# algorithm-for-finding-numerical-permutation-given-lexicographic-index

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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)) }`````` _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe
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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.ThanksTom DavieOn 3 Apr 2013, at 17:38, Lone Wolf <[hidden email]> 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[hidden email]http://www.haskell.org/mailman/listinfo/haskell-cafe_______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe
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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 a`bsf` 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 <[hidden email]> 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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## 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 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 a`bsf` 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 <[hidden email]> 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
[hidden email]
http://www.haskell.org/mailman/listinfo/haskell-cafe

_______________________________________________
Haskell-Cafe mailing list
[hidden email]
http://www.haskell.org/mailman/listinfo/haskell-cafe

_______________________________________________
Haskell-Cafe mailing list
[hidden email]
http://www.haskell.org/mailman/listinfo/haskell-cafe