Find the missing permutation: Difference between revisions

{{trans|C}}

‘DABC’, ‘BCAD’, ‘CADB’, ‘CDBA’, ‘CBAD’, ‘ABDC’, ‘ADBC’, ‘BDCA’,

‘DCBA’, ‘BACD’, ‘BADC’, ‘BDAC’, ‘CBDA’, ‘DBCA’, ‘DCAB’]

L.break

{{out}}

{{trans|BBC BASIC}}

Very compact version, thanks to the clever [[#Raku|Raku]] "xor" algorithm.

PERMMISX CSECT

USING PERMMISX,R15 set base register

MISS DC 4XL1'00',C' is missing' buffer

YREGS

{{out}}

<pre>DBAC is missing</pre>

=={{header|8080 Assembly}}==

puts: equ 9 ; CP/M print string

org 100h

db 'DABC','BCAD','CADB','CDBA','CBAD','ABDC','ADBC','BDCA'

db 'DCBA','BACD','BADC','BDAC','CBDA','DBCA','DCAB'

{{out}}

<pre>Missing permutation: DBAC</pre>

=={{header|8086 Assembly}}==

org 100h

mov si,perms ; Start of permutations

perms: db 'ABCD','CABD','ACDB','DACB','BCDA','ACBD','ADCB','CDAB'

db 'DABC','BCAD','CADB','CDBA','CBAD','ABDC','ADBC','BDCA'

{{out}}

<pre>Missing permutation: DBAC</pre>

=={{header|Action!}}==

DEFINE PTR="CARD"

DEFINE COUNT="23"

Print(missing)

{{out}}

[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Find_the_missing_permutation.png Screenshot from Atari 8-bit computer]

=={{header|Ada}}==

procedure Missing_Permutations is

subtype Permutation_Character is Character range 'A' .. 'D';

Ada.Text_IO.Put_Line ("Missing Permutation:");

Put (Missing_Permutation);

=={{header|Aime}}==

paste(record r, index x, text p, integer a)

{

return 0;

{{Out}}

<pre>DBAC</pre>

=={{header|ALGOL 68}}==

Uses the XOR algorithm of the Raku sample.

# the list to find the missing permutation of #

[]STRING list = ( "ABCD", "CABD", "ACDB", "DACB", "BCDA"

print( ( REPR ABS m ) )

OD

{{out}}

<pre>

the letter that occurs least.

{{out}}

perms⍪←↑'DABC' 'BCAD' 'CADB' 'CDBA' 'CBAD' 'ABDC' 'ADBC' 'BDCA'

perms⍪←↑'DCBA' 'BACD' 'BADC' 'BDAC' 'CBDA' 'DBCA' 'DCAB'

missing perms

=={{header|AppleScript}}==

Yosemite OS X onwards (uses NSString for sorting):

--------------- RAREST LETTER IN EACH COLUMN -------------

on |words|(s)

words of s

{{Out}}

<pre>"DBAC"</pre>

=={{header|Arturo}}==

"ABCD" "CABD" "ACDB" "DACB" "BCDA" "ACBD" "ADCB" "CDAB" "DABC"

"BCAD" "CADB" "CDBA" "CBAD" "ABDC" "ADBC" "BDCA" "DCBA" "BACD"

allPerms: map permutate split "ABCD" => join

{{out}}

=={{header|AutoHotkey}}==

CompleteList := Perm( "ABCD" )

}

return substr(L, 1, -1)

=={{header|AWK}}==

This reads the list of permutations as standard input and outputs the missing one.

split($1,a,"");

for (i=1;i<=4;++i) {

}

print s[1]s[2]s[3]s[4]

{{Out}}

=={{header|BBC BASIC}}==

{{works with|BBC BASIC for Windows}}

perms$() = "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", \

\ "CDAB", "DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", \

NEXT

PRINT $$^miss&(0) " is missing"

