196 algorithm code golf

Question

Write a short program for 196-algorithm. The algorithm starts from an integer, then adds its reverse to it until a palindrome is reached.

e.g.

input = 5280
5280 + 0825 = 6105
6105 + 5016 = 11121
11121 + 12111 = 23232
output = 23232

Input

an integer, which is not a lyrchrel number (that is, it does eventually yield a palindrome under this algorithm, rather than continuing infinitely)

Output

the palindrome reached.

Answer 1

APL (22 characters)

{a≡⌽a←⍕(⍎⍵)+⍎⌽⍵:a⋄∇a}⍞

This works in Dyalog APL. Here's an explanation, from right to left:

• { ... }⍞: Get input from the user as characters (⍞) and feed it to our function ({ ... }).
• Within the direct function (⋄ separates statements, so we look at them from left to right):
  • a≡⌽a←⍕(⍎⍵)+⍎⌽⍵ : a: Evaluate (⍎) the right argument's (⍵) reverse (⌽), and add that to the evaluated version of the right argument itself. Then, format the result (⍕; i.e., give its character representation), assign (←) that to the variable a, and finally test if a's reverse is equivalent to a (i.e., is a a palindrome?). If true, return a; otherwise...
  • ∇a: Feed a back into our function (∇ is implicit self-reference).

Example:

      {a≡⌽a←⍕(⍎⍵)+⍎⌽⍵:a⋄∇a}⍞
412371
585585

Answer 2

Python 2, 55 bytes

Following JPvdMerwe suggestion:

n=input()
while`n`!=`n`[::-1]:n+=int(`n`[::-1])
print n

Python 2, 62:

n=raw_input()
while n!=n[::-1]:n=`int(n)+int(n[::-1])`
print n

Answer 3

GolfScript, 29 chars

~]{{.`.-1%.@={;0}{~+1}if}do}%

---

Selected commentary

The meat of the program is the do loop, of course. So I'll just cover that.

1. .` copies the number and stringifies it.
2. .-1% copies that string version and reverses it.
3. .@ copies the reversed version, and brings the original non-reversed version to the front.

So, say, the number is 5280. At this stage, the stack is: 5280 "0825" "0825" "5280". The stage is set for the comparison. (After the comparison, the stack will be left at 5280 "0825" no matter what---the items to compare have been popped off.)

1. If the string and the reverse are the same, we don't care about the reversed string, so just pop it off (;) and return 0 (to end the do loop).
2. If they don't match, then evaluate (~) the reversed string (to make it a number), add (+) that to the original number, and return 1 (to continue the do loop).

Answer 4

Ruby — 56 chars

x,r=gets
x="#{x.to_i+r.to_i}"until x==r=x.reverse
puts x

Answer 5

Just exercising my Pyth skills, not a serious contender.

Pyth, 16 bytes

L?bqb_by`+vbi_bTyz

Equivalent to Python 3:

y=lambda b:b if b==b[::-1] else y(str(eval(b)+int(b[::-1],10)))
print y(input())

Answer 6

J 25 ^{27 31}

f=:(+g)^:(~:g=.|.&.":)^:_
e.g.
f 5280
23232

Answer 7

CJam, 22 21 bytes

CJam was created after this question was asked, so technically its an invalid submission. But I found the question interesting, so here goes:

r{__W%:X=0{~X~+s1}?}g

Explanation:

r{                 }g    "Read the input number as string and enter a while-do loop";
  __                     "Make 2 copies of the string number";
    W%:X                 "Reverse the second and store it in X";
        =                "Check if the number is already palindrome";
         0{      }?      "Put 0 on stack if it is palindrome, else, run the code block";
           ~             "Convert the string to int";
            X~           "Put the reverse string on stack and convert it to int too";
              +s         "Add the two numbers and covert back the result to string";

The core logic is that in each while-do iteration, you first check if palindrome is achieved or not. If not, add the reverse to the number. Pretty much what the algorithm is!

Try it online here

Answer 8

This is an actual contender, since J has been around for decades.

