Implement Bogosort

Question

Is solving Sudoku too hard? Even the brute force version? Here's a coding exercise that's a little easier. I hope. :-P

Write the shortest function to implement bogosort. In specific, your function should:

• Take an array (or your language's equivalent) as input
• Check if its elements are in sorted order; if so, return the array
• If not, shuffle the elements, and start again

The shortest entry wins. In the case of a tie, a function that supports a custom comparator (and/or pseudorandom number generator) is favoured. Any remaining ties are resolved by favouring the earlier submission.

---

Clarifications: You can use any element type you want, as long as there's some way to order them, of course. Also, the shuffling has to be uniform; none of this "I'll just quicksort it and call it shuffled" business. :-)

Answer 1

Perl 6: 23 chars

@s.=pick(*)until[<=] @s

Answer 2

APL(Dyalog), 20

{⍵≡⍵[⍋⍵]:⍵⋄∇⍵[?⍨⍴⍵]}

Explanation

⍵ is the (right) argument
⍵≡⍵[⍋⍵]: Checks if ⍵ sorted equals ⍵ itself
:⍵: If yes, then return ⍵
∇⍵[?⍨⍴⍵]: Else, generate an array of 1 to ⍴⍵ (length of ⍵) in random order, reorder ⍵ according to that (⍵[...]), and apply the function to it (∇)

---

Suddenly revisiting this problem and...

APL(Dyalog), 19

{∧/2≤/⍵:⍵⋄∇⍵[?⍨⍴⍵]}

Was just thinking about sorting an array in the check makes it kind of pointless (not saying that Bogosort is meaningful), a more accurate implementation would be ∧/2≤/⍵, and that happens to lower the char count.

Answer 3

APL (22)

{(⍳X←⍴⍵)≡⍋⍵:⍵⋄∇⍵[X?X]}

Usage:

    {(⍳X←⍴⍵)≡⍋⍵:⍵⋄∇⍵[X?X]} 3 2 1
1 2 3

Explanation:

• ⍋⍵: returns the indexes of the items in sorted order, so ⍋30 10 20 gives 2 1 3
• (⍳X←⍴⍵)≡⍋⍵:⍵ Store the length of input list in X. If range [1..X] is equal to the sorted index order, the list is sorted, so return it.
• ⋄∇⍵[X?X]: if this is not the case, recurse with shuffled array.

Answer 4

Ruby - 33 characters

g=->l{l.shuffle!!=l.sort ?redo:l}

Answer 5

Mathematica, 40 37

NestWhile[RandomSample,#,Sort@#!=#&]&

With whitespace:

NestWhile[RandomSample, #, Sort@# != # &] &

Answer 6

J - 34 27

f=:({~?~@#)^:(1-(-:/:~))^:_

eg:

f 5 4 1 3 2
1 2 3 4 5

f 'hello'
ehllo

The {~ ?~@# part shuffles the input:

({~ ?~@#) 1 9 8 4
4 8 9 1
({~ ?~@#) 'abcd'
bdca

Answer 7

Python 61

Sorts in place.

import random
def f(l):
 while l!=sorted(l):random.shuffle(l)

Answer 8

Python 94

from itertools import*
def f(a):return [x for x in permutations(a) if x==tuple(sorted(a))][0]

Other python answers use random.shuffle(). The documentation of the python random module states:

Note that for even rather small len(x), the total number of permutations of x is larger than the period of most random number generators; this implies that most permutations of a long sequence can never be generated.

Answer 9

K, 31 25

{while[~x~x@<x;x:x@(-#x)?#x];x}

{x@(-#x)?#x}/[{~x~x@<x};]

.

k){x@(-#x)?#x}/[{~x~x@<x};] 3 9 5 6 7 9 1
`s#1 3 5 6 7 9 9

.

k){x@(-#x)?#x}/[{~x~x@<x};] "ascsasd"
`s#"aacdsss"

Answer 10

C++11, 150 characters

#include<deque>
#include<algorithm>
void B(std::deque &A){while(!std::is_sorted(A.begin(),A.end())std::random_shuffle(myvector.begin(),myvector.end());}

Just.. made for fun.

Answer 11

Python (69 chars)

from random import*
def f(a):
 while a>sorted(a):shuffle(a)
 return a

Sorts integers in increasing numeric order. Note that recursive solutions, like

from random import*;f=lambda a:a>sorted(a)and(shuffle(a)or f(a))or a

will fail due to stack overflow for even small inputs (say N>5), because Python does not do tail-call optimisation.

