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A particular ranking of a permutation associates an integer with a particular ordering of all the permutations of a set of distinct items. For our purposes the ranking will assign integers $0..(n!-1)$ to an ordering of all the permutations of the integers $0..(n-1)$ .

For example, the permutations of the digits zero to 3 arranged lexicographically have the following rank:

  PERMUTATION      RANK
  (0, 1, 2, 3) ->  0
  (0, 1, 3, 2) ->  1
  (0, 2, 1, 3) ->  2
  (0, 2, 3, 1) ->  3
  (0, 3, 1, 2) ->  4
  (0, 3, 2, 1) ->  5
  (1, 0, 2, 3) ->  6
  (1, 0, 3, 2) ->  7
  (1, 2, 0, 3) ->  8
  (1, 2, 3, 0) ->  9
  (1, 3, 0, 2) -> 10
  (1, 3, 2, 0) -> 11
  (2, 0, 1, 3) -> 12
  (2, 0, 3, 1) -> 13
  (2, 1, 0, 3) -> 14
  (2, 1, 3, 0) -> 15
  (2, 3, 0, 1) -> 16
  (2, 3, 1, 0) -> 17
  (3, 0, 1, 2) -> 18
  (3, 0, 2, 1) -> 19
  (3, 1, 0, 2) -> 20
  (3, 1, 2, 0) -> 21
  (3, 2, 0, 1) -> 22
  (3, 2, 1, 0) -> 23

Algorithms exist that can generate a rank from a permutation for some particular ordering of permutations, and that can generate the same rank from the given individual permutation (i.e. given a rank of 17 produce (2, 3, 1, 0) in the example above).

One use of such algorithms could be in generating a small, random, sample of permutations of $n$ items without duplicates when the total number of permutations is large. Remember that the total number of permutations of $n$ items is given by $n!$ which grows large very quickly: A 32 bit integer can only hold $12!$ , a 64 bit integer only $20!$ . It becomes difficult to take the straight-forward approach of generating all permutations then taking a random sample of them.

A question on the Stack Overflow site asked how to generate one million random and indivudual permutations of 144 items.

Task

Create a function to generate a permutation from a rank.
Create the inverse function that given the permutation generates its rank.
Show that for $n=3$ the two functions are indeed inverses of each other.
Compute and show here 4 random, individual, samples of permutations of 12 objects.

Stretch goal

State how reasonable it would be to use your program to address the limits of the Stack Overflow question.

References

Ranking and Unranking Permutations in Linear Time by Myrvold & Ruskey. (Also available via Google here).
Ranks on the DevData site.
Another answer on Stack Overflow to a different question that explains its algorithm in detail.

C

C: Myrvold and Ruskey

This is algorithm #1 from the M&R paper. <lang c>#include <stdio.h>

include <stdlib.h>

define SWAP(a,b) do{t=(a);(a)=(b);(b)=t;}while(0)

void _mr_unrank1(int rank, int n, int *vec) {

   int t, q, r;
   if (n < 1) return;

   q = rank / n;
   r = rank % n;
   SWAP(vec[r], vec[n-1]);
   _mr_unrank1(q, n-1, vec);

}

int _mr_rank1(int n, int *vec, int *inv) {

   int s, t;
   if (n < 2) return 0;

   s = vec[n-1];
   SWAP(vec[n-1], vec[inv[n-1]]);
   SWAP(inv[s], inv[n-1]);
   return s + n * _mr_rank1(n-1, vec, inv);

}

/* Fill the integer array <vec> (of size <n>) with the

* permutation at rank <rank>.
*/

void get_permutation(int rank, int n, int *vec) {

   int i;
   for (i = 0; i < n; ++i) vec[i] = i;
   _mr_unrank1(rank, n, vec);

}

/* Return the rank of the current permutation of array <vec>

* (of size <n>).
*/

int get_rank(int n, int *vec) {

   int i, r, *v, *inv;

   v = malloc(n * sizeof(int));
   inv = malloc(n * sizeof(int));

   for (i = 0; i < n; ++i) {
       v[i] = vec[i];
       inv[vec[i]] = i;
   }
   r = _mr_rank1(n, v, inv);
   free(inv);
   free(v);
   return r;

}

int main(int argc, char *argv[]) {

   int i, r, tv[4];

   for (r = 0; r < 24; ++r) {
       printf("%3d: ", r);
       get_permutation(r, 4, tv);

       for (i = 0; i < 4; ++i) {
           if (0 == i) printf("[ ");
           else printf(", ");
           printf("%d", tv[i]);
       }
       printf(" ] = %d\n", get_rank(4, tv));
   }

} </lang>

Output:

  0: [ 1, 2, 3, 0 ] = 0
  1: [ 3, 2, 0, 1 ] = 1
  2: [ 1, 3, 0, 2 ] = 2
  3: [ 1, 2, 0, 3 ] = 3
  4: [ 2, 3, 1, 0 ] = 4
  5: [ 2, 0, 3, 1 ] = 5
  6: [ 3, 0, 1, 2 ] = 6
  7: [ 2, 0, 1, 3 ] = 7
  8: [ 1, 3, 2, 0 ] = 8
  9: [ 3, 0, 2, 1 ] = 9
 10: [ 1, 0, 3, 2 ] = 10
 11: [ 1, 0, 2, 3 ] = 11
 12: [ 2, 1, 3, 0 ] = 12
 13: [ 2, 3, 0, 1 ] = 13
 14: [ 3, 1, 0, 2 ] = 14
 15: [ 2, 1, 0, 3 ] = 15
 16: [ 3, 2, 1, 0 ] = 16
 17: [ 0, 2, 3, 1 ] = 17
 18: [ 0, 3, 1, 2 ] = 18
 19: [ 0, 2, 1, 3 ] = 19
 20: [ 3, 1, 2, 0 ] = 20
 21: [ 0, 3, 2, 1 ] = 21
 22: [ 0, 1, 3, 2 ] = 22
 23: [ 0, 1, 2, 3 ] = 23

D

Translation of: C

Currently this doesn't use BigInts. <lang d>import std.stdio, std.algorithm, std.range;

alias TRank = ulong;

TRank factorial(in uint n) pure nothrow {

   TRank result = 1;
   foreach (immutable i; 2 .. n + 1)
       result *= i;
   return result;

}

/// Fill the integer array <vec> with the permutation at rank <rank>. void computePermutation(size_t N)(ref uint[N] vec, TRank rank) pure nothrow if (N > 0 && N < 22) {

   N.iota.copy(vec[]);

   foreach_reverse (immutable n; 1 .. N + 1) {
       immutable size_t r = rank % n;
       rank /= n;
       swap(vec[r], vec[n - 1]);
   }

}

/// Return the rank of the current permutation. TRank computeRank(size_t N)(in ref uint[N] vec) pure nothrow if (N > 0 && N < 22) {

   uint[N] vec2, inv = void;

   TRank mrRank1(in uint n) nothrow {
       if (n < 2)
           return 0;

       immutable s = vec2[n - 1];
       swap(vec2[n - 1], vec2[inv[n - 1]]);
       swap(inv[s], inv[n - 1]);
       return s + n * mrRank1(n - 1);
   }

   vec2[] = vec[];
   foreach (immutable i; 0 .. N)
       inv[vec[i]] = i;
   return mrRank1(N);

}

void main() {

   import std.random;

   uint[4] items1 = void;
   immutable rMax1 = items1.length.factorial;
   for (TRank rank = 0; rank < rMax1; rank++) {
       items1.computePermutation(rank);
       writefln("%3d: %s = %d", rank, items1, items1.computeRank);
   }
   writeln;

   uint[21] items2 = void;
   immutable rMax2 = items2.length.factorial;
   foreach (immutable _; 0 .. 5) {
       immutable rank = uniform(0, rMax2);
       items2.computePermutation(rank);
       writefln("%20d: %s = %d", rank, items2, items2.computeRank);
   }

}</lang>

Output:

  0: [1, 2, 3, 0] = 0
  1: [3, 2, 0, 1] = 1
  2: [1, 3, 0, 2] = 2
  3: [1, 2, 0, 3] = 3
  4: [2, 3, 1, 0] = 4
  5: [2, 0, 3, 1] = 5
  6: [3, 0, 1, 2] = 6
  7: [2, 0, 1, 3] = 7
  8: [1, 3, 2, 0] = 8
  9: [3, 0, 2, 1] = 9
 10: [1, 0, 3, 2] = 10
 11: [1, 0, 2, 3] = 11
 12: [2, 1, 3, 0] = 12
 13: [2, 3, 0, 1] = 13
 14: [3, 1, 0, 2] = 14
 15: [2, 1, 0, 3] = 15
 16: [3, 2, 1, 0] = 16
 17: [0, 2, 3, 1] = 17
 18: [0, 3, 1, 2] = 18
 19: [0, 2, 1, 3] = 19
 20: [3, 1, 2, 0] = 20
 21: [0, 3, 2, 1] = 21
 22: [0, 1, 3, 2] = 22
 23: [0, 1, 2, 3] = 23