J (16 bytes)

(+^:~:|.&.":)^:_

This is a verb, so it can be assigned to a variable in a J session and used like so:

   f =. (+^:~:|.&.":)^:_
   f 5280
23232

How it works:

(+^:~:|.&.":)^:_
 +^:~:           add if unequal
      |.&.":     reverse under string format
 +^:~:|.&.":     add reverse unless a palindrome
(           )^:_ repeat until unchanged

Answer 9

Python: 66

n=input()
while 1:
 r=int(`n`[::-1])
 if n==r:break
 n+=r
print n

Answer 10

Brachylog, 8 bytes

↔?|↔;?+↰

Try it online!

Somewhat similar to one of the first Brachylog programs I saw and was intrigued by, from the Brachylog introduction video.

?↔?.|↔;?+↰.  (initial ? and the .s are implicit)

?↔?          % If the input and its reverse are the same,
   .         %   then the input is the output
    |↔;?+↰   % Or, add the input and its reverse, and call this predicate recursively
          .  % The result of that is the output

Answer 11

Scala 82

def p(n:Int):Int={val s=n.toString.reverse.toInt
if(n==s)n else p(n+s)}
p(readInt)

Answer 12

JAGL Alpha 1.2 - 19, 21 with stdin

Not contending, just getting some experience with my language
Expects a number from stdin

T~d{DddgCi+dgdC=n}uSP

Explanation

T~                       Get a line of input, and eval to an integer
  d                      Duplicate (for first round)
   {Ddd                  Drop last and duplicate twice
       gCi               Convert to string, reverse, and convert back to integer
          +d             Add to original and duplicate
            gdC          Convert to string, duplicate, reverse
               =n}       If it isn't a palindrome, keep going
                  uSP    Run until palindrome reached, then print output number

Answer 13

Perl, 40 chars

$_=<>;$_+=$r while$_!=($r=reverse);print

Answer 14

05AB1E, 7 bytes (non-competing)

Non-competing, since the language postdates the challenge.

Code:

[DÂQ#Â+

Explanation:

[        # Infinite loop.
 DÂ      # Duplicate and bifurcate (which duplicates it and reverses the duplicate).
   Q#    # If the number and the number reversed are equal, break.
     Â+  # Add the reversed number to the initial number.

Uses the CP-1252 encoding. Try it online!.

Answer 15

Jelly, 9 bytes (non competing)

A very simple answer, just for the challenge of coding in and esoteric language.

ṚḌ+µŒḂ¬$¿

ṚḌ        : take the argument, reverse (and vectorize) it
  +µ      : add both
    ŒḂ¬$¿ : while the result isn't a palindrome

Try it online!

Should this answer be unclear or wrong on any level, feel free to point it out.

Thanks to Dennis for helping me out with this first small piece of code.

Answer 16

In Q (39 characters)

f:{while[x<>g:"I"$reverse -3!x;x+:g];x}

Sample Usage:

q)f 5280
23232

Edit:

Down to 34 now, same usage:

{while[x<>g:"I"$(|) -3!x;x+:g];x} 5280

Answer 17

C# - 103 99 chars

public int P(int i)
{
    var r = int.Parse(new string(i.ToString().Reverse().ToArray())));
    return r == i ? i : P(i + r);        
}

C# never does very well in golf. Elegant, but verbose.

Answer 18

Bash (64)

X=`rev<<<$1|sed s/^0*//`;[ $1 = $X ]&&echo $1||. $0 $(($1+$X))

Call with: bash <filename> <number>

Answer 19

r=input()
while 1:
    r=`r`
    if r==r[::-1]:
      break
    else:
      r=int(r)+int(r[::-1])

print r

Answer 20

PHP - 54 48 characters

<?for($s=`cat`;$s!=$r=strrev($s);$s+=$r);echo$s;

Test:

$ php 196.php <<< 5280
23232

Answer 21

Add++, 57 bytes

L,BDbRBv
D,f,@,1]A+
D,g,@,1]A=
D,h,@,{f}A{g}Q{h}
L,{h}2/i

Try it online!

