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On Nov 30, 2008, at 10:17 AM, Luke Palmer wrote:
On Sun, Nov 30, 2008 at 9:06 AM, Jake Mcarthur <[EMAIL PROTECTED]>
wrote:
Seems a bit easy, I think.
Data.List.permutations . nub
That is not what he meant. Given:
[1,1,2,2]
The results should be:
[1,1,2,2]
[1,2,2,1]
[2,2,1,1]
[1,2,1,2]
[2,1,2,1]
[2,1,1,2]
Heh, after a couple more seconds of thought, reversing the two
composed functions fixes it:
nub . permutations
Of course, neither my previous nonsolution nor this solution are
efficient for long lists, but I think it serves as a decent reference
implementation at least.
- - Jake
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