Circuit and Communication Complexity

1. Circuit and Communication Complexity

2. Given The communication game Gf : • Alice gets s.t. f(x)=1 • Bob gets s.t. f(y)=0 • Goal: Find i s.t. Karchmer – Wigderson Games

3. has a boolean circuit C that computes it. The deterministic communication complexity of a game G: The minimal CC of a protocol for G. CC of a protocol: the maximal number of bits (over all inputs) exchanged when running the protocol Boolean circuits using AND,OR,NOT gates with fan-in=2 Depth(C): The longest path from input to output in a boolean circuit C Depth(f): The minimal depth of a circuit that computes f

4. - by induction on Depth(C) then f(z) is zi or zi for some i Base:Depth(C )=0 Alice and Bob can simply answer: i (No need to exchange bits) CC(Gf ) = Depth( f ) C : circuit with minimal depth that computes f Lemma 1: Proof:

5. CC(Gf ) = Depth(f) Induction step: look at C’s top gate f1, f2 : the functions computed by C1, C2 By inductive hypothesis: f(x)=1 f1(x) = f2(x) = 1 f1(y) = 0or f2(y) = 0 f(y)=0

6. f1(x) = f2(x) = 1 f1(y) = 0or f2(y) = 0 f(x)=1 f(y)=0 Bob and Alice can apply a protocol for Bob and Alice can apply a protocol for CC(Gf ) = Depth(f) Protocol for Gf: • Bob sends a value in {1,2} indicating whether • f1(y) = 0or f2(y) = 0 - If Bob sends 1, Alice and Bob know: f1(y) = 0 , f1(x) = 1 - If Bob sends 2, Alice and Bob know: f2(y) = 0 , f2(x) = 1

7. CC(Gf ) = Depth(f) Either way, we get a protocol for Gf with one additional bit f(x)=1 f1(x) = 1or f2(x) = 1 f1(y) = f2(y) = 0 f(y)=0 By the same idea, Alice sends a value in {1,2} indicating whether f1(x) = 1or f2(x) = 1

8. For any two disjoint sets: GA,B is: • Alice gets • Bob gets • Goal: Find i s.t. f -1(1)={x | f(x)=1} f -1(0)={y | f(y)=0} Gfis the same as CC(Gf ) = Depth(f) We’ll define a more general communication game:

9. Claim: if CC(GA,B )=d then there’s a function such that: • f(x)=1 for every • f(y)=0 for every , such a function is f itself For Hence, CC(Gf ) = Depth(f) f separates A from B We’ll prove the claim by induction on d =CC(GA,B )

10. So there’s i such that for every and every , The functions f=zi or f= satisfy the claim requirements. For both, Depth(f )=0. CC(Gf ) = Depth(f) We’ll prove the claim by induction on d =CC(GA,B ) Base:d=0 Bob and Alice don’t need to exchange bits

11. This bit partitions A into two disjoint sets If Alice sends 0, the rest of the protocol is a protocol for If Alice sends 1, the rest of the protocol is a protocol for For and we have protocols P1, P2 s.t. CC(Gf ) = Depth(f) Induction step: All x in A for which the first bit Alice sends is 0 We have a protocol P with CC(P)=d for GA,B Assume Alice sends the first bit. All x in A for which the first bit Alice sends is 1

12. f0(x)=1 for every • f1(x)=1 for every • f0(y)= f1(y)=0 for every We define Then: • for every , we have • for every , we have CC(Gf ) = Depth(f) By the inductive hypothesis, we have f0 and f1 that satisfy: f satisfies the requirements.

13. Similarly, if Bob sends the first bit, B is partitioned into two disjoint sets The rest of the protocol is either for or (depending on the bit Bob sent) • g0(y)=0 for every • g1(y)=0 for every • g0(x)= g1(x)=1 for every CC(Gf ) = Depth(f) By the inductive hypothesis, we have g0 and g1 that satisfy:

14. We define Then: • for every , we have • for every , we have We have seen: , CC(Gf ) = Depth(f) g satisfies the requirements. =>CC(Gf ) = Depth( f )

15. Example: • Alice gets a graphx on n vertices that contains • a clique of size n/2. • Bob gets a graph y on n vertices that doesn’t contain • a clique of size n/2 • Goal: Find an edge in x that doesn’t exist in y • (or vice versa) By lemma 1: the Communication Complexity of this game = the circuit depth of the (n/2)-Clique function. By proving lower bounds on the CC of such games, It would be possible to find lower bounds for certain functions’ circuit depth.

