Lone Wolf | 3 Apr 18:38 2013
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algorithm-for-finding-numerical-permutation-given-lexicographic-index

http://stackoverflow.com/questions/8940470/algorithm-for-finding-numerical-permutation-given-lexicographic-index

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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Tom Davie | 3 Apr 18:44 2013
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Re: algorithm-for-finding-numerical-permutation-given-lexicographic-index

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 <amslonewolf <at> gmail.com> wrote:

http://stackoverflow.com/questions/8940470/algorithm-for-finding-numerical-permutation-given-lexicographic-index

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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Lone Wolf | 10 Apr 00:17 2013
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Re: algorithm-for-finding-numerical-permutation-given-lexicographic-index

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 <tom.davie <at> gmail.com> 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 <amslonewolf <at> gmail.com> wrote:

http://stackoverflow.com/questions/8940470/algorithm-for-finding-numerical-permutation-given-lexicographic-index

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)) }


_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe <at> haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe


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mukesh tiwari | 10 Apr 07:54 2013
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Re: 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 <amslonewolf <at> gmail.com> 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 <tom.davie <at> gmail.com> 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 <amslonewolf <at> gmail.com> wrote:

http://stackoverflow.com/questions/8940470/algorithm-for-finding-numerical-permutation-given-lexicographic-index

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)) }


_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe <at> haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe



_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe <at> haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe


_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe <at> haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe

Gmane