Re: [Haskell-cafe] algorithm-for-finding-numerical-permutation-given-lexicographic-index

Hi Lone Just coded quickly so still there is lot of chance for improvement. {-# LANGUAGE MultiWayIf #-} import Data.List import Data.Word import Data.Bits builtin_ctz_hask :: Word32 -> Word32 builtin_ctz_hask x | x == 0 = 32 | otherwise = n' - ( ( .&. ) x' 1 ) where ( n' , x' ) = foldl' computeBit ( 1 , x ) [ ( 0xFFFF , 16 ) , ( 0xFF , 8 ) , ( 0xF , 4 ) , ( 0x3 , 2 ) ] computeBit :: ( Word32 , Word32 ) -> ( Word32 , Word32 ) -> ( Word32 , Word32 ) computeBit ( nt , xt ) ( yt , cnt ) = if | (.&.) xt yt == 0 -> ( nt + cnt , xt `shiftR` ( fromIntegral cnt ) ) | otherwise -> ( nt , xt ) bitPermutation :: Word32 -> Word32 bitPermutation v = w where t :: Word32 t = (.|.) v ( v - 1 ) t' :: Word32 t' = complement t w :: Word32 w = ( .|. ) ( t + 1 ) ( ( (.&.) t' ( -t' ) - 1 ) `shiftR` ( fromIntegral ( 1 + builtin_ctz_hask v ) ) ) allPermutation :: Word32 -> [ Word32 ] allPermutation n = iterate bitPermutation n wordtoBin :: Word32 -> [ Word32 ] wordtoBin 0 = [ 0 ] wordtoBin n | mod n 2 == 1 = wordtoBin ( div n 2 ) ++ [ 1 ] | otherwise = wordtoBin ( div n 2 ) ++ [ 0 ] Here is the output. *Main> map ( concat . map show . wordtoBin ) . take 10 . allPermutation $ 19 ["010011","010101","010110","011001","011010","011100","0100011","0100101","0100110","0101001"] Mukesh On Wed, Apr 10, 2013 at 3:47 AM, Lone Wolf wrote:

How could I use Data.Bits to implement the below C code in Haskell?

http://www-graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation

Compute the lexicographically next bit permutation

Suppose we have a pattern of N bits set to 1 in an integer and we want the next permutation of N 1 bits in a lexicographical sense. For example, if N is 3 and the bit pattern is 00010011, the next patterns would be 00010101 , 00010110, 00011001, 00011010, 00011100, 00100011, and so forth. The following is a fast way to compute the next permutation.

unsigned int v; // current permutation of bits unsigned int w; // next permutation of bits

unsigned int t = v | (v - 1); // t gets v's least significant 0 bits set to 1 // Next set to 1 the most significant bit to change, // set to 0 the least significant ones, and add the necessary 1 bits. w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));

The __builtin_ctz(v) GNU C compiler intrinsic for x86 CPUs returns the number of trailing zeros. If you are using Microsoft compilers for x86, the intrinsic is _BitScanForward. These both emit absf instruction, but equivalents may be available for other architectures. If not, then consider using one of the methods for counting the consecutive zero bits mentioned earlier. Here is another version that tends to be slower because of its division operator, but it does not require counting the trailing zeros.

unsigned int t = (v | (v - 1)) + 1; w = t | ((((t & -t) / (v & -v)) >> 1) - 1);

Thanks to Dario Sneidermanis of Argentina, who provided this on November 28, 2009.

On Wed, Apr 3, 2013 at 12:44 PM, Tom Davie wrote:

...

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 wrote:

...

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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