Constrained ear decompositions in graphs and digraphs
Statement
Finally, we conjecture that, for every set of integers, deciding whether a digraph has a handle decomposition with all handles of length in is NP-complete, unless there exists such that .
Context
Candidate 1 of the open problems stated in "Constrained ear decompositions in graphs and digraphs", extracted for the Scalable Mathematical Discovery run.
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Attempts
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.
NEW
Problem: Literal reconstruction: for every fixed set , let be the language of finite digraphs having a directed handle/ear decomposition in which every handle has length in . The conjecture asserts that is NP-complete unless for some . This reconstruction follows the quoted phrase “for every set of integers”; no finiteness or decidability condition on is stated.
Result: The literal conjecture is false.
Let be a nonrecursive set, for example a shifted halting set, and set . Then is not equal to any finite initial interval .
For , let be the directed cycle of length . Under the standard handle-decomposition convention, the first handle is a directed cycle and is constrained by . The only directed cycle contained in is itself. Hence
Indeed, if , the one-handle decomposition consisting of is valid; conversely, any valid decomposition of must start with the whole cycle , whose length is , so .
If were NP-complete, then in particular , hence would be decidable. But then membership in the nonrecursive set would be decidable by mapping to and deciding whether , 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 natural repair would restrict to finite or at least decidable sets of positive integers; if infinite sets are allowed, one should also treat as the infinite initial interval exception.
Citation: No literature counterexample is needed; the proof uses only the standard undecidability of the halting problem.
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Machine check · not human verification
machine: correctscope Full solution as submitted; SMD novelty classification TYPE1
PASS
The proposed argument validly disproves the conjecture as literally stated for arbitrary sets . Taking a nonrecursive , directed cycles encode membership: has such a handle decomposition iff . Hence would decide a nonrecursive set if it were decidable. Since every NP-complete language is in NP and therefore decidable, is not NP-complete, while is not an initial interval. This only refutes the literal arbitrary-set formulation, not a repaired finite/decidable- 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 , 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 , 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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