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complexity-theorynp-completenp

If P != NP, are there more P than non-P problems or vice versa?


If P != NP, are there then more Polynomial problems than SuperPolynomial problems, or vice versa?


Solution

  • From a formal languages perspective, there are only countably many problems in P and uncountably many problems not in P. Every problem in P can be solved by a deterministic, polynomial-time Turing machine, and since the number of TMs is countably infinite, the number of languages in P is countably infinite. On the other hand, the number of total languages is equal to the number of possible subsets of strings, so there are uncountably many languages not in P. This result is, interestingly enough, independent of whether P = NP.

    If you restrict "problems" to "decidable problems" (that is, problems that are solvable by computers with unbounded time and storage space), then we know that there are only countably many total decidable problems. Countably infinitely many of them are in P and, regardless of whether P = NP, there are countably infinitely many of them not in P.

    Hope this helps!