How it works

L,	; Create a function 'lambda 1'
	; Example argument:   [5280]
    BD	; Digits;     STACK = [[5 2 8 0]]
    bR	; Reverse;    STACK = [[0 8 2 5]]
    Bv	; Undigits;   STACK = [825]

D,f,@,	; Define a monadic function 'f'
	; Example argument:         [5280]
    1]	; Call 'lambda 1';  STACK = [825]
    A+	; Add the argument; STACK = [6105]

D,g,@,	; Define a monadic function 'g'
	; Example argument:          [5280]
    1]	; Call 'lambda 1';   STACK = [825]
    A=	; Equal to argument; STACK = [0]

D,h,@,	; Define a monadic function 'h'
	; Example argument:  [5280]
   {f}	; Call 'f';  STACK = [6105]
   A{g}	; If 'g'...
   Q	;   Return
   {h}	; Else call 'h'

L,	; Define a function, 'lambda 2'
	; Example argument: [5280]
   {h}	; Call 'h'; STACK = [46464]
   2/i	; Halve;    STACK = [23232]

Answer 22

Python. 85 characters:

s,i=str,int;rs=lambda q:s(q)[::-1];n=i(input());
while rs(n)!=s(n):n+=i(rs(n));print n

If you don't want output on each iteration:

s,i=str,int;rs=lambda q:s(q)[::-1];n=i(input());
while rs(n)!=s(n):n+=i(rs(n))
print n

(one less character)

Answer 23

GNU dc, 46 bytes

Requires GNU dc, min version 1.4 (for R command).

[O~3RO*+rd0<R]sR[+lfx]sg[ddO~rd0<R+d3R!=g]dsfx

Input and output are top-of-stack, as usual. It takes a surprising amount of code to reverse digits in dc (unless I'm missing something, which is far from impossible). It does have the numeric range to behave nicely with inputs such as these (which will overflow 32-bit unsigned arithmetic, for example):

• 89 ⇒ 8,813,200,023,188
• 8997 ⇒ 16,668,488,486,661
• 10677 ⇒ 4,668,731,596,684,224,866,951,378,664

Explanation

# Reverse digits, starting after first digit extracted
[O~3RO*+r d0<R]sR

# Recursion helper - add and recurse
[+ lfx]sg

# Main recursive function
[dd O~r d0<R+ d3R !=g]dsfx

Answer 24

R, 193 109 105 bytes

-84 bytes thanks to Giuseppe! -4 byes thanks to JayCe!

function(x){"-"=utf8ToInt
S=strtoi
"!"=rev
while({z=paste(S(x)+S(intToUtf8(!-x),10));any(-z!=!-z)})x=z
z}

Try it online!

Answer 25

Powershell, 63 62 bytes

-1 byte thanks to @AdmBorkBork

param($m)for(;$m-$s;$m=-join"$s"[-1..-"$s".Length]){$s+=$m};$s

Test script:

$f = {
param($m)for(;$m-$s;$m=-join"$s"[-1..-"$s".Length]){$s+=$m};$s
}

&$f 5280

Answer 26

Windows PowerShell (63)

for($a=+"$input";-join"$a"[99..0]-ne$a){$a+=-join"$a"[99..0]}$a

I still hate it that there is no easy way to reverse a string.

Answer 27

Haskell 89 87 chars

r=read.reverse.show
main=getLine>>=print.head.filter(\x->x==r x).iterate(\x->x+r x).read

Somewhat readable version:

myFind p = head . filter p
rev = read . reverse . show
isPalindrome x = x == rev x
next x = x + rev x
sequence196 = iterate next
palindrome196 = myFind isPalindrome . sequence196

main = getLine >>= print . palindrome196 . read

The golfed version was created by manual inlining and renaming the remaining functions to single character names.

Answer 28

befunge, 57 bytes

"KCTS"4(&:0\v
\T\a*+\:0`jv>:a%\a/
0:+_v#-TD2$<^\
  @.<

though the code is places in a 4x19 grid, so might call it 76.

• first line is initializeing, and reading input number
• second line reverse first number in stack and put it in the second stack position.
• and the third line checks if a number is palindrome.

