Gap-Hardness - PCPs and Inapproximability Gap-producing and...

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Unformatted text preview: PCPs and Inapproximability Gap-producing and Gap-Preserving Reductions My T. Thai 1 1 Hardness of Approximation Consider a maximization problem such as MAX-E3SAT. To show that it is NP-hard to approximation to an approximation ratio for all , we need to define the GAP- c,s where = s/c and prove that GAP- c,s is NP-hard. How to prove a gap problem is NP-hard? Basically, there are two main methods: (1) Gap-producing reduction and (2) Gap-preserving reduction. 2 Gap-Producing Reduction 2.1 Max-Clique In the previous lecture note, we saw an example of proving GAP-MAX- E3SAT c,s using a gap-producing reduction. Now, let us consider another example: Max-Clique problem. Definition 1 GAP-Max-Clique c,s : Given < s c 1 , a graph G and an positive integer k , the YES and NO instances of GAP-Max-Clique c,s are defined as follows: Output YES: if ( G ) ck , that is G has a clique of size ck Output NO: if ( G ) < sk , that is G has no clique of size sk . where ( G ) is the size of the maximum clique in G . Again, if GAP-Max-Clique c,s is NP-hard, then approximating Max-Clique to s/c is NP-hard for any . To show a gap-problem NP-hard via gap-producing reduction, we need to reduce from a NP-hard problem L to the gap-problem in polynomial time. From the PCP theorem, we know that for any NP-complete L , there exists an ( O (log n ) , O (1))-restricted verifier V . Use V as the sub-routing to construct a gap-producing reduction. For example, consider the GAP-Max-Clique c,s problem and consider an NP-hard problem L . We need to construct a reduction R such that for each input x L , R will construct an instance ( G, k ) of GAP-Max-Clique c,s satisfying the following two constraints: 2 If x L , then ( G ) ck If x / L , then ( G ) < sk Lemma 1 [1] If 3SAT PCP c,s [ r, q ] , then there exists a deterministic reduc- tion running in time poly (2 r + q ) reducing 3SAT to GAP-Max-Clique c,s Before proving Lemma 1, we would like to show the following construction. Consider E 3 SAT , let be on n variables x 1 , . . . , x n and m clauses c 1 , . . . , c m . We create m rows of vertices, one for each clause. For each row, there are 7 vertices corresponding to all the satisfying assignments on that respective clause. Note that there are 7 partial assignments on 3 variables that make a clause satisfiable (and 1 make it unsatisfiable). For example, consider the clause ( x 1 , x 2 , x 3 ), then the seven vertices correspond to the following assignments: 000, 010, 011, 100, 101, 110, 111. Now, let see how we assign edges on these vertices. Each row in the graph will be an independent set. Edges between rows are joining vertices whose assignments are consistent. The consistent assignments are defined as follows: (1) two vertices shared a variable which has the same assignment; (2) two vertices with disjoint set of variables. For example, consider clause C 1 = ( x 1 , x 2 , x 3 ) and clause...
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This note was uploaded on 05/20/2011 for the course CIS 6930 taught by Professor Staff during the Spring '08 term at University of Florida.

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Gap-Hardness - PCPs and Inapproximability Gap-producing and...

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