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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 |V|=n k pairs of vertices {s i,t i } i=1,…,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 isn’t 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,E Q ) Each side p=1,2 contains n=|Q|/2 vertices (or questions) labeled q 1 p, q 2 p, …, q n p Each edge (q i 1,q j 2 ) (called a question edge) is associated with a bijection b ij :[d]→[d] Each edge (q i 1,q j 2 ) has a nonnegative (normalized) weight w ij

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Unique Games (cont.) A solution is an answer 1≤A i p ≤d for each question q i p A solution satisfies an edge (q i 1,q j 2 ) if the answers A i 1 and A j 2 agree, i.e. A j 2 =b ij (A i 1 ) Goal: Find a solution with maximum value (total weight of satisfied edges)

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Unique Games Conjecture [Khot 2002] For every η,δ>0 there exists d=d(η,δ) 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- η), or at most δ

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A Little About Hypercubes A d-dimensional hypercube is a graph G=(V,E) where V={0,1} d 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 (0,0,0)(1,0,0) (0,1,0)(1,1,0) (0,0,1)(1,0,1) (0,1,1)(1,1,1) Dimension-1 edges

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The Reduction from Unique Games to Multicut For every vertex q i p construct a d-dimensional hypercube C i p. Let the edges of these 2n cubes (called hypercube edges) have cost 1. For each question edge (q i 1,q j 2 ) extend b ij to a bijection b’ ij :{0,1} d →{0,1} d defined by Connect each vertex with using and edge (called a cross edge) with cost w ij Λ, where Λ=n/η. Define the demand pairs to be the antipodal pairs.

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The Reduction from Unique Games to Multicut w 11 w 23 w 11 Λ w 23 Λ 1 1 1 1

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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-η, then there exists a multicut M for the Multicut instance such that c(M) ≤ 2 d+1 n Proof: Construct the following multicut M: For every answer A i p take the dimension- A i p cut in cube C i p. For every edge (q i 1,q j 2 ) that the solution A doesn’t satisfy, take all the cross edges between C i 1 and C j 2.

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The Yes Instance (cont.) Removing M disconnects all the demand pairs: For every vertex v in C i p, define f(v) to be the A i p -th coordinate of v. For every edge (u,v) left, f(u)≠f(v) The cost of M is at most 2 d+1 n: 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 I a f to be the fraction of dimension-a edges (u,v) for which f(u) ≠ f(v). For a cutset M in a hypercube, define I a M to be the fraction of dimension-a edges that belong to M. Observe that |M| = 2 d-1 Σ a I a M 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 I a M ≥I a g. 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 β fraction of the antipodal pairs in H. Then for every x>0, if Σ a I a M ≤ βx then there exists a dimension a* such that I a* M ≥ 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 β fraction of the vertices v, then for every dimension a we have |I a f – I a g | ≤ 2β.

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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 2 d nL (where L=c(log(1/(η+δ))) ) whose removal disconnects α≥7/8 fraction of the demand pairs, then there exists a solution for the unique 2-prover game whose value is larger than δ.

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The No Instance (cont.) Proof: Let M we such a cutset for the Multicut instance. Let I a p,i be the influence of M for each cube C i p. Construct a randomized solution A for the unique 2-prover game instance. For each vertex q i p, we choose A i p to be the answer a with probability I a p,i / Σ a I a p,i. The expected value of A is at least δ, 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 (q i 1,q j 2 ) ): E 1 – fewer than half the demand pairs in C i 1 are disconnected in G \ M E 2 – M contains more than 2 d+2 L hypercube edges in C i 1. E 3 – M contains more than 2 d+2 L hypercube edges in C j 2. E 4 – M contains more than 2 d / 2 96L+7 cross edges between C i 1 and C j 2. 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. I a* 1,I ≥ 2 -96L /27 (according to Lemma 2) I bij(a*) 2,j ≥ I a* 1,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-η, then the Multicut instance has a cutset M which disconnects all of the demand pairs with cost c(M) ≤ 2 d+1 n If the unique game has a value less than δ, then the disconnected 7/8 of the demand pairs in the Multicut instance costs (at least) 2 d Ln Therefore, the Unique Games Conjecture implies that it is NP-hard to approximate Multicut within a factor of any L>0.

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Proof of Lemma 2 Σ a I a M ≤ βx means I a* M ≥ 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 KKL’s lemma: Σ a I a f /α + Σ a (I a f ) 4/3 ≥ 2p*logα/α (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 β fraction of the vertices v, then for every dimension a we have |I a f – I a g | ≤ 2β. Proof: For all but at most β2 d of the edges, we have f(u)=g(u) and f(v)=g(v). Therefore, only a β2 d /2 d-1 =2β fraction of edges can contribute to the difference between I a f and I a g.

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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…

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