Generate a random derangement

Question

Challenge description

A "derangement" of a sequence is a permutation where no element appears in its original position. For example ECABD is a derangement of ABCDE, but CBEDA is not:

ABCDE
 | |   <- B and D are in their orignal positions
CBEDA

Given a sequence, generate a random derangement of it.

Notes

• You may take either a string as an input or an array/list of elements (integers, chars, objects...)

• Instead of returning a new object, you can modify an existing one by swapping its elements

• Each derangement should have an equal probability of being generated

• You may assume that there is more than one element in the sequence and none appear more than once

Answer 1

CJam, 14 bytes

q:X{mr_X.=:|}g

Try it online!

Keeps shuffling the input until it's a derangement.

Explanation

q:X   e# Read input and store it in X.
{     e# While the condition at the end of the loop is truthy...
  mr  e#   Shuffle the string.
  _X  e#   Duplicate it and push the input.
  .=  e#   Element-wise equality check.
  :|  e#   Reduce OR over the list, gives something truthy if any character
      e#   remained in its original position.
}g

Answer 2

Jelly, 6 bytes

Ẋ=³S$¿

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Explanation

Ẋ    ¿    Shuffle the given list while this is nonzero for it:
    $       A two-step process:
 =³           Element-wise equality of it and L (the original list)...
   S          Sum the ones in this binary array.

Jonathan Allan saved a byte.

Answer 3

Python, 85 bytes

Modifies the list passed to it (allowed by meta and in the question).

from random import*
def D(l):
 o=l[:]
 while any(x==y for x,y in zip(o,l)):shuffle(l)

Try it online here!

Answer 4

ES6 (Javascript), 71, 69 bytes

Input and output are arrays, should work with any element types (strings, numbers e.t.c.), as long as they can be compared with "==".

Golfed

F=s=>(r=[...s]).sort(_=>Math.random()-.5).some((e,i)=>s[i]==e)?F(s):r

Test

F=s=>(r=[...s]).sort(_=>Math.random()-.5).some((e,i)=>s[i]==e)?F(s):r

F(['A','B','C','D'])
Array [ "D", "C", "A", "B" ]

F(['A','B','C','D'])
Array [ "D", "A", "B", "C" ]

F(['A','B','C','D'])
Array [ "C", "D", "B", "A" ]

F(['A','B','C','D'])
Array [ "D", "C", "B", "A" ]

F(['A','B','C','D'])
Array [ "C", "D", "B", "A" ]

Interactive Snippet

F=s=>(r=[...s]).sort(_=>Math.random()-.5).some((e,i)=>s[i]==e)?F(s):r

function G() {
    console.log(F(T.value.split``).join``); 
}

<input id=T value="ABCDEF"><button id=G onclick="G()">GENERATE</button>

Answer 5

Perl 6, 33 bytes

{first (*Zne$_).all,.pick(*)xx *}

A lambda that takes a list of integers or characters as input, and returns a new list.

If it must support lists of arbitrary values, ne would have to be replaced with !eqv (+2 bytes).

(Try it online.)

Explanation:

• { }: Defines a lambda.
• .pick(*): Generates a random shuffle of the input list.
• .pick(*) xx *: Creates a lazy infinite sequence of such shuffles.
• (* Zne $_).all: A lambda that zips two lists (its argument *, and the outer lambda's argument $_) with the ne (negative string equality) operator, yielding a list of booleans, and then creates an all junction to collapse them to a single boolean state.
• first PREDICATE, SEQUENCE: Takes the first element from our infinite sequence of permutations that fulfills the "derangement" test.

Answer 6

MATL, 7 bytes

This is a translation of my Octave post (and similar to some of the other submissions here). I posted my first MATL post yesterday (CNR crack), so I guess this is not optimal, but it's the best I've got so far.

To be honest, I'm not entirely sure t is needed in there, but it's the only way I can get this to work. It's used so that I can compare the user input (retrieved with G), with the random permutation. I would think I could compare the two without it, but...?

Anyway, here goes:

`Z@tG=a

`          % Loop
 Z@        % Random permutation of input
   t       % Duplicating the stack
    G      % Paste from clipboard G (user input)
     =     % Comparing the random permutation with the input (retrieved from clipboard)
      a    % any(input == random permutation)
           % Implicit end and display

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Answer 7

Brachylog, 19 18 15 13 bytes

@~.:?z:#da;?&

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Explanation

@~.                Output is a shuffle of the input
  .:?z             Zip the output with the input
      :#da         All couples of integers of the zip must be different
          ;      Or
           ?&      Call recursively this predicate with the same input

Answer 8

Perl 6, 45 bytes

{(@^a,{[.pick(*)]}...{none @a Zeqv@$_})[*-1]}

{(@^a,{[.pick(*)]}...{!sum @a Zeqv@$_})[*-1]}

Try it

Input is an Array of anything.

Expanded:

{
  (

    @^a,          # declare parameter, and seed sequence generator

    {             # lambda with implicit parameter ｢$_｣
      [           # store into an array
        .pick(*)  # shuffle ｢$_｣
      ]
    }

    ...           # keep generating the sequence until

    {
      none        # none
      @a          # of the outer blocks input
      Z[eqv]      # is zip equivalent
      @$_         # with the current value being tested
    }

  )[ * - 1 ]      # return the last value
}

Answer 9

Actually, 13 bytes

;;WX╚│♀=ΣWX)X

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Explanation:

;;WX╚│♀=ΣWX)X
;;             make two copies of input
  WX╚│♀=ΣW     while top of stack is truthy:
   X             discard top of stack
    ╚            shuffle array
     │           duplicate entire stack
      ♀=         compare corresponding elements in shuffled and original for equality
        Σ        sum (truthy if any elements are in the same position, else falsey)
          X)X  discard everything but the derangement

Answer 10

Octave, 56 55 bytes

x=input('');while any(x==(y=x(randperm(nnz(x)))));end,y

We have to use input('') since this is not a function. Also, since I can choose to have the input as a string we can use the trick that nnz(x)==numel(x).