{{out}}

<pre>

=={{header|Burlesque}}==

ln"ABCD"r@\/\\

(Feed permutations via STDIN. Uses the naive method).

the letters with the lowest frequency:

ln)XXtp)><)F:)<]u[/v\[

=={{header|C}}==

#define N 4

return 0;

{{out}}

<pre>Missing: DBAC</pre>

===By permutating===

{{works with|C sharp|C#|2+}}

using System.Collections.Generic;

}

}

===By xor-ing the values===

{{works with|C sharp|C#|3+}}

using System.Linq;

Console.WriteLine(string.Join("", values.Select(i => (char)i)));

}

=={{header|C++}}==

#include <vector>

#include <set>

std::copy(missing.begin(), missing.end(), std::ostream_iterator<std::string>(std::cout, "\n"));

return 0;

=={{header|Clojure}}==

(use 'clojure.math.combinatorics)

(use 'clojure.set)

(def s1 (apply hash-set (permutations "ABCD")))

(def missing (difference s1 given))

Here's a version based on the hint in the description. ''freqs'' is a sequence of letter frequency maps, one for each column. There should be 6 of each letter in each column, so we look for the one with 5.

"DABC" "BCAD" "CADB" "CDBA" "CBAD" "ABDC" "ADBC" "BDCA"

"DCBA" "BACD" "BADC" "BDAC" "CBDA" "DBCA" "DCAB"])

(defn v->k [fqmap v] (->> fqmap (filter #(-> % second (= v))) ffirst))

=={{header|CoffeeScript}}==

missing_permutation = (arr) ->

# Find the missing permutation in an array of N! - 1 permutations.

console.log missing_permutation(arr)

{{out}}

=={{header|Common Lisp}}==

'("ABCD" "CABD" "ACDB" "DACB" "BCDA" "ACBD" "ADCB" "CDAB" "DABC" "BCAD" "CADB" "CDBA"

"CBAD" "ABDC" "ADBC" "BDCA" "DCBA" "BACD" "BADC" "BDAC" "CBDA" "DBCA" "DCAB"))

(cons (count letter occurs) letter))

letters))))))

{{out}}

<pre>ROSETTA> (missing-perm *permutations*)

=={{header|D}}==

import std.stdio, std.string, std.algorithm, std.range, std.conv;

perms[0][maxCode - code].write;

}

{{out}}

<pre>DBAC

=={{header|EchoLisp}}==

;; use the obvious methos

(lib 'list) ; for (permutations) function

(set-substract (make-set all-perms) (make-set perms))

→ { DBAC }

=={{header|Elixir}}==

def find_miss_perm(head, perms) do

all_permutations(head) -- perms

"CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"]

{{out}}

=={{header|Erlang}}==

The obvious method. It seems fast enough (no waiting time).

-module( find_missing_permutation ).

is_different( [_H] ) -> true;

is_different( [H | T] ) -> not lists:member(H, T) andalso is_different( T ).

{{out}}

<pre>

=={{header|ERRE}}==

PROGRAM MISSING

PRINT("Solution is: ";SOL$)

END PROGRAM

{{out}}

<pre>

=={{header|Factor}}==

Permutations are read in via STDIN.

{{out}}

<pre>

'''Method:''' Read the permutations in as hexadecimal numbers, exclusive ORing them together gives the answer.

(This solution assumes that none of the permutations is defined as a Forth word.)

ABCD CABD xor ACDB xor DACB xor BCDA xor ACBD xor

ADCB xor CDAB xor DABC xor BCAD xor CADB xor CDBA xor

BADC xor BDAC xor CBDA xor DBCA xor DCAB xor

cr .( Missing permutation: ) u.

