Still, lots of brute forcing, not much elegance.

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# dice problem

### #11

Posted 09 October 2013 - 08:42 PM

### #12

Posted 11 October 2013 - 05:34 PM

It seems that this puzzle is NP-hard and not NP-complete (if I understand this correctly), as there is not an efficient way to prove that the solution is minimal, without an exhaustive search for a better one. (Though a solution can be proved to be valid easily enough, just not minimal.)

Actually it's O(1).

Your comment got me thinking, I couldn't say that it's in NP because there is no

**n**in the problem, We have 130 as a starting point and we want to see if any of the binome(129,8) dice combinations have 120 different sum...

Unless you are referring to the problem with "N-sided Dice", then yeah that's crazy exponential...

**Edited by Anza Power, 11 October 2013 - 05:36 PM.**

### #13

Posted 18 October 2013 - 03:06 PM

It seems that this puzzle is NP-hard and not NP-complete (if I understand this correctly), as there is not an efficient way to prove that the solution is minimal, without an exhaustive search for a better one. (Though a solution can be proved to be valid easily enough, just not minimal.)

Actually it's O(1).

Your comment got me thinking, I couldn't say that it's in NP because there is nonin the problem, We have 130 as a starting point and we want to see if any of the binome(129,8) dice combinations have 120 different sum...

Unless you are referring to the problem with "N-sided Dice", then yeah that's crazy exponential...

If you assume a specific problem, with specific inputs, then yes, people get lazy and say O(1), because the problem itself is a constant. That's cheating.

NP-complete requires a polynomial method to prove that a given solution is valid for a problem in NP. And these are generalized problems, so we never use O() notation for the specific ones (what would be the point?).

To be NP-complete, verifying a solution to the "smallest max face value for distinct sums of (s=sides, d=dice)" would have to scale as some power of s and/or d. There seems to be no way to do this short of duplicating the search, and the search is exponential (perhaps (s**d)**k?)

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