Answer 29

C++ TMP (256 characters)

#include<cstdio>
#define Y(A,B,C,D)template<int N>struct A<C,D>{enum{v=B};};
#define Z(A)template<int N,int M>struct A{enum{v=
#define n 5194
Z(R)R<N/10,M*10+N%10>::v};};Y(R,N,0,N)Z(S)S<N+M,R<N+M,0>::v>::v};};Y(S,N,N,N)main(){printf("%d",S<n+R<n,0>::v,0>::v);}

This version could be shortened a bit, but a 256-character answer is hard to pass up. Here's an un-golfed version:

#include <iostream>

template<size_t N>
class Reverse
{
    template<size_t M, size_t R>
    struct Inner
    {
        enum { value = Inner<M/10, R*10 + M%10>::value };
    };

    template<size_t R>
    struct Inner<0, R>
    {
        enum { value = R };
    };

public:
    enum { value = Inner<N, 0>::value };
};

template<size_t N>
class OneNineSix
{
    template<size_t M, size_t R=Reverse<M>::value>
    struct Inner
    {
        enum { value = OneNineSix<M + R>::value };
    };

    template<size_t M>
    struct Inner<M, M>
    {
        enum { value = M };
    };

public:
    enum { value = Inner<N + Reverse<N>::value>::value };
};

int main()
{
    const size_t N = 4123;

    std::cout << OneNineSix<N>::value << std::endl;
}

Answer 30

Pyke, 13 bytes (noncompeting)

D`_b]D$XIsr)h

Try it here!

 `_b          -     int(reversed(str(num))
D   ]         -    [num, ^]
     D        -   _ = ^
      $       -  delta(^)
       XI     - if ^:
         s    -  num = sum(_)
          r   -  goto_start()
            h - _[0]

Answer 31

Java, 90 84 bytes

n->{for(long t;(t=new Long(new StringBuffer(n+"").reverse()+""))!=n;)n+=t;return n;}

-6 bytes thanks to @O.O.Balance.

Try it online.

Explanation:

n->{           // Method with long as both parameter and return-type
  for(long t;  //  Temp-long, starting uninitialized
      (t=new Long(new StringBuffer(n+"").reverse()+""))
               //  Before every iteration, reverse the temp-long
      !=n;)    //  And loop as long as it's not equal to `n` yet
    n+=t;      //   Add this temp-long to the input
  return n;}   //  Return the modified input as result

Answer 32

Japt, 11 10 bytes

@¶ìw}a@±ìw

Try it

Answer 33

PHP, 57 Bytes

Try it online!

Code, recursive function

function f($n){echo(strrev($n)!=$n)?f(strrev($n)+$n):$n;}

Explanation

function f($n){
    echo(strrev($n)!=$n)? #check if non-palindrome
        f(strrev($n)+$n): #true, call again with $n + reverse $n
        $n;               #false (is a palindrome) echo $n

}

Answer 34

Perl 6, 36 bytes

{($_,{$_+.flip}...{$_==.flip})[*-1]}

Try it online!

An anonymous code block that returns the last element of a sequence defined by:

• The first element is the first parameter
• The i+1th element is the ith element plus the reverse of itself
• And ends when the element is equal to its reverse

Answer 35

C (gcc), 132 bytes

t;r(char*s){char*e=s+strlen(s)-1;for(;e>s;*e--=*s,*s++=t)t=*e;}g(o,l,f){for(l=malloc(9);sprintf(l,"%d",o),r(l),f=atoi(l),f-o;)o+=f;}

Call with g(n) for an int n. Try it online here.

