Seminar on Communication Complexity

1. Seminar on Communication Complexity Rotem Zach November 1st, 2009

2. Quick Overview • A rectangle in X × Y is a subset R ⊆ X × Y such that R = A × B for some A ⊆ X and B ⊆ Y. • A rectangle R ⊆X×Y is called f-monochromatic if f is fixed on R. • Every protocol f partitions X × Y into f-monochromatic rectangles. • If every partition of X×Y into f-monochromatic rectangles requires at least t rectangles, then D(f) ≥ ceil(log2t). These are the 0-rectangles and the 1-rectangles.

3. Lower Bound • No function has a single “hard” input. Why? • Alice and Bob can each send a bit saying “we have the half of the hard input”. • A lower bound D(m) gives a set of inputs for which any protocol that computes f must use at least m bits of communication

4. Material • Finding lower bounds for the communication complexity of problems: • Fooling Sets • Weights and Distributions • Rank Lower Bound

5. Fooling Set (Definition) Let f: X × Y → {0,1} . A set S ⊆ X × Y is called a fooling set (for f) if there exists a value z∈{0,1} such that: • For every (x,y) ∈ S, f(x,y) = z • For every two pairs (x1,y1) ≠ (x2,y2) in S, either f(x1,y2)≠z or f(x2,y1)≠z

6. Example • EQ: 0 ≤ x , y < 22 • EQ(x,y) = 1 iff x = y • How many mono-rectangles are there? • Which fooling set can you find? • S = {(00,00),(01,01),(10,10),(11,11)}

7. Lemma 1 If f has a fooling S set of size t, then D(f) ≥ log2t Proof: • Claim: No x,y ∈ S can be in the same rectangle • Assume R contains (x1,y1) ≠ (x2,y2) (both belong to S) • From the definition, f(x1,y2) ≠ z or f(x2,y1) ≠ z • Thus, R is not monochromatic. • Therefore, at least t monochromatic rectangles are need to cover f Return

8. Exercise 1 • GT: 0 ≤ x , y < 2n • GT(x,y) = 1 iff x > y • Examples: • GT(5,6) = 0 • GT(6,5) = 1 • GT(6,6) = 0 • Prove D(GT) ≥ n

9. Solution 1 • S = {(x,x) | 0 ≤ x < 2n} • For every (x,x) ∈ S, GT(x,x) = 0 • For every (x,x) ≠ (y,y) ∈ S,GT(x,y) = 1 or GT(y,x) = 1 ⇒ S is a fooling set • |S| = 2n • From lemma 1 , D(GT) ≥ log2(2n) = n

10. We can do better! • What did we actually prove? • We proved: number of 0-rectangles is at least 2n • Because every member of fooling set is in a 0-rectangle, and we have at least 2ndifferent members • But, we have at least one 1-rectangle • Thus D(GT) ≥ log2(2n+ 1) = n + 1 • And because trivial upper bound is n+1: D(GT) = n + 1

11. Material • Finding lower bounds for the communication complexity of problems: • Fooling Sets ✓ • Weights and Distributions • Rank Lower Bound

12. Weighting Rectangles • Idea - proving that if every monochromatic rectangle is “small”, there must be many of them. • Weight: • µ: X × Y → [0,1] • Sum of µ(x,y) for every (x,y) ∈ X × Y = 1 • We choose the weights!

13. Weighted Fooling Sets • Our choice of weights: • Given fooling set S of size t • µ(x,y) = 0 for (x,y) ∉ S • µ(x,y) = 1/t for (x,y) ∈ S • If we can prove that S is a fooling set, only one weighted point can be in every rectangle. Thus µ(R) ≤ 1/t.

