ·

      scope Reproduction by the VibeMathed site

      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.

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