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?0,1n, Bob holds a
  string B?0,1n.
• Question is AB?
• How many bits are required?
• Upper Bound?
• Lower Bound?

4
Equality Lower Bound

• Denote an instance by (A,B).
• Lemma For each T?T ?0,1n, 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 ?0,1n, 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 M1,,m items for sale.
• Each bidder has a valuation function
• vi2M-gtR
• Standard assumptions
• Normalized v(?)0
• Monotonicity v(T)v(S), S??T
• 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.
• Sni1vi(M) Sivi(Oi) OPT.

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Lower Bound 2 Bidders

• Theorem For any egt0 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 Sltm/2
• 0/1 Sm/2
• 1 Sgtm/2
• Denote by vc the dual of v
• vc(Sc) 0 Sltm/2
• 1-v(S) Sm/2
• 1 Sgtm/2
• For every allocation MS?Sc, 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, Tm/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))112.
• ? 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 PNP

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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 egt0, 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 Ai?Aj ?

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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
000 Y N N N N N N N
001 N Y N N N N N N
010 N N Y N N N N N
011 N N N Y N N N N
100 N N N N Y N N N
101 N N N N N Y N N
110 N N N N N N Y N
110 N N N N N N N Y
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Proof (Approx. Disj.) Another Example for Matrix
A\B 000 001 010 011 100 101 110 111
000 Y Y N N N N N N
001 Y Y N N N N N N
010 N N N N N N N N
011 N N N Y Y ?? N N
100 N N N Y Y Y N N
101 N N N Y ?? Y N N
110 N N N N N N Y N
110 Y N N N N N N N
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Proof (Approx. Disj.) Rectangles

• Definition a (combinatorial) rectangle is a
  cartesian product R1Rn where each Ri?Ai.
• 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.

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Proof Approximate Disjointness

• There are (n1)t YES instances (for every i?j
  Ai?Aj ?).
• A YES instance is a partition between (n1)
  players.
• Lemma any rectangle which does not contain a NO
  instance can contain at most nt YES instances.
• Corollary there are at least (11/n)t
  monochromatic rectangles.
• Corollary the communication complexity of
  approximate-disjointness is at least
• log((11/n)t) t(log(11/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 FPss1t.
  Each Ps is a partition Ps1,,Psn of M.
• A set of partitions FPss1t 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
1
2
3
4
5
6
7
8
9
P2
1
2
3
4
5
6
7
8
9
P3
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Existence of the partitions set

• Lemma Such a set F exists with Ftem/2n2/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.
• Prj is not in both Psii and Psjj1-1/n2
• The probability that they do not intersect
• PrPsii?Psjj? (1-1/n2)m e-m/n2

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Existence cont.

• Previous slide PrPsii?Psjj? e-m/n2
• We have at most n2t2 choices of indices.
• Using the union bound
• Pr? pair of parts that dont intersect
  n2t2(e-m/n2)
• Choose t em/2n2/n2 exp(m/n2).
• Pr? pair of parts that dont intersect lt 1
• Prall pair of parts intersect gt 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 PsiAi1.
• Define the valuations as
• Vi(S) 1 ? ?T, T?Bi and T?S
• 0 otherwise

Suppose A1101 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
1
2
3
4
5
6
7
8
9
P2
1
2
3
4
5
6
7
8
9
P3
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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.

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Remarks

• We used strings of size tem/2n2/n2, thus the
  communication complexity is W(em/2n2-5log(n)).
• If n lt m1/2-e, the communication complexity is
  exponential.
• Corollary For any egt0, an m1/2-e-approximation
  requires exponential communication.
• An m1/2-approximation algorithm exists.

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Set Cover

• A universe of size Mm.
• 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.

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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
  r1 sets are needed (for rlog(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 tem/(r2r)
• 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)me-n/2r.
• There are at most (2t choose r) sets, so we need
  to make sure that
• (2t choose r)e-n/2rlt1
• We can choose tem/(r2r).

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The Reduction

• We reduce from the approximate disjointness
  problem with strings of size t.
• Alice will construct the collection DSiAi1.
• Bob will construct the collection ESiBi1.
• 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 r1 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 r1 sets
  are needed.