Lecture 4

*** comment on Theorem from last time. 

showed that IF rank r implies rectangle of size 2^{-c(r)}|X||Y|, then (essentially)

D(f) \le \approx c(r)

Note that if D(f) \le c(r), then there must be a rectangle of size 2^{-c(r)}|X||Y|.

So this says that an upper bound of D(f) in terms of rank is *equivalent* to proving 

an upper bound on rectangle size in terms of rank.


**** Nondeterministic CC. 

RECALL:let C^0(f) be the smallest *cover* of 0’s by combin. rects.
	respectively C^1(f) cover of 1’s

* Define N^1(f) = log_2 C^1(f) and N^0(f) = log_2 C^0(f)

Then N^1 is “non-deterministic CC” and N^0 is “co-non-deterministic CC” 
	(analogy w/ NP, coNP)

* example showing exponential gap: D(EQ) = n+1

	N^0(EQ) \le log_2 n + 1 [by announcing index i where differ]

	Related claim: N^0(EQ) \ge log_2 n.

	Proof: general statement: D(f) \le C^0(f) + 1 [also C^1(f) + 1]
		why? Alice sends list of 0-rectangles x is in
			Bob compares with list of 0-rectangles y is in
			f(x,y) = 0 iff some rectangle on both lists


* Comment: 0,1-fooling set and 0,1-rectangle size technique give lower bounds on C^0, C^1

	example: N^1(EQ) \ge n

* Comment: not rank method. This gives lower bounds on best *disjoint* cover
 
* Theorem: D(f) = O(N^0(f)*N^1(f))    “P = NP \cap coNP”

	Corollary: log_2(C^D(f)) \le D(f) \le O(log_2^2 (C^D(f)))

		where C^D is smallest *disjoint* cover. Proof? 

* proof of Theorem: 

	general idea: kill off 1/2 of remaining 0-rectangles in each phase

	Phase of protocol: A looks for 1-rectangle containing x that intersects
		\le 1/2 of remaining 0-rectangles, sends to Bob or else “couldn’t find one”

		if A couldn’t find one, B looks for 1-rectangle containing y that intersects
		\le 1/2 of remaining 0-rectangles, sends to Alice or else “couldn’t find”

		if either succeed, we reduced number of 0-rectangles by 1/2

		no 0 rectangles left? then output f(x,y) = 1.

		Both fail? then output f(x,y) = 0
		
		[if f(x,y) had been equal to 1, then rectangle containing (x, y)
			either intersects \le 1/2 0-rectangles in rows or in columns, 
			since otherwise it would intersect some 0 rectangle in both] 

	Each phase costs N^1(f) + O(1)

	At most N^0(f) = log_2 C^0(f) phases. 


* this result is tight. To show this, we study Disjointness variant Disj_k^n

	note: |X| = |Y| = n choose k

	1. N^0(Disj^n_k) \le log_2 n [why?]

	2. N^1(Disj^n_k) \le O(k + log log n)

	Proof: will show C^1(Disj^n_k) \le 2^{2k}ln((n choose k)^2), by prob. method.

		Choose random set S (each element in/out w prob 1/2 independently)
		
		Define R_S = {x : x \in S} x {y: y \in complement of S}

		Pr_S[(x,y) \in R_S] = 2^{-2k}
		
		so exists one that covers 2^{-2k} fraction of remaining

		repeat specified number of times to reduce remaining 1’s to < 1.

		note this critically uses non-disjointness of cover. 

	3. D(Disj^n_k) \ge log_2 (n choose k)

		Proof by showing matrix has full rank. Refer to picture. 

		(I | 0; M | -I(n-2k)) * (D^n_{k-1} | ? ; ? | 0) = (D^{n-1}_k | ?; 0 | (n-2k+1)D^{n-1}^{k-1})

		where M[x, z] = 1 if z of size k contains x\{n}, and 0 otherwise

		base cases: D^n_1 and D^n_{n/2} have full rank

		Consider x containing {n} vs. y omitting n: two cases (intersect or disjoint); both give 0
		
		Consider x containing {n} vs. y omitting n: x’ = x \ {n} and y’ = y \ {n}. two cases (x’, y’ intersect or disjoint)
		
			first case gives 0, second gives n - 2k + 1