The Communication Complexity of Approximate Set Packing and Covering

1. The Communication Complexity of Approximate Set Packing and Covering Noam Nisan Speaker: Shahar Dobzinski

2. Communication Complexity • n players, computationally unlimited. • Each player i holds some private input Ai. • The goal is to compute some function f(Ai,…,An). • We are counting only the number of bits transmitted the players. • Worst case analysis.

3. Communication Complexity – Equality • 2 players (Alice and Bob). • Input: Alice holds a string A{0,1}n, Bob holds a string B{0,1}n. • Question: is A=B? • How many bits are required? • Upper Bound? • Lower Bound?

4. Equality Lower Bound • Denote an instance by (A,B). • Lemma: For each T≠T’ {0,1}n, the sequence of bits for (T,T) is different than the sequence of bits for (T’,T’). • The answer for both (T,T) and (T’,T’) is YES. • Proof: Suppose that there are T,T’ such that the sequences are identical.

5. Equality Lower Bound – cont. • What happens when the instance is (T,T’)? • Alice sends the first bit. • Same bit in (T,T’) and (T,T) • Bob sends the same bit for T and for T’. • Same goes for Alice, in the next round. • Corollary: the sequence of bits is the same for (T,T’) and for (T,T). • But (T,T’) is a NO instance and (T,T) is a YES instance - a contradiction.

6. Equality Lower Bound • We proved that for each T≠T’ {0,1}n, the sequence of bits for (T,T) is different than the sequence of bits for (T’,T’). • There are 2n different such sequences. • Log(2n)=n is a lower bound for the number of bits needed.

7. Combinatorial Auctions • n bidders, a set of M={1,…,m} items for sale. • Each bidder has a valuation function vi:2M->R+ • Standard assumptions: • Normalized: v()=0 • Monotonicity: v(T)≥v(S), ST • Goal: a partition of M, S1,…,Sn, such thatSvi(Si) is maximized. • We will call Svi(Si) the total social welfare.

8. Combinatorial Auctions – cont. • Problem: input is “exponential” - we are interested in algorithms that are polynomial in n and m. • Two approaches: • Bidding langauges • Example: single minded bidders • Communication complexity

9. Upper Bound • Give all items to bidder i that maximizes vi(M). • Proposition: n-approximation to the optimal total social welfare. • Proof: denote the optimal allocation by O1,…,On. • Sni=1vi(M) ≥ Sivi(Oi) = OPT.

10. Lower Bound – 2 Bidders • Theorem: For any e>0 any (2-e)-approximation to the total social welfare requires exponential communication. • Two bidders with valuations v1 and v2. • The valuations will have the following form: v(S) = 0 |S|<m/2 0/1 |S|=m/2 1 |S|>m/2 • Denote by vc the “dual” of v: vc(Sc) = 0 |S|<m/2 1-v(S) |S|=m/2 1 |S|>m/2 • For every allocation M=SSc, v(S)+vc(Sc)=1.

11. Main Lemma • Lemma: Let v1 and v2 be two different valuations. The sequence of bits for (v1,vc1) is different than the sequence of bits for (v2,vc2). • Proof: Suppose the sequences are identical. Then the sequence of bits for (v1,vc2) is the same too. • Same reasoning as before. • The allocation produced for (v1,vc1), (v2,vc2), (v1,vc2), (v2,vc1) is the same.

12. Main Lemma – cont. • There is a bundle T, T=|m/2|, such that v1(T)≠v2(T). WLOG v1(T)=1 and v2(T)=0. • Thus v2c(Tc)=1, and the optimal solution for (v1,v2c) is 2. • The protocol generated an optimal allocation (S,Sc). So v1(S)+v2c(Sc)=2. • But ((v1(S)+v1c(Sc))+ (v2(S)+v2c(Sc))=1+1=2. •  v1c(Sc)+v2(S)=0. • A contradiction to the optimality of the protocol.

13. The Lower Bound – cont. • If v1≠v2 then the sequence of bits for (v1,vc1) is different than the sequence of bits for (v2,vc2). • The number of different valuations is 2(m choose m/2). • Since for each (v,vc) we have a different sequence of bits, the communication complexity is at least log(2(m choose m/2)) = (m choose m/2) = exp(m)

14. Corollaries • Optimal solution requires exponential communication. • An (2-e)-approximation of the total social welfare requires exponential communication. • tight for 2 bidders. • Unconditional lower bound • even if P=NP

15. Lower Bound – General Number of Bidders • Theorem: Any approximation of the optimal total social welfare to a factor better than min(n,m1/2-e), for any e>0, requires exponential communication. • This lower bound holds not only for deterministic communication, but also for randomized and non-deterministic setting.

16. Approximate Disjointness • n players, each holds a string of length t. • The string of player i specifies a subset Ai{1,…,t}. • The goal is to distinguish between the following two extreme cases: • NO: iAi ≠  • YES: for every i≠j AiAj = 

17. Approximate Disjointness – cont. • Theorem: The approximate disjointness requires communication complexity of at least W(t/n4). This lower bound also holds for the randomized and non-deterministic settings. (Alon-Matias-Szegedi) • Theorem: The approximate disjointness requires communication complexity of at least W(t/n). (Radhakrishnan-Srinivasan)

18. Proof (Approx. Disj.) – Equality Matrix

19. Proof (Approx. Disj.) – Another Example for Matrix

20. Proof (Approx. Disj.) – Rectangles • Definition: a (combinatorial) rectangle is a cartesian product R1*…*Rn where each RiAi. • Definition: a monochromatic rectangle is a rectangle which doesn’t contain both YES instances and NO instances. • Lemma: log(number of monochromatic rectangles) is a lower bound for the communication complexity. • we proved a special case before.

