⚠️ Warning: This is a draft ⚠️
This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.
{{task}}
ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB
Listed above are all of the permutations of the symbols '''A''', '''B''', '''C''', and '''D''', ''except'' for one permutation that's ''not'' listed.
;Task: Find that missing permutation.
;Methods:
-
Obvious method: enumerate all permutations of '''A''', '''B''', '''C''', and '''D''', and then look for the missing permutation.
-
alternate method: Hint: if all permutations were shown above, how many times would '''A''' appear in each position? What is the ''parity'' of this number?
-
another alternate method: Hint: if you add up the letter values of each column, does a missing letter '''A''', '''B''', '''C''', and '''D''' from each column cause the total value for each column to be unique?
;Related task:
- [[Permutations]])
360 Assembly
{{trans|BBC BASIC}} Very compact version, thanks to the clever [[#Perl 6|Perl 6]] "xor" algorithm.
* Find the missing permutation - 19/10/2015
PERMMISX CSECT
USING PERMMISX,R15 set base register
LA R4,0 i=0
LA R6,1 step
LA R7,23 to
LOOPI BXH R4,R6,ELOOPI do i=1 to hbound(perms)
LA R5,0 j=0
LA R8,1 step
LA R9,4 to
LOOPJ BXH R5,R8,ELOOPJ do j=1 to hbound(miss)
LR R1,R4 i
SLA R1,2 *4
LA R3,PERMS-5(R1) @perms(i)
AR R3,R5 j
LA R2,MISS-1(R5) @miss(j)
XC 0(1,R2),0(R3) miss(j)=miss(j) xor substr(perms(i),j,1)
B LOOPJ
ELOOPJ B LOOPI
ELOOPI XPRNT MISS,15 print buffer
XR R15,R15 set return code
BR R14 return to caller
PERMS DC C'ABCD',C'CABD',C'ACDB',C'DACB',C'BCDA',C'ACBD'
DC C'ADCB',C'CDAB',C'DABC',C'BCAD',C'CADB',C'CDBA'
DC C'CBAD',C'ABDC',C'ADBC',C'BDCA',C'DCBA',C'BACD'
DC C'BADC',C'BDAC',C'CBDA',C'DBCA',C'DCAB'
MISS DC 4XL1'00',C' is missing' buffer
YREGS
END PERMMISX
{{out}}
DBAC is missing
Ada
with Ada.Text_IO;
procedure Missing_Permutations is
subtype Permutation_Character is Character range 'A' .. 'D';
Character_Count : constant :=
1 + Permutation_Character'Pos (Permutation_Character'Last)
- Permutation_Character'Pos (Permutation_Character'First);
type Permutation_String is
array (1 .. Character_Count) of Permutation_Character;
procedure Put (Item : Permutation_String) is
begin
for I in Item'Range loop
Ada.Text_IO.Put (Item (I));
end loop;
end Put;
Given_Permutations : array (Positive range <>) of Permutation_String :=
("ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD",
"ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA",
"CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD",
"BADC", "BDAC", "CBDA", "DBCA", "DCAB");
Count : array (Permutation_Character, 1 .. Character_Count) of Natural
:= (others => (others => 0));
Max_Count : Positive := 1;
Missing_Permutation : Permutation_String;
begin
for I in Given_Permutations'Range loop
for Pos in 1 .. Character_Count loop
Count (Given_Permutations (I) (Pos), Pos) :=
Count (Given_Permutations (I) (Pos), Pos) + 1;
if Count (Given_Permutations (I) (Pos), Pos) > Max_Count then
Max_Count := Count (Given_Permutations (I) (Pos), Pos);
end if;
end loop;
end loop;
for Char in Permutation_Character loop
for Pos in 1 .. Character_Count loop
if Count (Char, Pos) < Max_Count then
Missing_Permutation (Pos) := Char;
end if;
end loop;
end loop;
Ada.Text_IO.Put_Line ("Missing Permutation:");
Put (Missing_Permutation);
end Missing_Permutations;
Aime
void
paste(record r, index x, text p, integer a)
{
p = insert(p, -1, a);
x.delete(a);
if (~x) {
x.vcall(paste, -1, r, x, p);
} else {
r[p] = 0;
}
x[a] = 0;
}
integer
main(void)
{
record r;
list l;
index x;
l.bill(0, "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB",
"CDAB", "DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC",
"BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB");
x['A'] = x['B'] = x['C'] = x['D'] = 0;
x.vcall(paste, -1, r, x, "");
l.ucall(r_delete, 1, r);
o_(r.low, "\n");
return 0;
}
{{Out}}
DBAC
AppleScript
{{Trans|JavaScript}} {{Trans|Haskell}} (Statistical versions) Taking the third approach from the task description, and composing with functional primitives:
Yosemite OS X onwards (uses NSString for sorting):
use framework "Foundation" -- ( sort )
-- RAREST LETTER IN EACH COLUMN ----------------------------------------------
on run
intercalate("", ¬
map(composeAll({¬
head, ¬
curry(minimumBy)'s |λ|(comparing(|length|)), ¬
group, ¬
sort}), ¬
transpose(map(chars, ¬
|words|("ABCD CABD ACDB DACB BCDA ACBD " & ¬
"ADCB CDAB DABC BCAD CADB CDBA " & ¬
"CBAD ABDC ADBC BDCA DCBA BACD " & ¬
"BADC BDAC CBDA DBCA DCAB")))))
--> "DBAC"
end run
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- chars :: String -> [String]
on chars(s)
characters of s
end chars
-- Ordering :: (-1 | 0 | 1)
-- compare :: a -> a -> Ordering
on compare(a, b)
if a < b then
-1
else if a > b then
1
else
0
end if
end compare
-- comparing :: (a -> b) -> (a -> a -> Ordering)
on comparing(f)
script
on |λ|(a, b)
tell mReturn(f) to compare(|λ|(a), |λ|(b))
end |λ|
end script
end comparing
-- composeAll :: [(a -> a)] -> (a -> a)
on composeAll(fs)
script
on |λ|(x)
script
on |λ|(f, a)
mReturn(f)'s |λ|(a)
end |λ|
end script
foldr(result, x, fs)
end |λ|
end script
end composeAll
-- curry :: (Script|Handler) -> Script
on curry(f)
script
on |λ|(a)
script
on |λ|(b)
|λ|(a, b) of mReturn(f)
end |λ|
end script
end |λ|
end script
end curry
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- foldr :: (b -> a -> a) -> a -> [b] -> a
on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to |λ|(item i of xs, v, i, xs)
end repeat
return v
end tell
end foldr
-- group :: Eq a => [a] -> [[a]]
on group(xs)
script eq
on |λ|(a, b)
a = b
end |λ|
end script
groupBy(eq, xs)
end group
-- groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
on groupBy(f, xs)
set mf to mReturn(f)
script enGroup
on |λ|(a, x)
if length of (active of a) > 0 then
set h to item 1 of active of a
else
set h to missing value
end if
if h is not missing value and mf's |λ|(h, x) then
{active:(active of a) & x, sofar:sofar of a}
else
{active:{x}, sofar:(sofar of a) & {active of a}}
end if
end |λ|
end script
if length of xs > 0 then
set dct to foldl(enGroup, {active:{item 1 of xs}, sofar:{}}, tail(xs))
if length of (active of dct) > 0 then
sofar of dct & {active of dct}
else
sofar of dct
end if
else
{}
end if
end groupBy
-- head :: [a] -> a
on head(xs)
if length of xs > 0 then
item 1 of xs
else
missing value
end if
end head
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- length :: [a] -> Int
on |length|(xs)
length of xs
end |length|
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- minimumBy :: (a -> a -> Ordering) -> [a] -> a
on minimumBy(f, xs)
if length of xs < 1 then return missing value
tell mReturn(f)
set v to item 1 of xs
repeat with x in xs
if |λ|(x, v) < 0 then set v to x
end repeat
return v
end tell
end minimumBy
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- sort :: [a] -> [a]
on sort(xs)
((current application's NSArray's arrayWithArray:xs)'s ¬
sortedArrayUsingSelector:"compare:") as list
end sort
-- tail :: [a] -> [a]
on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
-- transpose :: [[a]] -> [[a]]
on transpose(xss)
script column
on |λ|(_, iCol)
script row
on |λ|(xs)
item iCol of xs
end |λ|
end script
map(row, xss)
end |λ|
end script
map(column, item 1 of xss)
end transpose
-- words :: String -> [String]
on |words|(s)
words of s
end |words|
{{Out}}
"DBAC"
AutoHotkey
IncompleteList := "ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB"
CompleteList := Perm( "ABCD" )
Missing := ""
Loop, Parse, CompleteList, `n, `r
If !InStr( IncompleteList , A_LoopField )
Missing .= "`n" A_LoopField
MsgBox Missing Permutation(s):%Missing%
;-------------------------------------------------
; Shortened version of [VxE]'s permutation function
; http://www.autohotkey.com/forum/post-322251.html#322251
Perm( s , dL="" , t="" , p="") {
StringSplit, m, s, % d := SubStr(dL,1,1) , %t%
IfEqual, m0, 1, return m1 d p
Loop %m0%
{
r := m1
Loop % m0-2
x := A_Index + 1, r .= d m%x%
L .= Perm(r, d, t, m%m0% d p)"`n" , mx := m1
Loop % m0-1
x := A_Index + 1, m%A_Index% := m%x%
m%m0% := mx
}
return substr(L, 1, -1)
}
AWK
This reads the list of permutations as standard input and outputs the missing one.