16. is a monotone function if for every , implies iff For every i A vector corresponds to the set iff Monotone Complexity Monotone functions: This order (called Hamming partial order) can be thought of as the containment order between sets:

17. Let be a monotone function. • A minterm of f is s.t. f(x) = 1 and for every x’< x , f(x’) = 0 • A maxterm of f is s.t. f(y) = 0 and for every y’> y , f(y’) = 1

18. 0 f(x)= 1 otherwise Example: (0,0,…,0,1), (0,1,0,…,0) are maxterms (1,1,0,…,0), (1,0,…,0,1) are minterms

19. is monotone iff it can be computed by a monotone circuit Monotone boolean circuit: A circuit which uses only AND and OR gates Monotone circuits can’t compute all functions: it’s impossible to simulate NOT by using AND and OR Monotone circuits can compute all monotone functions. For example, this can be done by taking the DNF of all 1’s in all minterms, or the CNF of all 0’s in all maxterms

20. : For a monotone function For a monotone function f, • Mon-Size(f ): the minimal size of a monotone circuit for f • Mon-Depth(f ): the minimal depth of a monotone circuit for f

21. Several lower bounds were proved for the monotone size and the monotone depth of certain functions. For the non-monotone case, there are no non-trivial lower bounds. There is no lower bound better than linear for the circuit size of any explicit function, and no lower bound better than logarithmic for the circuit depth of any explicit function.

22. Given a monotone function The communication game Mf : • Alice gets s.t. f(x)=1 • Bob gets s.t. f(y)=0 • Goal: Find i s.t. The goal is always achievable, otherwise, which means Monotone Karchmer – Wigderson Games That is: Find i s.t. xi=1 and yi=0

23. Lemma 2: for every monotone function CC(Mf) = Mon-Depth(f ) Proof: similar to the non-monotone case with the following modifications: • When constructing a protocol from a given circuit: Base case: if Depth(f ) = 0, f(z) = zi. So xi=1, yi=0, and Alice and Bob always answer: i Induction step: When dividing the circuit into two sub-circuits, each is monotone and has a suitable protocol. Bob or Alice (depending on the top gate) send the first bit, and decide which of the two monotone games to solve.

24. When constructing a circuit from a given protocol: Base case: if there is no communication, Alice and Bob both know an i for which xi>yi, so the circuit is f(z)=zi Induction step: Each communication bit splits the game Into two sub-games of smaller CC. Since the original game is monotone, so are the two sub-games. The circuits for these games are monotone. Since we only add an AND or an OR gate, the entire circuit Remains monotone.

25. A lower bound of was proved for the monotone depth of the (n/2)-Clique function Example: • Alice gets a graph x on n vertices that contains • a clique of size n/2 • Bob gets a graph x on n vertices that doesn’t contain • a clique of size n/2 • Goal: find an edge in x that doesn’t exist in y By lemma 2: the Communication Complexity of this game = the monotone circuit depth of the (n/2)-Clique function.

26. Monotone KW Game, Given a monotone function The communication game : • Alice gets s.t. x is a minterm of f (so f(x)=1) • Bob gets s.t. y is a maxterm of f (so f(y)=0) • Goal: Find i s.t. That is: Find i s.t. xi=1 and yi=0

27. is a restriction of to a subset of inputs. Hence, any protocol for is also a protocol for So: In fact, for every monotone We have to prove

28. : Given a protocol for , we will construct a protocol for with the same communication complexity. Alice gets x s.t. f(x)=1, and finds a minimal x’ s.t. . And f(x’)=1 In the same way, Bob finds a maxterm This is done by successively changing 1’s to 0’s in x x’ is a minterm

29. Alice and bob run the protocol for on x’ and y’, and find i s.t. x’i = 1 and y’i = 0 Since and we have xi=1 and yi=0 We get: for every monotone

30. We will prove a lower bound of for the monotone depth of the Matching function. Match(x) = • 1 x contains k independent edges • 0 otherwise Lower Bound for Matching The function Match: Let n=3k, and x a graph on n vertices.

