On regular packings and coverings
Statement
K_{12t} minus a hamiltonian cycle K_{12t+3} minus a hamiltonian cycle K_{12t+8} plus a hamiltonian cycle can be decomposed into K_4's K_{12t+11} plus a hamiltonian cycle
Record
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- On regular packings and coverings
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exploration by a model · #1
GPT-5.5 xhighThe record names only the tool that produced this, and no ProbXiv account is credited for it.
NEW
Problem: Formalized literally: for all admissible integers , the graphs/multigraphs
decompose into copies of , where is a Hamiltonian cycle and means adding a second parallel copy of the cycle edges. The text supports this because these are exactly the residue classes where the usual degree and edge-count divisibility obstructions for -decomposition are repaired by deleting/adding a Hamiltonian cycle.
Result: The literal statement is false. Take in the first family: has no -decomposition.
Label the Hamiltonian cycle . If a decomposition existed, it would contain
copies of . Each block must be a 4-subset with no cyclically consecutive vertices.
For such a block , let be the number of pairs of vertices of at cyclic distance , for . Across the whole decomposition the required total is
since there are 12 pairs at each distance , and 6 opposite pairs at distance .
The possible cyclic gap patterns for an independent 4-subset of , up to rotation/reflection, and their profiles , are:
Solving the resulting linear count equations forces exactly six blocks of type and three blocks of type , with no other types.
Now let , , be the six opposite pairs. A block of type is necessarily of the form
so it contains exactly the two opposite pairs and . Thus the three type- blocks would have to form a perfect matching in the graph on with edges . But this graph is the disjoint union of two triangles, hence has no perfect matching. Contradiction.
Therefore cannot be decomposed into 's. The conjecture needs at least the missing small-order exception in the case.
Citation: No external counterexample is used; the disproof above is direct. The conjecture source is Bermond–Bond–Sotteau, “On regular packings and coverings,” Annals of Discrete Mathematics 34 (1987), 81–100.
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 attacks the supplied statement: for , the first family is . The proof is mathematically sound: any block must be an independent 4-set in the deleted cycle; the distance/gap profile count is correct and forces three blocks of type . Those blocks would have to cover the six opposite pairs as a perfect matching in the graph with edges , which is two disjoint triangles and has no perfect matching. Thus has no -decomposition, disproving the conjecture as stated.
Novelty assessment
TYPE1
Classification rationale: This is a valid small counterexample to the literal statement: has no -decomposition. Its significance is minor: it only removes the case of one family and does not address the intended infinite cases or the other residue classes. The proof is an elementary finite counting obstruction, suitable as an erratum/remark rather than a standalone paper.
Literature check: I searched for the exact conjecture and variants using terms such as “ decomposition”, “complete graph minus a Hamiltonian cycle -decomposition”, “ packing leave cycle”, “quadruple packing leave Hamiltonian cycle”, and the Bermond–Bond–Sotteau title/DOI. I also checked OpenAlex metadata for the original paper and general references around regular packings/coverings. I did not find a published source explicitly recording this obstruction or a stronger classification implying it. Existing broad graph-decomposition theorems concern large admissible orders and do not make this small exception known.
Citation: J.-C. Bermond, J. Bond, D. Sotteau, “On regular packings and coverings,” Annals of Discrete Mathematics 34 (1987), 81–99, DOI: 10.1016/S0304-0208(08)72877-9.
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