Hardness of Approximating Multicut

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Hardness of Approximating Multicut

• S. Chawla, R. Krauthgamer, R. Kumar, Y. Rabani,
  D. Sivakumar (2005)
• Presented by Adin Rosenberg

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Multicut

• Input
• An undirected graph G(V,E), where Vn
• k pairs of vertices si,tii1,,k, called demand
  pairs
• Optional a cost function c on E
• Goal
• A multicut a subset of edges M, whose removal
  disconnects all of the demand pairs.
• Of course, minimize c(M) (or M if c isnt
  defined)

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Multicut an example
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What are we going to prove?

• Assuming the Unique Games conjecture is true,
  Multicut is NP-hard to approximate within any
  constant factor L

How are we going to prove this?

• We will show a reduction from a UG instance to a
  Multicut instance.

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Unique Games

• Input
• A bipartite graph G(Q,EQ)
• Each side p1,2 contains nQ/2 vertices (or
  questions) labeled q1p, q2p, , qnp
• Each edge (qi1,qj2) (called a question edge) is
  associated with a bijection bijd?d
• Each edge (qi1,qj2) has a nonnegative
  (normalized) weight wij

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Unique Games (cont.)

• A solution is an answer 1Aipd for each question
  qip
• A solution satisfies an edge (qi1,qj2) if the
  answers Ai1 and Aj2 agree, i.e. Aj2bij(Ai1)
• Goal
• Find a solution with maximum value (total weight
  of satisfied edges)

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Unique Games Conjecture Khot 2002

• For every ?,dgt0 there exists dd(?,d) such that
  it is NP-hard to determine whether a unique
  2-prover game with answer set of size d has a
  value of
• at least (1- ?), or
• at most d

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A Little About Hypercubes

• A d-dimensional hypercube is a graph G(V,E)
  where V0,1d and there is an edge between two
  vertices if they differ in exactly one
  coordinate.
• An edge (u,v) is called a dimension-a edge if u
  and v differ in coordinate a.
• A dimension-a cut is the set of dimension-a
  edges.
• The antipodal of a vertex u is the vertex which
  differs from u in every coordinate.

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A Little About Hypercubes
(0,1,1)
(1,1,1)
(0,0,1)
(1,0,1)
Dimension-1 edges
(0,1,0)
(1,1,0)
(0,0,0)
(1,0,0)
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The Reduction from Unique Games to Multicut

• For every vertex qip construct a d-dimensional
  hypercube Cip. Let the edges of these 2n cubes
  (called hypercube edges) have cost 1.
• For each question edge (qi1,qj2) extend bij to a
  bijection bij0,1d?0,1d defined by
• Connect each vertex with
  using and edge (called a cross edge) with cost
  wij?, where ?n/?.
• Define the demand pairs to be the antipodal
  pairs.

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The Reduction from Unique Games to Multicut
w11?
w11
1
1
w23?
1
1
w23
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The Yes Instance

• Claim If there is a solution A for the unique
  2-prover game with value of at least 1-?, then
  there exists a multicut M for the Multicut
  instance such that c(M) 2d1n
• Proof Construct the following multicut M
• For every answer Aip take the dimension- Aip cut
  in cube Cip.
• For every edge (qi1,qj2) that the solution A
  doesnt satisfy, take all the cross edges between
  Ci1 and Cj2.

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The Yes Instance (cont.)

• Removing M disconnects all the demand pairs
• For every vertex v in Cip, define f(v) to be the
  Aip-th coordinate of v.
• For every edge (u,v) left, f(u)?f(v)
• The cost of M is at most 2d1n
• Let S be the set of question edges not satisfied
  by the solution A.

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A Little More About Hypercubes

• For a function f on the vertices of a hypercube,
  define Iaf to be the fraction of dimension-a
  edges (u,v) for which f(u) ? f(v).
• For a cutset M in a hypercube, define IaM to be
  the fraction of dimension-a edges that belong to
  M.
• Observe that M 2d-1SaIaM
• And now some lemmas

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Lemma 1

• Let M be a cutset in a hypercube, and let g be
  the function labeling each vertex with the index
  of the connected component it belongs to. Then
  IaMIag.
• Proof M contains every edge (u,v) for which g(u)
  ? g(v)

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Lemma 2

• Let M be a cutset in a hypercube H.
• Suppose M disconnects at least a ß fraction of
  the antipodal pairs in H.
• Then for every xgt0, if SaIaM ßx then there
  exists a dimension a such that IaM 2-6x/27

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Lemma 3

• Let f,g be two function on the vertices of a
  hypercube.
• If f(v) g(v) for all but a ß fraction of the
  vertices v, then for every dimension a we have
  Iaf Iag 2ß.

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The No Instance

• Claim There exists a constant c such that if the
  Multicut instance has a cutset of cost at most
  2dnL (where Lc(log(1/(?d))) ) whose removal
  disconnects a7/8 fraction of the demand pairs,
  then there exists a solution for the unique
  2-prover game whose value is larger than d.

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The No Instance (cont.)

• Proof
• Let M we such a cutset for the Multicut instance.
• Let Iap,i be the influence of M for each cube
  Cip.
• Construct a randomized solution A for the unique
  2-prover game instance.
• For each vertex qip, we choose Aip to be the
  answer a with probability Iap,i / Sa Iap,i.
• The expected value of A is at least d, and
  therefore there exist a solution with such a
  value.

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The No Instance (cont.)

• Bound the probability of the following bad
  events (for a choice of the question edge
  (qi1,qj2) )
• E1 fewer than half the demand pairs in Ci1 are
  disconnected in G \ M
• E2 M contains more than 2d2L hypercube edges
  in Ci1.
• E3 M contains more than 2d2L hypercube edges
  in Cj2.
• E4 M contains more than 2d / 296L7 cross edges
  between Ci1 and Cj2.
• All bad events do not occur with probability of
  at least 1/8.

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The No Instance (cont.)

• Assuming none of the bad events occur
• There exists a dimension a s.t. Ia1,I
  2-96L/27 (according to Lemma 2)
• Ibij(a)2,j Ia1,i 2-96L-6 2-96L/54 (Lemma
  3)
• The expected value of A is

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What have we seen?

• If the unique game has a value greater than 1-?,
  then the Multicut instance has a cutset M which
  disconnects all of the demand pairs with cost
  c(M) 2d1n
• If the unique game has a value less than d, then
  the disconnected 7/8 of the demand pairs in the
  Multicut instance costs (at least) 2dLn
• Therefore, the Unique Games Conjecture implies
  that it is NP-hard to approximate Multicut within
  a factor of any Lgt0.

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Proof of Lemma 2

• SaIaM ßx means IaM 2-6x/27 for some a
• Proof
• Convert M to a two-sided cut M
• Define a Boolean function f according to the
  connected components of M
• Use KKLs lemma SaIaf/a Sa(Iaf)4/3 2ploga/a
  (where f is a Boolean function on a hypercube and
  p ½ is the balance of f)

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Proof of Lemma 3

• If f(v) g(v) for all but a ß fraction of the
  vertices v, then for every dimension a we have
  Iaf Iag 2ß.
• Proof
• For all but at most ß2d of the edges, we have
  f(u)g(u) and f(v)g(v).
• Therefore, only a ß2d/2d-12ß fraction of edges
  can contribute to the difference between Iaf and
  Iag.

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Credits to

• The authors of the paper for giving me what to
  talk about.
• Kahn, Kalai and Linial for saving us the Fourier
  analysis.
• Sarai for taking care of the lights.
• And of course, thank you all for listening