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By Peter Cameron

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Example text

There is a (verifying) Turing machine associated with every member of NP, and the Turing machine can be made to stop in polynomial time on every input. Unlike the P case, however, the clue or guessed solution is part of that input. Thus, if the machine stops with yes, we know the answer to the underlying problem is yes. But if the machine stops without proving yes, we can conclude very little. Perhaps the correct answer is yes, and we have used a wrong clue. Perhaps it is no. All we can be sure of is that the problem encoding plus the given clue did not lead to a yes result.

5 Dealing with NP-Hard Problems 49 constraints of that linear programming relaxation may require more than polynomial effort to derive. 3). Therein lies the mechanism for identifying an "easier" subset of instances of an NP-Hard discrete optimization problem. When the optimal value question can be resolved over a polynomially derivable linear programming relaxation, the yes/no value question is in NP\ the correctness of the given candidate value can be verified by guessing a solution that achieves it together with the derivation of needed linear programming constraints.

3. Clique is NP-Complete Proof. That Clique is in NP is obvious. Any nondeterministic algorithm can guess a subset of vertices and check in polynomial time whether the size of the subset is correct and if every pair of vertices in the subset are connected by an edge. Following, we shall show that (SAT) <* Clique. Let an instance of (SAT) be given by variable set X = {xl9x29 . · . ,xn) and clause set C = { c 1 , c 2 , . . 9ck). }. Clearly, from a given C, the corresponding graph, G can be constructed in time bounded by a polynomial in the length of C since V has cardinality less than the length of C and the cardinality of EisO(\V\2).

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Finite geometry and coding theory by Peter Cameron

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