{
split($1,a,"");
for (i=1;i<=4;++i) {
t[i,a[i]]++;
}
}
END {
for (k in t) {
split(k,a,SUBSEP)
for (l in t) {
split(l, b, SUBSEP)
if (a[1] == b[1] && t[k] < t[l]) {
s[a[1]] = a[2]
break
}
}
}
print s[1]s[2]s[3]s[4]
}
{{Out}} DBAC
BBC BASIC
{{works with|BBC BASIC for Windows}}
DIM perms$(22), miss&(4)
perms$() = "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", \
\ "CDAB", "DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", \
\ "BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"
FOR i% = 0 TO DIM(perms$(),1)
FOR j% = 1 TO DIM(miss&(),1)
miss&(j%-1) EOR= ASCMID$(perms$(i%),j%)
NEXT
NEXT
PRINT $$^miss&(0) " is missing"
END
{{out}}
DBAC is missing
Burlesque
ln"ABCD"r@\/\\
(Feed permutations via STDIN. Uses the naive method).
Version calculating frequency of occurences of each letter in each row and thus finding the missing permutation by choosing the letters with the lowest frequency:
ln)XXtp)><)F:)<]u[/v\[
C
#include <stdio.h>
#define N 4
const char *perms[] = {
"ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB",
"DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA",
"DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB",
};
int main()
{
int i, j, n, cnt[N];
char miss[N];
for (n = i = 1; i < N; i++) n *= i; /* n = (N-1)!, # of occurrence */
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++) cnt[j] = 0;
/* count how many times each letter occur at position i */
for (j = 0; j < sizeof(perms)/sizeof(const char*); j++)
cnt[perms[j][i] - 'A']++;
/* letter not occurring (N-1)! times is the missing one */
for (j = 0; j < N && cnt[j] == n; j++);
miss[i] = j + 'A';
}
printf("Missing: %.*s\n", N, miss);
return 0;
}
{{out}}
Missing: DBAC
C++
#include <algorithm>
#include <vector>
#include <set>
#include <iterator>
#include <iostream>
#include <string>
static const std::string GivenPermutations[] = {
"ABCD","CABD","ACDB","DACB",
"BCDA","ACBD","ADCB","CDAB",
"DABC","BCAD","CADB","CDBA",
"CBAD","ABDC","ADBC","BDCA",
"DCBA","BACD","BADC","BDAC",
"CBDA","DBCA","DCAB"
};
static const size_t NumGivenPermutations = sizeof(GivenPermutations) / sizeof(*GivenPermutations);
int main()
{
std::vector<std::string> permutations;
std::string initial = "ABCD";
permutations.push_back(initial);
while(true)
{
std::string p = permutations.back();
std::next_permutation(p.begin(), p.end());
if(p == permutations.front())
break;
permutations.push_back(p);
}
std::vector<std::string> missing;
std::set<std::string> given_permutations(GivenPermutations, GivenPermutations + NumGivenPermutations);
std::set_difference(permutations.begin(), permutations.end(), given_permutations.begin(),
given_permutations.end(), std::back_inserter(missing));
std::copy(missing.begin(), missing.end(), std::ostream_iterator<std::string>(std::cout, "\n"));
return 0;
}
C#
By permutating
{{works with|C sharp|C#|2+}}
using System;
using System.Collections.Generic;
namespace MissingPermutation
{
class Program
{
static void Main()
{
string[] given = new string[] { "ABCD", "CABD", "ACDB", "DACB",
"BCDA", "ACBD", "ADCB", "CDAB",
"DABC", "BCAD", "CADB", "CDBA",
"CBAD", "ABDC", "ADBC", "BDCA",
"DCBA", "BACD", "BADC", "BDAC",
"CBDA", "DBCA", "DCAB" };
List<string> result = new List<string>();
permuteString(ref result, "", "ABCD");
foreach (string a in result)
if (Array.IndexOf(given, a) == -1)
Console.WriteLine(a + " is a missing Permutation");
}
public static void permuteString(ref List<string> result, string beginningString, string endingString)
{
if (endingString.Length <= 1)
{
result.Add(beginningString + endingString);
}
else
{
for (int i = 0; i < endingString.Length; i++)
{
string newString = endingString.Substring(0, i) + endingString.Substring(i + 1);
permuteString(ref result, beginningString + (endingString.ToCharArray())[i], newString);
}
}
}
}
}
===By xor-ing the values=== {{works with|C sharp|C#|3+}}
using System;
using System.Linq;
public class Test
{
public static void Main()
{
var input = new [] {"ABCD","CABD","ACDB","DACB","BCDA",
"ACBD","ADCB","CDAB","DABC","BCAD","CADB",
"CDBA","CBAD","ABDC","ADBC","BDCA","DCBA",
"BACD","BADC","BDAC","CBDA","DBCA","DCAB"};
int[] values = {0,0,0,0};
foreach (string s in input)
for (int i = 0; i < 4; i++)
values[i] ^= s[i];
Console.WriteLine(string.Join("", values.Select(i => (char)i)));
}
}
Clojure
(use 'clojure.math.combinatorics)
(use 'clojure.set)
(def given (apply hash-set (partition 4 5 "ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB" )))
(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.
(def abcds ["ABCD" "CABD" "ACDB" "DACB" "BCDA" "ACBD" "ADCB" "CDAB"
"DABC" "BCAD" "CADB" "CDBA" "CBAD" "ABDC" "ADBC" "BDCA"
"DCBA" "BACD" "BADC" "BDAC" "CBDA" "DBCA" "DCAB"])
(def freqs (->> abcds (apply map vector) (map frequencies)))
(defn v->k [fqmap v] (->> fqmap (filter #(-> % second (= v))) ffirst))
(->> freqs (map #(v->k % 5)) (apply str) println)
CoffeeScript
missing_permutation = (arr) ->
# Find the missing permutation in an array of N! - 1 permutations.
# We won't validate every precondition, but we do have some basic
# guards.
if arr.length == 0
throw Error "Need more data"
if arr.length == 1
return [arr[0][1] + arr[0][0]]
# Now we know that for each position in the string, elements should appear
# an even number of times (N-1 >= 2). We can use a set to detect the element appearing
# an odd number of times. Detect odd occurrences by toggling admission/expulsion
# to and from the set for each value encountered. At the end of each pass one element
# will remain in the set.
result = ''
for pos in [0...arr[0].length]
set = {}
for permutation in arr
c = permutation[pos]
if set[c]
delete set[c]
else
set[c] = true
for c of set
result += c
break
result
given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''
arr = (s for s in given.replace('\n', ' ').split ' ' when s != '')
console.log missing_permutation(arr)
{{out}}
> coffee missing_permute.coffee
DBAC
Common Lisp
(defparameter *permutations*
'("ABCD" "CABD" "ACDB" "DACB" "BCDA" "ACBD" "ADCB" "CDAB" "DABC" "BCAD" "CADB" "CDBA"
"CBAD" "ABDC" "ADBC" "BDCA" "DCBA" "BACD" "BADC" "BDAC" "CBDA" "DBCA" "DCAB"))
(defun missing-perm (perms)
(let* ((letters (loop for i across (car perms) collecting i))
(l (/ (1+ (length perms)) (length letters))))
(labels ((enum (n) (loop for i below n collecting i))
(least-occurs (pos)
(let ((occurs (loop for i in perms collecting (aref i pos))))
(cdr (assoc (1- l) (mapcar #'(lambda (letter)
(cons (count letter occurs) letter))
letters))))))
(concatenate 'string (mapcar #'least-occurs (enum (length letters)))))))
{{out}}
ROSETTA> (missing-perm *permutations*)
"DBAC"
D
void main() {
import std.stdio, std.string, std.algorithm, std.range, std.conv;
immutable perms = "ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC
BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD
BADC BDAC CBDA DBCA DCAB".split;
// Version 1: test all permutations.
immutable permsSet = perms
.map!representation
.zip(true.repeat)
.assocArray;
auto perm = perms[0].dup.representation;
do {
if (perm !in permsSet)
writeln(perm.map!(c => char(c)));
} while (perm.nextPermutation);
// Version 2: xor all the ASCII values, the uneven one
// gets flushed out. Based on Perl 6 (via Go).
enum len = 4;
char[len] b = 0;
foreach (immutable p; perms)
b[] ^= p[];
b.writeln;
// Version 3: sum ASCII values.
immutable rowSum = perms[0].sum;
len
.iota
.map!(i => to!char(rowSum - perms.transversal(i).sum % rowSum))
.writeln;
// Version 4: a checksum, Java translation. maxCode will be 36.
immutable maxCode = reduce!q{a * b}(len - 1, iota(3, len + 1));
foreach (immutable i; 0 .. len) {
immutable code = perms.map!(p => perms[0].countUntil(p[i])).sum;
// Code will come up 3, 1, 0, 2 short of 36.
perms[0][maxCode - code].write;
}
}
{{out}}
DBAC
DBAC
DBAC
DBAC
EchoLisp
;; use the obvious methos
(lib 'list) ; for (permutations) function
;; input
(define perms '
(ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB))
;; generate all permutations
(define all-perms (map list->string (permutations '(A B C D))))
→ all-perms
;; {set} substraction
(set-substract (make-set all-perms) (make-set perms))
→ { DBAC }
Elixir
defmodule RC do
def find_miss_perm(head, perms) do
all_permutations(head) -- perms
end
defp all_permutations(string) do
list = String.split(string, "", trim: true)
Enum.map(permutations(list), fn x -> Enum.join(x) end)
end
defp permutations([]), do: [[]]
defp permutations(list), do: (for x <- list, y <- permutations(list -- [x]), do: [x|y])
end
perms = ["ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA",
"CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"]
IO.inspect RC.find_miss_perm( hd(perms), perms )
{{out}}
["DBAC"]
Erlang
The obvious method. It seems fast enough (no waiting time).