{{out}}

<pre>Missing permutation: DBAC ok</pre>

=={{header|Fortran}}==

'''Work-around''' to let it run properly with some bugged versions (e.g. 4.3.2) of gfortran: remove the ''parameter'' attribute to the array list.

implicit none

write (*, *)

{{out}}

<pre>DBAC</pre>

=={{header|FreeBASIC}}==

===Simple count===

' compile with: fbc -s console

Print : Print "hit any key to end program"

Sleep

{{out}}

<pre>The missing permutation is : DBAC</pre>

===Add the value's===

' compile with: fbc -s console

Print : Print "hit any key to end program"

Sleep

<pre>output is same as the first version</pre>

===Using Xor===

' compile with: fbc -s console

Print : Print "hit any key to end program"

Sleep

<pre>Output is the same as the first version</pre>

=={{header|GAP}}==

L :=

[ "ABCD", "CABD", "ACDB", "DACB", "BCDA",

# convert back to letters

s := "ABCD";

=={{header|Go}}==

Alternate method suggested by task description:

import (

}

fmt.Println(string(b))

Xor method suggested by Raku contributor:

b := make([]byte, len(given[0]))

for _, p := range given {

}

fmt.Println(string(b))

{{out}} in either case:

<pre>

=={{header|Groovy}}==

Solution:

def missingPerms

missingPerms = {List elts, List perms ->

: missingPerms(elts - e, ePerms).collect { [e] + it }

}.sum()

Test:

def p = ['ABCD', 'CABD', 'ACDB', 'DACB', 'BCDA', 'ACBD', 'ADCB', 'CDAB', 'DABC', 'BCAD', 'CADB', 'CDBA',

'CBAD', 'ABDC', 'ADBC', 'BDCA', 'DCBA', 'BACD', 'BADC', 'BDAC', 'CBDA', 'DBCA', 'DCAB'].collect { it as List }

def mp = missingPerms(e, p)

{{out}}

====Difference between two lists====

{{works with|GHC|7.10.3}}

import Control.Monad (join)

main :: IO ()

{{Out}}

<pre>["DBAC"]</pre>

Another, more statistical, approach is to return the least common letter in each of the four columns. (If all permutations were present, letter frequencies would not vary).

import Data.Ord (comparing)

main :: IO ()

{{Out}}

<pre>"DBAC"</pre>

{{Trans|JavaScript}}

{{Trans|Python}}

import Data.Bits (xor)

main :: IO ()

{{Out}}

<pre>DBAC</pre>

=={{header|Icon}} and {{header|Unicon}}==

procedure main()

write("The difference is : ")

every write(!givens, " ")

The approach above generates a full set of permutations and calculates the difference. Changing the two commented lines to the three below will calculate on the fly and would be more efficient for larger data sets.

if member(givens,x) then delete(givens,x) # remove givens as they are generated

A still more efficient version is:

procedure main()

if not member(givens, p) then write(p)

{{libheader|Icon Programming Library}}

=={{header|J}}==

'''Solution:'''

'''Use:'''

<pre>data=: >;: 'ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA'

Or the above could be a single definition that works the same way:

Or the equivalent explicit (cf. tacit above) definition:

item=. {. y

y -.~ item A.~ i.! #item

Or, the solution could be obtained without defining an independent program:

Here, <code>'ABCD'</code> represents the values being permuted (their order does not matter), and <code>4</code> is how many of them we have.

Yet another alternative expression, which uses parentheses instead of the [http://www.jsoftware.com/help/dictionary/d220v.htm passive operator] (<code>~</code>), would be:

Of course the task suggests that the missing permutation can be found without generating all permutations. And of course that is doable:

However, that's actually a false economy - not only does this approach take more code to implement (at least, in J) but we are already dealing with a data structure of approximately the size of all permutations. So what is being saved by this supposedly "more efficient" approach? Not much... (Still, perhaps this exercise is useful as an illustration of some kind of advertising concept?)

