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Constrained ear decompositions in graphs and digraphs

Combinatorics · math.CO · posed by Frédéric Havet, Nicolas Nisse · open

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Statement

Finally, we conjecture that, for every set A\mathcal{A} of integers, deciding whether a digraph has a handle decomposition with all handles of length in A\mathcal{A} is NP-complete, unless there exists h∈Nh \in \mathbb{N} such that A={1,…,h}\mathcal{A} = \{1, \dots, h\}.

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  • Constrained ear decompositions in graphs and digraphs
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No person has examined this. Everything below was judged by machines. say whether it holds →

  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: Literal reconstruction: for every fixed set A⊆Z\mathcal A\subseteq \mathbb Z, let LAL_{\mathcal A} be the language of finite digraphs having a directed handle/ear decomposition in which every handle has length in A\mathcal A. The conjecture asserts that LAL_{\mathcal A} is NP-complete unless A={1,…,h}\mathcal A=\{1,\dots,h\} for some h∈Nh\in\mathbb N. This reconstruction follows the quoted phrase “for every set A\mathcal A of integers”; no finiteness or decidability condition on A\mathcal A is stated.

    Result: The literal conjecture is false.

    Let K⊆{3,4,… }K\subseteq \{3,4,\dots\} be a nonrecursive set, for example a shifted halting set, and set A=K\mathcal A=K. Then A\mathcal A is not equal to any finite initial interval {1,…,h}\{1,\dots,h\}.

    For n≥3n\ge 3, let CnC_n be the directed cycle of length nn. Under the standard handle-decomposition convention, the first handle is a directed cycle and is constrained by A\mathcal A. The only directed cycle contained in CnC_n is CnC_n itself. Hence

    Cn∈LA⟺n∈A.C_n\in L_{\mathcal A}\quad\Longleftrightarrow\quad n\in \mathcal A.

    Indeed, if n∈An\in\mathcal A, the one-handle decomposition consisting of CnC_n is valid; conversely, any valid decomposition of CnC_n must start with the whole cycle CnC_n, whose length is nn, so n∈An\in\mathcal A.

    If LAL_{\mathcal A} were NP-complete, then in particular LA∈NPL_{\mathcal A}\in\mathrm{NP}, hence LAL_{\mathcal A} would be decidable. But then membership in the nonrecursive set A\mathcal A would be decidable by mapping nn to CnC_n and deciding whether Cn∈LAC_n\in L_{\mathcal A}, a contradiction.

    Thus the conjecture, as literally stated for arbitrary sets of integers, is false.

    The defect is the missing effectiveness/finiteness hypothesis on A\mathcal A. A natural repair would restrict A\mathcal A to finite or at least decidable sets of positive integers; if infinite sets are allowed, one should also treat N>0\mathbb N_{>0} as the infinite initial interval exception.

    Citation: No literature counterexample is needed; the proof uses only the standard undecidability of the halting problem.

  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 proposed argument validly disproves the conjecture as literally stated for arbitrary sets A\mathcal A. Taking a nonrecursive A⊆{3,4,… }\mathcal A\subseteq\{3,4,\dots\}, directed cycles CnC_n encode membership: CnC_n has such a handle decomposition iff n∈An\in\mathcal A. Hence LAL_{\mathcal A} would decide a nonrecursive set if it were decidable. Since every NP-complete language is in NP and therefore decidable, LAL_{\mathcal A} is not NP-complete, while A\mathcal A is not an initial interval. This only refutes the literal arbitrary-set formulation, not a repaired finite/decidable-A\mathcal A version.

    Novelty assessment

    TYPE1

    Classification rationale: The resolution is genuinely a valid literal refutation, but it is only a definitional/effectivity loophole: choose a nonrecursive length set A\mathcal A, so the induced decision problem is undecidable and hence not in NP. This does not advance the graph-theoretic complexity classification for finite or decidable A\mathcal A, and would not support a standalone combinatorics paper.

    Literature check: I found no prior published counterexample or erratum making this exact observation. Searches of the original DMTCS/DOAJ records, open mirrors, DuckDuckGo/Yahoo results, and bibliographic data located only the Havet–Nisse paper and mirrors; OpenAlex lists no citing works. Targeted searches for phrases such as “A-handle-decomposition NP-complete”, “nonrecursive set handle decomposition”, and the conjecture wording did not reveal an existing resolution.

    Citation: Original source: Frédéric Havet and Nicolas Nisse, “Constrained ear decompositions in graphs and digraphs,” Discrete Mathematics & Theoretical Computer Science 21(4), 2019, DOI: 10.23638/DMTCS-21-4-3. The counterexample uses only the standard undecidability of the halting problem.

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