-module( find_missing_permutation ).
-export( [difference/2, task/0] ).
difference( Permutate_this, Existing_permutations ) -> all_permutations( Permutate_this ) -- Existing_permutations.
task() -> difference( "ABCD", existing_permutations() ).
all_permutations( String ) -> [[A, B, C, D] || A <- String, B <- String, C <- String, D <- String, is_different([A, B, C, D])].
existing_permutations() -> ["ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"].
is_different( [_H] ) -> true;
is_different( [H | T] ) -> not lists:member(H, T) andalso is_different( T ).
{{out}}
6> find_the_missing_permutation:task().
["DBAC"]
ERRE
PROGRAM MISSING
CONST N=4
DIM PERMS$[23]
BEGIN
PRINT(CHR$(12);) ! CLS
DATA("ABCD","CABD","ACDB","DACB","BCDA","ACBD","ADCB")
DATA("CDAB","DABC","BCAD","CADB","CDBA","CBAD","ABDC","ADBC")
DATA("BDCA","DCBA","BACD","BADC","BDAC","CBDA","DBCA","DCAB")
FOR I%=1 TO UBOUND(PERMS$,1) DO
READ(PERMS$[I%])
END FOR
SOL$="...."
FOR I%=1 TO N DO
CH$=CHR$(I%+64)
COUNT%=0
FOR Z%=1 TO N DO
COUNT%=0
FOR J%=1 TO UBOUND(PERMS$,1) DO
IF CH$=MID$(PERMS$[J%],Z%,1) THEN COUNT%=COUNT%+1 END IF
END FOR
IF COUNT%<>6 THEN
!$RCODE="MID$(SOL$,Z%,1)=CH$"
END IF
END FOR
END FOR
PRINT("Solution is: ";SOL$)
END PROGRAM
{{out}}
Solution is: DBAC
Factor
Permutations are read in via STDIN.
USING: io math.combinatorics sequences sets ;
"ABCD" all-permutations lines diff first print
{{out}}
DBAC
Forth
'''Tested with:''' GForth, VFX Forth, SwiftForth, Win32 Forth. Should work with any ANS Forth system.
'''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.)
hex
ABCD CABD xor ACDB xor DACB xor BCDA xor ACBD xor
ADCB xor CDAB xor DABC xor BCAD xor CADB xor CDBA xor
CBAD xor ABDC xor ADBC xor BDCA xor DCBA xor BACD xor
BADC xor BDAC xor CBDA xor DBCA xor DCAB xor
cr .( Missing permutation: ) u.
decimal
{{out}}
Missing permutation: DBAC ok
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.
program missing_permutation
implicit none
character (4), dimension (23), parameter :: list = &
& (/'ABCD', 'CABD', 'ACDB', 'DACB', 'BCDA', 'ACBD', 'ADCB', 'CDAB', &
& 'DABC', 'BCAD', 'CADB', 'CDBA', 'CBAD', 'ABDC', 'ADBC', 'BDCA', &
& 'DCBA', 'BACD', 'BADC', 'BDAC', 'CBDA', 'DBCA', 'DCAB'/)
integer :: i, j, k
do i = 1, 4
j = minloc ((/(count (list (:) (i : i) == list (1) (k : k)), k = 1, 4)/), 1)
write (*, '(a)', advance = 'no') list (1) (j : j)
end do
write (*, *)
end program missing_permutation
{{out}}
DBAC
FreeBASIC
Simple count
' version 30-03-2017
' compile with: fbc -s console
Data "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD"
Data "ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA"
Data "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD"
Data "BADC", "BDAC", "CBDA", "DBCA", "DCAB"
' ------=< MAIN >=------
Dim As ulong total(3, Asc("A") To Asc("D")) ' total(0 to 3, 65 to 68)
Dim As ULong i, j, n = 24 \ 4 ' n! \ n
Dim As String tmp
For i = 1 To 23
Read tmp
For j = 0 To 3
total(j, tmp[j]) += 1
Next
Next
tmp = Space(4)
For i = 0 To 3
For j = Asc("A") To Asc("D")
If total(i, j) <> n Then
tmp[i] = j
End If
Next
Next
Print "The missing permutation is : "; tmp
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
The missing permutation is : DBAC
===Add the value's===
' version 30-03-2017
' compile with: fbc -s console
Data "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD"
Data "ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA"
Data "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD"
Data "BADC", "BDAC", "CBDA", "DBCA", "DCAB"
' ------=< MAIN >=------
Dim As ULong total(3) ' total(0 to 3)
Dim As ULong i, j, n = 24 \ 4 ' n! \ n
Dim As ULong total_val = (Asc("A") + Asc("B") + Asc("C") + Asc("D")) * n
Dim As String tmp
For i = 1 To 23
Read tmp
For j = 0 To 3
total(j) += tmp[j]
Next
Next
tmp = Space(4)
For i = 0 To 3
tmp[i] = total_val - total(i)
Next
Print "The missing permutation is : "; tmp
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
output is same as the first version
Using Xor
' version 30-03-2017
' compile with: fbc -s console
Data "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD"
Data "ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA"
Data "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD"
Data "BADC", "BDAC", "CBDA", "DBCA", "DCAB"
' ------=< MAIN >=------
Dim As ULong i,j
Dim As String tmp, missing = chr(0, 0, 0, 0) ' or string(4, 0)
For i = 1 To 23
Read tmp
For j = 0 To 3
missing[j] Xor= tmp[j]
Next
Next
Print "The missing permutation is : "; missing
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output is the same as the first version
GAP
# our deficient list
L :=
[ "ABCD", "CABD", "ACDB", "DACB", "BCDA",
"ACBD", "ADCB", "CDAB", "DABC", "BCAD",
"CADB", "CDBA", "CBAD", "ABDC", "ADBC",
"BDCA", "DCBA", "BACD", "BADC", "BDAC",
"CBDA", "DBCA", "DCAB" ];
# convert L to permutations on 1..4
u := List(L, s -> List([1..4], i -> Position("ABCD", s[i])));
# set difference (with all permutations)
v := Difference(PermutationsList([1..4]), u);
# convert back to letters
s := "ABCD";
List(v, p -> List(p, i -> s[i]));
Go
Alternate method suggested by task description:
package main
import (
"fmt"
"strings"
)
var given = strings.Split(`ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB`, "\n")
func main() {
b := make([]byte, len(given[0]))
for i := range b {
m := make(map[byte]int)
for _, p := range given {
m[p[i]]++
}
for char, count := range m {
if count&1 == 1 {
b[i] = char
break
}
}
}
fmt.Println(string(b))
}
Xor method suggested by Perl 6 contributor:
func main() {
b := make([]byte, len(given[0]))
for _, p := range given {
for i, c := range []byte(p) {
b[i] ^= c
}
}
fmt.Println(string(b))
}
{{out}} in either case:
DBAC
Groovy
Solution:
def fact = { n -> [1,(1..<(n+1)).inject(1) { prod, i -> prod * i }].max() }
def missingPerms
missingPerms = {List elts, List perms ->
perms.empty ? elts.permutations() : elts.collect { e ->
def ePerms = perms.findAll { e == it[0] }.collect { it[1..-1] }
ePerms.size() == fact(elts.size() - 1) ? [] \
: missingPerms(elts - e, ePerms).collect { [e] + it }
}.sum()
}
Test:
def e = 'ABCD' as List
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)
mp.each { println it }
{{out}}
[D, B, A, C]
Haskell
=Difference between two lists=
{{works with|GHC|7.10.3}}
import Data.List ((\\), permutations, nub)
import Control.Monad (join)
missingPerm
:: Eq a
=> [[a]] -> [[a]]
missingPerm = (\\) =<< permutations . nub . join
deficientPermsList :: [String]
deficientPermsList =
[ "ABCD"
, "CABD"
, "ACDB"
, "DACB"
, "BCDA"
, "ACBD"
, "ADCB"
, "CDAB"
, "DABC"
, "BCAD"
, "CADB"
, "CDBA"
, "CBAD"
, "ABDC"
, "ADBC"
, "BDCA"
, "DCBA"
, "BACD"
, "BADC"
, "BDAC"
, "CBDA"
, "DBCA"
, "DCAB"
]
main :: IO ()
main = print $ missingPerm deficientPermsList
{{Out}}
["DBAC"]
=Character frequency in each column=
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.List (minimumBy, group, sort, transpose)
import Data.Ord (comparing)
missingPerm
:: Ord a
=> [[a]] -> [a]
missingPerm = ((head . minimumBy (comparing length) . group . sort) <$>) . transpose
deficientPermsList :: [String]
deficientPermsList =
[ "ABCD"
, "CABD"
, "ACDB"
, "DACB"
, "BCDA"
, "ACBD"
, "ADCB"
, "CDAB"
, "DABC"
, "BCAD"
, "CADB"
, "CDBA"
, "CBAD"
, "ABDC"
, "ADBC"
, "BDCA"
, "DCBA"
, "BACD"
, "BADC"
, "BDAC"
, "CBDA"
, "DBCA"
, "DCAB"
]
main :: IO ()
main = print $ missingPerm deficientPermsList
{{Out}}
"DBAC"
=={{header|Icon}} and {{header|Unicon}}==
link strings # for permutes
procedure main()
givens := set![ "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB", "DABC", "BCAD", "CADB",
"CDBA", "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"]
every insert(full := set(), permutes("ABCD")) # generate all permutations
givens := full--givens # and difference
write("The difference is : ")
every write(!givens, " ")
end
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.
