Code Golf: 6174 - Kaprekar's mythical constant

Question

Why is the number 6174 so interesting? As defined by Wikipedia

Take any four-digit number, using at least two different digits. (Leading zeros are allowed.)

Arrange the digits in ascending and then in descending order to get two four-digit numbers, adding leading zeros if necessary.

Subtract the smaller number from the bigger number.

Go back to step 2.

The above process, known as Kaprekar's routine, will always reach 6174 in at most 7 iterations. Once 6174 is reached, the process will continue yielding it.

Write a program which runs the Kaprekar's routine against a given four-digit number (see definition above) printing out each step of the routine.

Rules:

Submissions must be complete programs.
Input must be read from standard input. Piping from echo is OK.
Input should be in numeric form.
Printing out leading zeros is required. (See examples below.)
Last line should say how many iterations were needed. Punctuation is required.

Examples:

> 2607
7620 - 0267 = 7353
7533 - 3357 = 4176
7641 - 1467 = 6174
Iterations: 3.

> 1211
2111 - 1112 = 0999
9990 - 0999 = 8991
9981 - 1899 = 8082
8820 - 0288 = 8532
8532 - 2358 = 6174
Iterations: 5.

> 6174
7641 - 1467 = 6174
Iterations: 1.

Any programming language is welcome. Extra points for esoteric ones + a small bounty.

Update 1: There is already a similar question.

Update 2: Added example for 6174 as input. Thanks to Peter Taylor for the notice.

@lunohodov If you already have 6174 as input, do you really need one iteration to reach 6174? I only ask because I'll have to write a special case in my program if you do. :-( — Gareth, Jun 10, 2011 at 15:31
@Gareth Yes, as the task is, as defined above, to run the Kaprekar's routine against a given four-digit number and 6174 should be treated as such. — lunohodov, Jun 12, 2011 at 10:15
This is the exact same question I had to do in the NCSS challenge. If you're doing this to get other people to give you the answer in the NCSS challenge, then that's pretty bad and it defeats the purpose of a challenge. If you're not, then I apolagize. — user6674, Oct 17, 2012 at 4:19
This question was posted in the middle of last year - I very much doubt anyone was trying to cheat for some test or challenge. — Gareth, Oct 20, 2012 at 8:38

Community · Accepted Answer · 2020-06-17 09:04:33Z

10

Ruby 1.9, 122 characters

puts"Iterations: #{(1..7).find{s=$_.chars.sort*"";puts [r=s.reverse,?-,s,?=,$_="%04d"%(r.to_i-s.to_i)]*" ";~/6174/}}."

Example invocation:

$ echo 1211 | ruby -ln kaprekar.rb

I've counted the -ln flag as 4 characters (difference between the normal invocation ruby kaprekar.rb and ruby -ln kaprekar.rb).

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 7, 2011 at 10:40

Ventero

10.1k3 gold badges24 silver badges55 bronze badges

\$\begingroup\$ I saved this script as kaprekar.rb then invoked it with ruby -lp kaprekar.rb. Entered a number and pressed <Enter> but the output is the entered number itself. Clearly I am missing something... Please advise how to use the script. \$\endgroup\$
– lunohodov
Jun 16, 2011 at 12:52
\$\begingroup\$ @lunohodov: I've added an example invocation. It now also generates the correct output for 6174 as input, which unfortunately brings this solution up to 128 characters. \$\endgroup\$
– Ventero
Jun 16, 2011 at 21:54
\$\begingroup\$ Using echo 1234 | ruby kaprekar.rb raises a warning and ends with an error undefined method 'chars' for nil:NilClass (NoMethodError). Executing echo 1234 | ruby -lp kaprekar.rb issues only a warning and behaves as expected. The output is not as expected, as it contains a warning message kaprekar.rb:3: warning: regex literal in condition \$\endgroup\$
– lunohodov
Jun 17, 2011 at 9:12
\$\begingroup\$ @lunohodov: Fixed the warning and the example invocation. \$\endgroup\$
– Ventero
Jun 17, 2011 at 9:30

