196 algorithm code golf

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

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).

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