Code Golf: Happy Primes!

Question

It's Sunday, time for a round of code golf!

Challenge

Write the shortest source code by character count to determine if an input number is a "happy prime", "sad prime", "happy non-prime", or "sad non-prime."

Input

The input should be a integer that comes from a command line argument or stdin. Don't worry about handling big numbers, but do so if you can/want. Behavior would be undefined for input values less than 1, but 1 has a definite result.

Output

Output should print the type of number: "happy prime", "sad prime", "happy non-prime", or "sad non-prime." The trailing newline is optional.

Examples

$ happyprime 139
happy prime
$ happyprime 2
sad prime
$ happyprime 440
happy non-prime
$ happyprime 78
sad non-prime

Definitions

Just in case your brain needs a refresher.

Happy Number

From Wikipedia,

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).

For example,

• 139
• 1^2 + 3^2 + 9^2 = 91
• 9^2 + 1^2 = 82
• 8^2 + 2^2 = 68
• 6^2 + 8^2 = 100
• 1^2 + 0^2 + 0^2 = 1

Prime Number

A prime number is an integer greater than 1 and has precisely two divisors: 1 and itself.

Happy Prime

A happy prime, is therefore a number that is both happy and prime.

Answer Selection

Obviously the answer will be the shortest source code by character count that outputs the specified results in all cases that I test. I will mark the answer once the next (community decided) code golf challenge comes along, so we can focus all our energies on that one. :)

Decision

Well, it looks like the there is a new code golf in town and it has been about a week since this question was posted, so I've marked the shortest source code as the answer (gnibbler's 64 character Golfscript solution). That said, I enjoyed both the 99 character Mathematica solution by belisarius and the cryptic 107 character dc solution by Nabb.

To all others, great work! I've never had so many programming language environments on my computer. I hope everyone has learned some new, dirty tricks for their favorite language.

Reuse

I've re-published some of the code produced by this competition as an example for a script I wrote to test various programs against a reference implementation for auto-grading. The README in that directory explains where the source code comes from and states that all code is re-used under the CC BY-SA 2.5 license (as stated in SO's legal section). Each directory is labeled with your display name at the time of the submission.

If you have a problem with your code being re-used in this fashion or the attribution, let me know and I will correct the error.

Choler

Can we assume the input number is greater than 1? Greater than some X? – Manamanacle

Well, greater than or equal to 1. – Choler

Doctor Who Episode 184("42")? – Arciniega

nuclearsandwich: Indeed, that was the impetus for the question. – Choler

it's valid to do this on SO? Awesome... – Disturb

Corey Ogburn: According to meta they are. Though some people don't like them, I think it provides an interesting view of various language features, and I've seen many unexpected ways of finding the answer here. (All shorter than my reference implementation.) – Choler

@Choler I think the message warning in SO needs you to prepend an "@" sign before the name to work. As in @Corey . This comment will commit suicide in 3 .. 2 .. 1 – Kleinstein

@belisarius: that is correct, I just assumed it was leftover from too many people using twitter. – Choler

Made a web application out of it in case you're in dire need to find out whether your favourite number is happy, prime, or both! startwithabreak.com/happyprimeweb Will try and golf it but seems like it's a bit late in the game. – Querida

@Radu: It's never too late! :-) – Gildus

It's good to see a lot of people using the regex-prime test. Although I suspect that may not work with 85-digit numbers – Wingfield

Answer 1

GolfScript - 64 chars (works for 1)

~:@.{0\`{15&.*+}/}*1=!"happy sad "6/=@,{@\)%!},,2=4*"non-prime">

This program does n iterations to determine the happiness of the number, which is very wasteful for large numbers, but code-golf is not about conserving resources other than characters. The prime test is similarly inefficient - dividing n by all the values from 1 to n inclusive and checking that there are exactly two values with zero remainder. So while it is theoretically correct, running with really large numbers is not practical on real computers

GolfScript - 63 chars (fails for 1)

~:@9{0\`{15&.*+}/}*1=!"happy sad "6/=@,2>{@\%!},!4*"non-prime">

Oneiric

I get...odd behavior when the input is 1. Is that from a limitation of golfscript or something else? (It says ../lib/golfscript.rb:353:in select': private method select' called for nil:NilClass (NoMethodError).) – Choler

@mjschultz, darn - 1 is a special case in that solution. One char extra for a better primality test – Whittle

Well, let's run it on unreal computers then. – Dusk

Answer 2

dc - 98 chars

$ cat happyprimes
[happy][sad]?dsI[[I~d*rd0<H+]dsHxd4<h]dshx[r]sr1=rP[ ][ non-]_1lI[1-d2>rdlIr%0<p]dspx-2=rP[prime]p
$ echo 1  |dc happyprimes
happy non-prime
$ echo 139|dc happyprimes
happy prime
$ echo 2  |dc happyprimes
sad prime
$ echo 440|dc happyprimes
happy non-prime
$ echo 78 |dc happyprimes
sad non-prime

