The Communication Complexity
of Approximate Set Packing and
Covering
Noam Nisan
Speaker: Shahar Dobzinski
Communication Complexity
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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.
Communication Complexity –
Equality
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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?
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Upper Bound?
Lower Bound?
Equality Lower Bound
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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’).
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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.
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What happens when the instance is (T,T’)?
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Alice sends the first bit.
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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 Bound
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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.
Combinatorial Auctions
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n bidders, a set of M={1,…,m} items for sale.
Each bidder has a valuation function
vi:2M->R+
Standard assumptions:
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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.
Combinatorial Auctions – cont.
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Problem: input is “exponential” - we are
interested in algorithms that are
polynomial in n and m.
Two approaches:
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Bidding langauges
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Example: single minded bidders
Communication complexity
Upper Bound
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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.
Lower Bound – 2 Bidders
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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.
Main Lemma
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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.
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Same reasoning as before.
The allocation produced for (v1,vc1), (v2,vc2),
(v1,vc2), (v2,vc1) is the same.
Main Lemma – cont.
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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.
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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)
Corollaries
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Optimal solution requires exponential
communication.
An (2-e)-approximation of the total
social welfare requires exponential
communication.
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tight for 2 bidders.
Unconditional lower bound
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even if P=NP
Lower Bound – General
Number of Bidders
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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.
Approximate Disjointness
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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:
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NO: iAi ≠ 
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YES: for every i≠j AiAj = 
Approximate Disjointness –
cont.
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Theorem: The approximate disjointness
requires communication complexity of
at least W(t/n4). This lower bound also
holds for the randomized and nondeterministic settings. (Alon-Matias-Szegedi)
Theorem: The approximate disjointness
requires communication complexity of
at least W(t/n). (Radhakrishnan-Srinivasan)
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
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
Proof (Approx. Disj.) –
Rectangles
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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.
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we proved a special case before.
Proof – Approximate
Disjointness
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There are (n+1)t YES instances (for every i≠j AiAj =
).
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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))
Proof – Approximate
Disjointness
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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.
The Combinatorial Auction
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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 approximatedisjointness with strings of size t (to be
determined later).
The Partitions Set
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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≠.
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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
Existence of the partitions set
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Lemma: Such a set F exists with
|F|=t=em/2n^2/n2
Proof: using the probabilistic method.
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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
Existence – cont.
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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)
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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.
The Reduction
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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
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
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
The Reduction – cont.
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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.
Remarks
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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.
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Corollary: For any e>0, an m1/2-e-approximation
requires exponential communication.
An m1/2-approximation algorithm exists.
Set Cover
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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.
Lower Bound
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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:
The r-Covering Class
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A class C={(S1,S1c),…,(St,Stc)} has the rCovering property if every collection of
at most r sets, which does not contain a
set and its complementary, does not
cover all M.
Existence
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Lemma: For any given r≤ log(m) – O(loglog(m)),
there is a class C with t=em/(r2^r)
Proof: Probabalistic construction.
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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).
The Reduction
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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.