Programming Puzzles & Code Golf Stack Exchange is a question and answer site for programming puzzle enthusiasts and code golfers. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A happy number is defined by the following process. Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers, while those that do not end in 1 are unhappy numbers (or sad numbers). Given a number print whether it is happy or unhappy.

Sample Inputs

Sample Outputs

Note: Your program should not take more than 10 secs for any number below 1,000,000,000.

share|improve this question

22 Answers 22

up vote 6 down vote accepted

Golfscript - 34 chars


Basically the same as this and these.

The reason for 9 iterations is described in these comments (this theoretically returns correct values up to about 10^10^10^974 (A001273)).

share|improve this answer

Ruby, 77 characters

a=gets.to_i;a=eval"#{a}".gsub /./,'+\&**2'until a<5
puts a<2?:Happy: :Unhappy
share|improve this answer
Ok, so I kinda get how this works (Literally taking each number, splitting it and adding the square of each digit), but what's with the stop condition of (a < 5) and using (a < 2) to decide if it's happy or not? I don't question the validity, just the logic. – Mr. Llama Apr 25 '11 at 22:42
That is the same as a <= 4 and a <= 1. If the cycle has a 1 in it then it is happy, and if it has a 4 in it, then it is not happy. See the wikipedia section about the unhappy cycle. So once the value of a is 4 or less, he checks if a is -- the result of that is your answer. – Casey Apr 25 '11 at 23:57

C - 115

char b[1<<30];a;main(n){for(scanf("%d",&n);b[n]^=1;n=a)for

This uses a 230-byte (1GB) array as a bitmap to keep track of which numbers have been encountered in the cycle. On Linux, this actually works, and efficiently so, provided memory overcommitting is enabled (which it usually is by default). With overcommitting, pages of the array are allocated and zeroed on demand.

Note that compiling this program on Linux uses a gigabyte of RAM.

share|improve this answer
Why would you need anywhere close to that amount of memory for this problem? – Peter Olson Apr 25 '11 at 20:32
@Peter: I suppose the approach is to (naively) catch a cycle for any number in the allowed input range from 1 to 1,000,000,000. But I agree that in light of happy number theory, the only check necessary is if the number 4 is reached, because that's the only cycle that will ever occur. – mellamokb Apr 25 '11 at 20:51
I'm curious: why does compiling it require so much RAM? – Peter Taylor Apr 25 '11 at 21:22
Appears to work fine on Windows 7 with MSVC 10. Doesn't consume any notable amount of memory while compiling and only marks the array in the page file (something that sounds way safer than the story you linked about memory overcommitting suggests ;-)). – Joey Apr 25 '11 at 22:11
I love the naivety of this approach. And the abuse of for loops is beautiful. – dmckee Apr 27 '11 at 1:52

Haskell - 77

f 1="Happy"
f 4="Unhappy"
f n=f$sum[read[c]^2|c<-show n]
share|improve this answer

Golfscript, 49 43 41 40 39 chars


Every happy number converges to 1; every unhappy number converges to a cycle containing 4. Other than exploiting that fact, this is barely golfed at all.

(Thanks to Ventero, from whose Ruby solution I've nicked a trick and saved 6 chars).

share|improve this answer

eTeX, 153

\let~\def~\E#1{\else{\fi\if1#1H\else Unh\fi appy}\end}~\r#1?{\ifnum#1<5

Called as etex filename.tex 34*23 + 32/2 ? (including the question mark at the end). Spaces in the expression don't matter.

EDIT: I got down to 123, but now the output is dvi (if compiled with etex) or pdf (if compiled with pdfetex). Since TeX is a typesetting language, I guess that's fair.

\def~{\expandafter\r\the\numexpr}\def\r#1?{\ifnum#1<5 \if1#1H\else
Unh\fi appy\end\fi~\s#1{0?}}\def\s#1{+#1*#1\s}~\noexpand
share|improve this answer

Python - 81 chars

while n>4:n=sum((ord(c)-48)**2for c in`n`)

Some inspiration taken from Ventero and Peter Taylor.

share|improve this answer
better off doing a int(c) than ord(c)-48.... – st0le Apr 26 '11 at 4:32

Javascript (94 92 87 86)

do{n=0;for(i in a){n+=a[i]*a[i]|0}a=n+''}while(n>4);alert(['H','Unh'][n>1?1:0]+'appy')

Input is provided by setting a to the number desired.

