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

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 3

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 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

05AB1E, 7 bytes

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 10

Python: 66

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

Answer 11

Perl, 40 chars

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

Answer 12

PHP - 54 48 characters

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

Test:

$ php 196.php <<< 5280
23232

Answer 13

Scala 82

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

Answer 14

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 15

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 16

Bash (64)

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

Call with: bash <filename> <number>

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

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 19

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

print r

Answer 20

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 21

Java (JDK), 78 71 bytes

i->{for(int r=0,t;i!=r;)for(t=i+=r,r=0;t>0;t/=10)r=r*10+t%10;return i;}

Try it online!

-7 Thanks to Celingcat!

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

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 24

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 25

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 26

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 27

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 28

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 29

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 30

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