 5757702426915204486: [1, 4, 12, 10, 5, 13, 20, 9, 17, 8, 2, 19, 15, 3, 16, 11, 14, 18, 7, 6, 0] = 5757702426915204486
 9054497950101639559: [17, 3, 19, 0, 12, 9, 20, 1, 5, 7, 15, 2, 16, 10, 18, 4, 8, 11, 14, 6, 13] = 9054497950101639559
 6430238494482930297: [15, 7, 13, 3, 11, 0, 4, 2, 20, 5, 10, 16, 14, 6, 19, 18, 12, 1, 17, 8, 9] = 6430238494482930297
 6844249986266452118: [18, 20, 11, 19, 10, 12, 8, 9, 3, 13, 7, 15, 0, 1, 6, 5, 14, 17, 4, 16, 2] = 6844249986266452118
12804085840772788456: [8, 4, 14, 2, 5, 12, 19, 0, 9, 17, 11, 7, 16, 1, 20, 6, 10, 15, 18, 3, 13] = 12804085840772788456

Go

<lang go>package main

import (

   "fmt"
   "math/rand"

)

// returns permutation q of n items, using Myrvold-Ruskey rank. func MRPerm(q, n int) []int {

   p := ident(n)
   var r int
   for n > 0 {
       q, r = q/n, q%n
       n--
       p[n], p[r] = p[r], p[n]
   }
   return p

}

// returns identity permutation of n items. func ident(n int) []int {

   p := make([]int, n)
   for i := range p {
       p[i] = i
   }
   return p

}

// returns Myrvold-Ruskey rank of permutation p func MRRank(p []int) (r int) {

   p = append([]int{}, p...)
   inv := inverse(p)
   for i := len(p) - 1; i > 0; i-- {
       s := p[i]
       p[inv[i]] = s
       inv[s] = inv[i]
   }
   for i := 1; i < len(p); i++ {
       r = r*(i+1) + p[i]
   }
   return

}

// returns inverse of a permutation. func inverse(p []int) []int {

   r := make([]int, len(p))
   for i, x := range p {
       r[x] = i
   }
   return r

}

// returns n! func fact(n int) (f int) {

   for f = n; n > 2; {
       n--
       f *= n
   }
   return

}

func main() {

   n := 3
   fmt.Println("permutations of", n, "items")
   f := fact(n)
   for i := 0; i < f; i++ {
       p := MRPerm(i, n)
       fmt.Println(i, p, MRRank(p))
   }
   n = 12
   fmt.Println("permutations of", n, "items")
   f = fact(n)
   m := map[int]bool{}
   for len(m) < 4 {
       r := rand.Intn(f)
       if m[r] {
           continue
       }
       m[r] = true
       fmt.Println(r, MRPerm(r, n))
   }

}</lang>

Output:

permutations of 3 items
0 [1 2 0] 0
1 [2 0 1] 1
2 [1 0 2] 2
3 [2 1 0] 3
4 [0 2 1] 4
5 [0 1 2] 5
permutations of 12 items
340494881 [4 2 3 9 0 8 10 11 1 6 7 5]
469128647 [7 6 10 5 2 3 1 0 8 4 9 11]
460982459 [4 9 5 7 0 8 6 10 2 1 3 11]
432900481 [0 6 5 10 8 2 4 7 3 9 11 1]

Haskell

Without the random part. <lang haskell>fact n = foldr (*) 1 [1..n]

-- always assume elements are unique rankPerm [] _ = [] rankPerm list n = c:rankPerm (a++b) r where (q,r) = n `divMod` fact (length list - 1) (a,c:b) = splitAt q list

permRank [] = 0 permRank (x:xs) = length(filter (<x) xs) * fact (length xs) + permRank xs

main = mapM_ f [0..23] where f n = print (n, p, permRank p) where p = rankPerm [0..3] n</lang>

Output:

from rank to permutation back to rank:

(0,[0,1,2,3],0)
(1,[0,1,3,2],1)
(2,[0,2,1,3],2)
(3,[0,2,3,1],3)
(4,[0,3,1,2],4)
(5,[0,3,2,1],5)
(6,[1,0,2,3],6)
(7,[1,0,3,2],7)
(8,[1,2,0,3],8)
(9,[1,2,3,0],9)
(10,[1,3,0,2],10)
(11,[1,3,2,0],11)
(12,[2,0,1,3],12)
(13,[2,0,3,1],13)
(14,[2,1,0,3],14)
(15,[2,1,3,0],15)
(16,[2,3,0,1],16)
(17,[2,3,1,0],17)
(18,[3,0,1,2],18)
(19,[3,0,2,1],19)
(20,[3,1,0,2],20)
(21,[3,1,2,0],21)
(22,[3,2,0,1],22)
(23,[3,2,1,0],23)

J

The J primitive A. provides an effective solution to this task. Generating 4 random permutations of 144 items takes about 6 milliseconds so solving the Stack Overflow question ought to take about 100 minutes on that particular test machine.

<lang j> A. 2 0 1 NB. return rank of permutation 4

  4 A. i.3                 NB. return permutation of rank 4

2 0 1

  0 1 2 3 4 5 A. i. 3      NB. generate all 6 permutations for 3 items

0 1 2 0 2 1 1 0 2 1 2 0 2 0 1 2 1 0

  A. 0 1 2 3 4 5 A. i. 3   NB. ranks of each permuation

0 1 2 3 4 5

  ]ranks=: 4 ? !12         NB. 4 random numbers sampled from integers 0 to 12!

315645285 249293994 432230943 23060830

  ranks A. i.12            NB. 4 random samples of 12 items
7 10 11  8  4 0  2 3 9 5  6 1
6  2  8 11 10 0  5 9 7 1  3 4

10 9 1 2 0 3 8 6 7 5 11 4

0  7  4  6 11 5 10 3 9 8  1 2
  (4 ?@$ !144x) A. i.144   NB. 4 random samples of 144 items

117 36 129 85 128 95 27 14 15 119 45 60 21 98 135 106 18 64 132 97 79 84 35 139 101 75 59 13 141 99 86 40 10 140 23 92 125 6 68 41 69 20 56 12 127 65 142 116 71 54 1 5 121 8 78 73 48 30 80 131 111 57 66 100 138 77 37 124 136... 111 65 136 58 92 46 4 83 20 54 21 10 72 110 56 28 13 18 73 133 105 117 63 126 114 43 5 80 45 88 86 108 11 29 0 129 71 141 59 53 113 137 2 102 95 15 35 74 107 61 134 36 32 19 106 100 55 69 76 142 64 49 9 30 47 123 12 97 42...

64 76 139 122  37 127 57 143 32 108 46 17 126   9  51  59  1 74  23  89  42 124 132  19  93 137  70 86  14 112 83  91 63  39 73  18  90 120 53 103 140  87  43  55 131 40 142 102 107 111  80  65  61 34  66  75 88 92 13 138  50 117 97  20  44   7 56  94  41...

139 87 98 118 125 65 35 112 10 43 85 66 58 131 36 30 50 11 136 130 71 100 79 142 40 69 101 84 143 33 95 26 18 94 13 68 8 0 47 70 129 48 107 64 93 16 83 39 29 81 6 105 78 92 104 60 15 55 4 14 7 91 86 12 31 46 20 133 53...</lang>

Java

The basic approach used is to consider the rank a variable-base number, where the symbol set used to encode each digit is the set of all symbols available, minus the symbols already used by digits to the left. Because the symbol used and its numerical value are easily conflated, it may be easier to think of the digits using letters. If n = 5, we may have an ordered symbol set [A, B, C, D, E]. If the leftmost digits of the permutation used B and C, the set is reduced to [A, D, E]. So if E is encountered in the next digit, it has a numerical value of 2 (its index within the symbol set).

Regarding generating a random permutation, the code generates a random rank and converts it to a permutation and back to a rank, to demonstrate that it functions correctly. However, if random permutations were all that was needed, a simpler approach would be to generate random ranks that were already in variable-base form (int[] rather than a large BigInteger). This would reduce the CPU/memory requirements, as it wouldn't need to do arbitrary-precision arithmetic, and could instead do everything with 32-bit ints. The method would simply translate the int[] variable-base rank into an int[] permutation.

