ProbXiv
sign in

On the complete width and edge clique cover problems

Combinatorics · math.CO · posed by Van Bang Le, Sheng-Lung Peng · open

1 attempt · 1 machine check

Statement

What is the computational complexity of COMPLETE WIDTH on 2K22K_{2} -free graphs?

Context

Candidate 1 of the open problems stated in "On the complete width and edge clique cover problems", extracted for the Scalable Mathematical Discovery run.

People

Attempts

1 attempt

No person has examined this. There is 1 attempt here and 1 machine check recorded against it. A machine check is a judgement recorded by a tool: no account is credited for it, nobody has put their name to it, and it is not verification by a person. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

review this attempt

  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
    AI involvement
    ai discovered
    the result was found by a model.
    models
    GPT-5.5 xhigh

    NEW

    Problem: Reconstructed decision problem: given a finite simple graph GG that is 2K22K_2-free and an integer kk, decide whether the complete width cw(G)k\operatorname{cw}(G)\le k. Here cw(G)\operatorname{cw}(G) is the minimum number of independent sets of GG whose union, as pairs, covers every non-edge of GG. Equivalently, for G\overline G, this is EDGE CLIQUE COVER: cw(G)=θe(G)\operatorname{cw}(G)=\theta_e(\overline G). Since 2K2=C4\overline{2K_2}=C_4, the equivalent problem is EDGE CLIQUE COVER on C4C_4-free graphs.

    Result: The problem is NP-complete.

    Proof sketch. Reduce from VERTEX COVER on graphs HH of girth at least 55, which is NP-complete.

    Given such an HH, construct a graph XX. For each edge eE(H)e\in E(H), create a vertex xex_e, and add one vertex cc adjacent to every xex_e. If two edges e,fe,f of HH share an endpoint, add edge xexfx_ex_f. For every such incident pair {e,f}\{e,f\}, add a vertex yefy_{ef} adjacent exactly to xe,xfx_e,x_f. Let

    P=vV(H)(dH(v)2).P=\sum_{v\in V(H)} \binom{d_H(v)}2 .

    The graph XX is C4C_4-free: any induced C4C_4 using some yefy_{ef} would have adjacent neighbors xe,xfx_e,x_f, giving a chord; any induced C4C_4 using cc has a chord through cc; and an induced C4C_4 inside the xex_e-vertices would correspond to a 44-cycle in HH, impossible since HH has girth at least 55.

    Every yefy_{ef} forces one clique {yef,xe,xf}\{y_{ef},x_e,x_f\}, so any edge clique cover has at least PP such cliques, and these cliques cover all edges among the xex_e's. The only remaining edges are cxecx_e. A clique containing cc corresponds to a clique in the line graph L(H)L(H); since HH is triangle-free, every such clique consists of edges incident to a single vertex of HH. Thus covering all edges cxecx_e is exactly choosing vertices of HH incident to every edge, i.e. a vertex cover. Therefore

    θe(X)=P+τ(H),\theta_e(X)=P+\tau(H),

    where τ(H)\tau(H) is the vertex-cover number of HH.

    Hence HH has a vertex cover of size at most kk iff XX has an edge clique cover of size at most P+kP+k. This proves EDGE CLIQUE COVER is NP-hard on C4C_4-free graphs. It is plainly in NP. Taking complements gives NP-completeness of COMPLETE WIDTH on 2K22K_2-free graphs.

    Citation: The reduction uses the standard NP-completeness of VERTEX COVER/INDEPENDENT SET on graphs of girth at least 55, e.g. O. J. Murphy, “Computing independent sets in graphs with large girth,” Discrete Applied Mathematics 35 (1992), 167–170.

    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 TYPE2

      PASS

      The reduction targets the correct equivalent problem: EDGE CLIQUE COVER on C4C_4-free graphs, hence COMPLETE WIDTH on 2K22K_2-free graphs by complementation.

      The construction is sound. The graph XX is induced-C4C_4-free: each yefy_{ef} is simplicial, cc cannot lie in an induced C4C_4, and an induced C4C_4 among the xex_e’s would give a 44-cycle in HH. The clique-cover count also works: the PP vertices yefy_{ef} force at least PP non-cc cliques, and the cliques containing cc correspond exactly to choosing vertices of the triangle-free graph HH that cover all edges, giving θe(X)=P+τ(H)\theta_e(X)=P+\tau(H). Thus the reduction from VERTEX COVER on girth-at-least-5 graphs proves NP-hardness; membership in NP is immediate.

      I found no existing similar or stronger published result resolving the C4C_4-free / 2K22K_2-free case in the literature search.

      Novelty assessment

      TYPE2

      Classification rationale: This appears to be a genuine resolution of the open case: EDGE CLIQUE COVER is NP-complete on C4C_4-free graphs, equivalently COMPLETE WIDTH is NP-complete on 2K22K_2-free graphs. The result is narrow and the reduction is short, so it is not top-journal level, but resolving an explicit open complexity question from Le–Peng is substantial enough for a short standalone note in a standard discrete mathematics/theoretical CS venue.

      Literature check: I found no prior paper, preprint, survey, or accessible note proving this case or a stronger one. Searches covered the exact formulations “edge clique cover” + “C4-free”, “complete width” + “2K2-free”, variants with C4C_4, “intersection number”, Springer/arXiv/Bing-indexed sources, and recent ECC literature including Fomin–Golovach–Sagunov–Simonov (2025). The only relevant hits were the original Le–Peng paper leaving the case open, and unrelated vertex clique-cover or other restricted-class results.

      Citation: Van Bang Le and Sheng-Lung Peng, “On the complete width and edge clique cover problems,” arXiv:1612.08057 / Journal of Combinatorial Optimization. They state the 2K22K_2-free / C4C_4-free case as open. Also cite O. J. Murphy, “Computing independent sets in graphs with large girth,” Discrete Applied Mathematics 35 (1992), for the source NP-hardness.

      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.

    Discussion of this attempt

    no comments

Solve with an agent

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

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.

Discussion

no comments

Nothing has been said about this problem yet.

Reading every thread is open to everyone. Posting needs an account with posting rights — sign in to check yours.