Asymmetric Communication Complexity

1. Asymmetric Communication Complexity And its implications on Cell Probe Complexity Based on a paper of Peter Bro Miltersen, Noam Nisan, Muli Safra and Avi Wigderson Slides by Elad Verbin

2. Purpose We want to get Lower Bounds. Best known lower bounds: • Sorting is Ω(nlogn) in the comparison model • Trivial lower bounds. i.e. MAX is Ω(n) • What can we really do, i.e. for RAM?

3. Outline • Yao decided to strengthen the model – Considered the Cell Probe model. • Lower bounding Cell Probe is hard too. We strengthen even more – Communication Complexity

4. Outline • We show the relationship between Cell Probe and Communication Complexity. • We show how to get lower bounds for Communication Complexity using two techniques: • The Richness Technique • The Round Elimination Technique

5. Communication Complexity • The problem : f:XY{0,1} • Alice gets xX, Bob gets yY, their goal is to exchange messages to decide f(x,y). • A solution is a communication protocol that can compute f(x,y) for all x,y. f(x,y)

6. Asymmetric Communication Complexity • A Communication Protocol that computes function f in which Alice sends at most a bits and Bob sends at most b bits is called a [a,b]-protocol for f. f(x,y) Pink<=a Blue<=b

7. Randomized Protocols • If the protocol is allowed to flip (public) coins, and gives the correct answer with probability > 2/3 it is called a randomized protocol. • If it always correctly identifies a 0-instance it is called a one-sided error protocol.

8. Example • For any problem, there are trivial deterministic protocols: • [log|X|,1]-protocol • [1,log|Y|]-protocol.

9. The Problem DISJ(N,k,l) • We work on the universe U={0,1,…,N-1} • Alice gets x, a set of k elements • Bob gets y, a set of l elements • They must decide if x∩y=Ø x y x∩y=Ø?

10. A one-sided error randomized [O(k),O(l)]-protocol for DISJ(N,k,l) • If we say that x and y are disjoint then we want to have complete confidence. If we say they intersect, we want to be reasonably certain. • Flip public coins to get a sequence of random subsets of the universe: R1, R2, …

11. x y R3

12. x y R3 y=y∩R3 y

13. x y R3 y=y∩R3 R8 y x=x∩R8 AND SO ON….

14. A one-sided error randomized [O(k),O(l)]-protocol for DISJ(N,k,l) • We don’t really send the index of R8, we just send the distance from the last set (8-3=5). This means that the expected numbers of bits sent by a player is equal to the size of his set • If at some point one of the sets becomes empty, then the originals were disjoint – say so. Otherwise, after a long time, say that there is an intersection.

15. A one-sided error randomized [O(k),O(l)]-protocol for DISJ(N,k,l) • If x and y were indeed disjoint, the sizes of x and y decrease by a factor of 2 each round. Therefore the total communication is [O(k),O(l)]. • If the sets were disjoint, what is the chance that we say that there is an intersection? Very low.

16. Fixed-round protocols • If t alternating messages are sent and each message is of size a or b it is called a [t,a,b]-protocol. a b a t b f(x,y)

17. Static Data Structure Problems • A static data structure problem is a function f:DQR • D – the data • Q – the queries • R – Possible answers. Typically, R={0,1} DS query Data

18. The Problem MEMBERSHIP(N) • INPUT: a set S[N] • QUERIES: of the form “xS?” • D={S[N]}, |D|=2N • Q=[N] • R={0,1} • The trivial solution is optimal.

19. The Problem MEMBERSHIP(N,n) • INPUT: a set S[N] of size n • QUERIES: of the form “xS?” • D={S[N] | |S|=n}, |D|=choose(N,n) • Q=[N] • R={0,1}

20. The Cell Probe Model • Parameter w – word size • s cells, each containing w bits. • Each query probes at most t cells to get answer • A query is a decision tree of depth t and degree 2w

21. MEMBERSHIP(N,n) Solutions: • Keep every possible answer. s=N, t=1 (better – s=N/w, t=1) • Keep a nonredundant representation. s=log(choose(N,n)), t=log(choose(N,n))

22. MEMBERSHIP(N,n) Solutions: • Keep a sorted list of all elements. s=nlog(N)/w , t=log(n)*log(N)/w • There is a randomized solution with s=(n/w)c, t=O(1), for some constant c.

23. What is the connection between Cell Probe and Asymmetric Communication Complexity?

24. ACC <-> Cell Probe • The communication problem related to a static data structure problem f:DQ{0,1} if the problem where Alice gets a query, Bob gets the data, and they should decide if this is a “yes” instance or a “no” instance

25. Communication Problem MEM(N,n) • Alice gets x[N], Bob gets y[N], |y|=n, they should decide if xy. • Trivial protocols: [1,nlogN] , [logN,1]

26. Lemma CP->AAC If there is a solution to the data structure problem with word size w taking s cells and with query time t, then there is a [2t,log(s),w]-protocol for the communication problem Therefore a lower bound on ACC gives us a lower bound on Cell Probe

27. Finer points of CP->AAC • How is the communication complexity model stronger than the Cell Probe Model? • Answer: In its adaptivity • Which form of Cell Probe lower bounds can we get from the CP->AAC Lemma? • Answer: the bound on space is up to a polynomial

28. Restricted AAC->CP If there is a [O(1),a,b]-protocol for the communication problem then the data structure problem has a solution with word size w=b, t=O(1) and s=2O(a) Proof: The Data Structure for input y contains the message Bob should send next for every possible history of messages Alice can send, for any query.

