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**The Communication Complexity of Approximate Set Packing and Covering**

Noam Nisan Speaker: Shahar Dobzinski

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**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.

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**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?

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**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.

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**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.

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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.

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**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), ST Goal: a partition of M, S1,…,Sn, such that Svi(Si) is maximized. We will call Svi(Si) the total social welfare.

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**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

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**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.

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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) = |S|<m/2 0/1 |S|=m/2 |S|>m/2 Denote by vc the “dual” of v: vc(Sc) = |S|<m/2 1-v(S) |S|=m/2 |S|>m/2 For every allocation M=SSc, v(S)+vc(Sc)=1.

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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.

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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.

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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)

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**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

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**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.

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**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 AiAj =

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**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)

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

A\B 000 001 010 011 100 101 110 111 Y N

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

A\B 000 001 010 011 100 101 110 111 Y N

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**Proof (Approx. Disj.) – Rectangles**

Definition: a (combinatorial) rectangle is a cartesian product R1*…*Rn where each RiAi. 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.

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**Proof – Approximate Disjointness**

There are (n+1)t YES instances (for every i≠j 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 least log((1+1/n)t) = t(log(1+1/n))

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**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.

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**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).

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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, PsiiPsjj≠. i.e. every two parts from different partitions intersect. P1: 1 2 3 4 5 6 7 8 9 P2: 1 4 7 2 5 8 6 9 3 P3: 2 5 8 3 6 9 1 4 7

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**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[PsiiPsjj=] = (1-1/n2)m ≤ e-m/n^2

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**Existence – cont. Previous slide: Pr[PsiiPsjj=] ≤ 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.

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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, TBi and TS 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 P1: 1 2 3 4 5 6 7 8 9 P2: 1 4 7 2 5 8 6 9 3 P3: 2 5 8 3 6 9 1 4 7

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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 i≠j 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.

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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.

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**Set Cover A 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.

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**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:

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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.

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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).

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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 (AB ≠ ): there is some k AB. Alice holds Sk, and Bob holds Skc and these two sets cover the universe. YES instance (AB = ): 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.

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