Answer 12

D without custom comparator: 59 Characters

R f(R)(R r){while(!isSorted(r))r.randomShuffle();return r;}

More Legibly:

R f(R)(R r)
{
    while(!r.isSorted)
        r.randomShuffle();

    return r;
}

D with custom comparator: 69 Characters

R f(alias p,R)(R r){while(!isSorted!p(r))r.randomShuffle();return r;}

More Legibly:

R f(alias p, R)(R r)
{
    while(!isSorted!p(r))
        r.randomShuffle();

    return r;
}

Answer 13

Scala 73:

def s(l:Seq[Int]):Seq[Int]=if(l==l.sorted)l else s(util.Random.shuffle l)

In Scala, we can check whether the compiler did a tail-call optimization:

@annotation.tailrec
def s(l:Seq[Int]):Seq[Int]=if(l==l.sorted)l else s(util.Random shuffle l)

and yes, it did. However, for a short List of 100 values:

val rList = (1 to 100).map(x=>r.nextInt (500))
s(rList) 

took nearly 4 months to complete. ;)

Answer 14

C# (184 chars)

T[]S<T>(T[]i)where T:IComparable<T>{T l=default(T);while(!i.All(e=>{var r=e.CompareTo(l)>=0;l=e;return r;})){i=i.OrderBy(a=>Guid.NewGuid()).ToArray();l=default(T);}return i.ToArray();}

It's not really nice to do this in C#. You have to support generics to support both value and reference types. There is no array shuffle function or function to check if something is sorted.

Does anybody got any tips to make this better?

Edit Version that only sorts int (134 chars):

int[]S(int[]i){var l=0;while(!i.All(e=>{var r=e>=l;l=e;return r;})){i=i.OrderBy(a=>Guid.NewGuid()).ToArray();l=0;}return i.ToArray();}

Answer 15

GNU/BASH 65

b(){ IFS=$'\n';echo "$*"|sort -C&&echo "$*"||b $(shuf -e "$@");}

Answer 16

Javascript 291 characters

min

function f(e){var t=[].concat(e).sort();t.e=function(e){var n=true;t.forEach(function(t,r){if(t!=e[r])n=false});return n};while(!t.e(e.sort(function(){return Math.floor(Math.random()*2)?1:-1}))){console.log(e)}return e}

un-min

function f(a) {
var b = [].concat(a).sort();
b.e = function (z) {
    var l = true;
    b.forEach(function (v, i) {
        if (v != z[i]) l = false;
    });
    return l
};
while (!b.e(a.sort(function () {
    return Math.floor(Math.random() * 2) ? 1 : -1;
}))) {
    console.log(a);
}
return a;
}

Answer 17

C++14, 158 bytes

#include <algorithm>
#include <random>
[](int*a,int s){std::random_device r;for(std::knuth_b g(r());!std::is_sorted(a,a+s);std::shuffle(a,a+s,g));return a;};

Answer 18

Python - 61 chars

Recursive

from random import*;f=lambda l:l==sorted(l)or shuffle(l)>f(l)

Answer 19

Jelly, 6 bytes, language postdates challenge

ẊŒ¿’$¿

Try it online!

Explanation

ẊŒ¿’$¿
     ¿  While
 Œ¿’$     the input is not in its earliest possible permutation (i.e. sorted)
Ẋ       shuffle it

Œ¿ assigns a number to each permutation of a list; 1 is sorted, 2 has the last two elements exchanged, etc., up to the factorial of the list length (which is the list in reverse order). So for a sorted list, this has the value 1, and we can decrement it using ’ in order to produce a "not sorted" test that's usable as a Boolean in a while loop condition. The $ is to cause the condition to parse as a group.

Answer 20

PowerShell, 85 82 56 55 52 bytes

^{-26 bytes thanks to mazzy's suggestions
-1 byte thanks to AdmBorkBork
-3 bytes thanks to mazzy}

for($l=$args;"$l"-ne($l|sort)){$l=$l|sort{random}}$l

Try it online!

PowerShell does have a relatively cheap array comparison by casting them to strings and comparing that.

Answer 21

Go 1.20+, 95 bytes

import(."math/rand";."sort")
func f[T Interface](I T){for!IsSorted(I){Shuffle(I.Len(),I.Swap)}}

Attempt This Online!

Expects input to implement sort.Interface, which wants:

• Len() int: gets the number of elements.
• Less(i, j int) bool: checks whether the element at index i is less than the element at index j.
• Swap(i, j int): swaps elements at indexes i and j.

Why is this 1.20+? See below.

Go, 158 bytes, seeded ints (pre-1.20)

import(."math/rand";."sort";."time")
func f(I[]int)[]int{Seed(Now().Unix())
for!IntsAreSorted(I){Shuffle(len(I),func(i,j int){I[i],I[j]=I[j],I[i]})}
return I}

Attempt This Online!

Go, 131 bytes, unseeded ints (pre-1.20), truly random (post-1.20)

import(."math/rand";."sort")
func f(I[]int)[]int{for!IntsAreSorted(I){Shuffle(len(I),func(i,j int){I[i],I[j]=I[j],I[i]})}
return I}

Attempt This Online!

This solution is marked as both "unseeded" and "truly random" due to a change in Go v1.20 that changed the global, default random number generator to be seeded "randomly" rather than be default, deterministic rand.Seed(1) in pre-1.20 versions.

