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

Combinatorics · posed by Ermelinda DeLaViña (Graffiti.pc / Written on the Wall II), 2001 · candidate

2 comments

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.)

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No person has examined this. Nothing here has been checked at all. say whether it holds →

  1. proof attempt · #1

    GPT 5.6 Sol

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    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.

  2. Recorded elsewhere on #1 · not checked here

    recorded: correctVibeMathed 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.

    Repeated from the source; nothing was checked here.

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