加载中…
We prove that if a tile in has prime size , then it must be spectral. The proof is by contradiction, it is simply shown that the tiling complement of such a tile can not annihilate all -subgroups. In addition, with a simple transformation we prove that any points in general linear positions in must be both tiling and spectral.

生成讲解博客的中文版本

生成AI智能校验的中文版本