Credits to mellamokb.

share|improve this answer
Save 1 char: n==4?h="Unh":n==1?h="H":a=n+""}alert(h+"appy") – mellamokb Apr 26 '11 at 1:21
@mella Thanks. I also shaved another char off by changing || to |. – Peter Olson Apr 26 '11 at 1:28
Save 8 chars: Remove n==4?h.... Change to do...while loop with condition while(n>4). Then use this final statement instead: alert(["H","Unh"][n>1?1:0]+"appy") – mellamokb Apr 26 '11 at 1:43
@Mella Clever, I like it. – Peter Olson Apr 26 '11 at 2:01
@Mella n needs to be defined before the while loop, I'm trying to think of how to not repeat n=0; – Peter Olson Apr 26 '11 at 2:08

Python (98, but too messed up not to share)

f=lambda n:eval({1:'"H"',4:'"Unh"'}.get(n,'f(sum(int(x)**2for x in`n`))'))
print f(input())+"appy"

Way, way too long to be competitive, but perhaps good for a laugh. It does "lazy" evaluation in Python. Really quite similar to the Haskell entry now that I think about it, just without any of the charm.

share|improve this answer

dc - 47 chars


Brief description:

I~: Get the quotient and remainder when dividing by 10.
d*: Square the remainder.
0<H: If the quotient is greater than 0, repeat recursively.
+: Sum the values when shrinking the recursive stack.

4<h: Repeat the sum-of-squares bit while the value is greater than 4.

share|improve this answer

Befunge, 109

Returns correct values for 1<=n<=109-1.

v v              <   @,,,,,"Happy"<      >"yppahnU",,,,,,,@
share|improve this answer

J, 56


A verb rather than a standalone script since the question is ambiguous.


   happy =: 'Happy'"_`('Unhappy'"_)`([:$:[:+/*:@:"."0@":)@.(1&<+4&<)
happy =: 'Happy'"_`('Unhappy'"_)`([:$:[:+/*:@:"."0@":)@.(1&<+4&<)
   happy"0 (7 4 13)
happy"0 (7 4 13)
share|improve this answer

J (50)


I'm sure a more competent J-er than I can make this even shorter. I'm a relative newb.

New and improved:


Newer and even more improved, thanks to ɐɔıʇǝɥʇuʎs:

share|improve this answer
You can get a character by not splitting out 'appy'. I think you can also remove the parentheses aroundd ("."0) - adverbs bind tighter than conjunctions. – Jesse Millikan Apr 27 '11 at 5:36
I can't remove the parentheses around ("."0). That produces a rank error, but if I don't split 'Happy' and leave the result boxed, I can save a character. – Gregory Higley Apr 28 '11 at 20:12
The reason I can't leave out the parentheses around ("."0) is that conjunctions apply to the entire preceding train of verbs to which they're attached, which is not what I want. If I say +/@:("."0)@":, that is very different from +/@:"."0@:, which is actually (+/@:".)"0@:. – Gregory Higley May 1 '11 at 15:55
A massive necro, but you could save 4 chars by replacing 'Unhappy';'Happy' with Unhappy`Happy. – ɐɔıʇǝɥʇuʎs Dec 15 '14 at 7:39
@ɐɔıʇǝɥʇuʎs That works, but where is it documented that you can skip the quoting of strings with `? – Gregory Higley Dec 15 '14 at 10:50

Python (91 characters)

a=lambda b:b-1and(b-4and a(sum(int(c)**2for c in`b`))or"Unh")or"H";print a(input())+"appy"
share|improve this answer

Scala, 145 chars

def d(n:Int):Int=if(n<10)n*n else d(n%10)+d(n/10)
def h(n:Int):Unit=n match{
case 1=>println("happy")
case 4=>println("unhappy")
case x=>h(d(x))}
share|improve this answer
Wouldn't (n*n) be shorter as n*n , or does whitespace not suffice to separate an if-expression from the else? – Peter Taylor Apr 27 '11 at 11:50
Yes, I did so, Peter. – user unknown Apr 27 '11 at 12:07

