Given an integer partition of $n$, we consider the impartial combinatorial game LCTR in which moves consist of removing either the left column or top row of its Young diagram. We show that for both normal and mis\`ere play, the optimal strategy can consist mostly of mirroring the opponent's moves. We also establish that both LCTR and Downright are domestic as well as returnable, and on the other hand neither tame nor forced. For both games, those structural observations allow for computing the Sprague-Grundy value any position in $O(\log(n))$ time, assuming that the time unit allows for reading an integer, or performing a basic arithmetic operation. This improves on the previously known bound of $O(n)$ due to Ili\'c (2019). We also cover some other complexity measures of both games, such as state-space complexity, and number of leaves and nodes in the corresponding game tree.

翻译：给定整数$n$的一个分拆，我们考虑无偏组合博弈LCTR，其移动方式为移除杨图的最左列或最上一行。我们证明，在正常规则和反常规则下，最优策略均可主要由镜像对手的移动构成。还证明了LCTR与Downright既是Domestic又是Returnable，但既非Tame也非Forced。基于这些结构观察，对于这两个博弈，若时间单位允许读取整数或执行基本算术运算，则可在$O(\log(n))$时间内计算任意局面的Sprague-Grundy值。这一结果改进了Ili\'c（2019）先前得到的$O(n)$界。此外，我们还讨论了这两个博弈的其他复杂度度量，如状态空间复杂度及对应博弈树中叶节点与内部节点数量。