29. Lower Bounding The Communication Complexity

30. Communication Problem MEM(N,l) • Alice gets x[N], Bob gets y[N], |y|=l, they should decide if xy. • NONMEM(N,l) is the same problem, when Alice and Bob want to decide if xy • Trivial protocols: [1,l*logN] , [logN,1]

31. Problem <-> Matrix • We identify a communication problem f:X×Y{0,1} with a |X|×|Y| Matrix where M[x][y]=f(x,y). • The matrix of NONMEM(N,l) has N rows and columns. Each column has N-l 1-entries

32. Problem <-> Matrix • A problem (matrix) is (u,v)-rich if at least v columns contain at least u 1-entries. • NONMEM(N,l) is (N-l, )-rich. (4,3)-rich

33. The Richness Lemma • Let f be a communication problem that: • is (u,v)-rich • has a randomized one-sided error [a,b]-protocol. • Then f contains a u/2a+2 over v/2a+b+2 submatrix of 1-entries.

34. Randomized Lower Bound for MEM(N,l) • Say MEM(N,l) has a negative-one-sided error [a,b]-protocol. Let a<log(l), l<N/2. • Then NONMEM(N,l) has a one-sided error [a,b]-protocol

35. Randomized Lower Bound for NONMEM(N,l) • NONMEM(N,l) is (N-l, )-rich • Therefore it has a 1-submatrix of dimensions at least (N-l)/2a+2 over /2a+b+2

36. Randomized Lower Bound for NONMEM(N,l) • However, if there is a 1-submatrix of dimensions r on s then s≤ • By substituting for s and r, simplifying and bounding we get 2a(a+b)=Ω(l)

37. Randomized [O(a),O(l/2a)] Upper Bound for NONMEM(N,l) • On the other hand, NONMEM(N,l) has a [O(a),O(l/2a)]-protocol, for all a<log(l): • Alice sends Bob the first a indices of R’s that contain x. This allows Bob to reduce y to expected size l/2a. • Then Bob sends a couple indices that contain y. • If we are not yet sure that they are disjoint, we say that they intersect.

38. Tightness for NONMEM(N,l) • NONMEM(N,l) has a [O(a),O(l/2a)]-protocol, for all a<log(l) • 2a(a+b)= 2a(a+l/2a)=?O(l) Therefore the last result is tight? • There are constants c,c’>0, so that for any a, • b=l/2ca is enough. • b=l/2c’a is not enough.

39. The Richness Lemma • Let f be a communication problem that: • is (u,v)-rich • has a randomized one-sided error [a,b]-protocol. • Then f contains a u/2a+2 over v/2a+b+2 submatrix of 1-entries.

40. Proof of the Richness Lemma • First let us prove a weaker result: if f has a deterministic [a,b]-protocol then it contains a contains a u/2a over v/2a+b submatrix of 1-entries. • We prove this by induction on a+b:

41. Proof of the Richness Lemma • For a+b=0 – |X|≥u, |Y|≥v, and f(x,y)=1 for all x,y, so this is trivial. • Now, if Alice send the first bit: • X0 – inputs for which she sends 0 • X1 – inputs for which she sends 1 • Let f0, f1 be the restrictions of f to X0Y, X1Y.

42. Proof of the Richness Lemma • At least one of them is (u/2,v/2)-rich, and both have a [a-1,b] protocol. • By the induction it contains a (u/2)/2a-1 over (v/2)/2a+b-1 1-submatrix. • In the other case, Bob send the first bit. • Define Y0,Y1,f0,f1. At least one of them is (u,v/2)-rich, and proceed similarly.

43. Proof of the Richness Lemma • Now let us prove the general case: • Let S be the set of u*v rich-positions in the matrix • Let us look at some coin-flip sequence.

44. Proof of the Richness Lemma • Let X = #{1s in S} • E[X]>=2/3 * uv • => There exists such a sequence for which X>=2/3 * uv • Fix the sequence, to get a deterministic algorithm. This algorithm computes a function f’ that is close to f.

45. Proof of the Richness Lemma • By a counting argument, f’ is (u/4,v/4)-rich, and so it has a 1-submatrix of the required size ( u/2a+2 over v/2a+b+2 ) • This is a 1-submatrix in f too, because the error is one-sided. • Q.E.D.

46. A Richness Results for two-sided error • Let d,e>0, and let f:X×Y{0,1} be a communication problem with at least a d-fraction of 1s. If f has a randomized two-sided error [a,b]-protocol then f has a submatrix M of dimensions at least |X|/2O(a) over |Y|/2O(a+b) with at least a (1-e)-fraction of 1s.

47. The SPAN(n) Problem • In SPAN, Alice gets x{0,1}n and Bob gets a vector subspace y{0,1}n • y can be represented using a basis of k≤n vectors – O(n2) bits • Alice and Bob must decide if x∈y. • Trivial Protocols:[n,1] , [1,n2]

48. Lower bounds for SPAN • Let’s prove that in any [a,b] randomized one-sided error protocol for SPAN, either a=Ω(n), or b=Ω(n2) • We will assume that y is of dimension n/2. • We will prove that: • SPAN is (2n/2,2n2/4)-rich, and • SPAN does not contain a 1-submatrix of dimensions 2n/3 over 2n2/12

49. SPAN is (2n/2,2n2/4)-rich • Each subspace contains exactly 2n/2 vectors => each column contains 2n/2 1s. • How many subspaces of dimension n/2 are there? • Lets choose a basis: we have 2n-1 possibilities for the first vector, 2n-2 for the second, 2n-4 for the third, etc.

50. SPAN is (2n/2,2n2/4)-rich • We chose each basis (n/2)! times • How many basis does a subspace has? • We have 2n/2-1 options to choose the first vector, 2n/2-2 for the second, etc. • We again chose each basis (n/2)! times. • Thus, there are at least 2n2/4 subspaces of dimension n/2.