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Unformatted text preview: PCPs and Inapproximability Gapproducing and GapPreserving Reductions My T. Thai 1 1 Hardness of Approximation Consider a maximization problem such as MAXE3SAT. To show that it is NPhard 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 NPhard. How to prove a gap problem is NPhard? Basically, there are two main methods: (1) Gapproducing reduction and (2) Gappreserving reduction. 2 GapProducing Reduction 2.1 MaxClique In the previous lecture note, we saw an example of proving GAPMAX E3SAT c,s using a gapproducing reduction. Now, let us consider another example: MaxClique problem. Definition 1 GAPMaxClique c,s : Given < s c 1 , a graph G and an positive integer k , the YES and NO instances of GAPMaxClique 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 GAPMaxClique c,s is NPhard, then approximating MaxClique to s/c is NPhard for any . To show a gapproblem NPhard via gapproducing reduction, we need to reduce from a NPhard problem L to the gapproblem in polynomial time. From the PCP theorem, we know that for any NPcomplete L , there exists an ( O (log n ) , O (1))restricted verifier V . Use V as the subrouting to construct a gapproducing reduction. For example, consider the GAPMaxClique c,s problem and consider an NPhard problem L . We need to construct a reduction R such that for each input x L , R will construct an instance ( G, k ) of GAPMaxClique 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 GAPMaxClique 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.
 Spring '08
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