every x := permutes("ABCD") do # generate all permutations
if member(givens,x) then delete(givens,x) # remove givens as they are generated
else insert(givens,x) # add back any not given
A still more efficient version is:
link strings
procedure main()
givens := set("ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD",
"ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA",
"CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD",
"BADC", "BDAC", "CBDA", "DBCA", "DCAB")
every p := permutes("ABCD") do
if not member(givens, p) then write(p)
end
{{libheader|Icon Programming Library}} [http://www.cs.arizona.edu/icon/library/src/procs/strings.icn member 'strings' provides permutes(s) which generates all permutations of a string]
J
'''Solution:'''
permutations=: A.~ i.@!@#
missingPerms=: -.~ permutations @ {.
'''Use:'''
data=: >;: 'ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA'
data=: data,>;: 'CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'
missingPerms data
DBAC
Alternatives
Or the above could be a single definition that works the same way:
missingPerms=: -.~ (A.~ i.@!@#) @ {.
Or the equivalent explicit (cf. tacit above) definition:
missingPerms=: monad define
item=. {. y
y -.~ item A.~ i.! #item
)
Or, the solution could be obtained without defining an independent program:
data -.~ 'ABCD' A.~ i.!4
DBAC
Here, 'ABCD' represents the values being permuted (their order does not matter), and 4 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] (~), would be:
((i.!4) A. 'ABCD') -. data
DBAC
Of course the task suggests that the missing permutation can be found without generating all permutations. And of course that is doable:
'ABCD'{~,I.@(= <./)@(#/.~)@('ABCD' , ])"1 |:perms
DBAC
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:
,(~.#~2|(#/.~))"1|:data
DBAC
We could use arithmetic, as suggested in the task hints:
({.data){~|(->./)+/({.i.])data
DBAC
Java
'''optimized''' Following needs: [[User:Margusmartsepp/Contributions/Java/Utils.java|Utils.java]]
import java.util.ArrayList;
import com.google.common.base.Joiner;
import com.google.common.collect.ImmutableSet;
import com.google.common.collect.Lists;
public class FindMissingPermutation {
public static void main(String[] args) {
Joiner joiner = Joiner.on("").skipNulls();
ImmutableSet<String> s = ImmutableSet.of("ABCD", "CABD", "ACDB",
"DACB", "BCDA", "ACBD", "ADCB", "CDAB", "DABC", "BCAD", "CADB",
"CDBA", "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD", "BADC",
"BDAC", "CBDA", "DBCA", "DCAB");
for (ArrayList<Character> cs : Utils.Permutations(Lists.newArrayList(
'A', 'B', 'C', 'D')))
if (!s.contains(joiner.join(cs)))
System.out.println(joiner.join(cs));
}
}
{{out}}
DBAC
Alternate version, based on checksumming each position:
public class FindMissingPermutation
{
public static void main(String[] args)
{
String[] givenPermutations = { "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD",
"ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA",
"CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD",
"BADC", "BDAC", "CBDA", "DBCA", "DCAB" };
String characterSet = givenPermutations[0];
// Compute n! * (n - 1) / 2
int maxCode = characterSet.length() - 1;
for (int i = characterSet.length(); i >= 3; i--)
maxCode *= i;
StringBuilder missingPermutation = new StringBuilder();
for (int i = 0; i < characterSet.length(); i++)
{
int code = 0;
for (String permutation : givenPermutations)
code += characterSet.indexOf(permutation.charAt(i));
missingPermutation.append(characterSet.charAt(maxCode - code));
}
System.out.println("Missing permutation: " + missingPermutation.toString());
}
}
JavaScript
ES5
=Imperative=
The permute() function taken from http://snippets.dzone.com/posts/show/1032
permute = function(v, m){ //v1.0
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;
++i < l; x[2][i] = i, x[1][i] = x[0][i] = j /= --k);
for(r = new Array(q); ++p < q;)
for(r[p] = new Array(l), i = -1; ++i < l; !--x[1][i] && (x[1][i] = x[0][i],
x[2][i] = (x[2][i] + 1) % l), r[p][i] = m ? x[3][i] : v[x[3][i]])
for(x[3][i] = x[2][i], f = 0; !f; f = !f)
for(j = i; j; x[3][--j] == x[2][i] && (x[3][i] = x[2][i] = (x[2][i] + 1) % l, f = 1));
return r;
};
list = [ 'ABCD', 'CABD', 'ACDB', 'DACB', 'BCDA', 'ACBD', 'ADCB', 'CDAB',
'DABC', 'BCAD', 'CADB', 'CDBA', 'CBAD', 'ABDC', 'ADBC', 'BDCA',
'DCBA', 'BACD', 'BADC', 'BDAC', 'CBDA', 'DBCA', 'DCAB'];
all = permute(list[0].split('')).map(function(elem) {return elem.join('')});
missing = all.filter(function(elem) {return list.indexOf(elem) == -1});
print(missing); // ==> DBAC
=Functional=
(function (strList) {
// [a] -> [[a]]
function permutations(xs) {
return xs.length ? (
chain(xs, function (x) {
return chain(permutations(deleted(x, xs)), function (ys) {
return [[x].concat(ys).join('')];
})
})) : [[]];
}
// Monadic bind/chain for lists
// [a] -> (a -> b) -> [b]
function chain(xs, f) {
return [].concat.apply([], xs.map(f));
}
// a -> [a] -> [a]
function deleted(x, xs) {
return xs.length ? (
x === xs[0] ? xs.slice(1) : [xs[0]].concat(
deleted(x, xs.slice(1))
)
) : [];
}
// Provided subset
var lstSubSet = strList.split('\n');
// Any missing permutations
// (we can use fold/reduce, filter, or chain (concat map) here)
return chain(permutations('ABCD'.split('')), function (x) {
return lstSubSet.indexOf(x) === -1 ? [x] : [];
});
})(
'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}}
["DBAC"]
ES6
=Statistical=
==Using a dictionary==
(() => {
'use strict';
// transpose :: [[a]] -> [[a]]
let transpose = xs =>
xs[0].map((_, iCol) => xs
.map((row) => row[iCol]));
let xs = 'ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB' +
' DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA' +
' BACD BADC BDAC CBDA DBCA DCAB'
return transpose(xs.split(' ')
.map(x => x.split('')))
.map(col => col.reduce((a, x) => ( // count of each character in each column
a[x] = (a[x] || 0) + 1,
a
), {}))
.map(dct => { // character with frequency below mean of distribution ?
let ks = Object.keys(dct),
xs = ks.map(k => dct[k]),
mean = xs.reduce((a, b) => a + b, 0) / xs.length;
return ks.reduce(
(a, k) => a ? a : (dct[k] < mean ? k : undefined),
undefined
);
})
.join(''); // 4 chars as single string
// --> 'DBAC'
})();
{{Out}}
DBAC
==Composing functional primitives==
{{Trans|Haskell}}
(() => {
'use strict';
// MISSING PERMUTATION ---------------------------------------------------
// missingPermutation :: [String] -> String
const missingPermutation = xs =>
map(
// Rarest letter,
compose([
sort,
group,
curry(minimumBy)(comparing(length)),
head
]),
// in each column.