K, 43

share|improve this answer

Common Lisp 138

(format t"~Aappy~%"(do((i(read)(loop for c across(prin1-to-string i)sum(let((y(digit-char-p c)))(* y y)))))((< i 5)(if(= i 1)"H""Unh"))))

More readable:

(format t "~Aappy~%"
          ((i (read)
              (loop for c across (prin1-to-string i)
                    sum (let
                          ((y (digit-char-p c)))
                          (* y y)))))
          ((< i 5) (if (= i 1) "H" "Unh"))))

Would be shorter to just return "Happy" or "Unhappy" right from the (do), but arguably that wouldn't count as a whole program

share|improve this answer

Perl 5 - 77 Bytes

{$n=$_*$_ for split//,$u{$n}=$n;exit warn$/.'un'[$n==1].'happy'if$u{$n};redo}

$n is the input value

share|improve this answer

C++ 135 , 2 Lines

int n,i,j;int main(){for(std::cin>>n;n>1;n=++j&999?n*n+i:0)for(i=0;n/10;n/=10)i+=n%10*(n%10);std::cout<<(n?"H":"Unh")<<"appy";}

This is a modified version of the one I did here:

share|improve this answer
What is the &999 do? And how does it work if j is a garbage value? – David Grinberg Apr 11 '14 at 13:44
@Dgrin91, I wrote this 3 years ago, so i can't remember exactly how it works. I think the &999 makes the statement if(j==999){n = 0;}else{n=n*n +i;}, j shouldn't be a garbage value, globals are zero initialized. – Scott Logan Apr 14 '14 at 9:18

C: 1092 characters

#include <iostream>
using namespace std ;
int main ()
    int m , a[25] , kan=0 , y , z=0  , n , o=0, s , k=0 , e[25]  ;
    do {
m :
        for ( int j=1 ; j <10000 ; j++ )
            for (int i=0 ; j!=0 ; i++ )
                a[i]=j%10 ;
                j/=10 ;
                kan++ ;
            for ( int i=0 ; i<kan ; i++ )
                y=a[i]*a[i] ;
                z+=y ;
            k+=1 ;
            if (z==1)
               o++ ;

                 for (int f=0 ; f<k ; f++ )
                     e[f]=z ;
                 for ( int f=0 ; f=k-1 ; f++ )
                     for ( int p=f+1 ; p <k-1 ; p++ )
                             goto m ;
                         else { j=z ; goto n ; } 
    }while(o!=100) ;
    return 0 ;
share|improve this answer
Welcome to Programming Puzzles & Code Golf, @jannat. Please note that code golf is a challenge of writing the shortest code possible. That means, here we write unindented and almost unreadable code and force the limits of the language syntax to shorten our codes as possible. – manatwork Mar 9 '13 at 16:28 – aditsu Mar 11 '13 at 18:06

Yes, this challenge has three years; yes, it already has a winner answer; but since I was bored and done this for another challenge, thought I might put it up here. Surprise surprise, its long - and in...

Java - 280 264 bytes

import java.util.*;class H{public static void main(String[]a){int n=Integer.parseInt(new Scanner(,t;while((t=h(n))/10!=0)n=t;System.out.print(t==1?"":"");}static int h(int n){if(n/10==0)return n*n;else return(int)Math.pow(n%10,2)+h(n/10);}}


import java.util.*;

class H {

    public static void main(String[] a) {
        int n = Integer.parseInt(new Scanner(, t;
        while ((t = h(n)) / 10 != 0) {
            n = t;
        System.out.print(t == 1 ? "" : "");

    static int h(int n) {
        if (n / 10 == 0) {
            return n * n;
        } else {
            return (int) Math.pow(n % 10, 2) + h(n / 10);
share|improve this answer

C# 94 bytes

int d(int n)=>n<10?n*n:(d(n%10)+d(n/10));string h(int n)=>n==1?"happy":n==4?"unhappy":h(d(n));

For any given number (as int), h() will return the correct value. You can try the code on .NetFiddle.

Kudos to user unknown for the original algorithm.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.