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I edited that part out to be safe. But I'm curious. I was trying to explain why P problems were in NP and my off the top of my head reasoning was that determini
by dshipper 14y ago
I edited that part out to be safe. But I'm curious. I was trying to explain why P problems were in NP and my off the top of my head reasoning was that deterministic non-polynomial time solutions exist for P problems as well as deterministic polynomial time ones.
If that's not why P is contained in NP, what's the real reason?
- loudandclear 14y agoSince a problem in P is can be solved in polynomial time, then a solution to it can be verified in polynomial time. If it can be verified in polynomial time, then it is in NP. It's this simple. However, it has nothing to do with "non-polynomial" time solutions, as you originally stated. NP and non-polynomial doesn't have a correspondence by their definitions.
- dshipper 14y agoGot it. Thanks for the clarification.