We study the impartial game PAP (``permutations avoiding patterns''), in which players take turns choosing patterns to avoid. We define a set of length $k$ patterns, $B_k$, and show that it is the unique minimal monotone-forcing subset of $S_k$: every sufficiently long permutation that avoids $B_k$ is monotone, and every monotone-forcing subset of $S_k$ must contain $B_k$. We prove a quadratic upper bound for the monotone-forcing threshold, and determine the exact thresholds for $k=3,4,5,6$. We use properties of the sets $B_k$ to prove that a reverse-reply strategy wins PAP on $S_n$ when $k=4$ for all $n \geq 10$; for $k=3$, the same strategy can be analysed directly. We conjecture that it is a winning strategy for all $k$ and $n$ sufficiently large.

翻译：我们研究无偏博弈PAP（“避排列模式”），玩家轮流选择要避免的模式。定义长度为$k$的模式集合$B_k$，并证明它是$S_k$中唯一的最小单调强迫子集：每个充分长的避开$B_k$的排列都是单调的，且$S_k$中每个单调强迫子集必定包含$B_k$。我们给出单调强迫阈值的二次上界，并确定$k=3,4,5,6$时的精确阈值。利用$B_k$集合的性质证明：当$k=4$时，对所有$n \geq 10$，反转回应策略在$S_n$上的PAP博弈中获胜；对于$k=3$，同一策略可直接分析。我们猜想对所有充分大的$k$和$n$，该策略均为获胜策略。