admin
	[[schedule pushback]]
	ps1 returned 9/27
	ps2 due 10/02
	ps3 out 9/27
today
	defns of expanders
	existence
	comparing defns
vertex expander
	draw graph
		[[sparse graph, but well-connected]]
	G=(V,E) is a (K,A) vertex expander if
		S\subseteq V, |S|\le K
		|N(S)|\ge A*|S|
		N(S)={u: (u,v)\in E, v\in S}
	draw graph again
	interesting parameters
		D-regular graph
		N->\infty
		so get /family/ of graphs
		clearly |A|\le D
		want: (\Omega(N),1+\Omega(1)) O(1)-regular vertex expander
			O(1)-regular	hardest regime, most interesting
			\Omega(N)	this is as big as can be
			1+\Omega(1)	getting 1 is trivial, want to beat this by a constant as constant degree
	thm: D\ge 3 constant. \exists \alpha>0 such that random N-vertex D-regular graph is (\alpha N,D-1.01) vertex expander with probability \ge 1/2
	rmk
		D=2 does not give vertex expansion
			as get union of cycles and paths
			draw this
		A=D-1-\eps is the "right" answer
		this theorem is non-trivial:
			[[how to even sample a uniformly random D-regular graph?]]
	thm: D\ge 3, \exists \alpha>0 st random D-leftregular bipartite graph on 2N vertices is (\alpha N,D-2) vertex left-expander w/p \ge 1/2
		draw graph
	Pf
		via probabilistic method
			union bound + tail inequality
		random D-leftregular bipartite graph = each left vertex chooses D random neighbors on right side
		K\le \alpha N
		S\subseteq V, |S|=K
		choose DK neighbors of S in sequence v_1,\ldots,v_{DK}
		v_i is *repeat* if v_i=v_j for j<i
		Pr[v_i is repeat]=(i-1)/N\le KD/N
		Pr[|N(S)|\le (D-2)K]
			=\Pr[at least 2K repeats among v_1,\ldots,v_{KD}]
			\le (KD choose K/2) *(KD/N)^{2K}
		Pr[size K set does not expand]
			\le (N choose K)(KD choose 2K) *(KD/N)^{2K}
			\le (Ne/K)^k (KD\e/2K)^2K (KD/N)^{2K}
			=(e^3D^4K/4N)^K
		K\le \alpha N, choose \alpha=1/e^3D^4
			=1/4^K
		Pr[size \le K does not expand]\le \sum_i 1/4^k\le 1/3
			[[need to look at all sets \le K]]
	rmk:
		can get biregular graphs by considering unions of D perfect matchings
			[[lose independence, but it still works]]
		can translate results to uniformly random graphs via transfer theorems
	Q. explicit expander graphs?
		[[theme: don't just look at one type of expansion, but rather look at them all]]
spectral expanders
	G=(V,E) D-regular, random walk matrix M
	defn: \lambda(G)=\max_{x\perp u, ||x||=1} ||Mx||/||x||
		= magnitude of 2nd largest eigenvalue
	defn: G is a 1-\lambda(G)=\gamma(G) spectral expander
		want \gamma(G)\ge \Omega(1)
edge expander
	draw graph
	defn: G=(V,E) D-regular is a (K,\eps) edge expander if S\subset V |S|\le K, |cut(S,\bar{S}))\ge \eps D |S|
		[[\eps fraction of edges leave the set S]]
		[[no bottlenecks in the graph]]
	eg
		partitioning problems
			[[finding communities in social networks]]
		dynamic programming algorithms
			[[draw barbell graph, finding sparse cut enables to solve graph problem recursively]]
complexity of computing
	vertex		NP-hard
	edge		NP-hard
	spectral	P
relations
	goal: compare edge,vertex,spectral expansion
		[[can get approximations of vertex,edge expansion by using spectral expansion]]
	inherent loss: (N/4,1+1/4) vertex expander can have no spectral expansion
		graph can be union of two disjoint vertex expanders 
		hence disconnected
	optimal parameters for D-regular graphs
		vertex		(\alpha_D*N,D-1.1) vertex expansion
		spectral	\lambda(G)\le 2\sqrt{d-1}/D
				random graphs known to have this plus O(1)
				can actually meet this bound, called Ramanujan graph
					only recently shown to exist for all D
				BUT: exist Ramanujan graphs with vertex expansion \le D/2
					so optimal spectral does not imply optimal vertex expansion
spectral => vertex
	goal: show that spectral expander is also vertex expander
	defn: \pi probability distribution
		CP(\pi) is the collision probability
			[[draw two samples from \pi, what is the probability you get the same value]]
			[[measure of randomness of \pi]]
		=\sum \pi_i^2=<\pi,\pi>
	defn: Supp(\pi)=\{i:\pi_i>0\}
		[[also a measure of randomness]]
	eg
		uniform dist
			supp=n
			CP=1/n
		point dist
			supp=1
			CP=1
	lem: 
		CP(\pi)\ge 1/|Supp(\pi)|
	pf
		Recall cauchy schwartz
			|<x,y>|^2\le <x,x><y,y>
		S=Supp(\pi)
		1=|<1_Supp,\pi>|^2\le <1_S,1_S> <\pi,\pi> = |S|*CP(\pi)
	rmk: is equality iff \pi is uniform
		[[want *robust* equality]]
	lem:
		CP(\pi)=|\pi-u|^2+1/N
	Pf
		<\pi-u,\pi-u>
			=<pi,pi>-<u,pi>-<pi,u>+<u,u>
				<u,pi>=1/N for any pi
			=<pi,pi>-1/N
		[[implies previous lemma, but now does robustly]]
	thm: G D-regular graph with \lambga. Then for every \alpha, G is (\alpha N,1/((1-\alpha)\lambda^2+\alpha)) vertex expander
		[[RHS is constant if \lambda,alpha constants]]
		[[proof idea: random walk matrix makes distance to uniform smaller
			so this decreases CP
			so this increases support]]
	Pf
		let S be a subset of size \alpha N, take \pi=1_S/|S|
		|M\pi-u|^2\le \lambda^2 |pi-u|^2
			RHS = CP(\p)-1/N=1/S-1/N
			LHS = CP(M\pi) - 1/N \ge 1/|N(S)|-1/N
		solving gets N(S)\ge |S|\ge 1/(\lambda(1-\alpha)+\alpha)
	Rmk: if \alpha=1/2, get (N/2,1+\gamma) vertex expander
	Thm: for every \delta>0 D\ge 2. If G is  D-regular (N/2,1+\delta) vertex expander => \gamma(G)\ge \Omega((\delta/D)^2)
		[[won't prove this here]]
	Rmk: so (N/2,1+\Omega(1)) vertex expansion =~ \Omega(1) spectral expansion, for O(1)-degree graph
		[[contrast to non-equivalence in optimal parameter regimes]]
today
	defns of expanders
	existence
	comparing defns
next time
	comparing defns
	random walks on expanders