The 3 SAT Problem : Input : A 3
نویسندگان
چکیده
Following are the two main results of this paper, that the Radius and the Covering Radius of a code are hard to compute. Proof: First, note that this problem is in NP, as a center of a k-radius sphere which contains C is a linear time veriiable witness to the fact that R(C) k. To show that the problem is NP hard, we reduce 3SAT to MR. Given a 3CNF formula = ! 1 ^ : : : ^ ! t ; we deene the following code of length 2(n + 1): C 4 = Y 2(n+1) n c ! 1 ^00; : : : ; c ! t ^00 o : Observe that the code C is computable in polynomial time from the formula. We now prove the following validity claim of the reduction: is satissable () R(C) n + 1: Let v 2 f0; 1g n be a satisfying assignment for. Since (((v)^00) is doubled, Lemma 2 implies that
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