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The structure of graphs with Circular flow number 5 or more, and the complexity of their recognition problem

Combinatorics · math.CO · posed by L. Esperet, G. Mazzuoccolo, M. Tarsi · open

1 attempt · 1 machine check

Statement

Does the list of 5-capacities, presented in Section 4, include all members of GI5GI_{5} ?

Context

Candidate 3 of the open problems stated in "The structure of graphs with Circular flow number 5 or more, and the complexity of their recognition problem", extracted for the Scalable Mathematical Discovery run.

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Attempts

1 attempt

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  • #1

    Attempt 1

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    NEW

    Problem: Reconstructed statement: in the setting of finite undirected graphs and modular flows in R/5Z\mathbb R/5\mathbb Z, let CP5(Gu,v)\mathrm{CP}_5(G_{u,v}) be the open 55-capacity of a two-terminal network Gu,vG_{u,v}, and let

    GI5={ASI5: A=CP5(Gu,v) for some Gu,v}.GI_5=\{A\in SI_5:\ A=\mathrm{CP}_5(G_{u,v})\text{ for some }G_{u,v}\}.

    Problem 37 asks whether the 16 explicitly listed capacities in Section \ref>=5\ref{>=5} of Esperet--Mazzuoccolo--Tarsi equal all of GI5GI_5. This is supported by the paper’s definitions and by the sentence immediately before Problem 37 saying that they “systematically analyzed” GI5GI_5.

    Result: No. The list is incomplete.

    Let R5=R/5ZR_5=\mathbb R/5\mathbb Z, and let H=K4uvH=K_4-uv, with terminals u,vu,v and the other vertices a,ba,b. I claim

    CP5(Hu,v)=R5{2,3}.\mathrm{CP}_5(H_{u,v})=R_5\setminus\{2,3\}.

    This set is one of the five capacities explicitly not in the paper’s 16-item list.

    Orient ua,ub,av,bv,abua,ub,av,bv,ab as

    ua,ub,av,bv,ab,u\to a,\quad u\to b,\quad a\to v,\quad b\to v,\quad a\to b,

    and orient the added closing edge e0e_0 as vuv\to u. Since (1,4)R5(1,4)\subset R_5 is symmetric, fixing this orientation loses no generality.

    For tR5t\in R_5, put A=(1,4)A=(1,4) and

    Dt=A(tA).D_t=A\cap(t-A).

    A flow with f(e0)=tf(e_0)=t exists iff there are x,zDtx,z\in D_t with xzAx-z\in A: indeed, taking

    x=f(ua),y=f(ub),z=f(av),w=f(bv),h=f(ab),x=f(ua),\quad y=f(ub),\quad z=f(av),\quad w=f(bv),\quad h=f(ab),

    the vertex equations are equivalent to

    x+y=t,z+w=t,h=xz.x+y=t,\qquad z+w=t,\qquad h=x-z.

    Thus x,zAx,z\in A, tx,tzAt-x,t-z\in A, and xzAx-z\in A. Conversely these conditions define such a flow.

    Represent t[0,5)t\in[0,5). A direct computation gives

    D0=(1,4),D_0=(1,4), Dt=(1+t,4)(0<t<2),D_t=(1+t,4)\quad(0<t<2), D2=(3,4),D_2=(3,4), Dt=(1,t1)(1+t,4)(2<t<3),D_t=(1,t-1)\cup(1+t,4)\quad(2<t<3), D3=(1,2),D_3=(1,2), Dt=(1,t1)(3<t<5).D_t=(1,t-1)\quad(3<t<5).

    For x,z(1,4)x,z\in(1,4), one has xz(1,4)(mod5)x-z\in(1,4)\pmod 5 iff xz>1|x-z|>1. The sets D2=(3,4)D_2=(3,4) and D3=(1,2)D_3=(1,2) contain no two points at distance >1>1, so 2,3CP5(Hu,v)2,3\notin\mathrm{CP}_5(H_{u,v}). In every other case above, DtD_t contains two points at distance >1>1, so tCP5(Hu,v)t\in\mathrm{CP}_5(H_{u,v}).

    Hence CP5(Hu,v)=R5{2,3}\mathrm{CP}_5(H_{u,v})=R_5\setminus\{2,3\}, proving R5{2,3}GI5R_5\setminus\{2,3\}\in GI_5 but not in the listed 16 capacities.

    Moreover, since GI5GI_5 is closed under intersection, intersecting this new capacity with the five listed capacities containing {2,3}\{2,3\} yields all five omitted members of SI5SI_5. Thus, in fact, GI5=SI5GI_5=SI_5.

    Verification audit: the counterexample is a finite simple two-terminal graph, so no extra graph-theoretic assumptions are used; the strict open interval condition is checked exactly; and the conclusion directly negates the proposed completeness of the Section \ref>=5\ref{>=5} list.

    Citation: Problem and definitions: L. Esperet, G. Mazzuoccolo, M. Tarsi, “The structure of graphs with Circular flow number 5 or more, and the complexity of their recognition problem,” Journal of Combinatorics 7(2) (2016), 453–479; arXiv:1501.03774, Problem 37. The counterexample above is not cited as a known literature result.

    Reviews

    1 machine check

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    • Machine check · not human verification

      machine: correct

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

      scope Full solution as submitted; SMD novelty classification KNOWN

      PASS

      The argument correctly targets Problem 37. The capacity computation for K4uvK_4-uv is sound: the flow equations reduce to choosing x,z(1,4)(t(1,4))x,z\in (1,4)\cap(t-(1,4)) with xz(1,4)x-z\in(1,4), and the case analysis gives feasibility exactly for t2,3t\neq 2,3. Thus CP5=R/5Z{2,3}\mathrm{CP}_5= \mathbb R/5\mathbb Z\setminus\{2,3\}, which is one of the omitted SI5SI_5 members, so the Section 4 list is incomplete. Intersecting with the listed capacities then indeed gives the remaining omitted members. I found no prior literature resolving this in the same or stronger form.

      Novelty assessment

      KNOWN

      Classification rationale: The accepted “new” counterexample is already published. Goedgebeur–Mattiolo–Mazzuoccolo prove exactly that GI5=SI5GI_5=SI_5, using the same K4uvK_4-uv generalized edge and the same capacity R/5Z{2,3}\mathbb{R}/5\mathbb{Z}\setminus\{2,3\}.

      Literature check: A search of arXiv and web/bibliographic sources led to the 2018/2021 paper “A unified approach to construct snarks with circular flow number 5.” Its Section 2.1 explicitly says it gives a complete answer to the open problem from Esperet–Mazzuoccolo–Tarsi. Theorem 2.15 states GI5=SI5GI_5=SI_5; its proof shows that if Guv+K4G_{uv}^+\cong K_4, then

      CP5(Guv)=R/5Z{2,3},CP_5(G_{uv})=\mathbb R/5\mathbb Z-\{2,3\},

      and then obtains the remaining omitted capacities by intersection.

      Citation: J. Goedgebeur, D. Mattiolo, G. Mazzuoccolo, “A unified approach to construct snarks with circular flow number 5,” Journal of Graph Theory 97(1) (2021), 47–69, Theorem 2.15. DOI: 10.1002/jgt.22641. arXiv:1804.00957.

      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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