1. Rotem Zach November 1 st, 2009

2. 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(log 2 t). These are the 0-rectangles and the 1-rectangles.

3. 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. Finding lower bounds for the communication complexity of problems: – Fooling Sets – Weights and Distributions – Rank Lower Bound

5. 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 (x 1,y 1 ) ≠ (x 2,y 2 ) in S, either f(x 1,y 2 )≠z or f(x 2,y 1 )≠z 0001 1101 0101 1010

6. EQ: 0 ≤ x, y < 2 2 – 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)} EQ00011011 001000 010100 100010 110001

7. If f has a fooling S set of size t, then D(f) ≥ log 2 t Proof: Claim: No x,y ∈ S can be in the same rectangle – Assume R contains (x 1,y 1 ) ≠ (x 2,y 2 ) (both belong to S) – From the definition, f(x 1,y 2 ) ≠ z or f(x 2,y 1 ) ≠ z – Thus, R is not monochromatic. Therefore, at least t monochromatic rectangles are need to cover f Return

8. GT: 0 ≤ x, y < 2 n – GT(x,y) = 1 iff x > y Examples: – GT(5,6) = 0 – GT(6,5) = 1 – GT(6,6) = 0 Prove D(GT) ≥ n x\y1……2n2n 100…0 …10…… ………00 2n2n 1…10

9. S = {(x,x) | 0 ≤ x < 2 n } 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| = 2 n From lemma 1, D(GT) ≥ log 2 (2 n ) = n x\y1……2n2n 100…0 …10…… ………00 2n2n 1…10

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

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

12. 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. 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. GT x\y1……2n2n 100…0 …10…… ………00 2n2n 1…10 1……2n2n 1 12n12n 0…0 …0 12n12n …… ……… 12n12n 0 2n2n 0…0 12n12n Weights

15. Formalization of weights Definition: µ: 2 F → [0,1] µ (F) = 1 A ∩ 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) ≥ log 2 (1/ δ)

16. 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 AE

17. Define P = 2 ({1,2,3,…,n}) The weights: For every A,E ∈ P: µ(A,Ā) = 1/2 n (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. 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 ≤ 1/2 n

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

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

21. 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. For any function f: X × Y → {0,1}, D(f) ≥ log 2 rank(M)

23. For each leaf k where output is 1: – M k (x,y) = 1 iff (x,y) ∈ R k where R k is all the inputs leading to k – Claim: If f(x,y) = 1 there exists a single matrix where M k (x,y) = 1 Each input leads to exactly one leaf ⇒ for each input (x,y), there is exactly one k such that M k (x,y) = 1 – Thus, M = ∑ M k where k goes over all leaves D(f) ≥ log 2 rank(M) R1R1 … …R2R2 f = 10 00 M 1 = 00 01 M 2 =

24. M = ∑ M k where k goes over all leaves Because M k has a single rectangle, rank(M k )=1 Remember rank(A + B) ≤ rank(A) + rank(B) – We conclude, rank(M) ≤ ∑ rank(M k ) – rank(M) ≤ |Leaves of output 1| ≤ |Leaves| = |mono-rectangles| ⇒ D(f) ≥ log 2 rank(M) D(f) ≥ log 2 rank(M)

25. 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 = (j i,k ) s.t. ∀ j i,k = 1 M = J – M not ⇒ M = J + (–M not ) ⇒ rank(M) ≤ rank(M not ) + 1 ⇒ rank(M not ) ≥ rank(M) - 1 D(f) = log 2 (|Rectangles|) ≥ log 2 (rank(M) + rank(M not )) ≥ log 2 (2*rank(M) - 1)

26. 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. 00011011 000000 010001 100001 110111 rank(M) = 2 D(MAJ) ≥ log 2 (2*2-1) = 2 By previous means we could find better lower bound! M =

29. Is every partition a legal protocol? Suppose Alice starts: There are rectangles that aren’t contained in a any (non-trivial) partition. Same with Bob yy’y’’ x100 x’111 x’’001 NO!