31. By Lemma 2: The monotone KW game M0 for Match : • Alice gets a graph x on n vertices that contains a k-matching • Bob gets a graph y on n vertices that doesn’t contain • a k-matching • Goal: find an edge in x that doesn’t exist in y To prove a lower bound for Mon-Depth(Match) it’s enough to prove a lower bound for CC(M0)

32. Claim: The game M1: • Alice gets k independent edges x on n vertices (k-matching) • Bob gets a set y of k-1 vertices • Goal: find an edge in x that doesn’t touch • any of the vertices in y Proof: • Every k-matching is a minterm of Match • Every set of k-1 vertices is a maxterm of Match, by • considering all edges that touch y. • Every protocol for M0 can by applied on x,y to solve the game M1.

33. We get: To prove a lower bound for Mon-Depth(Match) it’s enough to prove a lower bound for CC(M1) P- a protocol for M1. We can assume that P outputs each possible answer with the same probability. Alice and Bob can use a common random string to permute the vertices before applying P

34. Claim: for any there exists a constant a s.t. The minimal CC of a probabilistic protocol P for which: Pr[P answers right] The game M2: • Alice gets a k-matching x • Bob gets a set y of k vertices • Goal: output 1 if there’s an edge in x that doesn’t touch • any of the vertices, and 0 otherwise.

35. Claim: for any there exists a constant a s.t. Bob has a set y of k vertices, and will randomly choose and remove it. Alice and Bob can apply P1 for M1 on (x,y’) and get That doesn’t touch any of the vertices in y’. Assume we have a deterministic protocol P1 for M1 Proof: We will construct a probabilistic protocol P2 for M2, with the same CC as P1 Bob is left with a set y’ of k-1 vertices.

36. If the edge edoesn’t touchv then P2 outputs 1. e doesn’t touch any vertex in y • If etouchesv then P2 outputs 0. P2assumes all edges in x touch y Analysis: If P2 outputs 1: e doesn’t touch v or any other vertex in y. In this case, there is no mistake If P2 outputs 0: there may be an edge e’ in x that doesn’t touch any of the vertices in y, P1 outputs e that touches v.

37. v e y e’ y’ Since the edges are independent, there’s at most one edge e that touches v. We are analyzing the case where there’s at least one edge e’ that doesn’t touch any of the vertices in y. An error occurs if e was output instead of any e’.

38. for any error prob. there exists a constant a s.t. The probability of that is at most 1/2 P1 outputs each answer with the same probability. If there are m answers, the probability is 1/m. To reduce the error probability, we can repeat this protocol any number of times.

39. We have: We’ll show: The communication complexity of this problem is The 3-Distinctness problem: • Alice and Bob each get a string of n letters from {a,b,c} • Goal: decide whether or not there’s i s.t. xi=yi. ? To prove the lower bound for M2, We’ll show a reduction from the 3-distinctness problem.

40. a c b b c a Alice and Bob convert their input from the 3-distinctness Problem to the following graph: For each coordinate construct an independent triangle with vertices labeled a,b,c. Denote each edge by the same letter as the vertex it doesn’t touch

41. a a a c b c b c b c b c b a a c b a 1 2 n a a a c b c b c b b c b c b c a a a 1 2 n • Alice interprets her n coordinates of x as thecorresponding • n edges in the n triangles. Example: Alice’s input is {a,c,…,b} • Bob interprets his n coordinates of y as the corresponding • n vertices in the n triangles. Example: Bob’s input is {b,b,…,a}

42. We’ve obtained a lower bound of for the Probabilistic communication complexity of M2 We’ll call Alice’s set of edges x’, and Bob’s set of vertices y’ There’s an edge in x’ that doesn’t touch y’ iff there’s a coordinate i such that xi=yi In each triangle, vertex d doesn’t touch edge d

43. CC(3-Distinctness) is By reduction from the Disjointness problem, which has CC : • Alice gets • Bob gets • Goal: return 1 iff there’s i s.t. xi=yi=1 ! In the disjointness problem: There’s i for which xi=yi=1 iff there’s i for which x’i=y’i To get a 3-distinctness problem: • Alice turns each 0 to b and each 1 to aand gets x’ • Bob turns each 0 to c and each 1 to aand gets y’