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Community · Accepted Answer · 2020-06-17 09:04:33Z

9

Perl - 147 143 134 130 129 126 129 128 126

for($_=<>;$_-6174+!$c;$c++){$_=reverse$d=join'',sort split//,"$_"
|$|x4;printf"$_ - $d = %04d\n",$_-=$d}die"Iterations: $c.\n"

EDIT: Now complies with 6174 case, at the cost of a few chars... run with echo -n <number> | perl kaprekar.pl

EDIT: Finally back to where I was before :D

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 7, 2011 at 5:23

swilliams

4862 silver badges4 bronze badges

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Martin Ueding · Accepted Answer · 2011-06-08 11:55:16Z

7

Python, 141 chars

n=input()
i=0
while n-6174:a=''.join(sorted("%04d"%n));b=a[::-1];n=int(b)-int(a);print"%s - %s = %04d"%(b,a,n);i+=1
print"Iterations: %d."%i

edited Jun 8, 2011 at 11:55

answered Jun 7, 2011 at 20:25

Martin Ueding

1716 bronze badges

\$\begingroup\$ +1 for slick padding with %04d. I learned something today! \$\endgroup\$
– arrdem
Jun 7, 2011 at 20:49
3

\$\begingroup\$ A few suggestions: put the whole loop on one line using ;s. while n-6174. No space between print and the quote. \$\endgroup\$
– Keith Randall
Jun 8, 2011 at 3:31
\$\begingroup\$ @keith-randall: thanks, got it down to 141 now. \$\endgroup\$
– Martin Ueding
Jun 8, 2011 at 11:55

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Community · Accepted Answer · 2020-06-17 09:04:33Z

6

+25

Golfscript, 74 characters

);:|;{0):0;|$:§-1%" - "§" = ""0"4$~§~-+-4>:|n|6174`=!}do"Iterations: "0"."

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 10, 2011 at 10:01

Ventero

10.1k3 gold badges24 silver badges55 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

5

><> - 268 308

</&4pff1
v>i86*-:n&1-:&?!
>ao&        v
<v&0pff+1gff
 >&1+:4=   ?v&:a%:}-a,
 v&8[4r::0}~<
 >&1-:?!v&:@@:@(?$}&:&3%1=?}
 v      >~:}}:}@:}$:}
 \:n}:n}:n}:n}' - 'ooo:n}:n}:n}:n}' = 'ooo
 \a*+a*+a*+}a*+a*+a*+-:0&\
 v?       =4&:+1&,a-}:%a:<
/\&~~rnnnnao:29777****=   ?v
voooooooooooo"Iterations: "/
\ffgna'.'oo;

Not much of a contender for golf, but it was fun to write. :)

~~Run with ./fish.py kaprekar.fish -v <number>~~
EDIT: Now takes input from STDIN.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 10, 2011 at 17:01

swilliams

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Rotsor · Accepted Answer · 2011-06-20 20:28:36Z

5

Haskell, 197 192 182 181 characters

import List
p=putStrLn.unwords
"6174"%k|k>0=p["Iterations:",shows k"."]
n%k=p[b,"-",a,"=",c]>>c%(k+1)where a=sort n;b=reverse a;c=take 4$shows(read b-read a)"0"
main=getLine>>=(%0)

edited Jun 20, 2011 at 20:28

Rotsor

3551 silver badge7 bronze badges

answered Jun 9, 2011 at 16:54

hammar

4,21222 silver badges26 bronze badges

\$\begingroup\$ Inlining r and s saves 2 characters. Also, "000" is redundant. "0" is enough. This brings us to 188 characters. I'm surprised interact does not help here. It usually does. \$\endgroup\$
– Rotsor
Jun 20, 2011 at 4:04
\$\begingroup\$ Replacing show x++s with shows x s gains 2 more bytes. 186 now. \$\endgroup\$
– Rotsor
Jun 20, 2011 at 4:28
\$\begingroup\$ By using pattern guards (|k>0) one can get rid of f. Further renaming g to % gets us to 182 chars. \$\endgroup\$
– Rotsor
Jun 20, 2011 at 5:18

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Casey Chu · Accepted Answer · 2011-06-18 06:26:35Z

4

JavaScript, 189 182 165 chars

Credit to DocMax:

for(n=prompt(i=o=e='');!i--|n-6174;o+=n+' - '+a+' = '+(n=(p=n-a+e)[3]?p:0+p)+'\n')
  b=n.split(e).sort(),n=b.reverse(a=b.join(e)).join(e);
alert(o+"Iterations: "+~i+'.')

