Lecture 2. 1 Nondeterministic Communication Complexity

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1 Communication Complexity 16:198:671 1/26/10 Lecture 2 Lecturer: Troy Lee Scribe: Luke Friedman 1 Nondeterministic Communication Complexity 1.1 Review D(f): The minimum over all deterministic protocols of the number of bits communicated between Alice and Bob to compute f in the worst case. C P (f): Protocol partition number. The minimum over all protocols of the number of leaves in the protocol tree. Equivalently the minimum size of a partition of the communication matrix into monochromatic rectangles by a protocol. C D (f): Partition number. The minimum size of a partition of the communication matrix into monochromatic rectangles, without the contstraint that the partition is derived from a protocol. We have the following inequalities: 1.2 Cover Number C D (f) C P (f) 2 D(f) The cover number for ones, denoted C 1 (f), is the minimum number of monochromatic-1 rectangles needed to cover all the 1s in the communication matrix, with overlaps between the rectangles allowed. Formally C 1 (f) = min R Each R i R is a 1-monochromatic rectangle For all (x, y) with f(x, y) = 1 there exists R i R : (x, y) R i Immediately we have that C 1 (f) C D (f), because in a 1-cover the rectangles are allowed to overlap and we don t need to worry about covering the 0 s. The following example shows that sometimes C 1 (f) can be significantly less than C D (f). Define the Set Intersection Problem (SI n ) as follows: n SI n (x, y) = (x i y i ) = (x i y i ) (x 2 y 2 )... (x n y n ) i=1 We can use each (x i y i ) to define a rectangle in a 1-cover, which shows that C 1 (SI n ) n. However, last time we saw that the rank of M SIn is 2 n 1, so C D (SI n ) 2 n 1. 1

2 1.3 Nondeterministic Communication Complexity and Relation to Cover Number The nondeterministic communication complexity of a function f, denoted N 1 (f), is the size of the smallest string that can be used to convince both Alice and Bob that f(x, y) = 1. Formally, N 1 (f) = min k : There exists predicates A, B such that for all (x, y) : f(x, y) = 1 z {0, 1} k : A(x, z) = 1 B(y, z) = 1 f(x, y) = 0 z {0, 1} k : A(x, z) = 0 B(y, z) = 0 There is an analogous definition for the co-nodeterministic communication complexity, N 0 (f), as well. The following theorem shows the relationship between the cover number and nondeterministic communication complexity. Theorem 1. N 1 (f) = log C 1 (f) Proof. First we show that N 1 (f) log C 1 (f). Let k = C 1 (f), and let C be a cover of the communication matrix by k monochromatic-1 rectangles. Suppose f(x, y) = 1 and let z be a string identifying the rectangle in C containing (x, y). The length of z will be log(c 1 (f)) bits. Alice and Bob can both verify that the rectangle contains the row x and the column y respectively. If f(x, y) = 0, then no rectangle contains both the row x and the column y, so either Alice or Bob will reject. This shows that N 1 (f) log C 1 (f). Now we show that log C 1 (f) N 1 (f). Suppose N 1 (f) = k and let A, B be the predicates from the definition of nondeterministic communication complexity. For a given z {0, 1} k, define a rectangle C D with C = {x : A(x, z) = 1} and D = {y : B(y, z) = 1}. The union of all such rectangles is a 1-cover of the communication matrix M f, and there are 2 k such rectangles. This shows that log C 1 (f) N 1 (f). The nice combinatorial representation of nondeterministic communication complexity in terms of covers is a large part of the justification for considering the measure. 1.4 Relationship Between Deterministic and Nondeterministic Communication Complexity The following theorem shows that the deterministic commmunication complexity of a function is not too much greater than the maximium of the nondeterminstic and co-nondeterministic communication complexity of the function. Theorem 2. (Aho, Ullman, Yannakakis 83) D(f) (N 0 (f) + 1)(N 1 (f) + 1) 2

3 Proof. Let C 0 be a minimal cover of the 0 values in M f of size at most 2 N 0 (f), and let C 1 be a minimal cover of the 1 values in M f of size 2 N 1 (f). Let C be a cover that is the union of all the rectangles in C 0 and C 1. Note that since the monochromatic-0 rectangles are disjoint from the monochromatic-1 rectangles in C, no monochromatic-0 rectangle and monochromatic-1 rectangle can share both a column and a row. We will give a deterministic protocol P that computes F and uses at most (N 0 (f) + 1)(N 1 (F ) + 1) bits of communication: Alice and Bob will assume f(x, y) = 1 and try to prove this by showing that no monochromatic-0 rectangle in C contains the input (x, y). At all times during the prototcol there is an active set of monochromatic-0 rectangles from C that could possibly still contain the input (x, y). Initially all monochromatic-0 rectangles in C are active. Alice and Bob then repeat the following steps until the protocol terminates: 1. If the active set of rectangles is empty, Alice and Bob output f(x, y) = 1 and the protocol ends. Otherwise, Alice looks for a monochromatic-1 rectangle which intersects x and overlaps in rows with at most half of the active 0 rectangles. If she finds such a rectangle r, she says the name of the rectangle, which takes N 1 (f) = log C 1 (f) bits. Alice and Bob then eliminate all the 0 rectangles that do not overlap in rows with r from the active group of rectangles, which cuts the set of active rectangles down by at least half. Otherwise, if Alice does not find such a rectangle r, she says no. 2. If Alice said no, then Bob looks for a monochromatic-1 rectangle which intersects y and overlaps with at most half of the active 0 rectangles. If he finds such a rectangle s, he says the name of the rectangle, which takes N 1 (f) = log C 1 (f) bits. Alice and Bob then eliminate all the 0 rectangles that do not overlap in columns with s from the active group. Otherwise, if Bob does not find such a rectangle s, he answers that f(x, y) = 0 and the protocol ends To show proof of correctness, let R C be a rectangle that contains the input (x, y). If f(x, y) = 1 then because R cannot intersect any monochromatic-0 rectangle on both a row and a column, it must be the case that either Alice or Bob successfully finds a rectangle with the specified properties in steps 1 and 2. Therefore the protocol will continue until all the 0 rectangles are eliminated, and they will correctly output that f(x, y) = 1. If f(x, y) = 0 then the rectangle R will never be eliminated and so as the process terminates in a finite number of steps at some point the condition in item (2) will fail and the players will output 0. During each round, at least half of the 0 rectangles are eliminated from the active set, so there can be at most log(c 0 (f)) + 1 = N 0 (f) + 1 rounds. During each round, at most N 1 (f) + 1 bits are communicated, so altogether the protocol takes (N 0 (f) + 1)(N 1 (f) + 1) bits to execute in the worst case As an example of the usefulness of this theorem, consider again the Set Intersection 3

