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Definition: A graph $G = (V, E)$ is called an $(n, d, c)$-expander if it has $n$ vertices, the maximum degree of a vertex is $d$, and, for every set of vertices $W \subset V$ of cardinality $|W| \le n∕2$, the inequality $|N(W)| \ge c|W|$ holds, where $N(W)$ denotes the set of all vertices in $V \backslash W$ adjacent to some vertex in $W$

By considering a random bipartite three-regular graph on $2n$ vertices obtained by picking three random permutations between the two color classes, prove that there is a $c > 0$ such that for every n there exists a $(2n, 3, c)$-expander.

This is question 9.6.1 from The Probabilistic Method and I am tempted to argue that for any $c\le 1$ the graph is trivially an expander.

In the question we are asked to consider the 3-regular bipartite graph obtained by taking the edge union of three permutations on $[n]$. A permutation is basically a perfect matching, therefore this construction contains a perfect matching and by Hall's theorem, every subset $W\subset[n]$ satistifes $|N(W)|\ge|W|\ge c|W|$, in particular for sets $|W|\le 1/2$.

(it is actually even simpler than that, no need for Hall's theorem: by 3-regularity $|N(W)|<|W|$ would imply $3|N(W)|<3|W|$ which in english means $\#\{\text{outedges of } |N(W)|\} < \#\{\text{outedges of } |W|\}$ which is obviously false).

But then these hints from this website point to a completely different approach. I would be interested in additional hints on how to compute the probability that $N(W)<(1+\epsilon)|W|$ for a given $W\subset[n]$ as they say in the hints.

And of course I'd like to know if taking $c\le 1$ is OK...

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  • $\begingroup$ So, why would it be obvious that it always holds for $c\leq 1$? $\endgroup$
    – Clement C.
    Jan 17 at 9:47
  • $\begingroup$ @ClementC. by what I wrote in the main paragraph, the neighborhood of $W$ has size at least $|W|$ $\endgroup$
    – H. Walter
    Jan 17 at 9:56
  • $\begingroup$ @HWalter: doesn't Hall's theorem imply that only for sets W contained entirely in one side of the bipartite (sub)graph? $\endgroup$
    – Clement C.
    Jan 17 at 10:40
  • $\begingroup$ Good point. Say $W=W_1\cup W_2$ is the union of $W_1\subset A$ and $W_2\subset B$, then $|N(W)|=|N(W_1)|+|N(W_2)|\ge |W_1|+|W_2|=|W|$ $\endgroup$
    – H. Walter
    Jan 17 at 12:08
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    $\begingroup$ As mentioned (among other things) in the current answer by Misha Lavrov, this is not quite true. If you don't count $W$ as part of the neighbours of $N(W)$ (and you shouldn't here), then what you just wrote isn't generally true. $\endgroup$
    – Clement C.
    Jan 17 at 21:05
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You might have been confused by the definition of $N(W)$. Usually this just means "the set of vertices adjacent to $W$". In the definition you've quoted from Alon and Spencer, we don't include vertices that are in $W$, which is the standard thing to do for measuring vertex expansion.

As a result, many cubic bipartite graphs are not expanders for $c =1$. For an extreme example, take two copies of any cubic bipartite graph, and let $W$ be the set of all vertices in one copy. Then $|W| = \frac n2$ and $|N(W)| = 0$, so the vertex expansion of such a graph is $0$.

For a family of connected examples with bad expansion, consider the prism graphs, such as the one below:

enter image description here

These are $3$-regular with $n=2k$ vertices, and when $k$ is even, they are bipartite. However, we can cut the prism in half vertically (if it is drawn as in the picture above) and take $W$ to be the vertices in one half. Then $|W| = \frac n2$ and $|N(W)|=4$, so at best this graph has vertex expansion $\frac 8n$, which goes to $0$ as $n \to \infty$.

So, no: not all bipartite cubic graphs are expanders.


The reason why you're seeing hints for proving that $|N(W)| \ge (1+\epsilon)|W|$ is not because $c$ can't be less than $1$. It's because in general, if $W = W_1 \cup W_2$ where $W_1$ is on one side of the bipartition and $W_2$ on the other, $$ N(W) = (N(W_1) \setminus W_2) \cup (N(W_2) \setminus W_1) $$ which leads us to $$|N(W)| \ge (|N(W_1)| - |W_2|) + (|N(W_2)| - |W_1|) = |N(W_1)| + |N(W_2)| - |W|.$$ if we can show that $|N(W_1)| \ge (1+\epsilon)|W_1|$ and $|N(W_2)| \ge (1+\epsilon)|W_2|$, then we conclude that $|N(W)| \ge \epsilon|W|$. If this always holds, then we have an $\epsilon$-expander.

We have to be careful because $|W| \le \frac12|V|$ doesn't imply that $|W_1| \le \frac12n$ and $|W_2| \le \frac12n$. But we also have $|N(W)| \ge |N(W_1)| - |W_2| \ge |W_1| - |W_2|$, and similarly $|N(W)| \ge |W_2| - |W_1|$, so if the two sides of $W$ are too unbalanced, we get vertex expansion immediately. That tells us that $|W_1|, |W_2| \le (\frac12 + \frac12\epsilon)n$ in all hard cases.

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