Code:

<lang java>import java.math.BigInteger; import java.util.*;

class RankPermutation {

 public static BigInteger getRank(int[] permutation)
 {
   int n = permutation.length;
   BitSet usedDigits = new BitSet();
   BigInteger rank = BigInteger.ZERO;
   for (int i = 0; i < n; i++)
   {
     rank = rank.multiply(BigInteger.valueOf(n - i));
     int digit = 0;
     int v = -1;
     while ((v = usedDigits.nextClearBit(v + 1)) < permutation[i])
       digit++;
     usedDigits.set(v);
     rank = rank.add(BigInteger.valueOf(digit));
   }
   return rank;
 }
 
 public static int[] getPermutation(int n, BigInteger rank)
 {
   int[] digits = new int[n];
   for (int digit = 2; digit <= n; digit++)
   {
     BigInteger divisor = BigInteger.valueOf(digit);
     digits[n - digit] = rank.mod(divisor).intValue();
     if (digit < n)
       rank = rank.divide(divisor);
   }
   BitSet usedDigits = new BitSet();
   int[] permutation = new int[n];
   for (int i = 0; i < n; i++)
   {
     int v = usedDigits.nextClearBit(0);
     for (int j = 0; j < digits[i]; j++)
       v = usedDigits.nextClearBit(v + 1);
     permutation[i] = v;
     usedDigits.set(v);
   }
   return permutation;
 }
 
 public static void main(String[] args)
 {
   for (int i = 0; i < 6; i++)
   {
     int[] permutation = getPermutation(3, BigInteger.valueOf(i));
     System.out.println(String.valueOf(i) + " --> " + Arrays.toString(permutation) + " --> " + getRank(permutation));
   }
   Random rnd = new Random();
   for (int n : new int[] { 12, 144 })
   {
     BigInteger factorial = BigInteger.ONE;
     for (int i = 2; i <= n; i++)
       factorial = factorial.multiply(BigInteger.valueOf(i));
     // Create 5 random samples
     System.out.println("n = " + n);
     for (int i = 0; i < 5; i++)
     {
       BigInteger rank = new BigInteger((factorial.bitLength() + 1) << 1, rnd);
       rank = rank.mod(factorial);
       int[] permutation = getPermutation(n, rank);
       System.out.println("  " + rank + " --> " + Arrays.toString(permutation) + " --> " + getRank(permutation));
     }
   }
 }

}</lang>

Output:

0 --> [0, 1, 2] --> 0
1 --> [0, 2, 1] --> 1
2 --> [1, 0, 2] --> 2
3 --> [1, 2, 0] --> 3
4 --> [2, 0, 1] --> 4
5 --> [2, 1, 0] --> 5
n = 12
  459460043 --> [11, 5, 7, 1, 3, 8, 10, 6, 2, 9, 4, 0] --> 459460043
  242238791 --> [6, 0, 9, 5, 11, 2, 8, 4, 10, 7, 3, 1] --> 242238791
  43886594 --> [1, 2, 0, 11, 6, 8, 7, 9, 3, 5, 4, 10] --> 43886594
  356431614 --> [8, 11, 2, 3, 0, 6, 10, 5, 4, 1, 7, 9] --> 356431614
  344630629 --> [8, 6, 11, 7, 3, 0, 5, 10, 4, 1, 9, 2] --> 344630629
n = 144
  4081840330208521662873206235509318516433197707528534675764597914562081781405735597462986112994867493592438058582576097677821221600974033669483476390594977992091821541791590292404824493917612378402336513544253687968370430594214991000754150383016996226 --> [105, 129, 133, 132, 96, 116, 91, 74, 128, 49, 29, 20, 118, 82, 61, 107, 63, 43, 115, 86, 25, 8, 102, 122, 18, 66, 67, 71, 47, 143, 1, 44, 109, 111, 131, 52, 58, 117, 130, 135, 48, 76, 101, 87, 142, 139, 33, 103, 31, 51, 136, 120, 55, 34, 78, 62, 77, 140, 9, 119, 99, 10, 27, 46, 2, 110, 80, 73, 53, 17, 68, 125, 35, 41, 6, 75, 92, 124, 16, 36, 26, 19, 7, 114, 59, 112, 108, 106, 90, 11, 70, 69, 113, 89, 134, 30, 84, 138, 22, 79, 100, 28, 94, 0, 57, 121, 141, 127, 39, 50, 15, 24, 64, 72, 60, 83, 81, 137, 42, 3, 14, 54, 98, 21, 4, 38, 65, 45, 123, 85, 95, 5, 93, 40, 56, 13, 23, 12, 97, 126, 37, 104, 32, 88] --> 4081840330208521662873206235509318516433197707528534675764597914562081781405735597462986112994867493592438058582576097677821221600974033669483476390594977992091821541791590292404824493917612378402336513544253687968370430594214991000754150383016996226
  2303393071343830907186150492848666018688104049718820526188554488347193604795830950201535244280363069858008773693396780501147779890336031158296935978114061601636771344904185209112272057234471902872428445042702453836810958756677409979103815382798559553 --> [59, 109, 108, 98, 87, 75, 50, 8, 51, 44, 78, 63, 19, 127, 140, 45, 115, 128, 35, 26, 6, 39, 2, 7, 94, 14, 66, 5, 9, 31, 36, 68, 142, 138, 124, 130, 71, 129, 97, 58, 12, 132, 40, 80, 139, 143, 118, 79, 46, 137, 116, 111, 117, 136, 103, 100, 43, 24, 65, 70, 73, 67, 76, 106, 102, 17, 134, 49, 54, 33, 0, 77, 89, 92, 56, 47, 42, 1, 57, 32, 30, 62, 25, 83, 29, 15, 126, 13, 55, 131, 4, 10, 85, 84, 91, 18, 121, 88, 104, 28, 95, 60, 72, 120, 114, 21, 133, 105, 37, 38, 3, 11, 122, 61, 74, 52, 16, 113, 96, 81, 99, 23, 101, 141, 69, 93, 110, 82, 34, 27, 123, 135, 86, 125, 90, 112, 48, 41, 53, 22, 64, 107, 119, 20] --> 2303393071343830907186150492848666018688104049718820526188554488347193604795830950201535244280363069858008773693396780501147779890336031158296935978114061601636771344904185209112272057234471902872428445042702453836810958756677409979103815382798559553
  848047014832080341751646538335377058839867620964996131754382188532880389234227596454312311520517451021055848535846150020108831781620073357956660780586648787723957832482692127095007493517434391511473891562806437927741246265623743953254070095131510626 --> [22, 0, 47, 4, 3, 10, 27, 40, 139, 30, 75, 21, 66, 68, 131, 7, 73, 79, 23, 135, 35, 126, 116, 62, 86, 140, 71, 113, 2, 114, 9, 76, 1, 132, 133, 48, 65, 78, 107, 12, 29, 16, 96, 8, 121, 56, 64, 108, 129, 50, 6, 119, 124, 13, 34, 33, 84, 111, 26, 141, 20, 120, 87, 142, 134, 55, 97, 25, 106, 39, 91, 49, 46, 72, 14, 19, 89, 63, 60, 54, 94, 11, 98, 31, 88, 101, 110, 37, 70, 95, 112, 143, 123, 41, 117, 92, 74, 90, 109, 115, 18, 130, 24, 5, 137, 138, 99, 77, 82, 118, 57, 43, 53, 102, 42, 105, 127, 61, 103, 59, 80, 45, 52, 15, 93, 83, 122, 100, 51, 128, 44, 32, 17, 38, 104, 81, 85, 36, 125, 67, 136, 28, 58, 69] --> 848047014832080341751646538335377058839867620964996131754382188532880389234227596454312311520517451021055848535846150020108831781620073357956660780586648787723957832482692127095007493517434391511473891562806437927741246265623743953254070095131510626
  3867119554188057320189830921237689379982446172130048814726708614069209339326961586538911571940960420075292018388619717817588596221122134526830668232334295443338743434633976186486468178683660450737612737202079329785249513121919049677640145391966805396 --> [100, 47, 42, 69, 16, 43, 66, 107, 73, 79, 12, 80, 41, 50, 9, 126, 95, 36, 26, 51, 123, 45, 52, 3, 93, 29, 83, 17, 82, 5, 81, 85, 131, 122, 113, 6, 75, 28, 59, 64, 138, 20, 74, 114, 27, 65, 105, 116, 62, 142, 141, 35, 115, 4, 49, 78, 2, 97, 130, 89, 110, 57, 90, 127, 72, 119, 44, 13, 99, 112, 118, 103, 77, 125, 92, 133, 104, 60, 76, 70, 23, 53, 55, 38, 108, 84, 96, 54, 128, 140, 25, 11, 24, 7, 124, 136, 8, 111, 46, 63, 31, 1, 102, 67, 94, 0, 132, 37, 91, 10, 101, 22, 34, 68, 14, 71, 21, 87, 30, 40, 129, 120, 18, 58, 61, 86, 56, 143, 32, 33, 15, 88, 137, 98, 106, 109, 135, 134, 48, 139, 121, 39, 19, 117] --> 3867119554188057320189830921237689379982446172130048814726708614069209339326961586538911571940960420075292018388619717817588596221122134526830668232334295443338743434633976186486468178683660450737612737202079329785249513121919049677640145391966805396
  4939977610738364532346788397924709243352263476618646232552639571823537393872611379891069437022354024663565745556550101222893478863124970842071831822047445469519025638779057663111504542116305321498448757808078281473228131641155000612722388784688217279 --> [128, 23, 97, 115, 131, 15, 21, 61, 90, 116, 32, 80, 59, 137, 7, 63, 43, 55, 11, 83, 27, 138, 114, 40, 122, 4, 132, 125, 54, 25, 95, 111, 72, 84, 17, 13, 31, 10, 16, 52, 126, 129, 91, 18, 47, 53, 34, 119, 57, 41, 110, 134, 108, 58, 127, 82, 66, 70, 33, 9, 98, 142, 100, 121, 30, 105, 36, 120, 48, 2, 28, 37, 5, 46, 44, 71, 107, 45, 19, 141, 86, 76, 109, 143, 118, 3, 130, 89, 73, 42, 56, 94, 35, 67, 136, 12, 74, 123, 24, 64, 93, 26, 14, 112, 88, 29, 77, 60, 85, 6, 0, 96, 103, 62, 50, 124, 8, 135, 22, 79, 68, 139, 78, 101, 20, 117, 51, 133, 81, 102, 39, 99, 113, 38, 104, 69, 106, 75, 92, 87, 49, 1, 140, 65] --> 4939977610738364532346788397924709243352263476618646232552639571823537393872611379891069437022354024663565745556550101222893478863124970842071831822047445469519025638779057663111504542116305321498448757808078281473228131641155000612722388784688217279