Original:

for(n=prompt(i=o=e='');n-6174;o+=(i++?n+"\n":e)+(n=(a=n.split(e).sort().join(e)).split(e).reverse().join(e))+' - '+a+' = ',n=n-a+e)while(!n[3])n=0+n
alert(o+n+"\nIterations: "+i+'.')

Ungolfed:

var i = 0;
var n = prompt();
var out = '';
while (n != 6174) {
    while ((n=''+n).length<4) n='0'+n // pad number
    if(i)out+=n+"\n"

    a = n.split('').sort().join('');
    n = a.split('').reverse().join('');

    out += n + ' - ' + a + ' = '
    n-=a
    i++;
}
console.log(out + "6174\nIterations: " + i + '.');

edited Jun 18, 2011 at 6:26

answered Jun 10, 2011 at 8:41

Casey Chu

1,78112 silver badges10 bronze badges

1

\$\begingroup\$ I think that you can change n != 6174 to n-6174 since it will return zero, which is false (at least in C and Python). \$\endgroup\$
– Martin Ueding
Jun 10, 2011 at 12:56
\$\begingroup\$ The credit should go to keith-randall who suggested it for my Python solution. \$\endgroup\$
– Martin Ueding
Jun 10, 2011 at 16:57
\$\begingroup\$ You can save 5 more characters by replacing while(n.length<4) with while(!n[3]). \$\endgroup\$
– DocMax
Jun 16, 2011 at 3:27
1

\$\begingroup\$ I can't stop staring at this one! The following a) fixes the output when n=6174, b) rearranges when n+'\n' is added to avoid the conditional and an extra \n, c) uses a temp to avoid a join-split-join sequence, d) takes advantage of the fact that we only ever have to add a single '0' for padding: for(n=prompt(i=0,o=e='');n-6174;i++,o+=(n=(b=n.split(e).sort(),a=b.join(e),b).reverse().join(e))+' - '+a+' = '+(n=('0'+(n-a)).slice(-4))+'\n');alert(o+"Iterations: "+i+'.') which should be 172 chars. \$\endgroup\$
– DocMax
Jun 17, 2011 at 6:31
1

\$\begingroup\$ Impressive! According to the spec above though, when n=6174, it has to go through at least one iteration, so I added a check if i is 0 (+4) but combined that with the i++. Unfortunately, that gives an off by one error, so I switched the increment to a decrement and then used a little bitwise trickery at the end (-1). Then I changed i=0,o=e='' to i=o=e='' (-2), reformatted the for loop to avoid extra parentheses (-1), expanded (b=...,a=...,b) bit (-2), and sneaked a=b.join inside the reverse() call (-1). So 169, not bad! \$\endgroup\$
– Casey Chu
Jun 17, 2011 at 8:00

| Show 3 more comments

Community · Accepted Answer · 2020-06-17 09:04:33Z

3

PowerShell, 125 128 130 131

for($a,$OFS=$input+'';$b-6174;++$i){$a=$b=+($c=''+($x="$a 000"[0..4]|sort)[4..0])-"$x"
"$c-$x = {0:d4}"-f$a}"Iterations: $i."

