1 Apr
2007
1 Apr
'07
7:41 p.m.
On Mon, Apr 02, 2007 at 09:12:17AM +1000, Duncan Coutts wrote:
On Sun, 2007-04-01 at 16:46 -0400, Paul Hudak wrote:
Here's a solution that I think is a bit more elegant.
-Paul
josephus n k = let loop xs = let d:r = drop (k-1) xs in d : loop (filter (/= d) r) in take n (loop (cycle [1..n]))
Lovely.
.. must.. resist ... urge to fuse ...
Resist -- there's little to gain from fusing an asymptotically slower algorithm. (This is O(n^2) compared with the original O(nk) and a possible O(n log k).)