A convex polyhedron is Rupert if a hole can be cut into it (making its genus $1$) such that an identical copy of the polyhedron can pass through the hole. Resolving a conjecture of Jerrard-Wetzel-Yuan, Steininger and Yurkevich recently constructed a convex polyhedron which is not Rupert. We propose a search for the simplest possible non-Rupert polyhedron and provide numerical evidence suggesting that a particular stellated tetrahedron is not Rupert. The computational techniques utilize linear program solvers to compute the largest possible scalings of polygons that can be translated to fit in other polygons. The relative simplicity of the stellated tetrahedron as compared to other polyhedra allows this more rudimentary check to be computationally tractable. In particular, we show that over 88% of a particular encoding of $\text{SO}(3) \times \text{SO}(3)$ equipped with the standard measure does not yield a Rupert passage.

翻译：一个凸多面体被称为鲁伯特的，如果能在其上挖出一个洞（使其亏格为$1$），且该多面体的一个全等副本能通过这个洞。为解决Jerrard-Wetzel-Yuan的一个猜想，Steininger和Yurkevich近期构造了一个非鲁伯特的凸多面体。我们提出寻找可能最简单的非鲁伯特多面体，并提供数值证据表明，一个特定的星形四面体是非鲁伯特的。该计算技术利用线性规划求解器计算多边形可平移放入其他多边形中的最大缩放比例。由于星形四面体相较于其他多面体更为简单，使得这种基础性检验在计算上易于处理。特别地，我们证明在$\text{SO}(3) \times \text{SO}(3)$配备标准测度的特定编码中，超过88%的区域不产生鲁伯特通道。