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Let d:V→Z0+d: V \to \mathbb{Z}_0^+ be a symmetric function that satisfies d(∅)=0d(\emptyset) = 0 and ∀X,Y⊆V\forall X, Y \subseteq V (19) and (20). Let R^:V→Z0+\hat{R}: V \to \mathbb{Z}_0^+ be an even valued, symmetric, skew-supermodular function. Suppose that R^(X)≤d(X)∀X⊆V\hat{R}(X) \le d(X) \quad \forall X \subseteq V. Then there exists a pairing MM on TdT_d that satisfies dM(X)≤d(X)−R^(X)∀X⊆V.d_M(X) \le d(X) - \hat{R}(X) \quad \forall X \subseteq V.

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  • Recent results on well-balanced orientations
  • FAR
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  1. exploration by a model · #1

    GPT-5.5 xhigh

    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.

    NEW

    Problem: Reconstructed statement: for a finite set VV, d:2V→Z≥0d:2^V\to\mathbb Z_{\ge0} is symmetric, d(∅)=0d(\emptyset)=0, and satisfies the standard connectivity-function inequalities

    d(X)+d(Y)≥d(X∩Y)+d(X∪Y),d(X)+d(Y)≥d(X∖Y)+d(Y∖X)d(X)+d(Y)\ge d(X\cap Y)+d(X\cup Y),\qquad d(X)+d(Y)\ge d(X\setminus Y)+d(Y\setminus X)

    for all X,Y⊆VX,Y\subseteq V. Let

    Td={v∈V:d({v}) is odd}.T_d=\{v\in V:d(\{v\})\text{ is odd}\}.

    A pairing MM on TdT_d is a perfect matching, and dM(X)d_M(X) counts matched pairs crossing XX. The question asks whether, for every even-valued symmetric skew-supermodular R^≤d\hat R\le d, there is MM with

    dM(X)≤d(X)−R^(X)∀X⊆V.d_M(X)\le d(X)-\hat R(X)\quad\forall X\subseteq V.

    Result: The statement is false.

    Let V=F23V=\mathbb F_2^3. For every nonzero a∈F23a\in\mathbb F_2^3 and ϵ∈F2\epsilon\in\mathbb F_2, let

    Ha,ϵ={x:a⋅x=ϵ}.H_{a,\epsilon}=\{x:a\cdot x=\epsilon\}.

    Define d:2V→Z≥0d:2^V\to\mathbb Z_{\ge0} by

    d(X)={0,∣X∣=0,8,1,∣X∣=1,7,2,∣X∣=2,6,3,∣X∣=3,5,2,∣X∣=4 and X=Ha,ϵ for some a,ϵ,4,∣X∣=4 and X is not such a hyperplane.d(X)= \begin{cases} 0,& |X|=0,8,\\ 1,& |X|=1,7,\\ 2,& |X|=2,6,\\ 3,& |X|=3,5,\\ 2,& |X|=4\text{ and }X=H_{a,\epsilon}\text{ for some }a,\epsilon,\\ 4,& |X|=4\text{ and }X\text{ is not such a hyperplane.} \end{cases}

    Then dd is symmetric, d(∅)=0d(\emptyset)=0, and d(X)≡∣X∣(mod2)d(X)\equiv |X|\pmod2. Hence Td=VT_d=V.

    The function dd satisfies the two displayed inequalities. Indeed, with h(X)=min⁡(∣X∣,8−∣X∣)h(X)=\min(|X|,8-|X|), hh is submodular and d=h−21{X hyperplane}d=h-2\mathbf 1_{\{X\text{ hyperplane}\}}. If neither of X,YX,Y is a hyperplane, submodularity follows from that of hh. If both are hyperplanes, they are equal, complementary, or meet in two points, and the inequality is immediate. If exactly XX is a hyperplane, put a=∣X∩Y∣a=|X\cap Y|, c=∣V∖(X∪Y)∣c=|V\setminus(X\cup Y)|; a direct check using g(t)=tg(t)=t for t≤3t\le3, g(4)=2g(4)=2, gives

    d(X∩Y)+d(X∪Y)=g(a)+g(c)≤2+h(Y)=d(X)+d(Y).d(X\cap Y)+d(X\cup Y)=g(a)+g(c)\le 2+h(Y)=d(X)+d(Y).

    Posimodularity follows from submodularity and symmetry.

    Now take R^≡0\hat R\equiv0, which is even-valued, symmetric, skew-supermodular, and satisfies R^≤d\hat R\le d.

    Let MM be any pairing of VV, with pairs {xi,yi}\{x_i,y_i\}, i=1,…,4i=1,\dots,4. Set zi=xi+yi≠0z_i=x_i+y_i\neq0. Since ∑v∈F23v=0\sum_{v\in\mathbb F_2^3}v=0, we have ∑izi=0\sum_i z_i=0. Every linear dependence among the ziz_i’s has even size, so there exists nonzero a∈F23a\in\mathbb F_2^3 with

    a⋅zi=1∀i.a\cdot z_i=1\quad\forall i.

    Thus the hyperplane H={x:a⋅x=0}H=\{x:a\cdot x=0\} contains exactly one endpoint of every matched pair, so

    dM(H)=4.d_M(H)=4.

    But HH is one of the special hyperplanes, so d(H)=2d(H)=2. Therefore

    dM(H)=4>2=d(H)−R^(H),d_M(H)=4>2=d(H)-\hat R(H),

    contradicting the required inequality. Hence no such pairing exists.

    Citation: Original question: Bernáth, Iwata, T. Király, Z. Király, Szigeti, “Recent results on well-balanced orientations,” Discrete Optimization 5 (2008), 663–676. The counterexample above is not taken from a known source.

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The counterexample targets the stated pairing assertion (with R^≡0\hat R\equiv0). The constructed dd is symmetric, nonnegative, has d(∅)=0d(\emptyset)=0, satisfies the required submodular/posimodular inequalities, and has Td=VT_d=V. The linear-algebra argument correctly shows that every perfect matching of V=F23V=\mathbb F_2^3 is crossed by some affine hyperplane HH, giving dM(H)=4d_M(H)=4, while d(H)=2d(H)=2, so the required inequality fails. I found no reliable prior source containing this same or stronger counterexample.

    Novelty assessment

    TYPE1

    Classification rationale: The counterexample appears genuinely new, but it is a small, elementary obstruction to a specialized abstract generalization. It answers a published question negatively, yet the proof is short and does not develop broader theory beyond the example, so I would not expect it to support a standalone standard-journal paper by itself.

    Literature check: I searched for the exact problem and variants: “feasible pairing for connectivity functions,” “symmetric crossing submodular pairing,” “Question 11/12 well-balanced orientations,” “odd-vertex pairing,” “admissible odd-vertex pairings,” “matroid connectivity function pairing,” and searches involving the Fano/affine-geometry construction. I found the original Bernáth–Iwata–Király–Király–Szigeti paper and related later work on admissibility/complexity of odd-vertex pairings and well-balanced orientations, but no published counterexample to this abstract connectivity-function pairing question or to the stronger version with R^≡0\hat R\equiv0.

    Citation: A. Bernáth, S. Iwata, T. Király, Z. Király, Z. Szigeti, “Recent results on well-balanced orientations,” Discrete Optimization 5 (2008), 663–676, Section 10, Question 12 in the technical-report version.

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