2 ms·
b!/(b-n)! only = b!/(b/C)! when C = 2, but interesting regardless. I'll revisit when I have the chance. In any case, I make the assertion that the algorithm p
by dllthomas 12y ago
b!/(b-n)! only = b!/(b/C)! when C = 2, but interesting regardless. I'll revisit when I have the chance.
In any case, I make the assertion that the algorithm proposed in the article is in fact O(b!/(b-n)!), with the intended definition of StateL.
- TheLoneWolfling 12y agoWhoops, started with C = 2 and generalized, incorrectly it would seem. I'll edit it. Edit: edited.
- dllthomas 12y agoThanks :) C = 2 was enough, of course. So O(n!) is between O(k^n) and O(n^n)?
- TheLoneWolfling 12y agoYep. Though we often weaken a big Theta of n! to a big O of n^n, as it's easier to deal with.
- dllthomas 12y agoThat makes sense, and probably contributed to my confusion.