Answer 22

JavaScript, 98 92 chars

More precisely ECMAScript 2015 (ES6)

Minified:

let b=r=>{for(;!r.every((o,t)=>o==[...r].sort()[t]);)r.sort(()=>Math.random()-.5);return r};

Invocation: console.log(b([3, 2, 1, 5, 67, 4, 6, 5, 6, 4]))

Not minified:

let bogoSort = (myArray) => {
  while (!myArray.every((v, i) => v == [...myArray].sort()[i])) {      
    myArray.sort(() => Math.random() -0.5)    
  }
  return myArray
}

Explanation:

• myArray.every() is an element-wise compare function. Comparing shuffled with sorted array
• someArray.sort() sorts in place so saving assignments, but also needs cloning for the sorted array
• [...myArray] clones the array (array destructuring)
• () => (Math.random() > .5) ? 1 : -1 is a custom compare function. Coin flip whether to order an element in front or behind

Answer 23

Python, 71 characters

from random import*
def b(l):
 while l!=sorted(l):shuffle(l)
 return(l)

Not the shortest one, but it works.

Ungolfed:

from random import *
def bogosort(list):
    while list != sorted(list):
        shuffle(list)

    return(list)

Usage:

Input: b([1, 7, 3, 9, 2])

Output: [1, 2, 3, 7, 9]

Answer 24

Matlab, 59 bytes

Relatively straight forward approach:

x=input('');while~issorted(x);x=x(randperm(numel(x)));end;x

Answer 25

J, 22 bytes

$:@({~?~@#)`]@.(-:/:~)

This is a recursive, tacit monad using an agenda. Here's how it works:

Let y be our list. First, the verb on the right of the agenda is -:/:~. This a verb graciously provided by Leaky Nun. It matches (-:) whether or not the input is sorted (/:~) using a monadic hook. ((f g) y = y f (g y)) This returns a one or a zero accordingly. The left hand side of the agenda is a gerund of two verbs: on the right is the identity verb ], and on the left is where the recursion takes place. The agenda selects either the identity verb at position 1 if the list is sorted, and the longer verb at position 0 if the list isn't sorted.

$:@({~?~@#) calls $: (the longest verb it is contained in) atop the result of {~?~@# on y. This shuffles the list, as ?~@# takes the permutations of the length of y, being randomly-sorted indices of y. {~, in a monadic hook, returns a list from y whose indices are the right arg. This shuffled list is then called again with the agenda, and repeats until it is sorted.

Answer 26

C++, 166 bytes

Meh.

#import<algorithm>
#import<random>
#define r b.begin(),b.end()
template<class v>
v f(v b){auto a=std::mt19937();while(!std::is_sorted(r))std::shuffle(r,a);return b;}

This should work on all STL containers that have begin() and end().

Ungolfed:

#include <algorithm>
#include <random>
template <class v>
v f(v b) {
    auto a = std::mt19937();
    while (!std::is_sorted(b.begin(),b.end()))
        std::shuffle(b.begin(),b.end(),a);

    return b;
}

Answer 27

APL (Dyalog Extended), 15 bytes

(?⍨∘≢⊇⊢)⍣{⍺≡∧⍺}

Try it online!

Answer 28

Brachylog, 5 bytes

∈&ṣ≤₁

Try it online!

When I first saw ais523's Brachylog answer (as opposed to his Jelly answer, because if I'm not mistaken user62131 was also him), I wondered, what if it used backtracking instead of recursion? So at first, I tried ṣ≤₁. Turns out, since choosing something at random doesn't produce multiple outputs so much as it just produces one output nondeterministically, the shuffle predicate ṣ can't be backtracked to, so running that will simply fail unless you're lucky enough to shuffle it right on the first try. After that, I tried pṣ≤₁, which worked most of the time, but since a finitely long list has finitely many permutations, it still failed at random sometimes. After having abandoned the goal of achieving length reduction, I finally came up with this:

         The input
∈        is an element of
         an unused implicit variable,
 &       and the input
  ṣ      shuffled randomly
   ≤₁    which is increasing
         is the output.

(Demonstration of randomness)

Although it actually can be a bit shorter if we take some liberties with I/O...

Brachylog, 4 bytes

⊆ṣ≤₁

Try it online!

In order for the output to be useful, the input must not contain any duplicate elements, because in addition to sorting the input, this bogosort predicate adds in a random number of duplicate elements and zeroes. (Hypothetically, it could add in anything, but it just kind of doesn't.) Ordinarily I wouldn't bother mentioning something so far from correctly functioning, but I feel it's in the spirit of the challenge.

⊆        An ordered superset of the input
 ṣ       shuffled randomly
  ≤₁     which is increasing
         is the output.

Answer 29

Perl 6, 28 bytes

{({.pick(*)}...~.sort).tail}

Try it online!

Anonymous code block that shuffles the list until it is sorted. Note that it sorts the list at least once, which is allowed. And no, the {.pick(*)} can't be replaced with *.pick(*)

Answer 30

Pyth, 11 bytes

Wn=Q.SQSQ;Q

Pretty happy with this, probably can be golfed a bit more

Explanation

Wn=Q.SQSQ;Q
W    While
  =Q.SQ    Variable Q (input variable) shuffled 
 n  Does not equal
       SQ    Variable Q sorted
             ;  Do nothing (loop ends)
              Q    And output variable Q

Try it online!