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From: Brown Bannister on 5 Jun 2010 09:48 / / Algorithm that gets NP-complete language [[sub-sum problem | Subset-SUM]]. / / / / This algorithm in polynomial time, if and only if'''P'''=''' NP'''. / / / / "In polynomial time," says it returns "yes" in polynomial time when / / The answer should be "Yes", and runs forever when she "no." / / / / Input: S = final group of integers / / Output: "Yes" if every subgroup of S amounts to 0. / / Production of the production did not win another. / / Note: "The number of program P 'is obtained by program / / Write binary complete in P, then / / Given that a string of bits to be / / Program. Every possible program can be / / Created in this way, although most do nothing / / Because of syntax errors. /> <Br = 1 to infinity ... N For p = 1 ... N Run a program P for a number of steps with input N S If the program outputs a list of distinct integers They are all complete AND S AND complete the amount to 0 /> <br Then Output "Yes" and stop If and only if ,''','' F.'''"= NP''', then it in polynomial time algorithm, to'''Language NP'''-complete." Receipt "This measure gives a" Yes "answers polynomial time, but is allowed to run forever, when the answer is "no." Note that this is not very practical, even if NP = P''''''''''''. If the shortest program that can solve Subset-SUM in polynomial time is''on''bits long, the algorithm above will try 2 <sup> b </ sup> -1 other programs first. Maybe we want to "solve" the problem Subset-SUM, and not just "get" the language Subset-SUM. That is, we want to stop and return algorithm always "yes" or "no" answer. If NP = P'''''''''''', So who does this algorithm in polynomial time, authentication method that uses polynomial time, that no amount of change algorithm above. Another algorithm is obtained by replacing the if statement above with this algorithm: If the program outputs to complete math proof At every stage of proof is legal The conclusion is that S does (or not) has sub total to 0 Then Output "Yes" (or "no") and a stop Posted by M.M.M.
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