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Average distance in weighted graphs

Combinatorics · math.CO · posed by Peter Dankelmann · open

1 attempt · 1 machine check

Statement

Let n ≥ 3 and let G be a 2-connected graph of order n with a nonnegative vertex weight function c. Then,

μc(G)≤{n4NN−1ifn is even,n4NN−1−N4n(N−1)ifn is odd.\mu_{c}(G)\leq \begin{cases}\frac{n}{4}\frac{N}{N-1}&\text{if}n \text{ is even,}\\ \frac{n}{4}\frac{N}{N-1}-\frac{N}{4n(N-1)}&\text{if}n \text{ is odd.}\end{cases}

Context

Candidate 1 of the open problems stated in "Average distance in weighted graphs", extracted for the Scalable Mathematical Discovery run.

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1 attempt

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    NEW

    Problem: Reconstructed conjecture: for a finite simple 2-connected graph GG of order n≥3n\ge 3, a nonnegative vertex weight function cc, total weight

    N=∑v∈V(G)c(v)>1,N=\sum_{v\in V(G)}c(v)>1,

    and weighted average distance

    μc(G)=1N(N−1)∑u,v∈V(G)c(u)c(v)dG(u,v),\mu_c(G)=\frac{1}{N(N-1)}\sum_{u,v\in V(G)}c(u)c(v)d_G(u,v),

    one has

    μc(G)≤{n4NN−1,n even,n4NN−1−N4n(N−1),n odd.\mu_{c}(G)\leq \begin{cases} \frac{n}{4}\frac{N}{N-1},& n \text{ even},\\[2mm] \frac{n}{4}\frac{N}{N-1}-\frac{N}{4n(N-1)},& n \text{ odd}. \end{cases}

    This is the standard weighted-distance convention and recovers the classical unweighted cycle bound when c≡1c\equiv 1.

    Result: The conjecture is false.

    Let G=K2,3G=K_{2,3}, with bipartition

    A={a1,a2,a3},B={b1,b2}.A=\{a_1,a_2,a_3\},\qquad B=\{b_1,b_2\}.

    Then GG has order n=5n=5 and is 2-connected: deleting any vertex leaves either K2,2K_{2,2} or K1,3K_{1,3}, both connected.

    Define positive integer weights

    c(ai)=3(i=1,2,3),c(bj)=1(j=1,2).c(a_i)=3\quad(i=1,2,3),\qquad c(b_j)=1\quad(j=1,2).

    Thus

    N=3+3+3+1+1=11.N=3+3+3+1+1=11.

    Distances in K2,3K_{2,3} are 11 across the bipartition and 22 between distinct vertices in the same part. Hence the ordered weighted distance sum is

    ∑u,vc(u)c(v)d(u,v)=6⋅3⋅3⋅2+12⋅3⋅1⋅1+2⋅1⋅1⋅2=108+36+4=148.\sum_{u,v}c(u)c(v)d(u,v) = 6\cdot 3\cdot 3\cdot 2 + 12\cdot 3\cdot 1\cdot 1 + 2\cdot 1\cdot 1\cdot 2 = 108+36+4 = 148.

    Therefore

    μc(G)=14811⋅10=7455.\mu_c(G)=\frac{148}{11\cdot 10}=\frac{74}{55}.

    The conjectured odd-nn bound for n=5,N=11n=5,N=11 is

    541110−114⋅5⋅10=3325.\frac54\frac{11}{10}-\frac{11}{4\cdot 5\cdot 10} = \frac{33}{25}.

    But

    7455−3325=7275>0.\frac{74}{55}-\frac{33}{25} = \frac{7}{275}>0.

    So

    μc(G)>541110−114⋅5⋅10,\mu_c(G)>\frac54\frac{11}{10}-\frac{11}{4\cdot 5\cdot 10},

    contradicting the conjecture.

    This is not a zero-weight or boundary-case failure: all weights above are positive integers and N=11>1N=11>1.

    Citation: No prior resolution is used; the counterexample above is explicit and self-contained.

    Reviews

    1 machine check

    No person has reviewed this attempt. 1 machine check below — a machine check is not human verification.

    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification TYPE1

      PASS

      The counterexample attacks the stated weighted 2-connected conjecture directly. K2,3K_{2,3} is indeed 2-connected, the weights are positive integers with N=11N=11, and the ordered weighted distance sum is correctly computed as 148148, giving μc=74/55\mu_c=74/55. For n=5,N=11n=5,N=11, the conjectured bound is 33/2533/25, and 74/55>33/2574/55>33/25. I found no prior comparable disproof in accessible literature/search results.

      Novelty assessment

      TYPE1

      Classification rationale: The counterexample is genuinely new as far as I could determine, but it is extremely small and elementary: a direct calculation on K2,3K_{2,3} with simple positive integer weights. It would be appropriate as an erratum/comment or as part of a note giving a corrected theorem, but not as a standalone combinatorics paper.

      Literature check: I found the original paper: Peter Dankelmann, “Average distance in weighted graphs,” Discrete Mathematics 312(1), 12–20 (2012), DOI 10.1016/j.disc.2011.02.010; Zbl 1238.05071. I searched for the exact title, “Dankelmann” with “weighted average distance,” “Conjecture 1,” “2-connected weighted graphs,” and K2,3K_{2,3}-based counterexample terms. arXiv searches found no exact-title or weighted-average-distance follow-up; broader Dankelmann average-distance hits concern proximity/remoteness and other unweighted distance parameters. DOAJ and GitHub repository/issue/discussion searches returned no relevant prior counterexample. OpenAlex metadata lists the paper and related works but no apparent correction or stronger known disproof. I found no accessible prior resolution of this conjecture.

      Citation: Peter Dankelmann, “Average distance in weighted graphs,” Discrete Mathematics 312(1):12–20, 2012. DOI: 10.1016/j.disc.2011.02.010.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

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