transpose(map(stringChars, xs))
)
.join('');
// GENERIC FUNCTIONAL PRIMITIVES -----------------------------------------
// transpose :: [[a]] -> [[a]]
const transpose = xs =>
xs[0].map((_, iCol) => xs.map(row => row[iCol]));
// sort :: Ord a => [a] -> [a]
const sort = xs => xs.sort();
// group :: Eq a => [a] -> [[a]]
const group = xs => groupBy((a, b) => a === b, xs);
// groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
const groupBy = (f, xs) => {
const dct = xs.slice(1)
.reduce((a, x) => {
const
h = a.active.length > 0 ? a.active[0] : undefined,
blnGroup = h !== undefined && f(h, x);
return {
active: blnGroup ? a.active.concat(x) : [x],
sofar: blnGroup ? a.sofar : a.sofar.concat([a.active])
};
}, {
active: xs.length > 0 ? [xs[0]] : [],
sofar: []
});
return dct.sofar.concat(dct.active.length > 0 ? [dct.active] : []);
};
// length :: [a] -> Int
const length = xs => xs.length;
// comparing :: (a -> b) -> (a -> a -> Ordering)
const comparing = f =>
(x, y) => {
const
a = f(x),
b = f(y);
return a < b ? -1 : a > b ? 1 : 0
};
// minimumBy :: (a -> a -> Ordering) -> [a] -> a
const minimumBy = (f, xs) =>
xs.reduce((a, x) => a === undefined ? x : (
f(x, a) < 0 ? x : a
), undefined);
// head :: [a] -> a
const head = xs => xs.length ? xs[0] : undefined;
// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f)
// compose :: [(a -> a)] -> (a -> a)
const compose = fs => x => fs.reduce((a, f) => f(a), x);
// curry :: ((a, b) -> c) -> a -> b -> c
const curry = f => a => b => f(a, b);
// stringChars :: String -> [Char]
const stringChars = s => s.split('');
// TEST ------------------------------------------------------------------
return missingPermutation(["ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD",
"ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC",
"BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"
]);
// -> "DBAC"
})();
{{Out}}
DBAC
jq
{{works with|jq|1.4}}
The following assumes that a file, Find_the_missing_permutation.txt, has the text exactly as presented in the task description.
To find the missing permutation, we can for simplicity invoke jq twice: jq -R . Find_the_missing_permutation.txt | jq -s -f Find_the_missing_permutation.jq
The first invocation simply converts the raw text into a stream of JSON strings; these are then processed by the following program, which implements the parity-based approach.
The program will handle permutations of any set of uppercase letters. The letters need not be consecutive. Note that the following encoding of letters is used: A => 0, B => 1, ....
'''Infrastructure''':
If your version of jq has transpose/0, the definition given here (which is the same as in [[Matrix_Transpose#jq]]) may be omitted.
def transpose:
if (.[0] | length) == 0 then []
else [map(.[0])] + (map(.[1:]) | transpose)
end ;
# Input: an array of integers (based on the encoding of A=0, B=1, etc)
# corresponding to the occurrences in any one position of the
# letters in the list of permutations.
# Output: a tally in the form of an array recording in position i the
# parity of the number of occurrences of the letter corresponding to i.
# Example: given [0,1,0,1,2], the array of counts of 0, 1, and 2 is [2, 2, 1],
# and thus the final result is [0, 0, 1].
def parities:
reduce .[] as $x ( []; .[$x] = (1 + .[$x]) % 2);
# Input: an array of parity-counts, e.g. [0, 1, 0, 0]
# Output: the corresponding letter, e.g. "B".
def decode:
[index(1) + 65] | implode;
# encode a string (e.g. "ABCD") as an array (e.g. [0,1,2,3]):
def encode_string: [explode[] - 65];
'''The task''':
map(encode_string) | transpose | map(parities | decode) | join("")
{{Out}}
$ jq -R . Find_the_missing_permutation.txt | jq -s -f Find_the_missing_permutation.jq
"DBAC"
Julia
{{works with|Julia|0.6}}
== Obvious method == Calculate all possible permutations and return the first not included in the array.
using BenchmarkTools, Combinatorics
function missingperm(arr::Vector)
allperms = String.(permutations(arr[1])) # revised for type safety
for perm in allperms
if perm ∉ arr return perm end
end
end
arr = ["ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB", "DABC", "BCAD",
"CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD", "BADC", "BDAC",
"CBDA", "DBCA", "DCAB"]
@show missingperm(arr)
{{out}}
missingperm(arr) = "DBAC"
== Alternative method 1 == {{trans|Python}}
function missingperm1(arr::Vector{<:AbstractString})
missperm = string()
for pos in 1:length(arr[1])
s = Set()
for perm in arr
c = perm[pos]
if c ∈ s pop!(s, c) else push!(s, c) end
end
missperm *= first(s)
end
return missperm
end
== Alternative method 2 == {{trans|Perl 6}}
function missingperm2(arr::Vector)
len = length(arr[1])
xorval = zeros(UInt8, len)
for perm in [Vector{UInt8}(s) for s in arr], i in 1:len
xorval[i] ⊻= perm[i]
end
return String(xorval)
end
@show missingperm(arr)
@show missingperm1(arr)
@show missingperm2(arr)
@btime missingperm(arr)
@btime missingperm1(arr)
@btime missingperm2(arr)
{{out}}
missingperm(arr) = "DBAC"
missingperm1(arr) = "DBAC"
missingperm2(arr) = "DBAC"
6.460 μs (148 allocations: 8.55 KiB)
6.780 μs (24 allocations: 2.13 KiB)
3.100 μs (50 allocations: 2.94 KiB)
K
split:{1_'(&x=y)_ x:y,x}
g: ("ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB")
g,:(" CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB")
p: split[g;" "];
/ All permutations of "ABCD"
perm:{:[1<x;,/(>:'(x,x)#1,x#0)[;0,'1+_f x-1];,!x]}
p2:a@(perm(#a:"ABCD"));
/ Which permutations in p are there in p2?
p2 _lin p
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1
/ Invert the result
~p2 _lin p
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
/ It's the 20th permutation that is missing
&~p2 _lin p
,20
p2@&~p2 _lin p
"DBAC"
Alternative approach:
table:{b@<b:(x@*:'a),'#:'a:=x}
,/"ABCD"@&:'{5=(table p[;x])[;1]}'!4
"DBAC"
Third approach (where p is the given set of permutations):
,/p2@&~(p2:{x@m@&n=(#?:)'m:!n#n:#x}[*p]) _lin p
Kotlin
// version 1.1.2
fun <T> permute(input: List<T>): List<List<T>> {
if (input.size == 1) return listOf(input)
val perms = mutableListOf<List<T>>()
val toInsert = input[0]
for (perm in permute(input.drop(1))) {
for (i in 0..perm.size) {
val newPerm = perm.toMutableList()
newPerm.add(i, toInsert)
perms.add(newPerm)
}
}
return perms
}
fun <T> missingPerms(input: List<T>, perms: List<List<T>>) = permute(input) - perms
fun main(args: Array<String>) {
val input = listOf('A', 'B', 'C', 'D')
val strings = listOf(
"ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB",
"DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA",
"DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"
)
val perms = strings.map { it.toList() }
val missing = missingPerms(input, perms)
if (missing.size == 1)
print("The missing permutation is ${missing[0].joinToString("")}")
else {
println("There are ${missing.size} missing permutations, namely:\n")
for (perm in missing) println(perm.joinToString(""))
}
}
{{out}}
The missing permutation is DBAC
Lua
Using the popular Penlight extension module - https://luarocks.org/modules/steved/penlight
local permute, tablex = require("pl.permute"), require("pl.tablex")
local permList, pStr = {
"ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB",
"DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA",
"DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"
}
for perm in permute.iter({"A","B","C","D"}) do
pStr = table.concat(perm)
if not tablex.find(permList, pStr) then print(pStr) end
end
{{out}}
DBAC
Maple
lst := ["ABCD","CABD","ACDB","DACB","BCDA","ACBD","ADCB","CDAB","DABC","BCAD","CADB","CDBA","CBAD","ABDC","ADBC","BDCA","DCBA","BACD","BADC","BDAC","CBDA","DBCA","DCAB"]:
perm := table():
for letter in "ABCD" do
perm[letter] := 0:
end do:
for item in lst do
for letter in "ABCD" do
perm[letter] += StringTools:-FirstFromLeft(letter, item):
end do:
end do:
print(StringTools:-Join(ListTools:-Flatten([indices(perm)], 4)[sort(map(x->60-x, ListTools:-Flatten([entries(perm)],4)),'output=permutation')], "")):
{{Out|Output}}
"DBAC"
=={{header|Mathematica}} / {{header|Wolfram Language}}==
ProvidedSet = {"ABCD" , "CABD" , "ACDB" , "DACB" , "BCDA" , "ACBD",
"ADCB" , "CDAB", "DABC", "BCAD" , "CADB", "CDBA" , "CBAD" , "ABDC",
"ADBC" , "BDCA", "DCBA" , "BACD", "BADC", "BDAC" , "CBDA", "DBCA", "DCAB"};
Complement[StringJoin /@ Permutations@Characters@First@#, #] &@ProvidedSet
->{"DBAC"}
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.
function perm = findMissingPerms(list)
permsList = perms(list(1,:)); %Generate all permutations of the 4 letters
perm = []; %This is the functions return value if the list is not missing a permutation
%Normally the rest of this would be vectorized, but because this is
%done on a vector of strings, the vectorized functions will only access
%one character at a time. So, in order for this to work we have to use
%loops.