We could use parity, as suggested in the task hints:

We could use arithmetic, as suggested in the task hints:

=={{header|Java}}==

Following needs: [[User:Margusmartsepp/Contributions/Java/Utils.java|Utils.java]]

import com.google.common.base.Joiner;

System.out.println(joiner.join(cs));

}

{{out}}

Alternate version, based on checksumming each position:

{

public static void main(String[] args)

System.out.println("Missing permutation: " + missingPermutation.toString());

}

=={{header|JavaScript}}==

The permute() function taken from http://snippets.dzone.com/posts/show/1032

for(var p = -1, j, k, f, r, l = v.length, q = 1, i = l + 1; --i; q *= i);

for(x = [new Array(l), new Array(l), new Array(l), new Array(l)], j = q, k = l + 1, i = -1;

missing = all.filter(function(elem) {return list.indexOf(elem) == -1});

====Functional====

// [a] -> [[a]]

'ABCD\nCABD\nACDB\nDACB\nBCDA\nACBD\nADCB\nCDAB\nDABC\nBCAD\nCADB\n\

CDBA\nCBAD\nABDC\nADBC\nBDCA\nDCBA\nBACD\nBADC\nBDAC\nCBDA\nDBCA\nDCAB'

{{Out}}

===ES6===

====Statistical====

=====Using a dictionary=====

'use strict';

// --> 'DBAC'

{{Out}}

=====Composing functional primitives=====

{{Trans|Haskell}}

'use strict';

// -> "DBAC"

{{Out}}

<pre>DBAC</pre>

====XOR====

Folding an xor operator over the list of character codes:

'use strict';

return main()

{{Out}}

<pre>DBAC</pre>

If your version of jq has transpose/0, the definition given here

(which is the same as in [[Matrix_Transpose#jq]]) may be omitted.

if (.[0] | length) == 0 then []

else [map(.[0])] + (map(.[1:]) | transpose)

# encode a string (e.g. "ABCD") as an array (e.g. [0,1,2,3]):

'''The task''':

{{Out}}

=={{header|Julia}}==

=== Obvious method ===

Calculate all possible permutations and return the first not included in the array.

function missingperm(arr::Vector)

"CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD", "BADC", "BDAC",

"CBDA", "DBCA", "DCAB"]

{{out}}

=== Alternative method 1 ===

{{trans|Python}}

missperm = string()

for pos in 1:length(arr[1])

return missperm

end

=== Alternative method 2 ===

{{trans|Raku}}

len = length(arr[1])

xorval = zeros(UInt8, len)

@btime missingperm1(arr)

@btime missingperm2(arr)

<pre>

missingperm(arr) = "DBAC"

=={{header|K}}==

g: ("ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB")

p2@&~p2 _lin p

Alternative approach:

table:{b@<b:(x@*:'a),'#:'a:=x}

,/"ABCD"@&:'{5=(table p[;x])[;1]}'!4

Third approach (where p is the given set of permutations):

,/p2@&~(p2:{x@m@&n=(#?:)'m:!n#n:#x}[*p]) _lin p

=={{header|Kotlin}}==

fun <T> permute(input: List<T>): List<List<T>> {

for (perm in missing) println(perm.joinToString(""))

}

{{out}}

=={{header|Lua}}==

Using the popular Penlight extension module - https://luarocks.org/modules/steved/penlight

local permList, pStr = {

"ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB",

pStr = table.concat(perm)

if not tablex.find(permList, pStr) then print(pStr) end

{{out}}

<pre>DBAC</pre>

=={{header|Maple}}==

perm := table():

for letter in "ABCD" do

end do:

end do:

{{Out|Output}}

<pre>"DBAC"</pre>

=={{header|Mathematica}} / {{header|Wolfram Language}}==

"ADCB" , "CDAB", "DABC", "BCAD" , "CADB", "CDBA" , "CBAD" , "ABDC",

"ADBC" , "BDCA", "DCBA" , "BACD", "BADC", "BDAC" , "CBDA", "DBCA", "DCAB"};

=={{header|MATLAB}}==

This solution is designed to work on a column vector of strings. This will not work with a cell array or row vector of strings.

