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Graffiti Conjecture 6

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graffiti-conjecture-6Combinatoricsposed by Graffiti, reported by Ermelinda DeLaViña, Siemion Fajtlowicz, and Bill Waller, 2002recorded: candidate

1 attempt · no person has looked

Statement

Every finite connected simple graph G satisfies

α(G)≥r(G)+ln⁡(ρ(G)),\alpha(G)\ge r(G)+\ln(\rho(G)),

where α(G)\alpha(G) is the independence number, r(G)r(G) is the radius, and ρ(G)\rho(G) is the minimum number of pairwise vertex-disjoint paths whose vertices cover V(G)V(G).

Context

Infinite family of counterexamples; mathematical argument internally checked, with external verification and novelty review pending.

Machine-generated (Graffiti); real but unfamous by construction.

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Attempts

1 attempt

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  • #1

    Attempt 1

    constructionGPT-5.6 Thinking with Jackson (prompter) ·
    AI involvement
    ai discovered
    — the result was found by a model.
    models
    GPT-5.6 Thinking
    people
    Jackson (prompter)

    GPT-5.6 Thinking produced and checked an infinite family of counterexamples. For each integer s >= 0, it considered a tree T_s formed from the path v_0v_1...v_{4s+7} by attaching leaves at v_{2s+2} and v_{2s+5}. It proved that

    α(Ts)=2s+5,r(Ts)=2s+4,ρ(Ts)=3.\alpha(T_s)=2s+5,\qquad r(T_s)=2s+4,\qquad \rho(T_s)=3.

    Since ln⁡3>1\ln 3>1, it follows that

    α(Ts)=2s+5<2s+4+ln⁡3=r(Ts)+ln⁡ρ(Ts).\alpha(T_s)=2s+5<2s+4+\ln 3=r(T_s)+\ln\rho(T_s).

    Thus every T_s is a counterexample, disproving the conjecture and providing infinitely many counterexamples. The AI also audited the final proof line by line.

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