Julia

Julia has native support for permutations. Depending upon its arguments, nthperm will either return the rank of a permutation or permute a vector according to the provided rank. randperm(n) returns a random permutation of n objects.

Note that, because Julia uses 1-based array indexing, permutations consists of lists of [1, ..., n] rather than the [0, ..., n-1] used for many of the other solutions to this task. Also, Julian partition ranks range from 1 to n! rather than from 0 to n!-1.

Unfortunately, as of the 0.3 release of Julian, this code can not be used to address the StackOverflow question. Although many of Julia's permutation built-in functions support large permutations, nthperm(p) is limited to partitions of no more than 20 objects. p = randperm(114) works fine, but nthperm(p) throws an OverflowError. Arguably, this is a bug, and it may be rectified in future releases. <lang Julia> nobjs = 4 a = collect(1:nobjs) println("All permutations of ", nobjs, " objects:") for i in 1:factorial(nobjs)

   p = nthperm(a, i)
   prank = nthperm(p)
   print(@sprintf("%5d => ", i))
   println(p, " (", prank, ")")

end

nobjs = 12 nsamp = 4 ptaken = Int[] println() println(nsamp, " random permutations of ", nobjs, " objects:") for i in 1:nsamp

   p = randperm(nobjs)
   prank = nthperm(p)
   while prank in ptaken
       p = randperm(nobjs)
       prank = nthperm(p)
   end
   push!(ptaken, prank)
   println("         ", p, " (", prank, ")")

end </lang>

Output:

All permutations of 4 objects:
    1 => [1,2,3,4] (1)
    2 => [1,2,4,3] (2)
    3 => [1,3,2,4] (3)
    4 => [1,3,4,2] (4)
    5 => [1,4,2,3] (5)
    6 => [1,4,3,2] (6)
    7 => [2,1,3,4] (7)
    8 => [2,1,4,3] (8)
    9 => [2,3,1,4] (9)
   10 => [2,3,4,1] (10)
   11 => [2,4,1,3] (11)
   12 => [2,4,3,1] (12)
   13 => [3,1,2,4] (13)
   14 => [3,1,4,2] (14)
   15 => [3,2,1,4] (15)
   16 => [3,2,4,1] (16)
   17 => [3,4,1,2] (17)
   18 => [3,4,2,1] (18)
   19 => [4,1,2,3] (19)
   20 => [4,1,3,2] (20)
   21 => [4,2,1,3] (21)
   22 => [4,2,3,1] (22)
   23 => [4,3,1,2] (23)
   24 => [4,3,2,1] (24)

4 random permutations of 12 objects:
         [5,9,4,1,11,12,6,8,10,2,7,3] (186192332)
         [10,12,4,3,2,9,7,1,8,6,11,5] (396717496)
         [8,1,2,5,10,3,11,9,12,7,6,4] (279524016)
         [1,10,4,12,6,5,8,9,3,2,7,11] (30095719)

Mathematica

Translation of: Haskell

<lang mathematica>fromrank[list_, 0] := list; fromrank[list_, n_] :=

 Prepend[fromrank[DeleteCases[list, #], 
     Mod[n, (Length@list - 1)!]], #] &@
  RankedMin[list, Quotient[n, (Length@list - 1)!] + 1];

rank[{}] = 0; rank[{x_, y___}] := Count[{y}, _?(# < x &)] Length@{y}! + rank[{y}];

Print /@ Table[{n, fromrank[{0, 1, 2, 3}, n],

   rank@fromrank[{0, 1, 2, 3}, n]}, {n, 0, 23}];

Do[Print@fromrank[Range[0, 12 - 1], RandomInteger[12!]], {4}]; Do[Print@fromrank[Range[0, 144 - 1], RandomInteger[144!]], {4}];</lang>

Output:

{0,{0,1,2,3},0}
{1,{0,1,3,2},1}
{2,{0,2,1,3},2}
{3,{0,2,3,1},3}
{4,{0,3,1,2},4}
{5,{0,3,2,1},5}
{6,{1,0,2,3},6}
{7,{1,0,3,2},7}
{8,{1,2,0,3},8}
{9,{1,2,3,0},9}
{10,{1,3,0,2},10}
{11,{1,3,2,0},11}
{12,{2,0,1,3},12}
{13,{2,0,3,1},13}
{14,{2,1,0,3},14}
{15,{2,1,3,0},15}
{16,{2,3,0,1},16}
{17,{2,3,1,0},17}
{18,{3,0,1,2},18}
{19,{3,0,2,1},19}
{20,{3,1,0,2},20}
{21,{3,1,2,0},21}
{22,{3,2,0,1},22}
{23,{3,2,1,0},23}
{8,1,10,2,3,6,4,5,9,0,7,11}
{9,0,11,2,3,6,8,10,1,5,7,4}
{6,4,7,0,3,10,5,11,1,9,2,8}
{7,11,0,3,1,6,10,2,4,5,8,9}
{117,128,29,8,33,81,15,54,132,64,80,6,20,111,22,44,49,28,96,40,55,66,58,126,16,97,26,41,57,32,39,138,71,83,3,76,110,107,103,31,88,141,5,50,67,19,122,127,35,47,93,23,106,84,116,108,69,14,4,139,123,62,133,78,87,131,102,134,74,13,1,104,0,135,72,100,37,113,61,89,34,105,82,25,129,46,43,95,109,65,121,70,73,85,48,120,143,51,98,79,101,56,68,52,24,119,140,17,18,45,11,60,91,7,90,10,27,59,53,21,99,36,115,63,2,12,38,92,118,125,136,112,137,124,30,42,9,77,75,86,114,142,130,94}
{77,39,109,18,142,36,22,6,13,12,139,27,92,134,47,30,88,107,21,85,137,70,9,7,26,86,10,63,143,119,116,130,74,72,81,19,108,136,122,104,1,58,24,73,42,45,135,121,8,76,83,14,99,65,141,128,82,124,79,75,87,91,15,95,96,114,11,52,50,129,34,29,125,89,84,71,20,4,140,49,118,16,64,46,68,110,100,62,123,3,131,53,5,113,37,43,48,44,51,105,32,67,115,111,132,38,102,55,0,93,94,23,25,17,80,133,112,78,56,103,138,61,97,2,54,31,60,120,40,41,126,66,117,127,28,35,98,59,57,69,106,33,101,90}
{118,52,6,67,24,8,105,3,55,29,99,111,14,21,0,48,45,80,131,63,76,16,68,25,125,72,47,98,126,75,28,30,85,143,129,90,32,54,119,117,116,88,115,73,123,11,7,78,141,87,114,91,96,37,106,46,31,133,100,20,17,27,42,36,134,138,120,9,66,74,112,33,77,101,104,12,64,121,40,58,43,136,61,135,132,79,81,49,69,19,82,4,53,1,38,84,108,56,70,86,10,89,107,44,15,35,26,5,110,137,62,140,92,113,103,142,97,51,93,65,128,18,95,50,102,23,2,57,13,34,71,130,83,94,39,60,59,109,122,41,127,22,124,139}
{100,48,125,13,85,95,54,128,111,79,10,103,83,52,143,78,16,114,133,105,43,53,104,37,12,5,17,26,68,61,73,141,34,14,138,140,59,21,77,99,29,117,108,62,39,41,45,91,116,25,131,31,118,94,90,46,49,98,51,137,7,101,44,56,50,87,110,120,142,27,135,18,58,113,69,75,42,107,130,86,80,127,123,15,3,1,122,47,134,106,119,92,136,20,55,74,36,65,67,72,81,0,23,96,32,88,126,97,19,64,4,102,76,66,35,132,139,60,129,6,30,124,57,70,71,82,22,112,24,9,115,63,8,33,89,93,84,38,28,11,109,121,2,40}

PARI/GP

The functions are built into GP already: numtoperm and permtonum <lang parigp>vector(3!,i,permtonum(numtoperm(3,i-1)))==vector(3!,i,i-1) vector(4,i,numtoperm(12,random(12!))) for(i=1,1e6,numtoperm(144,random(144!)))</lang>

Output:

%1 = 1
%2 = [[3, 11, 7, 9, 10, 5, 2, 12, 8, 1, 4, 6], [11, 9, 6, 1, 5, 3, 10, 2, 7, 4, 12, 8], [12, 3, 10, 6, 7, 4, 9, 11, 2, 8, 1, 5], [6, 10, 7, 2, 9, 12, 11, 4, 8, 3, 1, 5]]

Perl 6

It is similar to Haskell, but separate something like inversion vector. It is easy generate random inversion vector without BigInt.

