Construct a Permuter

Question

For this challenge you are going to make a function (your function may be a complete program) that takes a list as input and returns a permutation of that list. Your function must obey the following requirements.

• It must be deterministic.

• Composing your function with itself a variable number of times should be capable of getting a list to any of its permutations.

This is a code-golf question so answers will be scored in bytes, with less bytes being better.

Further rules

• You may take any type of list, ([Integer],[String],[[Integer]]) as long as it

  • Can be non empty
  • Can contain distinct objects with at least 16 possible values. (You can't use a Haskell [()] and claim your function is id)
  • Can contain duplicate objects (no sets)

• You may write a program or a function, but must obey standard IO.

Answer 1

CJam (11 bytes)

{_e!_@a#(=}

Online demo showing the full cycle for a four-element list with one duplicate element.

Dissection

{      e# Define a block
  _e!  e#   Find all permutations of the input. Note that if there are duplicate
       e#   elements in the input then only distinct permutations are produced.
       e#   Note also that the permutations are always generated in lexicographic
       e#   order, so the order is independent of the input.
  _@a# e#   Find the index of the input in the list
  (=   e#   Decrement and get the corresponding element of the list
       e#   Incrementing would also have worked, but indexing by -1 feels less
       e#   wrong than indexing by the length, and makes this more portable to
       e#   GolfScript if it ever adds a "permutations" built-in
}

Answer 2

Mathematica + Combinatorica (Built-in Package) 34 Bytes

19 bytes to load the package and 15 for the function.

<<"Combinatorica`";NextPermutation

Usage:

%@{c, b, a}

Without the built-in, 61 Bytes

Extract[s=Permutations[Sort@#],Mod[s~Position~#+1,Length@s]]&

Combinatorica is supposed to be fully incorporated into Mathematica, but I think the NextPermutation function was overlooked.

Answer 3

Python 3, 90 bytes

from itertools import*
def f(l):p=[*permutations(sorted(l))];return p[-~p.index(l)%len(p)]

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Answer 4

C++, 42 bytes

#include <algorithm>
std::next_permutation

This exact operation is a builtin in C++.

Answer 5

JavaScript (ES6), 145 139 137 134 108 bytes

Saved a whopping 25 bytes thanks to @Neil!

Takes input as an array of alphabetical characters. Returns the next permutation as another array.

a=>(t=x=y=-1,a.map((v,i)=>v<a[i+1]?(t=v,x=i):y=i>x&v>t?i:y),a[x]=a[y],a[y]=t,a.concat(a.splice(x+1).sort()))

How?

This is a generation in lexicographic order that processes the 4 following steps at each iteration:

1. Find the largest index X such that a[X] < a[X+1]

   a.map((v, i) => v < a[i + 1] ? (t = v, x = i) : ...)
   

2. Find the largest index Y greater than X such that a[Y] > a[X]

   a.map((v, i) => v < a[i + 1] ? ... : y = i > x & v > t ? i : y)
   

3. Swap the value of a[X] with that of a[Y]

   a[x] = a[y], a[y] = t
   

4. Sort the sequence from a[X + 1] up to and including the final element, in ascending lexicographic order

   a.concat(a.splice(x + 1).sort())
   

Example:

Demo

let f =

a=>(t=x=y=-1,a.map((v,i)=>v<a[i+1]?(t=v,x=i):y=i>x&v>t?i:y),a[x]=a[y],a[y]=t,a.concat(a.splice(x+1).sort()))

for(a = ["A", "B", "C", "D"], n = 0; n < 25; n++) {
  console.log(a.join(','));
  a = f(a);
}

Answer 6

Jelly, 6 bytes

Œ¿’œ?Ṣ

Cycles through the permutations in descending lexicographical order.

Try it online!

How it works

Œ¿’œ?Ṣ  Main link. Argument: A (array)

Œ¿      Compute the permutation index n of A, i.e., the index of A in the
        lexicographically sorted list of permutations of A.
  ’     Decrement the index by 1, yielding n-1.
     Ṣ  Sort A.
   œ?   Getthe (n-1)-th permutation of sorted A.

Answer 7

C, 161 bytes

Actual O(n) algorithm.

#define S(x,y){t=x;x=y;y=t;}
P(a,n,i,j,t)int*a;{for(i=n;--i&&a[i-1]>a[i];);for(j=n;i&&a[--j]<=a[i-1];);if(i)S(a[i-1],a[j])for(j=0;j++<n-i>>1;)S(a[i+j-1],a[n-j])}

Example usage:

int main(int argc, char** argv) {
    int i;
    int a[] = {1, 2, 3, 4};

    for (i = 0; i < 25; ++i) {
        printf("%d %d %d %d\n", a[0], a[1], a[2], a[3]);
        P(a, 4);
    }

    return 0;
}

Answer 8

Python 2, 154 bytes

x=input()
try:exec'%s=max(k for k in range(%s,len(x))if x[%s-1]<x[k]);'*2%tuple('i1kjii');x[i-1],x[j]=x[j],x[i-1];x[i:]=x[:i-1:-1]
except:x.sort()
print x

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Answer 9

Jelly, 10 bytes

ṢŒ!Q©i⁸‘ị®

Try it online!

Sort > all permutation > find input > add 1 > index into "all permutation

Answer 10

05AB1E, 7 bytes

œêD¹k>è

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Answer 11

PHP, 117 bytes

Takes input/output as string list of lower letters

$a=str_split($s=$argn);rsort($a);if(join($a)!=$s)for($n=$s;($c=count_chars)(++$n)!=$c($s););else$n=strrev($s);echo$n;

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