permsList = perms(list(1,:)); %Generate all permutations of the 4 letters

end %for

{{out}}

'CABD';

'ACDB';

ans =

=={{header|Nim}}==

{{trans|Python}}

proc missingPermutation(arr: openArray[string]): string =

CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB""".splitWhiteSpace()

{{out}}

<pre>DBAC</pre>

some utility functions:

let rec insert x = function

| [] -> [[x]]

(* convert a char list to a string *)

let string_of_chars cl =

resolve the task:

"ABCD";"CABD";"ACDB";"DACB";

"BCDA";"ACBD";"ADCB";"CDAB";

let results = List.filter (fun v -> not(List.mem v deficient_perms)) perms

Alternate method : if we had all permutations,

of the number of occurences of each letter.

The following program works with permutations of at least 3 letters:

let n = String.length s in

Array.init n (fun i -> int_of_char s.[i] - 65);;

find_missing deficient_perms;;

=={{header|Octave}}==

'BCDA';'ACBD';'ADCB';'CDAB'; ...

'DABC';'BCAD';'CADB';'CDBA'; ...

bits(there) = 0;

missing = dec2base(find(bits)-1,'ABCD')

=={{header|Oz}}==

Using constraint programming for this problem may be a bit overkill...

GivenPermutations =

["ABCD" "CABD" "ACDB" "DACB" "BCDA" "ACBD" "ADCB" "CDAB" "DABC" "BCAD" "CADB" "CDBA"

{System.showInfo "Missing: "#P}

end

=={{header|PARI/GP}}==

v=apply(u->permtonum(apply(n->n-64,Vec(Vecsmall(u)))),v);

t=numtoperm(4, binomial(4!,2)-sum(i=1,#v,v[i]));

{{out}}

<pre>%1 = "DBAC"</pre>

=={{header|Pascal}}==

like [[c]], summation, and [[Raku]] XORing

{$MODE DELPHI} //for result

writeln(CountOccurences,' is missing');

writeln(CheckXOR,' is missing');

DBAC is missing</pre>

the first missing rotation is the target.

my %hash; @hash{@_} = ();

for my $s (@_) { exists $hash{$_} or return $_

@perms = qw(ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA

CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB);

{{out}}

If the string XOR was of all the permutations, the result would be a string of nulls "\0",

since one is missing, it is the result of XOR of all the rest :)

or if you don't like eval...

Every permutation has a "reverse", just take all reverses and remove the "normals".

my %h = map { $_, '' } reverse =~ /\w+/g;

delete @h{ /\w+/g };

=={{header|Phix}}==

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #008080;">constant</span> <span style="color: #000000;">perms</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #008000;">"ABCD"</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">"CABD"</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">"ACDB"</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">"DACB"</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">"BCDA"</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">"ACBD"</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">"ADCB"</span><span style="color: #0000FF;">,</span> <span style="color: #008000;">"CDAB"</span><span style="color: #0000FF;">,</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%s\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">r</span><span style="color: #0000FF;">})</span>

<span style="color: #000080;font-style:italic;">-- (relies on brute force(!) - but this is the only method that could be made to cope with &gt;1 omission)</span>

{{out}}

<pre>

=={{header|PHP}}==

$finalres = Array();

function permut($arr,$result=array()){

permut($given);

print_r(array_diff($finalres,$givenPerms)); // Array ( [20] => DBAC )

=={{header|Picat}}==

All assume that the variables P1 and/or Perms has been defined:

"ADCB","CDAB","DABC","BCAD","CADB","CDBA",

"CBAD","ABDC","ADBC","BDCA","DCBA","BACD",

Perms = permutations("ABCD"),

% ...

===Very imperative===

Missing = _,

foreach(P in Perms, Missing = _)

end

end,

===Somewhat less imperative===

Missing2 = _,

foreach(P in Perms, Missing2 = _)

end

end,

===Using findall===

println(missing3=difference(Perms,P1)).

