You find a stain on the wall and decide to cover it with non-overlapping stickers of a single identical shape (rotation and reflection are allowed). Is it possible to find a sticker shape that fails to cover the stain? In this paper, we consider this problem under polyomino constraints and complete the classification of always-coverable stain shapes (polyominoes). We provide proofs for the maximal always-coverable polyominoes and construct concrete counterexamples for the minimal not always-coverable ones, demonstrating that such cases exist even among hole-free polyominoes. This classification consequently yields an algorithm to determine the always-coverability of any given stain. We also show that the problem of determining whether a given sticker can cover a given stain is $\NP$-complete, even though exact cover is not demanded. This result extends to the 1D case where the connectivity requirement is removed. As an illustration of the problem complexity, for a specific hexomino (6-cell) stain, the smallest sticker found in our search that avoids covering it has, although not proven minimum, a bounding box of $325 \times 325$.

翻译：你在墙上发现一处污渍，决定用单一相同形状的贴纸（允许旋转和反射）不重叠地覆盖它。是否存在某种贴纸形状无法覆盖该污渍？本文在多联骨牌约束下研究该问题，完成了始终可覆盖污渍形状（多联骨牌）的完整分类。我们证明了最大规模的始终可覆盖多联骨牌，并为最小非始终可覆盖情形构造了具体反例，表明此类情况即使在没有空洞的多联骨牌中依然存在。该分类结果进而产生了一种判定任意给定污渍是否始终可覆盖的算法。我们还证明：判定给定贴纸能否覆盖给定污渍的问题是 $\NP$ 完全的，尽管并不要求精确覆盖。此结果可推广至移除连通性要求的一维情形。为说明问题复杂性，针对某个特定六联骨牌（6单元格）污渍，我们在搜索中找到的能避免覆盖它的最小贴纸（虽未证明为理论最小）其边界框尺寸达到 $325 \times 325$。