# the communication complexity of approximate set packing and covering

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The Communication Complexity of Approximate Set Packing and Covering. Noam Nisan Speaker: Shahar Dobzinski. Communication Complexity. n players, computationally unlimited. Each player i holds some private input A i . The goal is to compute some function f(A i ,,A n ). - PowerPoint PPT Presentation

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• The Communication Complexity of Approximate Set Packing and CoveringNoam Nisan

Speaker: Shahar Dobzinski

• Communication Complexityn 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.

• Communication Complexity Equality2 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?

• Equality Lower BoundDenote an instance by (A,B).Lemma: For each TT {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.

• 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.

• Equality Lower BoundWe proved that for each TT {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.

• Combinatorial Auctionsn bidders, a set of M={1,,m} items for sale.Each bidder has a valuation functionvi:2M->R+Standard assumptions:Normalized: v()=0Monotonicity: v(T)v(S), ST Goal: a partition of M, S1,,Sn, such that Svi(Si) is maximized.We will call Svi(Si) the total social welfare.

• Combinatorial Auctions cont.Problem: input is exponential - we are interested in algorithms that are polynomial in n and m.Two approaches:Bidding langaugesExample: single minded biddersCommunication complexity

• Upper BoundGive 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.

• Lower Bound 2 BiddersTheorem: 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/2Denote by vc the dual of v: vc(Sc) = 0 |S|m/2For every allocation M=SSc, v(S)+vc(Sc)=1.

• Main LemmaLemma: 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.

• 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.

• The Lower Bound cont.If v1v2 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 leastlog(2(m choose m/2)) = (m choose m/2) = exp(m)

• CorollariesOptimal solution requires exponential communication.An (2-e)-approximation of the total social welfare requires exponential communication.tight for 2 bidders.Unconditional lower boundeven if P=NP

• Lower Bound General Number of BiddersTheorem: 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.

• Approximate Disjointnessn 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 ij AiAj =

• 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)

• Proof (Approx. Disj.) Equality Matrix

• Proof (Approx. Disj.) Another Example for Matrix

• Proof (Approx. Disj.) RectanglesDefinition: a (combinatorial) rectangle is a cartesian product R1**Rn where each RiAi.Definition: a monochromatic rectangle is a rectangle which doesnt 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.

• Proof Approximate DisjointnessThere are (n+1)t YES instances (for every ij AiAj = ).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 leastlog((1+1/n)t) = t(log(1+1/n))

• Proof Approximate DisjointnessLemma: 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.

• 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).

• The Partitions SetWe will use a set of partitions F={Ps|s=1t}. Each Ps is a partition Ps1,,Psn of M.A set of partitions F={Ps|s=1t} has the pair wise intersection property if for every choice of ij, and every sisj, PsiiPsjj.i.e. every two parts from different partitions intersect. 123456789123456789123456789P1:P2:P3:

• Existence of the partitions setLemma: Such a set F exists with |F|=t=em/2n^2/n2Proof: using the probabilistic method.for each partition, place each element independently at random in one part of the partition.Fix ij, sisj, and an item j. Pr[j is not in both Psii and Psjj]=1-1/n2 The probability that they do not intersect:Pr[PsiiPsjj=] = (1-1/n2)m e-m/n^2

• Existence cont.Previous slide: Pr[PsiiPsjj=] e-m/n^2We have at most n2t2 choices of indices.Using the union bound:Pr[ pair of parts that dont intersect] n2t2(e-m/n^2)Choose t = em/2n^2/n2 = exp(m/n2).Pr[ pair of parts that dont intersect] < 1Pr[all pair of parts intersect] > 0Such a set exists.

• The ReductionWe 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, TBi and TS 0 otherwise123456789123456789123456789P1:P2:P3:Suppose A1=101The first bidder values all bundles which contain {1,2,3} or {2,5,8} with 1, and the rest of the bundles with 0

• The Reduction cont.NO instance (iAi ): there is some kiAi. Assign Pki to bidder i, and the total social welfare is n.YES instance (for every ij AiAj = ): 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.

• RemarksWe 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.

• Set CoverA universe of size |M|=m.n players, each holds a collection Ai2M.Goal: find the minimum cardinality set cover.

Upper bound: the greedy algorithm is a ln(m) approximation.Lower bound a reduction from approximate disjointness.

• Lower Bound2 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:

• The r-Covering ClassA 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.

• ExistenceLemma: 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 .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)me-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
• The ReductionWe 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=