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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 A i. The goal is to compute some function f(A i,…,A n ). 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 2 n different such sequences. Log(2 n )=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 v i :2 M ->R + Standard assumptions: Normalized: v( )=0 Monotonicity: v(T)≥v(S), S T Goal: a partition of M, S 1,…,S n, such that v i (S i ) is maximized. We will call v i (S i ) 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 v i (M). Proposition: n-approximation to the optimal total social welfare. Proof: denote the optimal allocation by O 1,…,O n. n i=1 v i (M) ≥ i v i (O i ) = OPT.

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Lower Bound – 2 Bidders Theorem: For any >0 any (2- )-approximation to the total social welfare requires exponential communication. Two bidders with valuations v 1 and v 2. The valuations will have the following form: v(S) = 0 |S|

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Main Lemma Lemma: Let v 1 and v 2 be two different valuations. The sequence of bits for (v 1,v c 1 ) is different than the sequence of bits for (v 2,v c 2 ). Proof: Suppose the sequences are identical. Then the sequence of bits for (v 1,v c 2 ) is the same too. Same reasoning as before. The allocation produced for (v 1,v c 1 ), (v 2,v c 2 ), (v 1,v c 2 ), (v 2,v c 1 ) is the same.

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

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The Lower Bound – cont. If v 1 ≠v 2 then the sequence of bits for (v 1,v c 1 ) is different than the sequence of bits for (v 2,v c 2 ). The number of different valuations is 2 (m choose m/2). Since for each (v,v c ) 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- )-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,m 1/2- ), for any >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 A i {1,…,t}. The goal is to distinguish between the following two extreme cases: NO: i A i ≠ YES: for every i≠j A i A j =

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Approximate Disjointness – cont. Theorem: The approximate disjointness requires communication complexity of at least (t/n 4 ). 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 (t/n). (Radhakrishnan-Srinivasan)

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Proof (Approx. Disj.) – Equality Matrix A\B YNNNNNNN 001NYNNNNNN 010NNYNNNNN 011NNNYNNNN 100NNNNYNNN 101NNNNNYNN 110NNNNNNYN NNNNNNNY

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Proof (Approx. Disj.) – Another Example for Matrix A\B YYNNNNNN 001YYNNNNNN 010NNNNNNNN 011NNNYY NN 100NNNYYYNN 101NNNY YNN 110NNNNNNYN YNNNNNNN

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Proof (Approx. Disj.) – Rectangles Definition: a (combinatorial) rectangle is a cartesian product R 1 *…*R n where each R i A i. 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 A i A j = ). A YES instance is a partition between (n+1) players. Lemma: any rectangle which does not contain a NO instance can contain at most n t 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 n t YES instances. Reminder: a NO instance is i A i ≠ . 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” – n t.

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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={P s |s=1…t}. Each P s is a partition P s 1,…,P s n of M. A set of partitions F={P s |s=1…t} has the pair wise intersection property if for every choice of i≠j, and every s i ≠s j, P si i P sj j ≠ . i.e. every two parts from different partitions intersect P1:P1: P2:P2: P3:P3:

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

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Existence – cont. Previous slide: Pr[P si i P sj j = ] ≤ e -m/n^2 We have at most n 2 t 2 choices of indices. Using the union bound: Pr[ pair of parts that don’t intersect] ≤ n 2 t 2( e -m/n^2) Choose t = e m/2n^2 /n 2 = exp(m/n 2 ). 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 A i as input, constructs the collection B i = {P s i |A i =1}. Define the valuations as: V i (S) = 1 T, T B i and T S 0 otherwise P1:P1: P2:P2: P3:P3: Suppose A 1 =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

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The Reduction – cont. NO instance ( i A i ≠ ): there is some k i A i. Assign P k i to bidder i, and the total social welfare is n. YES instance (for every i≠j A i A j = ): 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=e m/2n^2 /n 2, thus the communication complexity is (e m/2n^2-5log(n) ). If n < m 1/2- , the communication complexity is exponential. Corollary: For any >0, an m 1/2- -approximation requires exponential communication. An m 1/2 -approximation algorithm exists.

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Set Cover A universe of size |M|=m. n players, each holds a collection A i 2 M. 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 2 M, and Bob holds a collection B 2 M. 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={(S 1,S 1 c ),…,(S t,S t c )} 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=e m/(r2^r) Proof: Probabalistic construction. put each element of the universe in the set S j 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=e m/(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={S i |A i =1}. Bob will construct the collection E={S i |B i =1}. NO instance (A B ≠ ): there is some k A B. Alice holds S k, and Bob holds S k c 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.

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