Given $n$-vertex simple graphs $X$ and $Y$, the friends-and-strangers graph $\mathsf{FS}(X, Y)$ has as its vertices all $n!$ bijections from $V(X)$ to $V(Y)$, where two bijections are adjacent if and only if they differ on two adjacent elements of $V(X)$ whose mappings are adjacent in $Y$. We consider the setting where $X$ and $Y$ are both edge-subgraphs of $K_{r,r}$: due to a parity obstruction, $\mathsf{FS}(X,Y)$ is always disconnected in this setting. Sharpening a result of Bangachev, we show that if $X$ and $Y$ respectively have minimum degrees $\delta(X)$ and $\delta(Y)$ and they satisfy $\delta(X) + \delta(Y) \geq \lfloor 3r/2 \rfloor + 1$, then $\mathsf{FS}(X,Y)$ has exactly two connected components. This proves that the cutoff for $\mathsf{FS}(X,Y)$ to avoid isolated vertices is equal to the cutoff for $\mathsf{FS}(X,Y)$ to have exactly two connected components. We also consider a probabilistic setup in which we fix $Y$ to be $K_{r,r}$, but randomly generate $X$ by including each edge in $K_{r,r}$ independently with probability $p$. Invoking a result of Zhu, we exhibit a phase transition phenomenon with threshold function $(\log r)/r$: below the threshold, $\mathsf{FS}(X,Y)$ has more than two connected components with high probability, while above the threshold, $\mathsf{FS}(X,Y)$ has exactly two connected components with high probability. Altogether, our results settle a conjecture and completely answer two problems of Alon, Defant, and Kravitz.

翻译：给定$n$顶点简单图$X$和$Y$，友人与陌生人图$\mathsf{FS}(X, Y)$以所有$n!$个从$V(X)$到$V(Y)$的双射作为顶点，两个双射相邻当且仅当它们在$V(X)$的两个相邻元素上取值不同，且这两个元素的像在$Y$中相邻。我们考虑$X$和$Y$均为$K_{r,r}$边子图的情形：由于奇偶性障碍，此时$\mathsf{FS}(X,Y)$总是非连通的。通过对Bangachev的结果进行精化，我们证明：若$X$和$Y$的最小度分别满足$\delta(X)$和$\delta(Y)$，且$\delta(X) + \delta(Y) \geq \lfloor 3r/2 \rfloor + 1$，则$\mathsf{FS}(X,Y)$恰好有两个连通分支。这表明$\mathsf{FS}(X,Y)$避免孤立顶点的阈值等于其恰好具有两个连通分支的阈值。我们还考虑一种概率设定：固定$Y=K_{r,r}$，但通过以概率$p$独立保留$K_{r,r}$中的每条边来随机生成$X$。借助Zhu的结果，我们揭示了以$(\log r)/r$为阈值函数的相变现象：低于阈值时$\mathsf{FS}(X,Y)$高概率具有多于两个连通分支，而高于阈值时$\mathsf{FS}(X,Y)$高概率恰好有两个连通分支。综合来看，我们的结果解决了一个猜想并完全回答了Alon、Defant和Kravitz提出的两个问题。