gpt 6 astra ultra attempt on CS-384

Balanced signing: an explicit Euler-tour flow and the denominator obstruction

Definitions and mathematical statement

Shared notation.

Standard number systems are denoted by $\mathbb N,\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$; index ranges, computational models and specialized notation follow the cited problem source and the definitions below.

For an undirected graph let $\delta(S)$ be the edges with exactly one endpoint in $S$. In the source's subtour cut relaxation, $x_e\geq0$ and $x(\delta(S))\geq2$ for every nonempty proper $S\subset V$. For an even terminal set $T$, a balanced signing is $s:T\to\{-1,1\}$ with $\sum_{v\in T}s(v)=0$. A capacitated transshipment routes these supplies and demands using edge capacities $x_e$; its cut feasibility condition is $|\sum_{v\in S\cap T}s(v)|\leq x(\delta(S))$ for every $S\subseteq V$.

Statement recorded in the supplied source.

For every point of the subtour-elimination polytope and every even terminal set, does there exist a balanced signing realizable by a capacitated transshipment? [S2]

Scope qualification.

The source passage uses the cut-based relaxation without degree equations. The displayed capacities are the coordinates of the given point; there is no additional scaling of capacities in this question.

Source status.

The supplied catalogue labels this record open. Its source-status metadata was checked against the supplied snapshot on 2026-10-11. This import does not independently establish current open status or mathematical correctness. The catalogue literature assessment is dated 2026-08-22.

Short English statement

Choose equally many source and sink terminals so their unit demands can be routed within the subtour point's edge capacities.

Sources

• S1 Ulam AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 30004910 (OWR-8415356-021). Catalogue curation and metadata: CC BY 4.0. The supplied source question is retained; definitions and the English summary are editorial additions. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Combinatorial Optimization $2022$, Oberwolfach Reports, DOI 10.4171/owr/2021/53, pp. 2893--2954 https://doi.org/10.4171/owr/2021/53.
• S3 Combinatorial Optimization, Oberwolfach Reports 53/2021, primary report PDF https://ems.press/content/serial-article-files/46931. Underlying Oberwolfach Reports material retains its source-specific licence, including CC BY-SA 4.0 where applicable; see THIRD_PARTY_NOTICES.md.

Verbatim supplied catalogue question

For every point of the subtour-elimination polytope and every even terminal set, does there exist a balanced signing realizable by a capacitated transshipment?

A constructive special case

The primary question requires capacities $x$ themselves. We prove an explicit signing and flow when $2x$ describes a connected Eulerian multigraph, and examine why clearing arbitrary denominators does not preserve the unit supplies. The source's broader half-integral discussion is not imported without checking its Eulerian hypothesis.

Euler-tour construction with exact capacities.

Assume $m_e=2x_e$ are nonnegative integers and the multigraph with $m_e$ copies of edge $e$ is connected on the relevant vertices and has even degree at every vertex. It has a closed walk using each edge copy once. For completeness, follow unused edges from any start until returning to it: at a different vertex, the number of used incidences upon an arrival is odd, so an unused edge remains unless all incidences there had odd total, contrary to even degree. This forms a closed trail. If edges remain, connectivity supplies a vertex of this trail incident to an unused component; form another closed trail there and splice it into the first. The number of unused edges decreases, so this process terminates with an Euler tour.

Let $T$ have even size. If $T$ is empty use zero flow. Otherwise mark one occurrence, for example the first occurrence, of every terminal on the cyclic tour. There are $2h$ marks. Give them alternating signs $+1,-1$ in tour order. On the tour segment from mark $i$ to mark $i+1$, send signed flow $+1/2$ in the tour direction if $i$ has positive sign, and $-1/2$ in the tour direction otherwise; indices are cyclic. At an unmarked occurrence the incoming and outgoing signed values coincide, so the divergence is zero. At a marked occurrence the outgoing minus incoming signed values equal its assigned sign. Even though a vertex may occur elsewhere on the tour, those unmarked occurrences contribute zero. Thus the resulting divergence is the required unit sign at each terminal and zero elsewhere. The number of positive and negative marks is equal, so the signing is balanced.

