4 ms·

There's an easy better method: assign variables starting from the lower order digits, greedily check correctness, and skip over all permutations with that prefi

There's an easy better method: assign variables starting from the lower order digits, greedily check correctness, and skip over all permutations with that prefix if it fails. Although I don't know how one would go about proving or disproving if this is asymptotically better or just a (substantial) constant-factor speedup.

dllthomas 11y ago
Oh! I had misread your original comment to be, "There are b! / (b - n)! ways to choose n of b unique numbers, [so] this solution is O(b^n)." You were contrasting! I believe that b!/(b-n)! is asymptotically the same class as b^n (whether we're varying b or n), hence my confusion. I had initially skimmed the article, and assumed he was already handling uniqueness. With the correct definition of StateL, the solution presented is equivalent to what you describe here, but apparently that definition is deferred to a future article. The idea would be that, for StateL, for_each would pick one value and pass that value to the lambda, interpreting the result in a context where the sel only sees the remaining values in the list.
- TheLoneWolfling 11y ago
  > I believe that b!/(b-n)! is asymptotically the same class as b^n Nope. It looks like it at first glance, but it isn't. Look at what happens when n == b/C: b^(b/C) / (b!/(b - b/C)!) == b^(b/C) / (b!/(b*(C-1)/C)!) == e^(log(b^(b/C)) + log((b/C)!) - log((b*(C-1)/C)!)) ~= e^(b/C log(b) + [b*(C-1)/C] log (b*(C-1)/C) - [b*(C-1)/C] - b log b + b) == e^((b+b (-1+C) log((-1+C)/C))/C) == e^(b/C* (1+(C-1)log((C-1)/C))) (I used Stirling's Approximation, namely the form ln(x!) ~= x ln x - x for large x) However, note that (1+(C-1)log((C-1)/C)) / C is greater than zero for all C > 1, as 1+(C-1)log((C-1)/C) > 0 for all C > 1. As such, this diverges for all C > 1.
  - dllthomas 11y 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 11y ago
      Whoops, started with C = 2 and generalized, incorrectly it would seem. I'll edit it. Edit: edited.
      - dllthomas 11y ago
        Thanks :) C = 2 was enough, of course. So O(n!) is between O(k^n) and O(n^n)?
        TheLoneWolfling 11y ago
        Yep. Though we often weaken a big Theta of n! to a big O of n^n, as it's easier to deal with.
        dllthomas 11y ago
        That makes sense, and probably contributed to my confusion.

dllthomas 11y ago

Oh! I had misread your original comment to be, "There are b! / (b - n)! ways to choose n of b unique numbers, [so] this solution is O(b^n)." You were contrasting! I believe that b!/(b-n)! is asymptotically the same class as b^n (whether we're varying b or n), hence my confusion. I had initially skimmed the article, and assumed he was already handling uniqueness. With the correct definition of StateL, the solution presented is equivalent to what you describe here, but apparently that definition is deferred to a future article. The idea would be that, for StateL, for_each would pick one value and pass that value to the lambda, interpreting the result in a context where the sel only sees the remaining values in the list.

TheLoneWolfling 11y ago
> I believe that b!/(b-n)! is asymptotically the same class as b^n Nope. It looks like it at first glance, but it isn't. Look at what happens when n == b/C: b^(b/C) / (b!/(b - b/C)!) == b^(b/C) / (b!/(b*(C-1)/C)!) == e^(log(b^(b/C)) + log((b/C)!) - log((b*(C-1)/C)!)) ~= e^(b/C log(b) + [b*(C-1)/C] log (b*(C-1)/C) - [b*(C-1)/C] - b log b + b) == e^((b+b (-1+C) log((-1+C)/C))/C) == e^(b/C* (1+(C-1)log((C-1)/C))) (I used Stirling's Approximation, namely the form ln(x!) ~= x ln x - x for large x) However, note that (1+(C-1)log((C-1)/C)) / C is greater than zero for all C > 1, as 1+(C-1)log((C-1)/C) > 0 for all C > 1. As such, this diverges for all C > 1.
- dllthomas 11y 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 11y ago
    Whoops, started with C = 2 and generalized, incorrectly it would seem. I'll edit it. Edit: edited.
    - dllthomas 11y ago
      Thanks :) C = 2 was enough, of course. So O(n!) is between O(k^n) and O(n^n)?
      - TheLoneWolfling 11y ago
        Yep. Though we often weaken a big Theta of n! to a big O of n^n, as it's easier to deal with.
        dllthomas 11y ago
        That makes sense, and probably contributed to my confusion.