Lecture 3 *** Rank lower bound method Definition 1.27 Lemma: D(f) \ge log_2 rank(f). Proof: write each 1-rectangle as rank 1 matrix, and sum Note, can do this for 0-rectangles via doing 1-rectangles of J - M_f So actually, Lemma: D(f) \ge log_2 (2rank(f) - 1) * Example: EQ, GT * Example: IP (look at square) * Example: DISJ (consider as tensor product of {1,1;1,0} ) Got this far last time. * Brief digression. Rank of matrices over different fields. If K subfield of F, then Rank_K(M) = Rank_F(M). [Follows because F is a vectorspace over K… omitting details to avoid algebra] Claim: rank_F(M) \le rank_R(M), for any field F Proof: By first lemma, can replace R with Q. Suppose we have \sum_i \lamda_i c_i = 0 so c_i’s are linearly dependent. Then clear denominators. key: can assume gcd{\lamda_i} = 1. Then replacing k with (1 + 1 + 1 … + 1) in field F, we get same sum. And not all the replaced \lamda_i are 1, because of GCD assumption Two other results: Exercise 1.31 (both use that Rank_{F_2} \le Rank_R) * Rank lower bound subsumes fooling sets for Boolean f: Lemma: fooling set of size t implies rank \ge sqrt(t) Proof: Let M be a matrix with fooling set of size t. WLOG restrict to fooling set cols and rows Note M[i,i]*M[i,i] = z^2 while M[j,i]*M[i,j] = 0 (for j \ne i) Thus t = rank(M \circ M^T) \le Rank(M \otimes M^T) = Rank(M)*Rank(M^T) = rank(M)*rank(M) * Log rank conjecture: exists a universal constant c, so that for all Boolean f, D(f) = (log rank(f))^c Two variants where this fails: * non-boolean f. Consider Intersection(x, y) = |x \cap y|. Rank(Intersection) = n (write as product of two matrices over R) But fooling set of size 2^n just as with Disjointness, so D(Inter) \ge n * rank over F_2. Rank_2(Inner-Product) = n. But we already showed D(IP) \ge n Formalizing that “upper bound in terms of rank depends on understanding size of monochromatic rectangles in terms of rank” * Thm: suppose that for any boolean fn f with Rank(M_f) = r, we have a monochromatic rectangle R with |R| \ge 2^{-c(r)}|X||Y|. Then any boolean fn of rank r has D(f) \le O(log^2 r) + O(\sum_{i=0}^{log r} c(r/2^i)) Note: rank \le 1 implies giant rectangle Note: c(r) = polylog(r) proves log-rank conjecture. Proof: Consider M_f or rank r. Write as M = ( R S; P Q) Note rank(R) \le 1. So rank(S) + rank(P) \le r+1. WLOG rank(S) \le r/2 + 1 (else transpose matrix) Alice sends 1 bit: is x in top or bottom [If in top, then rank decreases to rank(R S) \le r/2 + 1] [If in bottom, then matrix size decreases to (1 - 2^{-c(r)})|X||Y|] Analysis: phase 1 lasts until rank is reduced to \approx r/2 At most 2^{c(r)}ln|X||Y| many “leaves” in protocol tree so far [note it is long and skinny] Rebalancing: simulate with O(c(r) + log ln |X||Y|) bits No repeated rows: |X| \le 2^r. No repeated cols: |Y| \le 2^r. cost of phase 1: O(c(r) + log(r)) bits. Phase 2: reduce rank to \approx r/4… Total as claimed in Theorem. QED. **** Nondeterministic CC. all powerful prover speaks first to A, B trying to convince them f(x,y) = z, then A, B engage in protocol completeness: if f(x,y) = z, then prover can convince them soundness: if f(x,y) \ne z, then prover cannot convince them * Example: Non-equality. O(log n). Prover announces bit position where x_i \ne y_i Equivalent viewpoint for Boolean f. let C^0(f) be the smallest *cover* of 0’s by combin. rects. respectively C^1(f) cover of 1’s. * Example of cover that is not disjoint (p.17) [cover with 4, partition requires 5] * 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) * Lemma: best (prover bits plus communication of A, B) = N^1(f) [+/- 2 bits] Proof: <- prover sends name of 1-rectangle in cover. A, B check if (x,y) is in it. -> after Prover’s “proof” (p bits) A, B engage in b bits of communication implies cover of at most 2^p + 2^b rectangles So we adopt N^0 and N^1 based on C^0 and C^1 as our definitions of co-ND and ND CC