for i = (1:size(permsList,1))
found = false;
for j = (1:size(list,1))
if (permsList(i,:) == list(j,:))
found = true;
break
end
end
if not(found)
perm = permsList(i,:);
return
end
end %for
end %fingMissingPerms
{{out}}
list = ['ABCD';
'CABD';
'ACDB';
'DACB';
'BCDA';
'ACBD';
'ADCB';
'CDAB';
'DABC';
'BCAD';
'CADB';
'CDBA';
'CBAD';
'ABDC';
'ADBC';
'BDCA';
'DCBA';
'BACD';
'BADC';
'BDAC';
'CBDA';
'DBCA';
'DCAB']
list =
ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB
>> findMissingPerms(list)
ans =
DBAC
Nim
{{trans|Python}}
import strutils
proc missingPermutation(arr): string =
result = ""
if arr.len == 0: return
if arr.len == 1: return arr[0][1] & arr[0][0]
for pos in 0 .. <arr[0].len:
var s: set[char] = {}
for permutation in arr:
let c = permutation[pos]
if c in s: s.excl c
else: s.incl c
for c in s: result.add c
const 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()
echo missingPermutation(given)
{{out}}
DBAC
OCaml
some utility functions:
(* insert x at all positions into li and return the list of results *)
let rec insert x = function
| [] -> [[x]]
| a::m as li -> (x::li) :: (List.map (fun y -> a::y) (insert x m))
(* list of all permutations of li *)
let permutations li =
List.fold_right (fun a z -> List.concat (List.map (insert a) z)) li [[]]
(* convert a string to a char list *)
let chars_of_string s =
let cl = ref [] in
String.iter (fun c -> cl := c :: !cl) s;
(List.rev !cl)
(* convert a char list to a string *)
let string_of_chars cl =
String.concat "" (List.map (String.make 1) cl)
resolve the task:
let deficient_perms = [
"ABCD";"CABD";"ACDB";"DACB";
"BCDA";"ACBD";"ADCB";"CDAB";
"DABC";"BCAD";"CADB";"CDBA";
"CBAD";"ABDC";"ADBC";"BDCA";
"DCBA";"BACD";"BADC";"BDAC";
"CBDA";"DBCA";"DCAB";
]
let it = chars_of_string (List.hd deficient_perms)
let perms = List.map string_of_chars (permutations it)
let results = List.filter (fun v -> not(List.mem v deficient_perms)) perms
let () = List.iter print_endline results
Alternate method : if we had all permutations, each letter would appear an even number of times at each position. Since there is only one permutation missing, we can find where each letter goes by looking at the parity of the number of occurences of each letter. The following program works with permutations of at least 3 letters:
let array_of_perm s =
let n = String.length s in
Array.init n (fun i -> int_of_char s.[i] - 65);;
let perm_of_array a =
let n = Array.length a in
let s = String.create n in
Array.iteri (fun i x ->
s.[i] <- char_of_int (x + 65)
) a;
s;;
let find_missing v =
let n = String.length (List.hd v) in
let a = Array.make_matrix n n 0
and r = ref v in
List.iter (fun s ->
let u = array_of_perm s in
Array.iteri (fun i x -> x.(u.(i)) <- x.(u.(i)) + 1) a
) v;
let q = Array.make n 0 in
Array.iteri (fun i x ->
Array.iteri (fun j y ->
if y mod 2 != 0 then q.(i) <- j
) x
) a;
perm_of_array q;;
find_missing deficient_perms;;
(* - : string = "DBAC" *)
Octave
given = [ 'ABCD';'CABD';'ACDB';'DACB'; ...
'BCDA';'ACBD';'ADCB';'CDAB'; ...
'DABC';'BCAD';'CADB';'CDBA'; ...
'CBAD';'ABDC';'ADBC';'BDCA'; ...
'DCBA';'BACD';'BADC';'BDAC'; ...
'CBDA';'DBCA';'DCAB' ];
val = 4.^(3:-1:0)';
there = 1+(toascii(given)-toascii('A'))*val;
every = 1+perms(0:3)*val;
bits = zeros(max(every),1);
bits(every) = 1;
bits(there) = 0;
missing = dec2base(find(bits)-1,'ABCD')
Oz
Using constraint programming for this problem may be a bit overkill...
declare
GivenPermutations =
["ABCD" "CABD" "ACDB" "DACB" "BCDA" "ACBD" "ADCB" "CDAB" "DABC" "BCAD" "CADB" "CDBA"
"CBAD" "ABDC" "ADBC" "BDCA" "DCBA" "BACD" "BADC" "BDAC" "CBDA" "DBCA" "DCAB"]
%% four distinct variables between "A" and "D":
proc {Description Root}
Root = {FD.list 4 &A#&D}
{FD.distinct Root}
{FD.distribute naiv Root}
end
AllPermutations = {SearchAll Description}
in
for P in AllPermutations do
if {Not {Member P GivenPermutations}} then
{System.showInfo "Missing: "#P}
end
end
PARI/GP
v=["ABCD","CABD","ACDB","DACB","BCDA","ACBD","ADCB","CDAB","DABC","BCAD","CADB","CDBA","CBAD","ABDC","ADBC","BDCA","DCBA","BACD","BADC","BDAC","CBDA","DBCA","DCAB"];
v=apply(u->permtonum(apply(n->n-64,Vec(Vecsmall(u)))),v);
t=numtoperm(4, binomial(4!,2)-sum(i=1,#v,v[i]));
Strchr(apply(n->n+64,t))
{{out}}
%1 = "DBAC"
Pascal
like [[c]], summation, and [[Perl 6]] XORing
program MissPerm;
{$MODE DELPHI} //for result
const
maxcol = 4;
type
tmissPerm = 1..23;
tcol = 1..maxcol;
tResString = String[maxcol];
const
Given_Permutations : array [tmissPerm] of tResString =
('ABCD', 'CABD', 'ACDB', 'DACB', 'BCDA', 'ACBD',
'ADCB', 'CDAB', 'DABC', 'BCAD', 'CADB', 'CDBA',
'CBAD', 'ABDC', 'ADBC', 'BDCA', 'DCBA', 'BACD',
'BADC', 'BDAC', 'CBDA', 'DBCA', 'DCAB');
chOfs = Ord('A')-1;
var
SumElemCol: array[tcol,tcol] of NativeInt;
function fib(n: NativeUint): NativeUint;
var
i : NativeUint;
Begin
result := 1;
For i := 2 to n do
result:= result*i;
end;
function CountOccurences: tresString;
//count the number of every letter in every column
//should be (colmax-1)! => 6
//the missing should count (colmax-1)! -1 => 5
var
fibN_1 : NativeUint;
row, col: NativeInt;
Begin
For row := low(tmissPerm) to High(tmissPerm) do
For col := low(tcol) to High(tcol) do
inc(SumElemCol[col,ORD(Given_Permutations[row,col])-chOfs]);
//search the missing
fibN_1 := fib(maxcol-1)-1;
setlength(result,maxcol);
For col := low(tcol) to High(tcol) do
For row := low(tcol) to High(tcol) do
IF SumElemCol[col,row]=fibN_1 then
result[col]:= chr(row+chOfs);
end;
function CheckXOR: tresString;
var
row,col: NativeUint;
Begin
setlength(result,maxcol);
fillchar(result[1],maxcol,#0);
For row := low(tmissPerm) to High(tmissPerm) do
For col := low(tcol) to High(tcol) do
result[col] := chr(ord(result[col]) XOR ord(Given_Permutations[row,col]));
end;
Begin
writeln(CountOccurences,' is missing');
writeln(CheckXOR,' is missing');
end.
{{out}}
DBAC is missing
DBAC is missing
Perl
Because the set of all permutations contains all its own rotations, the first missing rotation is the target.
sub check_perm {
my %hash; @hash{@_} = ();
for my $s (@_) { exists $hash{$_} or return $_
for map substr($s,1) . substr($s,0,1), (1..length $s); }
}
# Check and display
@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);
print check_perm(@perms), "\n";
{{out}}
DBAC
Perl 6
my @givens = <ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB>;
my @perms = <A B C D>.permutations.map: *.join;
.say when none(@givens) for @perms;
{{out}}
DBAC
Of course, all of these solutions are working way too hard, when you can just xor all the bits, and the missing one will just pop right out:
say [~^] <ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB>;
{{out}}
DBAC
Phix
constant perms = {"ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB",
"DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA",
"DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"}
-- 1: sum of letters
sequence r = repeat(0,4)
for i=1 to length(perms) do
r = sq_add(r,perms[i])
end for
r = sq_sub(max(r)+'A',r)
puts(1,r&'\n')
-- based on the notion that missing = sum(full)-sum(partial) would be true,
-- and that sum(full) would be like {M,M,M,M} rather than a mix of numbers.