14. Weighted Fooling Sets (Example) GT Weights

15. Probability Distribution • Formalization of weights • Definition: µ: 2F → [0,1]µ (F) = 1A ∩ B = ∅ ⇒ µ (A∪B) = µ (A) + µ (B) • Proposition: Let µ be a probability distribution of X × Y. If every monochromatic rectangle R has measure µ(R) ≤ δ, then D(f) ≥ log2 (1/ δ)

16. Exercise 2 • DISJ: x, y ⊆ {1,2,…,n} • DISJ(x,y) = 1 iff x ∩ y = ∅ • Examples: • DISJ({1,2,3},{3,4}) = 0 • DISJ({3,7,9},{1,2}) = 1 • Prove D(DISJ) ≥ n A E

17. Solution 2 • Define P = 2({1,2,3,…,n}) • The weights: For every A,E ∈ P: µ(A,Ā) = 1/2n (A ≠ Ē) µ(A,E) = 0 • Claim: Every rectangle contains at most one pair (A,Ā) • We will prove DISJ(A,Ā) ≠ DISJ(A, Ē). Why is this sufficient?

18. Solution 2 (continued) • Claim: Every rectangle contains at most one pair (A,Ā) • For every A ≠ E ∈ P, without loss of generality, exists x ∈ A and x ∉ E. ⇒ A ∩ Ē ≠ ∅ ⇒ DISJ(A,Ē) = 0 • DISJ(A,Ā) = 1 and DISJ(A, Ē) = 0 ⇒ Every rectangle contains at most one pair (A,Ā), has size ≤ 1/2n

19. Solution 2 (final) • Definition: For every A,E ∈ P: µ(A,Ā) = 1/2n(A ≠ Ē) µ(A,E) = 0 • We proved: Every rectangle contains at most one pair (A,Ā) ⇒ every rectangle has size ≤ 1/2n • From the proposition, D(f) ≥ log2 [1/(1/2n)] = n • (S = {(A,Ā) | A ∈ P} is a fooling set)

20. Material • Finding lower bounds for the communication complexity of problems: • Fooling Sets ✓ • Weights and Distributions ✓ • Rank Lower Bound

21. Rank Lower Bound • Algebraic tool to calculating lower bound • Create matrix M of size |X| * |Y| • Indexed by elements of X and Y • M(x,y) = f(x,y)

22. Lemma 2 • For any function f: X × Y → {0,1}, D(f) ≥ log2 rank(M)

23. Proof D(f) ≥ log2 rank(M) • For each leaf k where output is 1: • Mk(x,y) = 1 iff (x,y) ∈ Rk where Rk is all the inputs leading to k • Claim: If f(x,y) = 1 there exists a single matrix where Mk(x,y) = 1 • Each input leads to exactly one leaf ⇒ for each input (x,y), there is exactly one k such that Mk (x,y) = 1 • Thus, M = ∑ Mk where k goes over all leaves f = M1 = M2 =

24. Proof (continued) D(f) ≥ log2 rank(M) • M = ∑ Mk where k goes over all leaves • Because Mk has a single rectangle, rank(Mk)=1 • Remember rank(A + B) ≤ rank(A) + rank(B) • We conclude, rank(M) ≤ ∑ rank(Mk) • rank(M) ≤ |Leaves of output 1| ≤ |Leaves| = |mono-rectangles| ⇒ D(f) ≥ log2 rank(M)

25. Improvement • We actually found a bound for the number of 1 rectangles, 0 rectangles are symmetric • Look at not(f), instead of f, and exact same proof • Define J = (ji,k) s.t. ∀ji,k = 1 • M = J – Mnot⇒ M = J + (–Mnot) ⇒ rank(M) ≤ rank(Mnot) + 1 ⇒ rank(Mnot) ≥ rank(M) - 1 • D(f) = log2 (|Rectangles|) ≥ log2(rank(M) + rank(Mnot)) ≥ log2(2*rank(M) - 1)

26. Example 3 • MAJ: x,y ∈ {0,1}2 • MAJ(x,y) = 1 iff |0| < |1| (x and y combined) • If |0| = |1|, we return 0 • Examples: • MAJ(00,10) = 0 • MAJ(01,11) = 1 • MAJ(10,01) = 0 • Find D(MAJ)

27. Solution 3 M = • rank(M) = 2 • D(MAJ) ≥ log2(2*2-1) = 2 • By previous means we could find better lower bound!

28. Questions?

29. Food for Thought NO! • Is every partition a legal protocol? • Suppose Alice starts:There are rectangles that aren’tcontained in a any (non-trivial) partition. • Same with Bob