<lang perl6>use v6;

sub rank2inv ( $rank, $n = * ) {

   $rank.polymod( 1 ..^ $n );

}

sub inv2rank ( @inv ) {

   [+] @inv Z* [\*] 1, 1, * + 1 ... *

}

sub inv2perm ( @inv, @items is copy = ^@inv.elems ) {

   my @perm;
   for @inv.reverse -> $i {
       @perm.append: @items.splice: $i, 1;
   }
   @perm;

}

sub perm2inv ( @perm ) { #not in linear time

   (
       { @perm[++$ .. *].grep( * < $^cur ).elems } for @perm;  
   ).reverse;

}

for ^6 {

   my @row.push: $^rank;
   for ( *.&rank2inv(3) , &inv2perm, &perm2inv, &inv2rank )  -> &code {
       @row.push: code( @row[*-1] );
   }
   say @row;

}

my $perms = 4; #100; my $n = 12; #144;

say 'Via BigInt rank'; for ( ( ^([*] 1 .. $n) ).pick($perms) ) {

   say $^rank.&rank2inv($n).&inv2perm;

};

say 'Via inversion vectors'; for ( { my $i=0; inv2perm (^++$i).roll xx $n } ... * ).unique( with => &[eqv] ).[^$perms] {

   .say;

};

say 'Via Perl 6 method pick'; for ( { [(^$n).pick(*)] } ... * ).unique( with => &[eqv] ).head($perms) {

   .say

}; </lang>

Output:

[0 (0 0 0) [0 1 2] (0 0 0) 0]
[1 (0 1 0) [0 2 1] (0 1 0) 1]
[2 (0 0 1) [1 0 2] (0 0 1) 2]
[3 (0 1 1) [1 2 0] (0 1 1) 3]
[4 (0 0 2) [2 0 1] (0 0 2) 4]
[5 (0 1 2) [2 1 0] (0 1 2) 5]
Via BigInt rank
[4 3 1 8 6 2 0 7 9 11 5 10]
[0 8 11 4 9 3 7 5 2 6 10 1]
[5 7 9 11 10 6 4 1 2 3 0 8]
[9 11 8 6 3 5 7 2 4 0 1 10]
Via inversion vectors
[9 0 3 1 8 2 4 5 11 7 10 6]
[7 3 1 10 0 6 4 11 2 9 8 5]
[9 8 5 11 1 10 0 7 4 6 2 3]
[10 8 6 5 4 9 0 2 11 7 1 3]
Via Perl 6 method pick
[11 0 7 10 9 4 1 8 6 5 2 3]
[4 5 8 3 2 1 7 9 11 0 10 6]
[11 7 9 4 0 8 10 1 5 2 6 3]
[11 10 0 3 4 6 7 9 8 5 1 2]

Python

Python: Myrvold & Ruskey

This is based on the work shown in the paper by Myrvold & Ruskey and has algorithms for the two types of ordering of permutations that they show.
Their algorithm is efficient and pythons transparent use of arbitrary precision integers allows the Stack Overflow questions limits to be used. (Chopped at four rather than a million random samples although function get_random_ranks(144, 1e6) doesn't take long).

<lang python>from math import factorial as fact from random import randrange from textwrap import wrap

def identity_perm(n):

   return list(range(n))

def unranker1(n, r, pi):

   while n > 0:
       n1, (rdivn, rmodn) = n-1, divmod(r, n)
       pi[n1], pi[rmodn] = pi[rmodn], pi[n1]
       n = n1
       r = rdivn
   return pi

def init_pi1(n, pi):

   pi1 = [-1] * n
   for i in range(n): 
       pi1[pi[i]] = i
   return pi1

def ranker1(n, pi, pi1):

   if n == 1: 
       return 0
   n1 = n-1
   s = pi[n1]
   pi[n1], pi[pi1[n1]] = pi[pi1[n1]], pi[n1]
   pi1[s], pi1[n1] = pi1[n1], pi1[s]
   return s + n * ranker1(n1, pi, pi1)

def unranker2(n, r, pi):

   while n > 0:
       n1 = n-1
       s, rmodf = divmod(r, fact(n1))
       pi[n1], pi[s] = pi[s], pi[n1]
       n = n1
       r = rmodf
   return pi

def ranker2(n, pi, pi1):

   if n == 1: 
       return 0
   n1 = n-1
   s = pi[n1]
   pi[n1], pi[pi1[n1]] = pi[pi1[n1]], pi[n1]
   pi1[s], pi1[n1] = pi1[n1], pi1[s]
   return s * fact(n1) + ranker2(n1, pi, pi1)

def get_random_ranks(permsize, samplesize):

   perms = fact(permsize)
   ranks = set()
   while len(ranks) < samplesize:
       ranks |= set( randrange(perms) 
                     for r in range(samplesize - len(ranks)) )
   return ranks

def test1(comment, unranker, ranker):

   n, samplesize, n2 = 3, 4, 12
   print(comment)
   perms = []
   for r in range(fact(n)):
       pi = identity_perm(n)
       perm = unranker(n, r, pi)
       perms.append((r, perm))
   for r, pi in perms:
       pi1 = init_pi1(n, pi)
       print('  From rank %2i to %r back to %2i' % (r, pi, ranker(n, pi[:], pi1)))
   print('\n  %i random individual samples of %i items:' % (samplesize, n2))
   for r in get_random_ranks(n2, samplesize):
       pi = identity_perm(n2)
       print('    ' + ' '.join('%2i' % i for i in unranker(n2, r, pi)))
   print()

def test2(comment, unranker):

   samplesize, n2 = 4, 144
   print(comment)
   print('  %i random individual samples of %i items:' % (samplesize, n2))
   for r in get_random_ranks(n2, samplesize):
       pi = identity_perm(n2)
       print('    ' + '\n      '.join(wrap(repr(unranker(n2, r, pi)))))
   print()

if __name__ == '__main__':

   test1('First ordering:', unranker1, ranker1)
   test1('Second ordering:', unranker2, ranker2)
   test2('First ordering, large number of perms:', unranker1)</lang>

Output:

First ordering:
  From rank  0 to [1, 2, 0] back to  0
  From rank  1 to [2, 0, 1] back to  1
  From rank  2 to [1, 0, 2] back to  2
  From rank  3 to [2, 1, 0] back to  3
  From rank  4 to [0, 2, 1] back to  4
  From rank  5 to [0, 1, 2] back to  5

  4 random individual samples of 12 items:
     1  4  5 10  7  3  2  6  9 11  0  8
     0  4  9 11  8 10  7  2  6  5  1  3
    11  8  2  3 10  7  4  6  1  0  5  9
    11  6  7  1  9  5  4  0  2  8  3 10

Second ordering:
  From rank  0 to [1, 2, 0] back to  0
  From rank  1 to [2, 1, 0] back to  1
  From rank  2 to [2, 0, 1] back to  2
  From rank  3 to [0, 2, 1] back to  3
  From rank  4 to [1, 0, 2] back to  4
  From rank  5 to [0, 1, 2] back to  5

  4 random individual samples of 12 items:
     8  9  4 11  7  0  3  6  1 10  2  5
     1  2  0 11  8  6  3  9  4 10  7  5
     9  1  4  8  2 10  3  7  0  6 11  5
     4  7 11  2  5  0  3  8  9  6 10  1