===findall approach as a one-liner===

===Using ordsets===

The module <code>ordsets</code> must be imported,

% ...

===List comprehension===

List comprehension with <code>membchk/1</code> for the check)

===Using maps===

Map = new_map(),

foreach(P in P1) Map.put(P,1) end,

==="Merge sort" variants===

"Merge sort" variants, using sorted lists. <code>zip/2</code> requires that the length of the two lists are the same, hence the "dummy".

PermsSorted = Perms.sort(),

P1Sorted = P1.sort(),

% shorter

===Constraint modelling===

The <code>cp</code> module must be imported.

% ...

solve(Missing3),

ABCD2 = "ABCD",

===Matrix approach===

PermsLen = Perms.length,

P1Len = P1.length,

A2[I,J] := 1

end,

===Xor variant===

{{trans|Raku}}

===Count occurrences===

Count the character with the least occurrence (=5) for each positions (1..4). Some variants.

{{trans|K}}

println(missing14=[[O:O=5 in Occ]:Occ in [occurrences([P[I]:P in P1]):I in 1..4]]),

% sort a map according to values

sort2(Map) = [K=V:_=(K=V) in sort([V=(K=V): K=V in Map])]

Running all these snippets:

=={{header|PicoLisp}}==

(mapcar chop

(quote

(rot L) )

(unless (member Lst *PermList) # Check

{{out}}

<pre>DBAC</pre>

{{works with|PowerShell|4.0}}

function permutation ($array) {

function generate($n, $array, $A) {

)

$perm | where{-not $find.Contains($_)}

<b>Output:</b>

<pre>

=={{header|PureBasic}}==

Define.i i, j

Define.s a

Data.s "DABC","BCAD","CADB","CDBA","CBAD","ABDC","ADBC","BDCA"

Data.s "DCBA","BACD","BADC","BDAC","CBDA","DBCA","DCAB"

Based on the [[Permutations#PureBasic|Permutations]] task,

the solution could be:

NewList a.s()

findPermutations(a(), "ABCD", 4)

Print(#CRLF$ + "Press ENTER to exit"): Input()

=={{header|Python}}==

===Python: Calculate difference when compared to all permutations===

{{works with|Python|2.6+}}

given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA

allPerms = [''.join(x) for x in permutations(given[0])]

===Python:Counting lowest frequency character at each position===

i.e. it never needs to generate the full set of expected permutations.

def missing_permutation(arr):

"Find the missing permutation in an array of N! - 1 permutations."

print missing_permutation(given)

===Python:Counting lowest frequency character at each position: functional===

Uses the same method as explained directly above,

but calculated in a more functional manner:

>>> given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA

CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''.split()

>>> ''.join(Counter(x).most_common()[-1][0] for x in zip(*given))

'DBAC'

;Explanation

created by the call to <code>most_common()</code>

is the least common character.

>>> pp(list(zip(*given)), width=120)

[('A', 'C', 'A', 'D', 'B', 'A', 'A', 'C', 'D', 'B', 'C', 'C', 'C', 'A', 'A', 'B', 'D', 'B', 'B', 'B', 'C', 'D', 'D'),

>>> ''.join([Counter(x).most_common()[-1][0] for x in zip(*given)])

'DBAC'

===Python:Folding XOR over the set of strings===

Surfacing the missing bits:

{{Trans|JavaScript}}

from functools import reduce

[0, 0, 0, 0]

)

{{Out}}

<pre>DBAC</pre>

Credit to [[#Raku|Raku]] for the method, and noting that the strings are valid hexadecimal numbers.

ADCB CDAB DABC BCAD CADB CDBA

CBAD ABDC ADBC BDCA DCBA BACD

0 swap witheach ^

number$ echo$

{{out}}

=={{header|R}}==

This uses the "combinat" package, which is a standard R package:

library(combinat)