permutationIndex :: Int → [Int] → [Int]
permutationIndex [] = []
permutationIndex xs =
let len = length xs
max = fac len
divisor = max / len
i = index / divisor
el = xs !! i
in permutationIndex (index - divisor * i) (filter (!= el) xs)
Of course, this is not very efficient, because you're using lists, and attempting to index into them and measure their lengths. Perhaps a different data structure is in order.
Thanks
Tom Davie
On 3 Apr 2013, at 17:38, Lone Wolf
http://stackoverflow.com/questions/8940470/algorithm-for-finding-numerical-p...
How would you rewrite this into Haskell? The code snippet is in Scala.
/** example: index:=15, list:=(1, 2, 3, 4) */ def permutationIndex (index: Int, list: List [Int]) : List [Int] = if (list.isEmpty) list else { val len = list.size // len = 4 val max = fac (len) // max = 24 val divisor = max / len // divisor = 6 val i = index / divisor // i = 2 val el = list (i) el :: permutationIndex (index - divisor * i, list.filter (_ != el)) }
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