First ordering, large number of perms:
  4 random individual samples of 144 items:
    [92, 32, 141, 93, 97, 45, 70, 134, 60, 5, 99, 4, 13, 80, 68, 100, 77,
      115, 116, 34, 50, 117, 26, 31, 88, 128, 14, 35, 106, 129, 114, 73,
      101, 142, 29, 11, 1, 86, 17, 38, 130, 140, 84, 51, 81, 110, 111, 64,
      24, 3, 16, 6, 139, 104, 103, 8, 75, 62, 43, 113, 137, 48, 22, 53, 30,
      125, 33, 67, 69, 143, 83, 121, 123, 138, 102, 87, 57, 49, 58, 82, 20,
      109, 89, 59, 96, 56, 19, 10, 90, 28, 41, 94, 107, 108, 95, 74, 21, 66,
      40, 25, 46, 78, 44, 112, 124, 36, 135, 42, 132, 79, 37, 63, 15, 2, 61,
      47, 85, 72, 0, 119, 39, 52, 55, 65, 136, 23, 18, 127, 27, 126, 131,
      133, 71, 54, 7, 98, 9, 12, 118, 122, 120, 91, 76, 105]
    [36, 141, 8, 114, 28, 42, 32, 40, 75, 134, 72, 106, 78, 107, 43, 83,
      0, 13, 41, 48, 66, 4, 98, 136, 34, 35, 46, 92, 15, 135, 17, 79, 22,
      118, 95, 137, 81, 121, 60, 74, 110, 33, 80, 14, 62, 125, 3, 56, 115,
      49, 27, 1, 7, 84, 20, 64, 47, 140, 124, 119, 143, 23, 93, 87, 25, 12,
      69, 105, 68, 112, 10, 44, 101, 138, 99, 51, 127, 52, 122, 61, 21, 108,
      117, 39, 2, 85, 53, 130, 120, 18, 16, 109, 126, 103, 113, 5, 82, 19,
      73, 45, 58, 102, 129, 54, 96, 31, 57, 97, 65, 133, 59, 132, 88, 30,
      89, 86, 9, 116, 71, 142, 77, 94, 104, 63, 37, 139, 70, 131, 11, 111,
      123, 128, 24, 76, 90, 29, 6, 38, 100, 26, 55, 91, 50, 67]
    [32, 8, 35, 70, 140, 136, 59, 16, 15, 93, 118, 26, 132, 98, 71, 6,
      107, 19, 65, 11, 42, 97, 78, 85, 28, 96, 119, 82, 44, 2, 125, 58, 117,
      21, 27, 39, 48, 52, 92, 139, 142, 49, 38, 54, 79, 57, 127, 45, 109,
      17, 75, 95, 101, 86, 14, 122, 108, 131, 31, 141, 110, 128, 53, 84, 30,
      23, 66, 41, 13, 37, 36, 130, 113, 46, 50, 47, 115, 7, 100, 51, 137,
      134, 10, 55, 64, 43, 99, 102, 135, 24, 20, 73, 9, 90, 60, 111, 104,
      143, 1, 103, 3, 129, 12, 138, 33, 67, 123, 22, 112, 68, 120, 81, 133,
      91, 61, 25, 124, 56, 40, 121, 4, 18, 114, 126, 80, 106, 83, 5, 94,
      116, 88, 105, 63, 87, 77, 89, 72, 62, 74, 0, 29, 69, 34, 76]
    [28, 68, 101, 27, 40, 79, 73, 2, 75, 97, 135, 23, 17, 95, 58, 26, 80,
      93, 49, 31, 140, 39, 25, 35, 83, 114, 1, 13, 48, 29, 134, 122, 70, 0,
      91, 62, 77, 37, 14, 44, 24, 50, 126, 9, 42, 67, 100, 104, 99, 59, 55,
      21, 123, 138, 33, 116, 127, 115, 82, 61, 117, 130, 43, 86, 22, 12, 56,
      60, 47, 78, 121, 131, 36, 38, 51, 4, 139, 142, 128, 98, 84, 92, 111,
      5, 53, 106, 34, 6, 3, 20, 72, 112, 11, 136, 94, 32, 71, 132, 88, 124,
      85, 110, 108, 137, 30, 89, 74, 120, 8, 102, 41, 81, 129, 107, 63, 118,
      66, 7, 133, 65, 125, 64, 90, 141, 109, 57, 18, 69, 103, 76, 113, 16,
      52, 15, 19, 46, 96, 45, 10, 54, 105, 143, 87, 119]

Python: Iterative

Functions ranker1 and ranker2 can be changed from the recursive form as used in the paper to iterative versions if they are replaced with the following code that yields the same results: <lang python>def ranker1(n, pi, pi1):

   if n == 1: 
       return 0
   n1 = n-1
   s = pi[n1]
   pi[n1], pi[pi1[n1]] = pi[pi1[n1]], pi[n1]
   pi1[s], pi1[n1] = pi1[n1], pi1[s]
   return s + n * ranker1(n1, pi, pi1)

def ranker2(n, pi, pi1):

   result = 0
   for i in range(n-1, 0, -1):
       s = pi[i]
       pi[i], pi[pi1[i]] = pi[pi1[i]], pi[i]
       pi1[s], pi1[i] = pi1[i], pi1[s]
       result += s * fact(i)
   return result</lang>

Racket

Mimicking the Haskell style. <lang racket>

lang racket (require math)

(define-syntax def (make-rename-transformer #'match-define-values))

(define (perm xs n)

 (cond [(empty? xs) '()]
       [else (def (q r) (quotient/remainder n (factorial (sub1 (length xs)))))
             (def (a (cons c b)) (split-at xs q))
             (cons c (perm (append a b) r))]))

(define (rank ys)

 (cond [(empty? ys) 0]
       [else (def ((cons x0 xs)) ys)
             (+ (rank xs)
                (* (for/sum ([x xs] #:when (< x x0)) 1)
                   (factorial (length xs))))]))

</lang>

Output:

(for/list ([n 24])
  (define p (perm '(0 1 2 3) n))
  (list n p (rank p)))

'((0 (0 1 2 3) 0)
  (1 (0 1 3 2) 1)
  (2 (0 2 1 3) 2)
  (3 (0 2 3 1) 3)
  (4 (0 3 1 2) 4)
  (5 (0 3 2 1) 5)
  (6 (1 0 2 3) 6)
  (7 (1 0 3 2) 7)
  (8 (1 2 0 3) 8)
  (9 (1 2 3 0) 9)
  (10 (1 3 0 2) 10)
  (11 (1 3 2 0) 11)
  (12 (2 0 1 3) 12)
  (13 (2 0 3 1) 13)
  (14 (2 1 0 3) 14)
  (15 (2 1 3 0) 15)
  (16 (2 3 0 1) 16)
  (17 (2 3 1 0) 17)
  (18 (3 0 1 2) 18)
  (19 (3 0 2 1) 19)
  (20 (3 1 0 2) 20)
  (21 (3 1 2 0) 21)
  (22 (3 2 0 1) 22)
  (23 (3 2 1 0) 23))

REXX

The permsets subroutine (actually, a function) is a modified version of a REXX subroutine used elsewhere in Rosetta code.

This modified version starts the permute numbers with 0 (zero) instead of 1.

Since this REXX program generates permutations without recursive calls, testing for the limit for a stack overflow wouldn't be possible using this REXX program. <lang rexx>/*REXX program displays permutations of N number of objects (1, 2, 3, ···). */ parse arg N seed . /*obtain optional arguments from the CL*/ if N== | N=="," then N=4 /*Not specified? Then use the default.*/ if datatype(seed,'W') then call random ,,seed /*can make RANDOM numbers repeatable. */ permutes=permSets(N) /*returns N! (number of permutations).*/ w=length(permutes) /*used for aligning the SAY output. */

     do what=0  to permutes-1                   /*traipse through each of the permutes.*/
     z=permSets(N, what)                        /*get which of the  permutation  it is.*/
     say N 'items, permute' right(what,w)   "="   z  '  rank=' permSets(N,,z)
     end   /*what*/

say N=12

     do 4;     ?=random(0, N**4)                /*REXX has a  100k  RANDOM range.      */
     say  N   'items, permute'    right(?,6)     " is "     permSets(N,?)
     end   /*rand*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ permSets: procedure expose @. #; #=0; parse arg x,r,c; c=space(c); xm=x-1

                 do j=1  for x;    @.j=j-1;   end /*j*/
         _=0;    do u=2  for xm;   _=_ @.u;   end /*u*/
         if r==#  then return _;           if c==_  then return #

                 do  while .permSets(x,0); #=#+1; _=@.1
                    do v=2  for xm;   _=_  @.v;   end /*v*/
                 if r==#  then return _;   if c==_  then return #
                 end   /*while···*/
         return #+1

/*──────────────────────────────────────────────────────────────────────────────────────*/ .permSets: procedure expose @.; parse arg p,q; pm=p-1

                 do k=pm  by -1  for pm;   kp=k+1;  if @.k<@.kp  then do; q=k; leave; end
                 end   /*k*/

                 do j=q+1  while j<p;   parse  value  @.j  @.p   with   @.p  @.j;   p=p-1
                 end   /*j*/
           if q==0  then return 0
                 do p=q+1  while @.p<@.q;   end  /*p*/
           parse  value   @.p  @.q   with   @.q  @.p
           return 1</lang>

output when using the default inputs:

4 items, permute  0 = 0 1 2 3   rank= 0
4 items, permute  1 = 0 1 3 2   rank= 1
4 items, permute  2 = 0 2 1 3   rank= 2
4 items, permute  3 = 0 2 3 1   rank= 3
4 items, permute  4 = 0 3 1 2   rank= 4
4 items, permute  5 = 0 3 2 1   rank= 5
4 items, permute  6 = 1 0 2 3   rank= 6
4 items, permute  7 = 1 0 3 2   rank= 7
4 items, permute  8 = 1 2 0 3   rank= 8
4 items, permute  9 = 1 2 3 0   rank= 9
4 items, permute 10 = 1 3 0 2   rank= 10
4 items, permute 11 = 1 3 2 0   rank= 11
4 items, permute 12 = 2 0 1 3   rank= 12
4 items, permute 13 = 2 0 3 1   rank= 13
4 items, permute 14 = 2 1 0 3   rank= 14
4 items, permute 15 = 2 1 3 0   rank= 15
4 items, permute 16 = 2 3 0 1   rank= 16
4 items, permute 17 = 2 3 1 0   rank= 17
4 items, permute 18 = 3 0 1 2   rank= 18
4 items, permute 19 = 3 0 2 1   rank= 19
4 items, permute 20 = 3 1 0 2   rank= 20
4 items, permute 21 = 3 1 2 0   rank= 21
4 items, permute 22 = 3 2 0 1   rank= 22
4 items, permute 23 = 3 2 1 0   rank= 23

12 items, permute  10243  is  0 1 2 3 6 4 7 8 11 5 10 9
12 items, permute   4231  is  0 1 2 3 4 10 11 6 7 5 9 8
12 items, permute    422  is  0 1 2 3 4 5 9 8 10 7 6 11
12 items, permute   1212  is  0 1 2 3 4 6 10 5 9 7 8 11

Ruby

<lang ruby>class Permutation

 include Enumerable
 attr_reader :num_elements, :size
 
 def initialize(num_elements)
   @num_elements = num_elements
   @size = fact(num_elements)
 end
 
 def each
   return self.to_enum unless block_given?
   (0...@size).each{|i| yield unrank(i)}
 end
 
 def unrank(r)  # nonrecursive version of Myrvold Ruskey unrank2 algorithm.
   pi = (0...num_elements).to_a
   (@num_elements-1).downto(1) do |n|
     s, r = r.divmod(fact(n))
     pi[n], pi[s] = pi[s], pi[n]
   end
   pi
 end
 
 def rank(pi)  # nonrecursive version of Myrvold Ruskey rank2 algorithm.
   pi = pi.dup
   pi1 = pi.zip(0...pi.size).sort.map(&:last)
   (pi.size-1).downto(0).inject(0) do |memo,i|
     pi[i], pi[pi1[i]] = pi[pi1[i]], (s = pi[i])
     pi1[s], pi1[i] = pi1[i], pi1[s]
     memo += s * fact(i)
   end
 end
 
 private
 def fact(n)
   n.zero? ? 1 : n.downto(1).inject(:*)
 end

end</lang> Demo: <lang ruby>puts "All permutations of 3 items from and back to rank." perm = Permutation.new(3) (0...perm.size).each{|num| puts "#{num} --> #{prm=perm.unrank(num)} --> #{perm.rank(prm)}"}

puts "\n4 random samples of 12 items from and back to rank." perm = Permutation.new(12) 4.times{ puts "%9d --> %s --> %9d" % [r=rand(perm.size), prm=perm.unrank(r), perm.rank(prm)]}

puts "\n4 random uniq samples of 144 items:" perm, rands = Permutation.new(144), {}

generate 1_000_000 unique random numbers in the range (0...144!) (takes about 2.5 seconds)

rands[rand(perm.size)] = true until rands.size == 1_000_000

random_perms = rands.each_key.lazy{|k| perm.unrank(k)}

random_perms is lazy. Generate permutations one by one:

4.times do

 p r = random_perms.next
 p prm = perm.unrank(r)
 p perm.rank(prm) == r

end</lang>

Output:

All permutations of 3 items from and back to rank.
0 --> [1, 2, 0] --> 0
1 --> [2, 1, 0] --> 1
2 --> [2, 0, 1] --> 2
3 --> [0, 2, 1] --> 3
4 --> [1, 0, 2] --> 4
5 --> [0, 1, 2] --> 5

4 random samples of 12 items from and back to rank.
 99442243 --> [7, 6, 8, 3, 10, 1, 9, 11, 0, 4, 5, 2] -->  99442243
337326172 --> [7, 6, 10, 0, 2, 3, 11, 1, 5, 9, 4, 8] --> 337326172
  4778098 --> [9, 6, 7, 5, 2, 8, 11, 4, 10, 3, 1, 0] -->   4778098
468353447 --> [9, 4, 2, 3, 6, 10, 1, 7, 5, 0, 8, 11] --> 468353447

4 random uniq samples of 144 items:
2764575360456493218947191346199563786344679440083424317164174964729286291440941279012888793089017925404664271586714233946361449878956286173028220761623980334868793458099951877456608056440904453668056576598920423670911869425117449937938198123808117853
[98, 65, 31, 138, 68, 69, 10, 35, 143, 17, 49, 90, 100, 58, 79, 37, 89, 94, 11, 122, 5, 119, 47, 123, 128, 41, 67, 75, 33, 13, 8, 55, 80, 99, 85, 81, 45, 126, 115, 59, 133, 44, 140, 113, 54, 142, 125, 127, 50, 20, 70, 36, 60, 72, 52, 86, 134, 18, 96, 61, 114, 56, 1, 112, 109, 26, 2, 62, 97, 88, 57, 16, 83, 22, 76, 64, 129, 117, 121, 136, 73, 34, 27, 48, 3, 137, 104, 132, 101, 32, 7, 23, 25, 30, 4, 78, 6, 120, 28, 19, 42, 111, 124, 43, 135, 118, 131, 14, 40, 29, 51, 92, 93, 77, 95, 116, 106, 139, 91, 66, 102, 63, 84, 130, 21, 12, 38, 105, 141, 74, 46, 0, 39, 24, 108, 53, 15, 87, 107, 9, 82, 110, 103, 71]
true
581378746619134421610590787114075883604639790081640446988853702372476235480735354011614748882838959943069972039765230962978435387565720009660107060797009355093335736922722607963964082214822692777188121231234716922053393200924081454086807468556348101
[19, 48, 16, 89, 72, 109, 46, 34, 3, 52, 66, 122, 56, 119, 47, 84, 67, 88, 135, 113, 142, 4, 103, 124, 75, 31, 86, 81, 143, 2, 96, 117, 127, 39, 41, 33, 71, 106, 54, 17, 140, 83, 80, 68, 77, 125, 28, 38, 1, 139, 53, 35, 78, 65, 22, 10, 123, 20, 116, 23, 62, 32, 108, 29, 120, 128, 59, 111, 82, 64, 43, 42, 141, 70, 85, 57, 91, 44, 7, 61, 137, 45, 130, 132, 100, 112, 26, 58, 50, 55, 131, 27, 0, 74, 114, 115, 104, 95, 76, 73, 79, 5, 51, 93, 101, 24, 94, 8, 110, 138, 133, 92, 69, 134, 25, 87, 118, 121, 60, 126, 40, 36, 18, 98, 9, 99, 12, 129, 30, 107, 90, 63, 21, 97, 14, 49, 37, 13, 102, 105, 6, 136, 11, 15]
true
3862828872557168655028927772314213364838836589595169123363755483846504171095746597867391615687467779932466719611353020149271086793608853169788146783569713115890942229488492262312988677553070567823077055419333248270326769623765923370471995234511383513
[11, 29, 50, 68, 131, 139, 39, 84, 16, 143, 125, 62, 56, 120, 42, 80, 38, 83, 81, 49, 47, 116, 57, 34, 23, 124, 40, 137, 117, 135, 127, 43, 119, 18, 72, 33, 69, 141, 104, 102, 74, 133, 21, 0, 132, 78, 51, 98, 3, 7, 15, 26, 9, 122, 115, 103, 140, 85, 61, 130, 96, 87, 123, 5, 92, 22, 24, 93, 4, 88, 107, 121, 41, 136, 118, 65, 6, 77, 109, 60, 36, 105, 2, 30, 94, 10, 82, 110, 89, 106, 113, 32, 17, 111, 126, 71, 129, 46, 53, 20, 108, 55, 12, 28, 142, 37, 44, 45, 63, 73, 70, 134, 75, 99, 97, 14, 64, 90, 114, 8, 101, 66, 48, 27, 76, 13, 128, 112, 91, 95, 52, 25, 59, 35, 67, 86, 19, 58, 79, 138, 1, 54, 31, 100]
true
2918302898022044898081980708265879018867198180047026696353345520178885195108521785334274488901945168232326479535713828754624213355248635860483646511664861481065235873861496511507890774391370483865270125811094164531568140755985423978860285012352192610
[66, 140, 109, 101, 20, 62, 13, 60, 0, 116, 107, 1, 87, 99, 110, 38, 64, 25, 130, 2, 17, 84, 123, 36, 31, 94, 134, 51, 55, 103, 81, 30, 106, 15, 61, 21, 59, 142, 71, 92, 54, 79, 48, 132, 111, 7, 67, 65, 82, 49, 115, 10, 105, 19, 32, 22, 141, 9, 68, 53, 3, 63, 74, 56, 42, 23, 136, 129, 57, 126, 77, 39, 137, 50, 121, 117, 93, 5, 46, 45, 89, 96, 41, 122, 72, 47, 37, 11, 58, 91, 70, 128, 43, 127, 76, 95, 138, 100, 90, 8, 119, 120, 143, 14, 6, 33, 83, 28, 44, 27, 135, 88, 35, 78, 86, 18, 69, 73, 85, 112, 29, 80, 4, 104, 52, 108, 12, 114, 133, 118, 97, 26, 34, 24, 113, 40, 98, 124, 139, 125, 131, 16, 102, 75]
true

Really generating one million unique random permutations of 144 elements would take an estimated 11 hours on one core of my machine.