-- the final step is equivalent to eg {1528,1530,1531,1529}
-- max-r[i] -> { 3, 1, 0, 2}
-- to chars -> { D, B, A, C}
-- (but obviously both done in one line)
-- 2: the xor trick
r = repeat(0,4)
for i=1 to length(perms) do
r = sq_xor_bits(r,perms[i])
end for
puts(1,r&'\n')
-- (relies on the missing chars being present an odd number of times, non-missing chars an even number of times)
-- 3: find least frequent letters
r = " "
for i=1 to length(r) do
sequence count = repeat(0,4)
for j=1 to length(perms) do
count[perms[j][i]-'A'+1] += 1
end for
r[i] = smallest(count,1)+'A'-1
end for
puts(1,r&'\n')
-- (relies on the assumption that a full set would have each letter occurring the same number of times in each position)
-- (smallest(count,1) returns the index position of the smallest, rather than it's value)
-- 4: test all permutations
for i=1 to factorial(4) do
r = permute(i,"ABCD")
if not find(r,perms) then exit end if
end for
puts(1,r&'\n')
-- (relies on brute force(!) - but this is the only method that could be made to cope with >1 omission)
{{out}}
DBAC
DBAC
DBAC
DBAC
PHP
<?php
$finalres = Array();
function permut($arr,$result=array()){
global $finalres;
if(empty($arr)){
$finalres[] = implode("",$result);
}else{
foreach($arr as $key => $val){
$newArr = $arr;
$newres = $result;
$newres[] = $val;
unset($newArr[$key]);
permut($newArr,$newres);
}
}
}
$givenPerms = Array("ABCD","CABD","ACDB","DACB","BCDA","ACBD","ADCB","CDAB","DABC","BCAD","CADB","CDBA","CBAD","ABDC","ADBC","BDCA","DCBA","BACD","BADC","BDAC","CBDA","DBCA","DCAB");
$given = Array("A","B","C","D");
permut($given);
print_r(array_diff($finalres,$givenPerms)); // Array ( [20] => DBAC )
PicoLisp
(setq *PermList
(mapcar chop
(quote
"ABCD" "CABD" "ACDB" "DACB" "BCDA" "ACBD" "ADCB" "CDAB"
"DABC" "BCAD" "CADB" "CDBA" "CBAD" "ABDC" "ADBC" "BDCA"
"DCBA" "BACD" "BADC" "BDAC" "CBDA" "DBCA" "DCAB" ) ) )
(let (Lst (chop "ABCD") L Lst)
(recur (L) # Permute
(if (cdr L)
(do (length L)
(recurse (cdr L))
(rot L) )
(unless (member Lst *PermList) # Check
(prinl Lst) ) ) ) )
{{out}}
DBAC
PowerShell
{{works with|PowerShell|4.0}}
function permutation ($array) {
function generate($n, $array, $A) {
if($n -eq 1) {
$array[$A] -join ''
}
else{
for( $i = 0; $i -lt ($n - 1); $i += 1) {
generate ($n - 1) $array $A
if($n % 2 -eq 0){
$i1, $i2 = $i, ($n-1)
$temp = $A[$i1]
$A[$i1] = $A[$i2]
$A[$i2] = $temp
}
else{
$i1, $i2 = 0, ($n-1)
$temp = $A[$i1]
$A[$i1] = $A[$i2]
$A[$i2] = $temp
}
}
generate ($n - 1) $array $A
}
}
$n = $array.Count
if($n -gt 0) {
(generate $n $array (0..($n-1)))
} else {$array}
}
$perm = permutation @('A','B','C', 'D')
$find = @(
"ABCD"
"CABD"
"ACDB"
"DACB"
"BCDA"
"ACBD"
"ADCB"
"CDAB"
"DABC"
"BCAD"
"CADB"
"CDBA"
"CBAD"
"ABDC"
"ADBC"
"BDCA"
"DCBA"
"BACD"
"BADC"
"BDAC"
"CBDA"
"DBCA"
"DCAB"
)
$perm | where{-not $find.Contains($_)}
Output:
DBAC
PureBasic
Procedure in_List(in.s)
Define.i i, j
Define.s a
Restore data_to_test
For i=1 To 3*8-1
Read.s a
If in=a
ProcedureReturn #True
EndIf
Next i
ProcedureReturn #False
EndProcedure
Define.c z, x, c, v
If OpenConsole()
For z='A' To 'D'
For x='A' To 'D'
If z=x:Continue:EndIf
For c='A' To 'D'
If c=x Or c=z:Continue:EndIf
For v='A' To 'D'
If v=c Or v=x Or v=z:Continue:EndIf
Define.s test=Chr(z)+Chr(x)+Chr(c)+Chr(v)
If Not in_List(test)
PrintN(test+" is missing.")
EndIf
Next
Next
Next
Next
PrintN("Press Enter to exit"):Input()
EndIf
DataSection
data_to_test:
Data.s "ABCD","CABD","ACDB","DACB","BCDA","ACBD","ADCB","CDAB"
Data.s "DABC","BCAD","CADB","CDBA","CBAD","ABDC","ADBC","BDCA"
Data.s "DCBA","BACD","BADC","BDAC","CBDA","DBCA","DCAB"
EndDataSection
Based on the [[Permutations#PureBasic|Permutations]] task, the solution could be:
If OpenConsole()
NewList a.s()
findPermutations(a(), "ABCD", 4)
ForEach a()
Select a()
Case "ABCD","CABD","ACDB","DACB","BCDA","ACBD","ADCB","CDAB","DABC"
Case "BCAD","CADB","CDBA","CBAD","ABDC","ADBC","BDCA","DCBA","BACD"
Case "BADC","BDAC","CBDA","DBCA","DCAB"
Default
PrintN(A()+" is missing.")
EndSelect
Next
Print(#CRLF$ + "Press ENTER to exit"): Input()
EndIf
Python
Python: Calculate difference when compared to all permutations
{{works with|Python|2.6+}}
from itertools import permutations
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()
allPerms = [''.join(x) for x in permutations(given[0])]
missing = list(set(allPerms) - set(given)) # ['DBAC']
Python:Counting lowest frequency character at each position
Here is a solution that is more in the spirit of the challenge, 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."
# We won't validate every precondition, but we do have some basic
# guards.
if len(arr) == 0: raise Exception("Need more data")
if len(arr) == 1:
return [arr[0][1] + arr[0][0]]
# Now we know that for each position in the string, elements should appear
# an even number of times (N-1 >= 2). We can use a set to detect the element appearing
# an odd number of times. Detect odd occurrences by toggling admission/expulsion
# to and from the set for each value encountered. At the end of each pass one element
# will remain in the set.
missing_permutation = ''
for pos in range(len(arr[0])):
s = set()
for permutation in arr:
c = permutation[pos]
if c in s:
s.remove(c)
else:
s.add(c)
missing_permutation += list(s)[0]
return missing_permutation
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()
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:
from collections import Counter
>>> 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
It is rather obfuscated, but can be explained
by showing these intermediate results and noting
that zip(*x) transposes x;
and that at the end of the list
created by the call to most_common()
is the least common character.
from pprint import pprint as pp
>>> 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'),
('B', 'A', 'C', 'A', 'C', 'C', 'D', 'D', 'A', 'C', 'A', 'D', 'B', 'B', 'D', 'D', 'C', 'A', 'A', 'D', 'B', 'B', 'C'),
('C', 'B', 'D', 'C', 'D', 'B', 'C', 'A', 'B', 'A', 'D', 'B', 'A', 'D', 'B', 'C', 'B', 'C', 'D', 'A', 'D', 'C', 'A'),
('D', 'D', 'B', 'B', 'A', 'D', 'B', 'B', 'C', 'D', 'B', 'A', 'D', 'C', 'C', 'A', 'A', 'D', 'C', 'C', 'A', 'A', 'B')]
>>> pp([Counter(x).most_common() for x in zip(*given)])
[[('C', 6), ('B', 6), ('A', 6), ('D', 5)],
[('D', 6), ('C', 6), ('A', 6), ('B', 5)],
[('D', 6), ('C', 6), ('B', 6), ('A', 5)],
[('D', 6), ('B', 6), ('A', 6), ('C', 5)]]
>>> pp([Counter(x).most_common()[-1] for x in zip(*given)])
[('D', 5), ('B', 5), ('A', 5), ('C', 5)]
>>> pp([Counter(x).most_common()[-1][0] for x in zip(*given)])
['D', 'B', 'A', 'C']
>>> ''.join([Counter(x).most_common()[-1][0] for x in zip(*given)])
'DBAC'
>>>
R
This uses the "combinat" package, which is a standard R package:
permute.me <- c("A", "B", "C", "D") perms <- permn(permute.me) # list of all permutations perms2 <- matrix(unlist(perms), ncol=length(permute.me), byrow=T) # matrix of all permutations perms3 <- apply(perms2, 1, paste, collapse="") # vector of all permutations
incomplete <- c("ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB")
setdiff(perms3, incomplete)
{{out}}
```txt
[1] "DBAC"
Racket
#lang racket
(define almost-all
'([A B C D] [C A B D] [A C D B] [D A C B] [B C D A] [A C B D] [A D C B]
[C D A B] [D A B C] [B C A D] [C A D B] [C D B A] [C B A D] [A B D C]
[A D B C] [B D C A] [D C B A] [B A C D] [B A D C] [B D A C] [C B D A]
[D B C A] [D C A B]))
;; Obvious method:
(for/first ([p (in-permutations (car almost-all))]
#:unless (member p almost-all))
p)
;; -> '(D B A C)
;; For permutations of any set
(define charmap
(for/hash ([x (in-list (car almost-all))] [i (in-naturals)])
(values x i)))
(define size (hash-count charmap))
;; Illustrating approach mentioned in the task description.
