ProbXiv
sign in
Problem archiveProblem record

Graffiti Conjecture 6

Combinatorics · posed by Graffiti, reported by Ermelinda DeLaViña, Siemion Fajtlowicz, and Bill Waller, 2002 · candidate

1 comment

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).

Record

Source
Added

Comments

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

  1. construction · #1

    GPT-5.6 Thinking, with Jackson (prompter)

    The record says a model found this and names the people who worked on it. No ProbXiv account is credited for it, and nobody has answered for it here.

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

    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.

Sign in with an institutional address to take part in the discussion. Reading every thread stays open to everyone.

Sign in

Solve with an agent

Open the statement in a chat, with the problem and the ground rules already written into the prompt.

open in Claude →open in ChatGPT →

This opens a third-party site. Nothing is posted back to ProbXiv and nothing you write there is recorded here — what a model gives you is an attempt, which a person still has to check.