Hamiltonian Tetrahedralizations

Research-attempt provenance

Definitions and mathematical statement

Let $P\subset\mathbb R^3$ be a convex three-dimensional polytope. A tetrahedralization $\mathcal T$ partitions $P$ into finitely many tetrahedra meeting face-to-face, with vertices drawn only from the vertices of $P$. Subdivision of a nonsimplicial boundary face by diagonals is allowed; insertion of new vertices is not.

The dual graph $G_{\mathcal T}$ has one vertex for each tetrahedron, with two vertices adjacent precisely when the corresponding tetrahedra share a triangular face. It has a Hamiltonian path if its tetrahedra can be ordered $T_1,\ldots,T_m$ so that every tetrahedron occurs once and $T_i,T_{i+1}$ share a face for each $i<m$.

Does every convex $P$ admit a tetrahedralization with this property? The subdivision itself may be chosen to make the path possible; the claim is not that every tetrahedralization works. The path is in the dual graph, not a path through the original polytope vertices.

Allowing new Steiner points gives a different problem and is known to permit additional constructions. The source's unrestricted vertex-preserving existence question is the one retained here. No bound on the number of tetrahedra beyond finiteness is part of the main assertion.

Short English statement

Can every convex three-dimensional polytope be tetrahedralized, using its existing vertices, so that all tetrahedra can be visited once in an order crossing a shared face at each step?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 5500029 (AMR-054-0029), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problems Project - Computational Geometry, Problem 29. Original source indexed by AMR-054-0029. https://topp.openproblem.net/p29.

Literature and context check

The consulted TOPP Problem 29 page labels the question open and records obstructions for particular pulling triangulations. The supplied prior narrative asserting a general positive solution is internally inconsistent with that source; this attempt does not adopt it. The argument below is independent of that literature and tests a specific shelling-based shortcut.

Research attempt: Shellability does not supply a Hamiltonian dual path

A necessary condition for a finite graph $G$ to possess a Hamiltonian path is \[ \#\operatorname{components}(G-S)\le |S|+1 \quad\hbox{for every vertex set }S. \] Indeed, deleting $|S|$ vertices from a path leaves at most $|S|+1$ nonempty path pieces. Each component of $G-S$ contains at least one such piece because the path visited every vertex; graph edges may merge pieces but cannot increase the number of components.

Now start with a tetrahedron $T$ and stack one tetrahedron on each of three distinct boundary faces, with no further stacking. This is geometrically realizable as a convex polytope with a vertex-preserving tetrahedralization. Here is a justification of the geometric step often omitted from the graph picture. At any chosen triangular facet of a convex polytope, its barycenter satisfies strictly all other facet inequalities. Moving it a sufficiently small positive distance along the outward normal keeps those finitely many strict inequalities and violates only the chosen facet inequality. The convex hull with this new vertex adds exactly the pyramid over that facet; the visible boundary consists of that facet alone. The other two original facets remain facets. Apply this operation successively on the three selected original facets. Each new vertex is extreme, so all tetrahedra use only vertices of the resulting convex polytope.

The resulting four-tetrahedron dual graph is $K_{1,3}$: each new tetrahedron shares a triangle with $T$, while two new ones meet, if at all, along an edge or vertex of $T$. The order consisting of $T$ followed by its three attachments is a shelling in the direct sense that each attachment meets the preceding union in a single triangular facet. Yet the dual has no Hamiltonian path: deleting its center leaves three components, violating the bound $3\le2$.

Thus shelling, connectivity, and even convex realizability of every intermediate stacking step do not turn an arbitrary traversal into a face-adjacent Hamiltonian traversal. A depth-first walk would return through the central tetrahedron repeatedly. The problem asks for existence of some triangulation, however; the demonstrated failure of this triangulation leaves alternative tetrahedralizations entirely available. Any repair must modify the subdivision to eliminate such dual separation obstructions, rather than merely reorder the given shelling.

Verification and adversarial audit

The construction uses three separate facet stackings, and the strict inequality argument verifies convexity and extreme vertices without placing Steiner vertices in the interior. The dual's four vertices and three edges were counted explicitly. One or two attached tetrahedra instead have path duals, confirming the smallest branching obstruction in this family.

Audit corrections.

The prior unsupported solved narrative was rejected. Failure of one shellable triangulation was not confused with failure of the existential original target.

External computation.

None; symbolic arguments and boundary checks only.

Scope and remaining gaps

There is no proof that every convex polytope has a triangulation whose dual avoids all separating-set obstructions, and avoiding those obstructions alone would still be only necessary for Hamiltonicity. No negative example for all tetrahedralizations of one polytope is claimed.

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 calculations below the preserved source statement investigate a concrete route, but establish no substantiated new progress toward the entire original target. Known special cases and failed sufficient conditions are not credited as a solution.

References

The original source references and catalogue attribution are preserved in the Sources section above. The following URLs were actually consulted for this attempt; retention of an additional original reference does not claim fresh consultation of it.

• https://topp.openproblem.net/p29 (consulted 2026-10-11).

Model and contributor disclosure

OpenAI gpt-6-astra, requested reasoning configuration ultra, generated this research attempt on 2026-10-11. The exact serving revision is not exposed. Codex GPT-6 performed orchestration and repository-format review; its exact serving revision is also not exposed. The mathematical route, boundary checks and final adversarial self-audit followed contribution prompts A then B in docs/contribution_prompts.md, after reading CONTRIBUTING.md. Python 3.14.2 assembled the standalone documents; any mathematical computation is specified in Verification. No theorem prover was used. No human mathematical authorship, substantial human proof correction, or independent human proof checking is asserted. Model-assisted review, source retrieval, and structural checks do not establish correctness or novelty. Full private reasoning transcripts are unavailable as reproducibility artifacts; the complete arguments, inputs and outcomes actually used are stated here or in the companion batch audit.

Final status

Unresolved. This is a research attempt with an explicit failed route or restricted calculation, not a full solution, not a current-status certification, and not a claim that the elementary material is new. The catalogue's original source-status label is preserved independently of this attempt status.