Aggregate signed traversals on each original undirected edge, using an arbitrary reference direction. The absolute aggregate is at most the sum of traversal magnitudes, namely $m_e/2=x_e$. Cancellation can only decrease it. Therefore this is a transshipment within precisely the original capacities, with no appeal to a separate cut-sufficiency theorem. It proves the desired conclusion for this explicitly stated Eulerian special case.

What arbitrary denominator clearing actually gives.

For rational $x$, choose an integer $D$ with $Dx$ integral. The multigraph with multiplicities $2Dx_e$ has even degrees and is connected when the positive-capacity support is connected. The same construction gives unit supplies, but its capacity bound is $Dx_e$. Dividing the flow by $D$ restores capacities $x_e$ and simultaneously divides every nonzero supply by $D$. Those supplies are $\pm1/D$, not the required $\pm1$. Thus denominator clearing alone does not extend the proof with a capacity factor independent of $D$. The missing step is to coordinate sufficiently many terminal markings without increasing edge usage, rather than merely rescale the Euler tour.

Averaging feasible signings does not repair the problem.

The balance and flow constraints are linear once the signs are fixed, but the requirement $s(v)\in\{-1,1\}$ is not convex. If $(s,f)$ is a valid nonempty-terminal construction, reversing all signs and flows gives $(-s,-f)$ within the same capacities. Their average is $(0,0)$, which has no allowed terminal signs. Hence a convex decomposition of capacities into convenient cases, together with an arbitrary signing for each case, does not produce a valid signing by averaging. This does not prove that compatible choices of signs cannot exist; their compatibility is precisely an additional unresolved requirement of that route.

Audit

The primary signing problem and Euler-walk remark on printed page 2946 were inspected visually and in text. The audit checked repeated vertex visits, orientation cancellation, the last-to-first tour segment, empty $T$, and conservation at unmarked occurrences. It also made the even-degree condition explicit: integrality of $2x$ alone does not force every degree to be even after degree equations have been dropped. No experiment or general splitting-off theorem was used. The proof is a known Eulerian mechanism and does not settle arbitrary fractional subtour capacities.

Process and contributions

Mathematics and model-assisted adversarial self-audit: OpenAI gpt-6-astra, ultra reasoning, as dispatched for this research batch; exact serving revision not exposed. Codex (GPT-6; exact serving revision not exposed) materially assisted with orchestration and format review. No human mathematical authorship, mathematical edits, or independent human proof checking is asserted. Contribution prompt A was applied to the full definition and supplied source row, then prompt B was applied adversarially to the derived results. The item-specific audit above records the scope checks and rejected extrapolations. Local assembly and structural checks do not establish mathematical correctness. The source catalogue status is preserved in TOP_PROBLEM and is separate from this attempt status.

Integration review.

Codex (GPT-6; exact serving revision not exposed) reviewed the written arguments and expanded layout and source-reference wrappers. Repository metadata and the production web renderer were checked using Python 3.13 and JavaScript with MathJax 3.2.2. This model-assisted review is not independent human proof verification; no PDF compilation or formal proof checking is asserted.

Completion Estimate

COMPLETION: 0% The proved material is a rederivation of known or auxiliary facts rather than a substantiated new advance on the original target. The exact Eulerian special-case flow is proved, but arbitrary denominators and incompatible sign choices remain obstacles to extending it to every subtour point.

References

The complete original Sources subsection above is retained, including catalogue attribution and licence notices. The additional list records primary sources actually consulted for this attempt on 11 October 2026; uncited source links reproduced above were not all rechecked.

• Oberwolfach Report 53/2021, Combinatorial Optimization, original question and surrounding discussion read from the primary PDF text on 11 October 2026. Historical statements are not a current-status survey. https://ems.press/content/serial-article-files/46931.

Final status

Unresolved in this attempt. The exact Eulerian special-case flow is proved, but arbitrary denominators and incompatible sign choices remain obstacles to extending it to every subtour point.