gpt 6 astra ultra attempt on Math-1079

Strong 5-cycle double cover conjecture

Definitions and mathematical statement

Let $G=(V,E)$ be a finite simple cubic graph with no bridge, and fix a circuit $C$ of $G$. Here cubic means degree three at every vertex, a bridge is an edge whose removal increases the number of connected components, and a circuit is one connected simple cycle.

For this problem a covering cycle means an edge set $F\subseteq E$ for which every vertex of $(V,F)$ has degree zero or two. Thus its nontrivial components are disjoint circuits, and $F$ may be empty. A five-cycle double cover is a list $F_1,\ldots,F_5$ of such edge sets satisfying \[ \sum_{i=1}^5\mathbf1_{\{e\in F_i\}}=2 \qquad(e\in E). \] Repetitions and empty members are allowed, so the phrase "five" means at most five nonempty covering members.

The strong conjecture requires that for every pair $(G,C)$ one can choose the cover with $E(C)\subseteq F_j$ for some index $j$. Since the degrees inside $F_j$ are at most two, the given circuit then occurs as a whole component of $F_j$; additional disjoint circuits may occur there too.

It is essential that the five members need not themselves be connected. The statement does not ask for five individual circuits, and merely ensuring that the edges of $C$ appear somewhere in the double cover is insufficient. All edges of the specified circuit must belong to one common member.

Short English statement

Can every prescribed circuit in a bridgeless cubic graph be a component of one member of a five-cycle double cover?

Research attempt: an even-factor criterion and a Petersen certificate

Scope and current literature.

GPT-6 Astra Ultra, 1 October 2026. The prescribed circuit must occur as a whole component of one common covering member; covering its edges somewhere is insufficient. The source problem [S2] records this stronger condition. Oum's 2026 exposition [S3] presents the announced ordinary cycle double cover proof and separately discusses the five-member conjecture. That ordinary theorem imposes neither the five-member bound nor the prescribed-component requirement. The present attempt proves a sufficient matching condition and verifies a non-three-edge-colourable example with an odd prescribed circuit.

1. A useful sufficient condition.

Suppose $G$ has a perfect matching $M$ disjoint from $E(C)$ such that every component of the complementary two-factor $F=E(G)\setminus M$ has even length. Since every vertex of $C$ already has its two $C$-edges in $F$, the circuit $C$ is a whole component of $F$. Alternately colour each component of $F$ with colours one and two, and colour all matching edges with colour three. This is a proper three-edge-colouring: each vertex has one incident edge of each colour.

Let $F_{12},F_{13},F_{23}$ be the unions of the indicated colour classes. Each has degree two at every vertex, so each is an allowed covering member. Each edge belongs to exactly two of these three members. Moreover $F_{12}=F$ contains $C$ as a whole component. Appending two empty members gives a five-cycle double cover satisfying the prescribed-circuit condition. This proves the strong target whenever the stated matching and even-factor condition can be certified.

In particular, if $C$ is Hamiltonian, the unused edges of the cubic graph form a perfect matching. A cubic graph has an even number of vertices, by the degree sum $3|V|=2|E|$, so this Hamilton cycle has even length. The condition applies and proves the target for every prescribed Hamilton circuit. For a nonspanning even circuit, its parity alone does not guarantee the required matching or the parities of the other components.

2. An explicit odd-circuit example.

Use Petersen vertices $0,\ldots,9$ and index its edges as follows: \[ \begin{array}{c|rrrrrrrr} j&0&1&2&3&4&5&6&7\\ e_j&01&12&23&34&04&05&16&27\\[2pt] j&8&9&10&11&12&13&14&\\ e_j&38&49&57&68&79&58&69& \end{array} \] Here $ab$ denotes the unordered edge with endpoints $a,b$. Prescribe the outer five-cycle $C$ with edge indices $0,1,2,3,4$. The following five sets are a certificate: \[ \begin{aligned} F_1&=\{0,1,2,3,4\},\\ F_2&=\{0,2,5,6,7,8,10,11\},\\ F_3&=\{3,8,9,10,12,13\},\\ F_4&=\{1,6,7,12,14\},\\ F_5&=\{4,5,9,11,13,14\}. \end{aligned} \] Interpreting each index as its listed edge, every vertex has degree zero or two in each $F_i$. Counting the occurrences of every index $0,\ldots,14$ gives exactly two, and $F_1$ is precisely the prescribed circuit. The script [S4] checks these statements directly, and also enumerates all even edge sets and reproduces a search for such a certificate. The example demonstrates that the three-colouring criterion is sufficient but not necessary.

3. A genuine limit of the colouring approach.

A component using only two colours in a proper edge-colouring must alternate them and therefore have even length. An odd prescribed circuit can never be a component of any of the three bicoloured members produced in paragraph 1. Furthermore the same script checks that all six perfect matchings of Petersen have complements consisting of two five-cycles, ruling out a proper three-edge-colouring there. The explicit five-member certificate overcomes both limitations on this one input, without supplying a general construction for odd circuits.

Verification and first remaining gap.

The edge list fixes all labelling conventions and the certificate contains no implicit graphical choices. The search is finite and uses exact parity and incidence counts; it establishes only this specified Petersen pair $(G,C)$. For an arbitrary prescribed circuit, no matching with the required even complement is guaranteed, and no argument here bounds a more general repair procedure by five members while keeping $C$ intact.

Final status

UNRESOLVED. The even-factor criterion, all prescribed Hamilton circuits, and the displayed odd Petersen circuit are handled. The universal strong five-cycle double cover conjecture is not proved. Completion estimate: no defensible global percentage.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3173 (OPG-37241), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Strong 5-cycle double cover conjecture. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/strong_5_cycle_double_cover_conjecture.
• S3 S.-i. Oum, A proof of the cycle double cover conjecture by OpenAI: An exposition, arXiv:2607.16356v3 $2026$, especially Section 9.2. Checked 1 October 2026. https://arxiv.org/abs/2607.16356.
• S4 GPT-6 Astra Ultra, exact Petersen certificate and exhaustive finite search, 1 October 2026: scripts/check_attempt_batch02_connectivity_2026_10_01.py in this repository; run with Python 3.