Passes all test cases from the question.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 10, 2011 at 9:29

Joey

13k2 gold badges43 silver badges73 bronze badges

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pimvdb · Accepted Answer · 2011-06-07 09:12:46Z

2

JavaScript, 260 bytes

function z(c){for(u=c+y;u.length<4;)u=0+u;return u}for(p=prompt(i=0,r=y="");;)
if(s=(p+y).split(y).sort(),t=s.concat().reverse(),a=s.join(y),b=t.join(y),q=a<b?b:a,
w=a<b?a:b,p=z(q-w),i++,r+=z(q)+" - "+z(w)+" = "+p+"\n",p==6174)break;alert(r+
"Iterations: "+i+".")

answered Jun 7, 2011 at 9:12

pimvdb

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kotarak · Accepted Answer · 2011-06-08 14:01:22Z

2

Clojure, 256 characters

(let[i #(Integer/parseInt%)f #(format"%04d"%)a #(->>% f sort(apply str)i)d #(->>% f sort reverse(apply str)i)k #(let[u(d %)l(a %)n(- u l)](println(f u)"-"(f l)"="(f n))n)](while true(println"Iterations:"(count(take-while #(not=% 6174)(iterate k(read)))))))

answered Jun 8, 2011 at 14:01

kotarak

1211 bronze badge

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Gareth · Accepted Answer · 2011-06-12 10:32:06Z

2

Scala 2.9, 194 characters

object K extends App{var(c,s)=(0,args(0));do{var d=s.sorted;var e=d.reverse.toInt-d.toInt;s="%04d".format(e);println(d.reverse+" - "+d+" = "+s);c+=1}while(s!="6174");print("Iterations: "+c+".")}

Makes use of the App trait from Scala 2.9.

Edit: gives correct output for initial input of 6174.

edited Jun 12, 2011 at 10:32

answered Jun 8, 2011 at 18:05

Gareth

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rintaun · Accepted Answer · 2011-06-15 08:32:21Z

2

PHP, 215 259 276 characters

<?php echo">";$n=str_split(str_pad(trim(fgets(STDIN)),4,0,0));for($k=0,$z=0;$k-6174;$z++){sort($n);$a=implode($n);$b=strrev($a);$k=str_pad($b-$a,4,0,0);echo"$b - $a = $k\n";$n=str_split($k);}echo"Iterations: $z\n";

Ungolfed:

<?php
echo ">";
$n = str_split(str_pad(trim(fgets(STDIN)),4,0,0));
for($k=0, $z=0; $k-6174; $z++) {
    sort($n);
    $a = implode($n);
    $b = strrev($a);
    $k = str_pad($b-$a,4,0,0);
    echo "$b - $a = $k\n";
    $n = str_split($k);
}
echo "Iterations: $z\n";

edited Jun 15, 2011 at 8:32

answered Jun 14, 2011 at 20:59

rintaun

7814 silver badges6 bronze badges

\$\begingroup\$ I don't think you need the abs, max and min functions, since the sort will always mean that $b is greater than $a. That could save you 20ish characters. Also, I think that putting the sort inside the loop at the top will mean you only need to have it in your code once which will save you another 9. \$\endgroup\$
– Gareth
Jun 15, 2011 at 8:05
\$\begingroup\$ Wow, I guess I got distracted by the "subtract the smaller number from the bigger number" instruction. Thanks. \$\endgroup\$
– rintaun
Jun 15, 2011 at 8:25
\$\begingroup\$ <?function k($c){echo"> $c\n";$n=str_split(str_pad($c,4,0,0));for(;$k-6174;$z++){sort($n);$a=join($n);$b=strrev($a);$k=str_pad($b-$a,4,0,0);echo"$b - $a = $k\n";$n=str_split($k);}echo"Iterations: $z\n";} You can save 12 characters by changing your for statement, calling this as a function and using join instead of implode. \$\endgroup\$
– TwoScoopsofPig
Oct 17, 2012 at 19:02
\$\begingroup\$ Also, I hate mini-markdown. \$\endgroup\$
– TwoScoopsofPig
Oct 17, 2012 at 19:03

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Jonas Elfström · Accepted Answer · 2011-06-17 12:21:49Z

2

CoffeeScript, 233 225 characters

o=e='';i=0;n=prompt()
while n!=6174
  n=e+n;q=(n='0'+n if !n[3]) for x in [0..2];n?=q;o+=n+"\n" if i;a=n.split(e).sort().join(e);n=a.split(e).reverse().join(e);o+=n+' - '+a+' = ';n-=a;i++
alert(o+"6174\nIterations: "+i+'.')

