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From: Brown Bannister on
/ / 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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