Randomized Algorithms (Spring 2010)/Expander graphs and rapid mixing random walks

Graph Expansion

Expander graphs

We consider undirected graphs [math]\displaystyle{ G(V,E) }[/math]. In addition, we allow multiple edges between two vertices.

Some notations:

• For [math]\displaystyle{ S,T\subset V }[/math], let [math]\displaystyle{ E(S,T)=\{uv\in E\mid u\in S,v\in T\} }[/math].
• The Edge Boundary of a set [math]\displaystyle{ S\subset V }[/math], denoted [math]\displaystyle{ \partial S\, }[/math], is [math]\displaystyle{ \partial S = E(S, \bar{S}) }[/math].

Definition (Graph expansion) The expansion ratio of [math]\displaystyle{ G }[/math], is defined as [math]\displaystyle{ \phi(G)=\min_{S:|S|\le\frac{n}{2}}\frac{|\partial S|}{|S|}. }[/math]

Expander graphs are [math]\displaystyle{ d }[/math]-regular (multi)graphs with [math]\displaystyle{ d=O(1) }[/math] and [math]\displaystyle{ \phi(G)=\Omega(1) }[/math].

Existence

We will show the existence of expander graphs by the probabilistic method. In order to do so, we need to generate random regular graphs.

Suppose that [math]\displaystyle{ d }[/math] is even. We can generate a random [math]\displaystyle{ d }[/math]-regular graph [math]\displaystyle{ G(V,E) }[/math] as follows:

• Let [math]\displaystyle{ V }[/math] be the vertex set. Uniformly and independently choose [math]\displaystyle{ \frac{d}{2} }[/math] cycles of [math]\displaystyle{ V }[/math].
• For each vertex [math]\displaystyle{ v }[/math], for every cycle, assuming that the two neighbors of [math]\displaystyle{ v }[/math] in that cycle is [math]\displaystyle{ w }[/math] and [math]\displaystyle{ u }[/math], add two edges [math]\displaystyle{ wv }[/math] and [math]\displaystyle{ uv }[/math] to [math]\displaystyle{ E }[/math].

The resulting [math]\displaystyle{ G(V,E) }[/math] is a multigraph. That is, it may have multiple edges between two vertices. We will show that [math]\displaystyle{ G(V,E) }[/math] is an expander graph with high probability. Formally, for some constant [math]\displaystyle{ d }[/math] and constant [math]\displaystyle{ \alpha }[/math],

[math]\displaystyle{ \Pr[\phi(G)\ge \alpha]=1-o(1) }[/math].

By the probabilistic method, this shows that there exist expander graphs. In fact, the above probability bound shows something much more stronger: it shows that almost every regular graph is an expander.

Recall that [math]\displaystyle{ \phi(G)=\min_{S:|S|\le\frac{n}{2}}\frac{|\partial S|}{|S|} }[/math]. We call such [math]\displaystyle{ S\subset V }[/math] that [math]\displaystyle{ \frac{|\partial S|}{|S|}\lt \alpha }[/math] a "bad [math]\displaystyle{ S }[/math]". Then [math]\displaystyle{ \phi(G)\lt \alpha }[/math] if and only if there exists a bad [math]\displaystyle{ S }[/math] of size at most [math]\displaystyle{ \frac{n}{2} }[/math]. Therefore,

[math]\displaystyle{ \begin{align} \Pr[\phi(G)\lt \alpha] &= \Pr\left[\min_{S:|S|\le\frac{n}{2}}\frac{|\partial S|}{|S|}\lt \alpha\right]\\ &= \sum_{k=1}^\frac{n}{2}\Pr[\,\exists \mbox{bad }S\mbox{ of size }k\,]\\ &\le \sum_{k=1}^\frac{n}{2}\sum_{S\in{V\choose k}}\Pr[\,S\mbox{ is bad}\,] \end{align} }[/math]

Let [math]\displaystyle{ R\subset S }[/math] be the set of vertices in [math]\displaystyle{ S }[/math] which has neighbors in [math]\displaystyle{ \bar{S} }[/math], and let [math]\displaystyle{ r=|R| }[/math]. It is obvious that [math]\displaystyle{ |\partial S|\ge r }[/math], thus, for a bad [math]\displaystyle{ S }[/math], [math]\displaystyle{ r\lt \alpha k }[/math]. Therefore, there are at most [math]\displaystyle{ \sum_{r=1}^{\alpha k}{k \choose r} }[/math] possible choices such [math]\displaystyle{ R }[/math]. For any fixed choice of [math]\displaystyle{ R }[/math], the probability that an edge picked by a vertex in [math]\displaystyle{ S-R }[/math] connects to a vertex in [math]\displaystyle{ S }[/math] is at most [math]\displaystyle{ k/n }[/math], and there are [math]\displaystyle{ d(k-r) }[/math] such edges. For any fixed [math]\displaystyle{ S }[/math] of size [math]\displaystyle{ k }[/math] and [math]\displaystyle{ R }[/math] of size [math]\displaystyle{ r }[/math], the probability that all neighbors of all vertices in [math]\displaystyle{ S-R }[/math] are in [math]\displaystyle{ S }[/math] is at most [math]\displaystyle{ \left(\frac{k}{n}\right)^{d(k-r)} }[/math]. Due to the union bound, for any fixed [math]\displaystyle{ S }[/math] of size [math]\displaystyle{ k }[/math],

[math]\displaystyle{ \begin{align} \Pr[\,S\mbox{ is bad}\,] &\le \sum_{r=1}^{\alpha k}{k \choose r}\left(\frac{k}{n}\right)^{d(k-r)} \le \alpha k {k \choose \alpha k}\left(\frac{k}{n}\right)^{dk(1-\alpha)} \end{align} }[/math]

Therefore,

[math]\displaystyle{ \begin{align} \Pr[\phi(G)\lt \alpha] &\le \sum_{k=1}^\frac{n}{2}\sum_{S\in{V\choose k}}\Pr[\,S\mbox{ is bad}\,]\\ &\le \sum_{k=1}^\frac{n}{2}{n\choose k}\alpha k {k \choose \alpha k}\left(\frac{k}{n}\right)^{dk(1-\alpha)} \\ &\le \sum_{k=1}^\frac{n}{2}\left(\frac{en}{k}\right)^k\alpha k \left(\frac{ek}{\alpha k}\right)^{\alpha k}\left(\frac{k}{n}\right)^{dk(1-\alpha)}&\quad (\mbox{Stirling formula }{n\choose k}\le\left(\frac{en}{k}\right)^k)\\ &\le \sum_{k=1}^\frac{n}{2}\exp(O(k))\left(\frac{k}{n}\right)^{k(d(1-\alpha)-1)}. \end{align} }[/math]

The last line is [math]\displaystyle{ o(1) }[/math] when [math]\displaystyle{ d\ge\frac{2}{1-\alpha} }[/math]. Therefore, [math]\displaystyle{ G }[/math] is an expander graph with expansion ratio [math]\displaystyle{ \alpha }[/math] with high probability for suitable choices of constant [math]\displaystyle{ d }[/math] and constant [math]\displaystyle{ \alpha }[/math].