Try it here or with instructions here.

edited Jun 17, 2011 at 12:21

answered Jun 16, 2011 at 11:47

Jonas Elfström

4231 gold badge4 silver badges10 bronze badges

\$\begingroup\$ My browser freezes - had to cancel the execution of the script. \$\endgroup\$
– lunohodov
Jun 16, 2011 at 17:31
\$\begingroup\$ What number did you enter? I tried it in Firefox and Chrome for 4711 and 1 and a couple of others. \$\endgroup\$
– Jonas Elfström
Jun 16, 2011 at 20:19
\$\begingroup\$ Using 0 (as suggested by the prompt) or clicking the cancel button causes Safari to freeze. \$\endgroup\$
– lunohodov
Jun 17, 2011 at 9:17
\$\begingroup\$ I don't know why it suggested that. You have to enter a number between 1 and 9998 whose digits are not all identical. 0 is the same as 0000 and will cause an infinite loop. It seems most of the solutions here skipped validating input to keep the number of characters down. \$\endgroup\$
– Jonas Elfström
Jun 17, 2011 at 11:43
\$\begingroup\$ See i56.tinypic.com/bhhoqe.png Your output also ends with "It took 5 iterations to reach Kaprekar's constant." which does not conform to the requirements. \$\endgroup\$
– lunohodov
Jun 17, 2011 at 12:12

| Show 1 more comment

user unknown · Accepted Answer · 2011-06-22 09:11:55Z

Scala 276

object o{var i=0;def a(v:String){val c=v.toList.sortWith(_>_).mkString;val b=c.reverse;val d=c.toInt-b.toInt;val e="0"*(4-(d+"").length)+d;val p=c+" - "+b+" = "+e;if(d!=6174){println(p);i=i+1;a(e)}else{println(p+"\nIterations: "+(i+1)+".")}};def main(s:Array[String])=a(s(0))}

Scala 283

object o{var i=0;def a(v:String){val c=v.toList.sortWith(_>_).mkString;val b=c.reverse;val d=c.toInt-b.toInt;val e="0"*(4-(d+"").length)+d;val p=c+" - "+b+" = "+e;if(d!=6174){println(p);i=i+1;a(e)}else{println(p);println("Iterations: "+(i+1)+".")}};def main(s:Array[String])=a(s(0))}

diff:

else{println(p);println("Iterations: "+(i+1)+".")}};
// to
else{println(p+"\nIterations: "+(i+1)+".")}};

Dan Andreatta · Accepted Answer · 2011-06-26 07:51:47Z

2

GAWK - 152 chars

This is a GNU awk version. It may not work with other non-gnu versions.

{for(z=$1;z-6174+!c;++k){split(z,a,"");asort(a);for(b=c=i=0;i<4;z=c-b){c+=a[i+1]*10^i;b=b*10+a[++i]}printf c" - %.4d = "z"\n",b}print"Iterations: "k"."}

$ awk -f k.awk <<< 9992
2999 - 9992 = 6993
3699 - 9963 = 6264
2466 - 6642 = 4176
1467 - 7641 = 6174
Iterations: 4

edited Jun 26, 2011 at 7:51

answered Jun 23, 2011 at 15:40

Dan Andreatta

2312 silver badges2 bronze badges

\$\begingroup\$ I receive awk: calling undefined function asort. Awk version is 20070501 running on OSX 10.6.7. Don't forget the . after the number of iterations. \$\endgroup\$
– lunohodov
Jun 25, 2011 at 11:40
\$\begingroup\$ lunohodov@: Added missing point. Also, I have used gnu awk (gawk), and that may explain the missing function. \$\endgroup\$
– Dan Andreatta
Jun 26, 2011 at 7:53
\$\begingroup\$ The subtraction numbers are the wrong way round: e.g. should be 9992 - 2999 = 6993 \$\endgroup\$
– Chris Degnen
Jan 12, 2014 at 21:10