21. Proof – Approximate Disjointness • There are (n+1)t YES instances (for every i≠j AiAj = ). • A YES instance is a partition between (n+1) players. • Lemma: any rectangle which does not contain a NO instance can contain at most nt YES instances. • Corollary: there are at least (1+1/n)t monochromatic rectangles. • Corollary: the communication complexity of approximate-disjointness is at least log((1+1/n)t) = t(log(1+1/n))

22. Proof – Approximate Disjointness • Lemma: any rectangle which does not contain a NO instance can contain at most nt YES instances. • Reminder: a NO instance is iAi ≠ . • Proof: • Fix such rectangle R. • For each item j there must a player i such that never gets j. • Otherwise, we have a NO instance. • Upper bound to the number of YES instances: all allocations between the rest of the (n-1) players and “unallocated” – nt.

23. The Combinatorial Auction • We will prove that it requires exponential communication to distinguish between the case the total social welfare is 1 and the case that it is n. • We will reduce from the approximate-disjointness with strings of size t (to be determined later).

24. The Partitions Set • We will use a set of partitions F={Ps|s=1…t}. Each Ps is a partition Ps1,…,Psn of M. • A set of partitions F={Ps|s=1…t} has the pair wise intersection property if for every choice of i≠j, and every si≠sj, PsiiPsjj≠. • i.e. every two parts from different partitions intersect. 1 2 3 4 5 6 7 8 9 P1: P2: 1 4 7 2 5 8 6 9 3 2 5 8 3 6 9 1 4 7 P3:

25. Existence of the partitions set • Lemma: Such a set F exists with |F|=t=em/2n^2/n2 • Proof: using the probabilistic method. • for each partition, place each element independently at random in one part of the partition. • Fix i≠j, si≠sj, and an item j. Pr[j is not in both Psii and Psjj]=1-1/n2 • The probability that they do not intersect: Pr[PsiiPsjj=] = (1-1/n2)m ≤ e-m/n^2

26. Existence – cont. • Previous slide: Pr[PsiiPsjj=] ≤ e-m/n^2 • We have at most n2t2 choices of indices. • Using the union bound: Pr[ pair of parts that don’t intersect] ≤ n2t2(e-m/n^2) • Choose t = em/2n^2/n2 = exp(m/n2). • Pr[ pair of parts that don’t intersect] < 1 • Pr[all pair of parts intersect] > 0 • Such a set exists.

27. The Reduction • We reduce the approximate-disjointness problem to a combinatorial auction (m items, n bidders). • Each player i who got Ai as input, constructs the collection Bi = {Psi|Ai=1}. • Define the valuations as: Vi(S) = 1  T, TBi and TS 0 otherwise Suppose A1=101 The first bidder values all bundles which contain {1,2,3} or {2,5,8} with 1, and the rest of the bundles with 0 1 2 3 4 5 6 7 8 9 P1: P2: 1 4 7 2 5 8 6 9 3 2 5 8 3 6 9 1 4 7 P3:

28. The Reduction – cont. • NO instance (iAi ≠ ): there is some kiAi. Assign Pki to bidder i, and the total social welfare is n. • YES instance (for every i≠j AiAj = ): the total social welfare is at most 1. • Corollary: It requires exponential communication to distinguish between the case the total social welfare is 1, and the case that it is n.

29. Remarks • We used strings of size t=em/2n^2/n2, thus the communication complexity is W(em/2n^2-5log(n)). • If n < m1/2-e, the communication complexity is exponential. • Corollary: For any e>0, an m1/2-e-approximation requires exponential communication. • An m1/2-approximation algorithm exists.

30. Set Cover • A universe of size |M|=m. • n players, each holds a collection Ai2M. • Goal: find the minimum cardinality set cover. • Upper bound: the greedy algorithm is a ln(m) approximation. • Lower bound – a reduction from approximate disjointness.

31. Lower Bound • 2 players (Alice and Bob). • Alice holds a collection A 2M, and Bob holds a collection B 2M. • We will prove that it requires exponential communication to distinguish between the case 2 sets are needed to cover M, and the case at least r+1 sets are needed (for r=log(m)-O(loglog(m))). • We will require the following class of subsets of M:

32. The r-Covering Class • A class C={(S1,S1c),…,(St,Stc)} has the r-Covering property if every collection of at most r sets, which does not contain a set and its complementary, does not cover all M.

33. Existence • Lemma: For any given r≤ log(m) – O(loglog(m)), there is a class C with t=em/(r2^r) • Proof: Probabalistic construction. • put each element of the universe in the set Sj with probability ½. • For a random collection of r sets, the probability that a single element j is in their union is 1-2-r. • For a random collection of r sets, the probability that their union is M is (1-2-r)m≤e-n/2^r. • There are at most (2t choose r) sets, so we need to make sure that (2t choose r)e-n/2^r<1 • We can choose t=em/(r2^r).

34. The Reduction • We reduce from the approximate disjointness problem with strings of size t. • Alice will construct the collection D={Si|Ai=1}. • Bob will construct the collection E={Si|Bi=1}. • NO instance (AB≠ ): there is some k AB. Alice holds Sk, and Bob holds Skc and these two sets cover the universe. • YES instance (AB= ): at least r+1 sets are needed to cover the world. • Corollary: It requires exponential communication to distinguish between the case 2 sets cover the universe, and between the case at least r+1 sets are needed.