Ungolfed:

t; // temporary varaiable fpr swapping characters
r(char *s) { // helper function to reverse a string: takes a string argument and modifies it
    char *e = s + strlen(s) - 1; // find the end of the string
    for(; e > s; // move the end pointer and the start pointer towards each other until they meet
        *e-- = *s, *s++ = t) // swap the characters under the start and end pointers ...
        t = *e; // ... using the temporary variable
}
g(o, // function for the 196 algorithm: takes an int argument and returns an int
  l, f) { // abusing the argument list to declare to extra variables: one is a string for reversing the number, the other is an int
    for(l = malloc(9); // allocate space for the string
        sprintf(l, "%d", o), // convert the number to a string
        r(l), // reverse that string
        f = atoi(l), // convert it back to an int
        f - o; ) // loop until the two numbers are equal
        o += f; // add the two numbers
}

Answer 36

MATL, 18 bytes

`Vt2&P=?5M.}1MvUsT

Try it on MATL Online

         % (implicit input)
`        % do-while loop
  Vt     % convert the number into a string, duplicate it
  2&P    % flip the copy left-to-right
  =      % are they equal? 
    ?5M. % if yes, push the number back on the stack and exit
         % (implicit output display)
    }1M  % else, push the number and its reverse (as strings) again on to the stack
    vUs  % convert them to numbers and add them
    T    % "True" value to continue loop, this time with the sum as the input number
         % (implicit loop end)

Answer 37

Husk, 7 bytes

¤Ω`S↔=+

Try it online!

Alternatively we could use ΩS=↔S+↔ for the same amount of bytes, but I like the one above more.

Explanation

¤Ω`S↔=+
¤        -- compose the arguments of
 Ω       -- | iterate second function until first is truthy
         -- with
  `S     -- | flipped S: applying binary function to itself and
    ↔    -- | | itself reversed
         -- first function: check if palindrome (\x-> x == reverse x)
         -- second function: "196ify" (\x-> x + reverse x)

Answer 38

Ruby, 55 bytes

l=->x{x.reverse==x ? x:l[(x.to_i+x.reverse.to_i).to_s]}

Different approach than the other Ruby one, ended up golfing it down to one byte fewer

Explanation: makes a lambda l that recursively calls itself, each time adding the number's reverse, until the string is a palindrome

Slightly less golfed version:

func = ->x do
  (x.reverse == x) ? x : func[(x.to_i + x.reverse.to_i).to_s]
end

Answer 39

Excel VBA, 49 bytes

An immediate window function which takes input from range [A1] and outputs to the VBE immediate window.

n=[A1]:Do:n=n+r:r=StrReverse(n):Loop While n-r:?n

Answer 40

Lua 5.3.3, 57 bytes

Takes i as input, prints output:

r=0while i~=r do i=i+r|0r=0+(i..""):reverse()end print(i)

More readable version:

r=0
while i~=r do
  i=i+r|0
  r=0+(i..""):reverse()
end
print(i)

Super simple. Just repeatedly adds i to its reverse until i is equal to its reverse. Then it just prints the new value of i.

Answer 41

Actually, 16 bytes

;WX;$R≈+;$;R=YWX

Try it online!

Explanation:

;WX;$R≈+;$;R=YWX
;                 push a copy of n
 W            W   while top of stack is truthy:
  X                 pop and discard
   ;$R≈+            copy n, cast to string, reverse, cast to int, add
        ;$;R=Y      not palindrome (cast to string, copy, reverse, check inequality)
               X  pop and discard

Answer 42

J, 27 bytes

(+&.".|.)^:(-.@-:|.)^:_&.":

Try it online!

Answer 43

Python 2, 53 bytes

f=lambda x,y=0:x if x==y else f(x+y,int(`x+y`[::-1]))

Try it online!

Answer 44

Dyalog APL, 17 bytes

{⍵+⍎⌽⍕⍵}⍣{⍺≡⍎⌽⍕⍺}

{⍵+⍎⌽⍕⍵} add argument and its reverse...
⍣ ... until...
{⍺≡⍎⌽⍕⍺} ... the result is a palindrome.

Answer 45

Javascript ES6, 56 bytes

f=a=>(g=_=>[...""+a].reverse().join``)()==a?a:f(a+ +g())

Answer 46

Wren, 102 bytes

A super long program.

Fn.new{|x|
while(x!=x[-1..0])x=(Num.fromString(x)+Num.fromString(x[-1..0])).toString
System.write(x)
}

Try it online!