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Joaquim Rendeiro · Accepted Answer · 2011-07-02 20:09:53Z

2

Ruby, 179 chars but posting anyway

s=gets.chomp
n=0
begin
  s=s.to_s.chars.sort.reduce{|s,c|s+c}.rjust(4,'0')
  a=s.reverse
  puts"#{a} - #{s} = #{'%04d'%(s=a.to_i-s.to_i)}"
  n+=1
end while s!=6174
puts"Iterations: #{n}."

answered Jul 2, 2011 at 20:09

Joaquim Rendeiro

1312 bronze badges

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tmartin · Accepted Answer · 2012-10-17 11:20:01Z

2

K, 104

{b::();{b,:,k," = ",r:"0"^(-:4)$$. k:(x@>x)," - ",x@<x;r}\[$x];-1'c,,"Iterations: ",$#c:$[1=#b;b;-1_b];}

Test cases

k){b::();{b,:,k," = ",r:"0"^(-:4)$$. k:(x@>x)," - ",x@<x;r}\[$x];-1'c,,"Iterations: ",$#c:$[1=#b;b;-1_b];}'2607 1211 6174;
7620 - 0267 = 7353
7533 - 3357 = 4176
7641 - 1467 = 6174
Iterations: 3
2111 - 1112 = 0999
9990 - 0999 = 8991
9981 - 1899 = 8082
8820 - 0288 = 8532
8532 - 2358 = 6174
Iterations: 5
7641 - 1467 = 6174
Iterations: 1

answered Oct 17, 2012 at 11:20

tmartin

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Aman ZeeK Verma · Accepted Answer · 2011-06-14 17:38:46Z

1

PERL

chomp($n=<STDIN>);
    do{
       $t++;
       $desc=join('',reverse sort split(//,$n));
       $asc=join('', sort split(//,$n));
       $n=($desc - $asc);
       for($i=4;$i>length $n;$i--){
          $n="0".$n;
       }
       print $desc." - ".$asc." = ".$n."\n";
       $n="6174" if $n eq "0000";
    }while($n ne "6174");
    print "Iterations: $t.\n";

answered Jun 14, 2011 at 17:38

Aman ZeeK Verma

6394 silver badges11 bronze badges

\$\begingroup\$ Thats ~310 chars... \$\endgroup\$
– Aman ZeeK Verma
Jun 14, 2011 at 17:40

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Chris Degnen · Accepted Answer · 2019-03-23 10:59:43Z

Mathematica, 314 291 characters

This is the program, kaprekar.m :-

SetOptions[$Output,FormatType->OutputForm];
x=$ScriptCommandLine[[2]];
f[x_]:=(a=Characters@x;
b=Sort@ToExpression@a;
c=Sort[FromDigits/@{#,Reverse@#}&@b];
{c,{b,a}}=IntegerString[{#2-#&@@c,c},10,4];
Print[a," - ",b," = ",c];c)
x=f@x;
e=NestWhileList[f,x,#!="6174"&];
Print["Iterations: ",N@Length@e]

Setting the path prior to running :-

$ PATH=${PATH}:/Applications/Mathematica.app/Contents/MacOS ; export PATH

Running the program :-

$ MathematicaScript -script kaprekar.m 2607
7620 - 0267 = 7353
7533 - 3357 = 4176
7641 - 1467 = 6174
Iterations: 3.
$ MathematicaScript -script kaprekar.m 1211
2111 - 1112 = 0999
9990 - 0999 = 8991
9981 - 1899 = 8082
8820 - 0288 = 8532
8532 - 2358 = 6174
Iterations: 5.
$ MathematicaScript -script kaprekar.m 6174
7641 - 1467 = 6174
Iterations: 1.

640KB · Accepted Answer · 2019-03-21 17:44:43Z

0

PHP, 160 bytes

function k($n,$d=1){$o=str_split($n);sort($o);echo$q=strrev($r=join($o))," - $r = ",$n=str_pad($q-$r,4,0,0),"
",$n==6174?"Iterations: $d.":k($n,++$d);}k($argn);

Try it online!

