Bipartite friends and strangers walking on bipartite graphs

Abstract

Given n-vertex simple graphs X and Y, the friends-and-strangers graph $\mathrm{FS}(X,Y)$ has as its vertices all n! bijections from $V\left(X\right)$ to $V\left(Y\right)$, where two bijections are adjacent if and only if they differ on two adjacent elements of $V\left(X\right)$ 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, $\mathrm{FS}(X,Y)$ is always disconnected in this setting. Modestly improving a result of Bangachev, we show that if X and Y respectively have minimum degrees $\delta \left(X\right)$ and $\delta \left(Y\right)$ and they satisfy $\delta \left(X\right)+\delta \left(Y\right)\ge \lfloor 3r/2\rfloor +1$, then $\mathrm{FS}(X,Y)$ has exactly two connected components. This proves that the cutoff for $\mathrm{FS}(X,Y)$ to avoid isolated vertices is equal to the cutoff for $\mathrm{FS}(X,Y)$ 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 $(\log r)/r$. More precisely, below the threshold, $\mathrm{FS}(X,K_{r,r})$ has more than two connected components with high probability, while above the threshold, $\mathrm{FS}(X,K_{r,r})$ has exactly two connected components with high probability. Altogether, our results settle a conjecture and completely answer two problems of Alon, Defant, and Kravitz.