Hexahedral Meshing

Definitions and mathematical statement

Let $P\subset\mathbb R^3$ be a polyhedral solid homeomorphic to a closed three-ball. This makes the connected-boundary domain convention explicit; simply connected solids with separate cavity boundaries would enlarge the class. The boundary is given a conforming quadrilateral mesh: faces are planar quadrilaterals with disjoint relative interiors, meet edge-to-edge, and cover $\partial P$. Assume the number $b$ of boundary quadrilaterals is even.

A geometric hexahedral mesh is a finite face-to-face subdivision of $P$ into embedded polyhedral cells combinatorially equivalent to a cube. Every cell has six planar quadrilateral faces; cells need not be congruent, rectangular, or aligned with coordinate axes. New interior vertices are allowed, but the induced boundary mesh must be exactly the prescribed one, without subdividing its faces.

Does every such boundary mesh extend to a geometric hexahedral mesh? The parity condition is necessary: if $h$ is the number of cells and $i$ the number of interior faces, then $6h=2i+b$.

The source reports extensions by topological cells and by cells with straight edges but possibly nonplanar faces. The unresolved formulation retained here requires planar faces. Consequently, a topological extension alone does not answer it, and "hexahedron" does not mean an axis-aligned rectangular box.

Short English statement

Can every even quadrilateral mesh on the boundary of a polyhedral ball be filled by cube-shaped combinatorial cells whose faces are all planar?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 5500027 (AMR-054-0027), 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 27. Original source indexed by AMR-054-0027. https://topp.openproblem.net/p27.