Tcl

Translation of: D

<lang tcl># A functional swap routine proc swap {v idx1 idx2} {

   lset v $idx2 [lindex $v $idx1][lset v $idx1 [lindex $v $idx2];subst ""]

}

Fill the integer array <vec> with the permutation at rank <rank>

proc computePermutation {vecName rank} {

   upvar 1 $vecName vec
   if {![info exist vec] || ![llength $vec]} return
   set N [llength $vec]
   for {set n 0} {$n < $N} {incr n} {lset vec $n $n}
   for {set n $N} {$n>=1} {incr n -1} {

set r [expr {$rank % $n}] set rank [expr {$rank / $n}] set vec [swap $vec $r [expr {$n-1}]]

}

Return the rank of the current permutation.

proc computeRank {vec} {

   if {![llength $vec]} return
   set inv [lrepeat [llength $vec] 0]
   set i -1
   foreach v $vec {lset inv $v [incr i]}
   # First argument is lambda term
   set mrRank1 {{f n vec inv} {

if {$n < 2} {return 0} set s [lindex $vec [set n1 [expr {$n - 1}]]] set vec [swap $vec $n1 [lindex $inv $n1]] set inv [swap $inv $s $n1] return [expr {$s + $n * [apply $f $f $n1 $vec $inv]}]

   }}
   return [apply $mrRank1 $mrRank1 [llength $vec] $vec $inv]

}</lang> Demonstrating: <lang tcl>proc factorial {n} {

   for {set result 1; set i 2} {$i <= $n} {incr i} {

set result [expr {$result * $i}]

   }
   return $result

}

set items1 [lrepeat 4 ""] set rMax1 [factorial [llength $items1]] for {set rank 0} {$rank < $rMax1} {incr rank} {

   computePermutation items1 $rank
   puts [format "%3d: \[%s\] = %d" \

$rank [join $items1 ", "] [computeRank $items1]] } puts "" set items2 [lrepeat 21 ""] set rMax2 [factorial [llength $items2]] foreach _ {1 2 3 4} {

   # Note that we're casting to (potential) bignum, so entier() not int()
   set rank [expr {entier(rand() * $rMax2)}]
   computePermutation items2 $rank
   puts [format "%20lld: \[%s\] = %lld" \

$rank [join $items2 ", "] [computeRank $items2]] }</lang>

Output:

  0: [1, 2, 3, 0] = 0
  1: [3, 2, 0, 1] = 1
  2: [1, 3, 0, 2] = 2
  3: [1, 2, 0, 3] = 3
  4: [2, 3, 1, 0] = 4
  5: [2, 0, 3, 1] = 5
  6: [3, 0, 1, 2] = 6
  7: [2, 0, 1, 3] = 7
  8: [1, 3, 2, 0] = 8
  9: [3, 0, 2, 1] = 9
 10: [1, 0, 3, 2] = 10
 11: [1, 0, 2, 3] = 11
 12: [2, 1, 3, 0] = 12
 13: [2, 3, 0, 1] = 13
 14: [3, 1, 0, 2] = 14
 15: [2, 1, 0, 3] = 15
 16: [3, 2, 1, 0] = 16
 17: [0, 2, 3, 1] = 17
 18: [0, 3, 1, 2] = 18
 19: [0, 2, 1, 3] = 19
 20: [3, 1, 2, 0] = 20
 21: [0, 3, 2, 1] = 21
 22: [0, 1, 3, 2] = 22
 23: [0, 1, 2, 3] = 23

22386545476695773184: [16, 8, 18, 19, 10, 6, 13, 9, 20, 3, 2, 7, 17, 11, 5, 14, 1, 15, 12, 4, 0] = 22386545476695773184
16971674357521238016: [2, 10, 8, 17, 4, 16, 7, 1, 14, 19, 15, 5, 13, 3, 6, 0, 9, 12, 11, 20, 18] = 16971674357521238016
 2200782205637410304: [12, 2, 19, 5, 6, 1, 8, 0, 16, 20, 18, 3, 9, 7, 11, 4, 13, 17, 15, 10, 14] = 2200782205637410304
49795340002029010944: [16, 20, 10, 3, 12, 6, 5, 1, 4, 8, 19, 0, 7, 14, 17, 18, 11, 13, 2, 15, 9] = 49795340002029010944

The above code is not a particularly good use of a random number generator; the Tcl PRNG does not generate enough bits of true randomness per call to rand() for it to be useful in such a simplistic approach. However, it will do for the purposes of demonstration. A more sophisticated RNG is this one: <lang tcl>proc uniform {limit} {

   set bits [expr {ceil(log($limit)/log(2))+10}]
   for {set b [set r 0]} {$b < $bits} {incr b 16} {

incr r [expr {2**$b * int(rand() * 2**16)}]

   }
   return [expr {$r % $limit}]

}</lang>

Addressing the extra credit

The technique above can generate any random permutation of a list, but an equally viable approach (and one especially suitable to the SO question where the number of required permutations is small with respect to the number of possible permutations) would be to use a Knuth shuffle on a list and just use that with a quick check for repetitions, which could be implemented using a simple hashing check. (On the other hand, building a hash table of a million strings of 144 characters would not be too awkward on modern hardware.)

zkl

Translation of: D

<lang zkl>fcn computePermutation(items,rank){ // in place permutation of items

  foreach n in ([items.len()..1,-1]){
     r:=rank%n;
     rank/=n;
     items.swap(r,n-1);
  }
  items

}</lang> <lang zkl> /// Return the rank of a permutation. fcn computeRank(perm){

  N,p2,inv := perm.len(), perm.copy(), List.createLong(N,0);
  foreach n in (N){ inv[perm[n]] = n; }
  fcn(n,p2,inv){
     if(n<2) return(0);
     i,s:= n-1, p2[i];
     p2.swap(i,inv[i]);
     inv.swap(s,i);
     s + n*self.fcn(i,p2,inv);
  }(N,p2,inv);

}</lang> <lang zkl> // do some random permutations of 4 fcn factorial(n){ (1).reduce(n,'*) } items,max:=(4).toList(),factorial(items.len()); // (0,1,2,3), read only foreach rank in ((5).pump(List,(0).random.fp(max)).sort()){

  p:=computePermutation(items.copy(),rank);
  println("%3d: %s = %d".fmt(rank, p, computeRank(p)));

} println();

   // Permutations of 12 to compare to other solutions

items:=(12).toList(); // (0,1,2,3,..11) foreach rank in (T(340494881,469128647,460982459,432900481)){

  p:=computePermutation(items.copy(),rank);
  println("%12d: %s = %d".fmt(rank,p,computeRank(p)));

}</lang>

Output:

 16: L(3,2,1,0) = 16
 16: L(3,2,1,0) = 16
 16: L(3,2,1,0) = 16
 17: L(0,2,3,1) = 17
 23: L(0,1,2,3) = 23

   340494881: L(4,2,3,9,0,8,10,11,1,6,7,5) = 340494881
   469128647: L(7,6,10,5,2,3,1,0,8,4,9,11) = 469128647
   460982459: L(4,9,5,7,0,8,6,10,2,1,3,11) = 460982459
   432900481: L(0,6,5,10,8,2,4,7,3,9,11,1) = 432900481

Addressing the extra credit

computePermutation works fine with BigInts but there is no current way to generate a random number in range to 250! If a 63 bit range was OK, then (but don't hold your breath, this takes a while): <lang zkl>var [const] BN=Import("zklBigNum"); // libGMP one44Bang:=BN(144).factorial(); // 250 digits // Needed: random number [0,one44Bang) do(0d1_000_000){

  p:=computePermutation((144).toList().copy(),(0).random((0).MAX));

}</lang> and the TCL idea of shuffles (List.shuffle()) makes more sense.