4 Problem (SI n ), where SI n (x, y) = n x i y i We have seen previously that D(SI n ) = n + 1. and N 1 (SI n ) (log n). From that we can immediately conclude that N 0 (SI n ) n/ log(n). (In fact it turns out that N 0 (SI n ) n), as we will see next. i=1 1.5 Fooling Set Method A 1-fooling set for a function f is a set of inputs that map to 1 but such that any pair from the set cannot mutually lie in a 1-monochromatic rectangle. Formally for a fooling set F we have that (x, y) F : f(x, y) = 1 (x, y), (x, y ) F : F (x, y ) = 0 F (x, y) = 0 A fooling set can be used to get a lower bound on the nondeterministic communication complexity of f: immediately for any fooling set F for f we have that C 1 (f) F. One can analogously define a 0-fooling set to show lower bounds on C 0 (f). Let us use this to show that N 0 (SI n ) n. Consider the set of pairs F = {(x, x) : x {0, 1} n }. Here x denotes the bitwise complement of x. Clearly, SI n (x, x) = 0 for all x {0, 1} n. Now consider two distinct pairs (x, x), (y, ȳ). As x y we may assume, renaming and reordering as necessary, that x 1 = 0, y 1 = 1. But then ( x, y) intersect on the first bit and so SI n ( x, y) = 1. Thus we find C 0 (SI n ) 2 n and N 0 (SI n ) n. 2 Linear Programming Bounds for N 1 (f) Linear programming is an important tool in computer science, and we can use it to help derive bounds in communication complexity. The following program is an alternative characterization of the measure C 1 (f), where we assume that there is a variable W r for every 1-monochromatic rectangle r in the communication matrix M f : C 1 (f) = min Wr {W r} such that (x, y) : f(x, y) = 1 W r 1 r W r {0, 1} 4

5 Suppose we relax the constraint W r {0, 1} to W r 0. Then we have a linear program, which we denote by C 1 (f). Since C 1 is a relaxation, immediately we have that C 1 (f) C 1 (f). The following theorem shows that C 1 (f) is in fact a good approximation to C 1 (f). Theorem 3. C 1 (f) 2n C 1 (f) Proof. Let f be a function f : {0, 1} n {0, 1} n {0, 1} and consider the following dual program: C 1 (f) = max {α x,y} f=1 α x,y monochromatic 1 rectangles r : α x,y 0 α x,y 1 Lemma 4. C 1 (f) C 1 (f) Proof. Let {α x,y } be a feasible solution to the dual and {W r } an optimal solution to the primal. α x,y α x,y f=1 f=1 = r W r r α x,y W r r W r = C 1 (f) Thus we have C 1 (f) C 1 (f) C 1 (f). To finish the proof of the theorem, we will show that C 1 (f) C 1 (f) 2n. We will do this by constructing in a greedy way a cover and a feasible solution to the dual, and bounding the ratio of these solutions. In each stage of the greedy algorithm we choose R i which covers the largest number of currently uncovered elements and add it to the covering. Let R i be the number of elements 5

6 in R i not yet covered before this round. We set α x,y = 1 R i b for all (x, y) R i that have not yet been set, with b a constant to be determined later. In each round the size of the covering increases by 1 and the value of the dual solution increases by 1/b, so their ratio will be at most b. All that remains is to show that we can choose b to be at most 2n without violating the constraints of the dual. The next key claim bounds the weight in any rectangle r. Lemma 5. α x,y 1 b + 1 2b r b ln( r ) + O(1) b Proof. Consider the kth to last element (x, y) of r that is assigned during the greedy algorithm. We have that α x,y 1, since at this point R has at least k many elements uncovered, k b and the rectangle greedily chosen at this step thus must have at least k many elements. Let R be the size of the largest rectangle in M f. By Lemma 5, we can satisfy the constraints of the dual by choosing b such that ln(r)/b 1. Since R 2 2n, we can safely choose b to be 2n. 6