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My question concerns the sensitivity of maximum-size matchings (and more generally maximum-size $k$-cycle collections) to deletion of an edge in the graph.

Given a graph $G$, let a $k$-cycle be a cycle in the graph of length at most $k$. Let a $k$-allocation be a collection of disjoint $k$-cycles. For an allocation $A$ and node $v$, let $A(v)$ be the cycle containing $v$, where $A(v) = \varnothing$ if $v$ is not in a cycle. Let $G\setminus e$ be the graph obtained by deleting the edge $e$. Let $M^k(G)$ be the set of maximum-size $k$-allocations.

Consider an Erdos-Renyi random graph $G(m,p(m))$ with $m$ nodes, where between each node an undirected edge exists with probability $p(m)$ (and $p$ is a non-decreasing function). My question concerns the following: suppose we generate a graph $G(m,p(m))$, and then randomly draw a node $v$, edge $e \in G$, and allocations $A \in M^k(G)$ and $A' \in M^k(G\setminus e)$. What is the probability that $A(v) \neq A'(v)$? Denote this probability by $S(m)$. I'm really interested in the following question: Under what conditions does $S(m) \rightarrow 0$ as $m \rightarrow \infty$, and more importantly, what is the rate of convergence?

Yoshida and Zhou (2020) study a similar question, but do not discuss random graphs.

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  • $\begingroup$ This seems like a strange question to ask if you want to understand sensitivity to edge deletion. If you just picked two random allocations $A, A' \in M^k(G)$, I would expect that's already often enough to have $A(v) \ne A'(v)$ with high probability, and we don't even need to delete the edge. $\endgroup$ Commented May 27 at 7:14
  • $\begingroup$ You may well be correct. Do you have a reference for the claim? If that's the case then I'd need to modify the question. One alternative would be to fix an algorithm for finding (perhaps approximate) maximum-size $k$-allocations, and then ask about the effect of deleting an edge on the output of this algorithm. This is closer in spirit to Yoshida and Zhou (2020). $\endgroup$ Commented May 27 at 17:20
  • $\begingroup$ Actually, let me be more precise: I expect $A(v) \ne A'(v)$ to hold whp for dense enough graphs only. Here is some of the calculation behind that. Let $p^*(n)$ be the threshold for $G(n,p)$ to have a nearly-perfect $k$-allocation (with fewer than $k$ vertices uncovered). Then for $p \gg p^*(n) \log n$, say, we can write $G(n,p)$ as the disjoint union of $\log n$ independent copies of $G(n,p')$ with $p' \gg p^*(n)$; each $G(n,p')$ whp has a nearly-perfect $k$-allocation, they are nearly disjoint, and so for most $v$, $\Pr[A(v)=A'(v)] \le \frac1{\log n}$. $\endgroup$ Commented May 27 at 18:33
  • $\begingroup$ So that is the story for dense-enough random graphs. (Actually, there are probably many more maximum $k$-allocations, so the true probability is probably much lower; this is just an upper bound.) On the other hand, consider triangles (for simplicity), and take $n^{-1} \ll p(n) \ll n^{-5/6}$. Here, there are many triangles in $G(n,p)$, but whp all of them are vertex-disjoint, and $M^3(G)$ actually contains only one allocation; conditioned on this whp event, $\Pr[A(v)=A'(v)] = 1$. $\endgroup$ Commented May 27 at 18:33

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