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A SAT instance in CNF with $n$ variables has at most $2^n$ resolvents, therefore the resolution method is not in polynomial time.

Considering a 3-SAT instance, we have at most $n^3 + n^2 + n$ many resolvents. Therefore, we can decide 3-SAT in polynomial time by resolution.

I am sure, my second statement is wrong, because otherwise we would have $P = NP$. But I can not see my mistake. Can someone bring clarification here?

John Threepwood
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  • Your second statement is wrong, but I don't see how we can explain your mistake if you don't explain how you arrived at it. – Eric Tressler Jul 04 '13 at 12:42
  • @EricTressler Roughly: All 1-Literal clauses: $n$ many. All 2-Literal clauses $n^2$ many. All 3-Literal clauses $n^3$ many. Calculate all clauses (polynomial many) and check for an empty clause. – John Threepwood Jul 04 '13 at 12:49
  • The problem is that to determine if a CNF instance is satisfiable, you have to assign boolean values to each of the variables $x_i$ so that all of the existing clauses are satisfied simultaneously. You can't deal with them one by one, so you still have $2^n$ possible assignments of the $x_i$. – Eric Tressler Jul 04 '13 at 13:38
  • @EricTressler Thank you for the reply. Resolution is refutation complete: A SAT instance is satisfiable iff the empty clause can not be derived as resolvent. So building all resolvents is sufficient to test for satisfiability. – John Threepwood Jul 04 '13 at 13:50

3 Answers3

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There are only that many clauses but that only gives an upper bound for the time to test a single candidate solution. There are still $2^n$ candidates to test.

Dan Brumleve
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Your statement is not true because you only counted the clauses containing 3 & 2 & 1 literals but some times two clause of size 3 can produce a clause of size 4 and also two clause of size 4 can led to a resolvent clause of size 6. Thus there is not a well-known bound on the number of clauses that may produce.

bummi
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Kavosh Havaledarnejad
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Using some algorithms like DPLL or CDCL SAT solvers leads to much less than $2^n$ candidate testing but still an exponential function.

Also, in local search algorithms best bounds are about $1.333^n$ by Schöning's 1999 paper.

Alex M.
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Kavy
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    This question is very old and already has good answers. It is probably better to spend your effort to write very good answers for unanswered questions. I personally do not see what this contributes to the answers already provided. – TravisJ Sep 19 '15 at 16:44