Explanation:

x=input('')            % Self-explanatory
while any(x==y)        % Loop until x==y has only 0s (i.e. no elements are equal)
y=x(randperm(nnz(x)))  % Continue to shuffle the indices and assign x(indices) to y
end                    % End loop
y                      % Display y

Thanks to Luis for noticing that the input can be a string, thus I could use nnz instead of numel saving two bytes.

Answer 11

MATL, 13 bytes

This is a joint effort of @LuisMendo and me. In contrast to many other answers here this one is deterministic in the sense that it does not sample random permutations until it gets a derangement, but it generates all derangements and chooses one randomly.

Y@tG-!Af1ZrY)

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Explanation

Y@tG-!Af1ZrY)
Y@             generate all permutatoins
  t            create a duplicate
   G-!A        find the (logical) indices of all valid derangements (where no character of the string is in the same position as the original string)
       f       convert logical to linear indices
        1Zr    choose one of those indices randomly
           Y)  get the derangement (from the ones we generated earlier) at this index

Answer 12

Pyth - 10 9 bytes

This keeps on shuffling the input while any of the characters equal the characters at their index in the input.

.WsqVHQ.S

Try it online here.

.W           Iterate while
 s           Sum, this is works as any() on a boolean list
  qV         Vectorized equality
   H         The lambda variable for the check step
   Q         The input
 .S          Shuffle
  (Z)        Lambda variable, implicit
 (Q)         Start .W with input, implicit

Answer 13

Mathematica, 57 bytes

#/.x_:>RandomChoice@Select[Permutations@x,FreeQ[#-x,0]&]&

Unnamed function taking a list of whatevers as input and outputting a list. After generating all permutations # of the input x, we keep only the ones for which the set #-x of element-wise differences doesn't contain a 0; then we make a (uniformly) random choice from that set.

Answer 14

PHP, 85 bytes

for($a=$b=str_split($argv[1]);array_diff_assoc($a,$b)!=$a;)shuffle($b);echo join($b);

Copies the string argument to two arrays, shuffles one of them until the difference between them (also comparing indexes of the elements) equals the other one. Run with -r.

Answer 15

R, 59 bytes

z=x=1:length(y<-scan(,""));while(any(x==z))z=sample(x);y[z]

Reads a list of elements to STDIN, takes the length of the list and starts sampling ranges from 1 to the length, until it finds one that shares no places with the ordered list. Then prints that list.

Answer 16

Wonder, 32 bytes

f\@[/>#I zip#=[#0a\shuf#0]?f a?a

Usage:

f\@[/>#I zip#=[#0a\shuf#0]?f a?a];f[1 2 3 4 5]

Explanation

More readable:

f\@[
  some #I zip #= [#0; a\ shuf #0]
    ? f a
    ? a
]

Recursive function f. Does an element-wise comparison between f's input list and and a shuffled version of the input list. If the comparison yields any equal values, then f is called on the shuffled list. Otherwise, we simply return the shuffled list.

Answer 17

Ruby, 67 bytes

def f a
while (a.zip(o=a.shuffle).map{|x,y|x-y}.index 0);end
o
end

Answer 18

Octave, 54 53 bytes

@(a)((p=perms(a))(L=!any(p==a,2),:))(randi(sum(L)),:)

Generate all permutations of a and select randomly a row that doesn't have a common element with a .

note: It is accidentally the same as @flawr MATL answer!

Answer 19

Clojure, 94 90 79 bytes

#(let[s(shuffle %)](if(not(some(fn[[x y]](= x y))(map vector % s)))s(recur %)))

-4 bytes by changing the conditional inside the reduction to an and, and inlining done?.

-11 bytes by converting the reduction to some.

WOOT! Beat PHP.

Brute-force method. Shuffles the list while it's invalid. This finishes stupid fast considering it's a brute force method that does nothing to prevent duplicate tries. It found 1000 dearangments of a 1000 element long list in less than a second.

Ungolfed:

(defn dearang [ls]
  (let [s (shuffle ls)
        bad? (some (fn [[x y]] (= x y))
                (map vector ls s))]
    (if (not bad?) s (recur ls))))

Answer 20

Clojure, 56 bytes

#(let[s(shuffle %)](if((set(map = % s))true)(recur %)s))

Note that a string cannot be shuffled, must be passed through seq or vec.

Originally I tried #(first(remove(fn[s]((set(map = % s))true))(iterate shuffle %))) but recur approach is indeed shorter than iterate.

The magic is that (set(map = % s)) returns either a set of false, set of true or set of true and false. This can be used as a function, if it contains true then the answer is true, otherwise falsy nil. = is happy to take two input arguments, no need to wrap it with something.

((set [false]) true)
nil

Maybe there is even shorter way to check if any of the values is true?

Answer 21

APL, 11 bytes.

With the string in the right argument:

⍵[⍋(⍴⍵)?⍴⍵]

Explanation

ρ⍵ gets the length (or shape) of the right argument.

? returns a random array of (⍴⍵) of these numbers.

⍋ returns the order of them in order to ensure no duplicates.

⍵[..] represents the random assortment of the string using this index.