;; For each position, character with odd parity at that position.
(require data/bit-vector)
(for/list ([i (in-range size)])
(define parities (make-bit-vector size #f))
(for ([permutation (in-list almost-all)])
(define n (hash-ref charmap (list-ref permutation i)))
(bit-vector-set! parities n (not (bit-vector-ref parities n))))
(for/first ([(c i) charmap] #:when (bit-vector-ref parities i))
c))
;; -> '(D B A C)
RapidQ
Dim PList as QStringList
PList.addItems "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD", "ADCB", "CDAB"
PList.additems "DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC", "BDCA"
PList.additems "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB"
Dim NumChar(4, 65 to 68) as integer
Dim MPerm as string
'Create table with occurences
For x = 0 to PList.Itemcount -1
for y = 1 to 4
Inc(NumChar(y, asc(PList.Item(x)[y])))
next
next
'When a char only occurs 5 times it's the missing one
for x = 1 to 4
for y = 65 to 68
MPerm = MPerm + iif(NumChar(x, y)=5, chr$(y), "")
next
next
showmessage MPerm
'= DBAC
REXX
/*REXX pgm finds one or more missing permutations from an internal list & displays them.*/
list = 'ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA',
'CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'
@.= /* [↓] needs to be as long as THINGS.*/
@abcU = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' /*an uppercase (Latin/Roman) alphabet. */
things = 4 /*number of unique letters to be used. */
bunch = 4 /*number letters to be used at a time. */
do j=1 for things /* [↓] only get a portion of alphabet.*/
$.j=substr(@abcU,j,1) /*extract just one letter from alphabet*/
end /*j*/ /* [↑] build a letter array for speed.*/
call permSet 1 /*invoke PERMSET subroutine recursively*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
permSet: procedure expose $. @. bunch list things; parse arg ?
if ?>bunch then do; _=
do m=1 for bunch /*build a permutation. */
_=_ || @.m /*add permutation──►list.*/
end /*m*/
/* [↓] is in the list? */
if wordpos(_,list)==0 then say _ ' is missing from the list.'
end
else do x=1 for things /*build a permutation. */
do k=1 for ?-1
if @.k==$.x then iterate x /*was permutation built? */
end /*k*/
@.?=$.x /*define as being built. */
call permSet ?+1 /*call subr. recursively.*/
end /*x*/
return
'''output'''
DBAC is missing from the list.
Ring
list = "ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB"
for a = ascii("A") to ascii("D")
for b = ascii("A") to ascii("D")
for c = ascii("A") to ascii("D")
for d = ascii("A") to ascii("D")
x = char(a) + char(b) + char(c)+ char(d)
if a!=b and a!=c and a!=d and b!=c and b!=d and c!=d
if substr(list,x) = 0 see x + " missing" + nl ok ok
next
next
next
next
Output:
DBAC missing
Ruby
{{works with|Ruby|2.0+}}
given = %w{
ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB
}
all = given[0].chars.permutation.collect(&:join)
puts "missing: #{all - given}"
{{out}}
missing: ["DBAC"]
Run BASIC
list$ = "ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB"
for a = asc("A") to asc("D")
for b = asc("A") to asc("D")
for c = asc("A") to asc("D")
for d = asc("A") to asc("D")
x$ = chr$(a) + chr$(b) + chr$(c)+ chr$(d)
for i = 1 to 4 ' make sure each letter is unique
j = instr(x$,mid$(x$,i,1))
if instr(x$,mid$(x$,i,1),j + 1) <> 0 then goto [nxt]
next i
if instr(list$,x$) = 0 then print x$;" missing" ' found missing permutation
[nxt] next d
next c
next b
next a
{{out}}
DBAC missing
Scala
{{libheader|Scala}} {{works with|Scala|2.8}}
def fat(n: Int) = (2 to n).foldLeft(1)(_*_)
def perm[A](x: Int, a: Seq[A]): Seq[A] = if (x == 0) a else {
val n = a.size
val fatN1 = fat(n - 1)
val fatN = fatN1 * n
val p = x / fatN1 % fatN
val (before, Seq(el, after @ _*)) = a splitAt p
el +: perm(x % fatN1, before ++ after)
}
def findMissingPerm(start: String, perms: Array[String]): String = {
for {
i <- 0 until fat(start.size)
p = perm(i, start).mkString
} if (!perms.contains(p)) return p
""
}
val perms = """ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB""".stripMargin.split("\n")
println(findMissingPerm(perms(0), perms))
Scala 2.9.x
{{works with|Scala|2.9.1}}
println("missing perms: "+("ABCD".permutations.toSet
--"ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB".stripMargin.split(" ").toSet))
Seed7
$ include "seed7_05.s7i";
const func string: missingPermutation (in array string: perms) is func
result
var string: missing is "";
local
var integer: pos is 0;
var set of char: chSet is (set of char).EMPTY_SET;
var string: permutation is "";
var char: ch is ' ';
begin
if length(perms) <> 0 then
for key pos range perms[1] do
chSet := (set of char).EMPTY_SET;
for permutation range perms do
ch := permutation[pos];
if ch in chSet then
excl(chSet, ch);
else
incl(chSet, ch);
end if;
end for;
missing &:= min(chSet);
end for;
end if;
end func;
const proc: main is func
begin
writeln(missingPermutation([] ("ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD",
"ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA", "CBAD", "ABDC", "ADBC",
"BDCA", "DCBA", "BACD", "BADC", "BDAC", "CBDA", "DBCA", "DCAB")));
end func;
{{out}}
DBAC
Sidef
{{trans|Perl}}
func check_perm(arr) {
var hash = Hash()
hash.set_keys(arr...)
arr.each { |s|
{
var t = (s.substr(1) + s.substr(0, 1))
hash.has_key(t) || return t
} * s.len
}
}
var perms = %w(ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB)
say check_perm(perms)
{{out}}
DBAC
Tcl
{{tcllib|struct::list}}
package require struct::list
set have { \
ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC \
ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB \
}
struct::list foreachperm element {A B C D} {
set text [join $element ""]
if {$text ni $have} {
puts "Missing permutation(s): $text"
}
}
Ursala
The permutation generating function is imported from the standard library below and needn't be reinvented, but its definition is shown here in the interest of comparison with other solutions.
permutations = ~&itB^?a\~&aNC *=ahPfatPRD refer ^C/~&a ~&ar&& ~&arh2falrtPXPRD
The ~&j operator computes set differences.
#import std
#show+
main =
~&j/permutations'ABCD' -[
ABCD
CABD
ACDB
DACB
BCDA
ACBD
ADCB
CDAB
DABC
BCAD
CADB
CDBA
CBAD
ABDC
ADBC
BDCA
DCBA
BACD
BADC
BDAC
CBDA
DBCA
DCAB]-
{{out}}
DBAC
VBScript
Uses the 3rd method approach by adding the columns.
arrp = Array("ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD",_
"ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA",_
"CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD",_
"BADC", "BDAC", "CBDA", "DBCA", "DCAB")
Dim col(4)
'supposes that a complete column have 6 of each letter.
target = (6*Asc("A")) + (6*Asc("B")) + (6*Asc("C")) + (6*Asc("D"))
missing = ""
For i = 0 To UBound(arrp)
For j = 1 To 4
col(j) = col(j) + Asc(Mid(arrp(i),j,1))
Next
Next
For k = 1 To 4
n = target - col(k)
missing = missing & Chr(n)
Next
WScript.StdOut.WriteLine missing
{{Out}}
DBAC
XPL0
The list of permutations is input by using a command line like this: missperm <missperm.txt
code HexIn=26, HexOut=27;
int P, I;
[P:= 0;
for I:= 1 to 24-1 do P:= P xor HexIn(1);
HexOut(0, P);
]
{{out}}
0000DBAC
zkl
Since I just did the "generate the permutations" task, I'm going to use it to do the brute force solution.
var data=L("ABCD","CABD","ACDB","DACB","BCDA","ACBD","ADCB","CDAB",
"DABC","BCAD","CADB","CDBA","CBAD","ABDC","ADBC","BDCA",
"DCBA","BACD","BADC","BDAC","CBDA","DBCA","DCAB");
Utils.Helpers.permute(["A".."D"]).apply("concat").copy().remove(data.xplode());
Copy creates a read/write list from a read only list. xplode() pushes all elements of data as parameters to remove. {{out}}
L("DBAC")
ZX Spectrum Basic
10 LET l$="ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB"
20 LET length=LEN l$
30 FOR a= CODE "A" TO CODE "D"
40 FOR b= CODE "A" TO CODE "D"
50 FOR c= CODE "A" TO CODE "D"
60 FOR d= CODE "A" TO CODE "D"
70 LET x$=""
80 IF a=b OR a=c OR a=d OR b=c OR b=d OR c=d THEN GO TO 140
90 LET x$=CHR$ a+CHR$ b+CHR$ c+CHR$ d
100 FOR i=1 TO length STEP 5
110 IF x$=l$(i TO i+3) THEN GO TO 140
120 NEXT i
130 PRINT x$;" is missing"
140 NEXT d: NEXT c: NEXT b: NEXT a