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The algorithm needs to generate all possible combinations from a given list (empty set excluded).

list          =>         [1, 2, 3]
combinations  =>         [{1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}]

This algorithm would take O(2n) time complexity to generate all combinations. However, I'm not sure if an improved algorithm can bring this time complexity down. If an improved algorithm exists, please do share your knowledge!

In the case that it takes O(2n) which is exponential, I would like some insight regarding which class this algorithm belongs to P, NP, NP-Complete, or NP-Hard. Thanks in advance :)

Ashera
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    If the goal of the code is to produce a list of all possible combinations, then it's not possible to do better than O(2^n). The complexity of an algorithm can never be less than the size of the output. – user3386109 Oct 20 '20 at 04:12

2 Answers2

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P, NP, NP-complete, and NP-hard are all classes of decision problems, none of which contain problems that involve non-binary output (such as this enumeration problem).

Often people refer colloquially to problems in FNP as being in NP. This problem is not in FNP either because the length of the output string for the relation must be bounded by some polynomial function of the input length. It might be FNP-hard, but we're getting into the weeds that even a graduate CS education doesn't cover. Worth asking on the CS Stack Exchange if you care enough.

David Eisenstat
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This problem is in none of them except, arguably, NP-hard.

It is not in P because there is no polynomial time algorithm to do it. You cannot generate an exponential number of things in polynomial time.

It is not in NP because there is no polynomial time algorithm to validate the answer. You cannot process an exponential number of things in polynomial time.

It is not in NP-complete because everything in NP-complete must be in NP and it is not.

The argument for it being in NP-hard goes like this. You can say anything that you want about the members of the empty set. Including that they make monkeys fly out of your nose and can solve any problem in NP in polynomial time. So if we could find a polynomial solution, we can solve any NP problem fast, and therefore it meets the definition of NP-hard. But uselessly so - we know that no polynomial solution exists.

btilly
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  • I don't think the last point is correct. Even if we ignore for a moment that NP-hard concerns decision problems, an O(1) oracle for a problem like described in the question will not be of any help for solving NP problems polynomially. – ADdV Oct 20 '20 at 13:01
  • @ADdV An O(1) oracle would also not help make monkeys fly out of your nose. The point is that you can say anything that you want about the members of the empty set. – btilly Oct 20 '20 at 15:13
  • My point, of course, is that a problem is NP-hard if and only if an oracle machine with an oracle for the problem can solve any problem in NP in polynomial time. Nothing in the definition of NP-hardness concerns monkeys and noses. – ADdV Oct 20 '20 at 19:24