admin
	Problem 2.9 part 6 (pset2 question 1) is optional
today
	3-sat
	k-wise independence
	chebyschev inequality
	averaging samplers
	[[end of chapter 3]]
next time
	expanders [[chapter 4]]
exact 3-sat
	recall max-cut
		[[given a graph, want to partition into two parts to maximize number of edges across cut]]
		1/2-approx [[we saw randomized algo, derand via conditional expectations, pairwise independence]]
		.878-approx known, some evidence is best possible [[ie, doing better might be NP-hard]]
	defn
		(=3)-CNF is an AND\circ \OR^3
			each clause has *exactly* 3 literals on distinct variables (vars or negated vars)
			[[special case of 3CNF, which has at most 3 literals]]
	Max-Exact-3-SAT: given (=3)-CNF, compute the maximum number of simultaneously satisfiable clauses
	Rmk: NP-hard to compute exactly
	thm: randomized 7/8-approx algo
	thm[Hastad]: any constant \eps, 7/8+\eps approx is NP-hard
		[[can even get NP-hardness for subconstant \eps]]
	Pf:
		[[guess for the algorithm]]
		pick assignment \vx randomnly
		E[# satisfying clauses]
			=\sum_i \Pr[clause i is satisfied]
				u \or \not v \or w	has unique unsat assignment, so satisfied w/p 7/8
			=(# clauses) * 7/8
		rand algo: try this many times
		derandom:
			conditional expectations
			3-wise independence
k-wise independence
	defn 
		f:[N]\times [H]\to [M] is a k-wise independent hash function if
		for distinct i_1,\ldots,i_k\in[N]
		(f(i_1,h),\ldots,f(i_k,h))=(U_M)^k, when h chosen uniformly from H
	recall:
		h_{a,b((x)=ax+b is pairwise independent
		[[proven using facts about interpolating lines through points]]
	lem:
		for any (x_1,y_1),\ldots,(x_k,y_k) there exists a unique polynomial of degree <k passing through these points
	pf
		uniqueness:
			say f and g are two such polynomials
			then f-g is zero on all x_i
			so k roots, but degree <k => f-g is zero => f=g
		existence
			define L_i=\frac{\prod_{j\ne i} x-x_j}{\prod_{j\ne i} x_i-x_j}
				degree k-1
				L_i(x_i)=1
				L_i(x_j)=0
			define f=\sum_i y_i* L_i(x)
				degree k-1
				passes through desired points
	prop:
		h:\F\times\F^k\to\F
		h_{a_0,\ldots,a_k-1}(x)=a_0+a_1*x+\cdots + a_k x^{k-1} is k-wise independent
	Pf:
		for distinct x_1,\ldots,x_k\in\F
		any y_1,\ldots,y_k\in\F
		want \Pr_{a_i}[h(x_i)=y_i \for all i]=1/\F^k
			=(# degree <k polynomials passing through ((x_i,y_i))_i)/(# degree <k polynomials)
			=1/F^k
	Cor: deterministic 7/8-approx to max-exact-3-sat
averaging samplers
	meta-thm: 
		f(x)=\sum (k-local functions)
		E[f] cannot distinguish between true randomness and k-wise independence
	Q. what else is k-wise independence good for?
	Q. iid vars => chernoff tail bounds
		pairwise independent => ?
	defn: Var(X)=\E[(X-\mu)^2], \mu=E[X]
	Prop[Chebyschev]
		X random var
		Pr[|X-E[X]|\ge \eps]\le \var(X)/\eps^2
	Pr
		\Pr[|X-\mu|\ge \eps]
		=\Pr[|X-\mu|^2\ge \eps^2]
		now apply markov
	lem: X_1,\ldots,X_n pairwise independent. X=\sum_i X_i. Var(X)=\sum_i Var(X_i)
	Pf.
		Var(X)
		=\E[(\sum_i (X_i-\mu_i)^2]
		=\E[\sum_{i,j} (X_i-\mu_i)(X_j-\mu_j)]
		=\E[\sum_i (X_i-\mu_i)^2]
			+\E[\sum_{i\ne j} (X_i-\mu_i)(X_j-\mu_j)]
		=\E[\sum_i (X_i-\mu_i)^2]+0
		=\sum_i \Var(X_i)
	Cor. X_1,\ldots,X_n pairwise independent in [0,1]. X=\sum_i X_i/n
		Pr[|X-E[X]|\ge \eps]\le 1/(n\eps^2)
	Pf.
		Var(X)
			=\sum Var(X_i/n)
			=\sum Var(X_i)/n^2
			\le 1/n^2 \cdot \sum 1
				as |X_i-\mu_i|\le 1
			=1/n
		Chebyschev => result
	[[this means that pairwise independence is more than just fooling 2-local functions]]
	defn:
		a *(\delta,\eps) averaging sampler* is a function Samp:[N]\to[M]^t
		such that for every f:[M]\to[0,1]
		\Pr_{z_1,\ldots,z_t\from \Samp(U_{[N]})} [1/t \sum_i f(z_i)>\mu(f)+\eps] \le \delta
		*boolean sampler* if only for f:[M]\to\bits
		*explicit* if t-th output computable in poly(log N, log t) time
	rmk:
		general sampler
			compute estimate \hat{f} from samples in arbitrary way
			averaging samplers are more useful
		one-sided error guarantee
			is better for connections to other pseudorandom objects
			=> two-sided via considering 1-f
	parameters of avg samples	#samples			#rand bits
		independent samples	1/\eps^2*\log(1/\delta)		m*1/\eps^2*\log(1/\delta)
			chernoff for t samples in \bits^m
				requires t*m random bits
				gives \eps(-\eps^2t/4) error
		pairwise independent	1/\eps^2*1/delta		O(m+\log(1/\eps)+\log(1/\delta))
			t samples from \bits^m takes 2\ell random bits
				\ell=\max{\log t,m}=O(m+\log t)
				embed [t] and \bits^m into \F_{2^\ell}
			gives error 1/(t\eps^2)
		optimal			1/\eps^2*\log(1/delta)		O(m+\log(1/\eps)+\log(1/\delta))	[[open]]
	BPP error reduction: 
		error <1/3 -> error -> 1/2^{-k}
			so \eps=1/3 in the above
			\delta=2^{-k}
		independent samples					O(k)			O(km)
		pairwise independent					2^O(k)			O(m+k)
		optimal							O(k)			O(m+k)		[[we'll see!]]
skipped topics
	universal traversal sequences
	hash tables
today
	3-sat
	k-wise independence
	chebyschev inequality
	averaging samplers
	[[end of chapter 3]]
next time
	expanders [[chapter 4]]