$RD^d$ is bipartite graph in which the vertices are $(d−1)$-element subsets and $d$-element subsets of a $(2d−1)$-element ground set. Two vertices are adjacent if one of the corresponding sets is a subset of the other.

My question is why any edge together with its incident edges defines an antimatching of $2d-1$ edges? And why $2d-1$?

Answer probably is related to the fact that this graph is $d$-regular but I have problem to justify it.


Let $G$ be a $d$-regular bipartite graph. And let $e_{uv}$ be an edge of $G$.

Let $S$ the set of $e_{uv}$ and all its adjacent edges. Then the size of $S$ is $2d-1$ : They are $d$ edges connected at the vertex $u$, and $d$ connected at vertex $v$. As we double count $e_{uv}$, we reduce the count by one, and $|S|=2d-1$.

Recall that an antimatching $A$ of $G$ is a set of edges such that two edges are at most at distance 2, and that the distance between two edges is defined as the distance between corresponding vertices in $L(G)$ (the Line graph of $G$).

As all edges from $S$ are connected to $e_{uv}$, then the maximum distance between edges of $S$ is 2.

Therefore $S$ is an antimatching of size $2d-1$.

I think that's all to proove. (I never studied antimatching before, I might be missing something).

Note An antimatching can also be defined (equivalently) as a subgraph with no induced matching of size greater than 1. This is also clearly the case for $S$ :

Suppose there exist an induced matching of size 2 in $S$. Then it must contains one edge connected to $u$ and one connected to $v$. But then in order to be induced we would need to include $e_{uv}$, and would not be a matching. Hence a contradiction. Therefore the only induced matching of $S$ have size 1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.