Complete program, input is STDIN, run with php -nF.

Output

> echo 2607|php -nF kap.php
7620 - 0267 = 7353
7533 - 3357 = 4176
7641 - 1467 = 6174
Iterations: 3.

> echo 1211|php -nF kap.php
2111 - 1112 = 0999
9990 - 0999 = 8991
9981 - 1899 = 8082
8820 - 0288 = 8532
8532 - 2358 = 6174
Iterations: 5.

> echo 6174|php -nF kap.php
7641 - 1467 = 6174
Iterations: 1.

edited Mar 21, 2019 at 17:44

answered Mar 21, 2019 at 17:16

640KB

11.8k2 gold badges35 silver badges57 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Rust - 375 bytes

use std::io::{self,BufRead};fn main() {let mut x=io::stdin().lock().lines().next().unwrap().unwrap().parse::<i16>().unwrap();let mut n=0;println!("Iterations: {}.",loop {let mut v=[x/1000%10,x/100%10,x/10%10,x%10];v.sort();let j=v.iter().fold(0,|a,i|a*10+i);let k=v.iter().rev().fold(0,|a,i|a*10+i);x=k-j;n+=1;println!("{:04} - {:04} = {:04}",k,j,x);if x==6174{break n};});}

I present this as a possible "upper bound", I challenge anyone to find a language where a reasonable implementation of this is longer - as in there is nothing superfluous, but also nothing even remotely obvious that would shrink it significantly. The thing about Rust is that it takes about 120 characters just to read from stdin and parse into an integer. "Oh but then just use the string representation"... but im 99% confident that would be even longer

Jo King · Accepted Answer · 2019-03-22 05:50:45Z

Perl 6 -n flag, 105 bytes

say "Iterations: "~+.&{{{say $!=.flip~" - $_"," = ",($/=$!.EVAL.fmt("%04d"));$/}([~] .comb.sort)}...6174}

Try it online!

I finally got to use my {}...* trick, since we need to have at least one iteration for 6174. I'm not sure why I need the extra wrapping .&{ } around the sequence though, which kinda sucks.

Explanation:

    .&{                         } # Call a function on the input
       {                }...6174  # Create a sequence that goes until 6174
        {           }([~] .comb.sort) # Get the sorted digits of the number
         say $!=.flip~" - $_"," = "~($/=$!.EVAL.fmt("%04d"))  # Print the iteration
                        ;$/  # Return the result
say "Iterations: "~+.&{      }     # Print the number of iterations

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Code Golf: 6174 - Kaprekar's mythical constant

22 Answers 22

Ruby 1.9, 122 characters

Perl - 147 143 134 130 129 126 129 128 126

Python, 141 chars

Golfscript, 74 characters

><> - 268 308

Haskell, 197 192 182 181 characters

JavaScript, 189 182 165 chars

PowerShell, 125 128 130 131

JavaScript, 260 bytes

Clojure, 256 characters

Scala 2.9, 194 characters

PHP, 215 259 276 characters

CoffeeScript, 233 225 characters

GAWK - 152 chars

Ruby, 179 chars but posting anyway

K, 104

Mathematica, 314 291 characters

PHP, 160 bytes

Rust - 375 bytes

Perl 6 -n flag, 105 bytes

Explanation:

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Code Golf: 6174 - Kaprekar's mythical constant

22 Answers 22

Ruby 1.9, 122 characters

Perl - 147 143 134 130 129 126 129 128 126

Python, 141 chars

Golfscript, 74 characters

><> - 268 308

Haskell, 197 192 182 181 characters

JavaScript, 189 182 165 chars

PowerShell, 125 128 130 131

JavaScript, 260 bytes

Clojure, 256 characters

Scala 2.9, 194 characters

PHP, 215 259 276 characters

CoffeeScript, 233 225 characters

GAWK - 152 chars

Ruby, 179 chars but posting anyway

K, 104

Mathematica, 314 291 characters

PHP, 160 bytes

Rust - 375 bytes

Perl 6 -n flag, 105 bytes

Explanation:

Your Answer

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