Devitt

About dc: Wikipedia, manpage. – Gildus

Code golfing is the third use for dc, the other two are doing modular exponentiation and implementing bc on top of dc (and using bc instead of dc ;P ). – Intermingle

Answer 3

Mathematica 115 108 107 102 100 99 91 87 Chars

---

87 characters

Print[If[Nest[Tr[IntegerDigits@#^2]&,#,9]>1,Sad,Happy],If[PrimeQ@#," "," non-"],prime]&

-- Mr.Wizard

---

Da monkey learnt a few tricks (91 chars)

 Print[
       If[Nest[Plus@@(IntegerDigits@ #^2) &, #, 9] > 1, Sad, Happy ],
       If[PrimeQ@#, " ", " non-"], prime
      ] &

Invoke with %[7]

Edit 5 - 99 Chars/

Nine iterations is enough. Thanks @Nabb, @mjschultz

h = Print[
    If[Nest[Plus @@ (IntegerDigits@#^2) &, #, 9] > 1, "Sad ", "Happy "]
   , If[PrimeQ@#, "", "non-"], "prime"] &

Edit 4 - 100 Chars/

Same as Edit 3, replacing 10^2 by 99 (allowing 84 digits for input values) ... Thanks, @Greg

h = Print[
    If[Nest[Plus @@ (IntegerDigits@#^2) &, #, 99] > 1, "Sad ", "Happy "]
   , If[PrimeQ@#, "", "non-"], "prime"] &

Edit 3 - 102 Chars/

Reworked the loop again.

It is interesting that the recursion depth until eventually reaching 1 is bounded by (15 + Number of digits of the argument). See here

So for numbers with less than 85 digits (I think this limit is pretty well into the OP's "Don't worry about handling big numbers" consideration) the following code works

h = Print[
    If[Nest[Plus @@ (IntegerDigits@#^2) &, #, 10^2] > 1, "Sad ", "Happy "]
   , If[PrimeQ@#, "", "non-"], "prime"] &

I changed the "NestWhile" for the shorter "Nest", and so, instead of specifying a stop condition for the recursion, is enough to hardcode the desired recursion depth (10^2).

It is not very efficient, but that's the golfer's life :D

Edit 2 - 107 Chars/

Reworked the Sad/Happy assignment

h = Print[
     If[NestWhile[Plus @@ (IntegerDigits@#^2) &, #, # > 4 &] > 1,"Sad ","Happy "]
    ,If[PrimeQ@#, "", "non-"]
    , "prime"] &

All spaces/newlines except on literals are optional and added for readability

Explanation:

The

    NestWhile[Plus @@ (IntegerDigits@#^2) &, #, # > 4 &]

Recurses applying "function" [Add up sum of digits squared] until the result is 4 or less. The function has the property that it stagnates at "1", or enters the cycle {4, 16, 37, 58, 89, 145, 42, 20, 4, ...}.

So, when the outcome is "1", the number is "Happy" and when the outcome is "4", it is "Sad".

If the result is "2", the number is also SAD, because it'll enter the SAD cycle in the next iteration (2^2 = 4).

If the result is 3, the cycle is 3->9->81->65->61->37->58->89->145-> .... (Enters the SAD loop).

So, we can stop the recursion when the result is 4 or less, knowing that only a result of "1" will lead to a Happy number.

Perhaps other solutions may take advantage of this fact.

In fact, the results 5 and 6 also lead to SAD numbers, but that gives us only an efficiency boost and not a golfing advantage (I guess).

Edit 1 - 108 Chars/

Reworked the Loop Control logic

    h = Print[
        NestWhile[Plus@@(IntegerDigits@#^2) &, #, #>4 &] /.{1 →"Happy ",_→"Sad "}
          , If[PrimeQ@#, "", "non-"]
          , "prime"] &

Original - 115 Chars/

h = Print[
    If[NestWhile[Plus @@ (IntegerDigits@#^2) &, #, Unequal, All] == 1
        ,"Happy ", "Sad "],      
    If[PrimeQ@#, "", "non-"], "prime"] &

The statement

NestWhile[Plus @@ (IntegerDigits@#^2) &, #, Unequal, All]

performs the recursive application of the sums of the digits squared, until some value repeats. The "Unequal,All" part takes care of the comparison across the preceding values list. Finally returns the repeated value, which is "1" for Happy Numbers.

Sample run

h[7]
Happy prime
h[2535301200456458802993406410753]
Sad non-prime

Looping (Slightly changing the Print statement)

1 Happy non-prime
2 Sad prime
3 Sad prime
4 Sad non-prime
5 Sad prime
6 Sad non-prime
7 Happy prime
8 Sad non-prime
9 Sad non-prime
10 Happy non-prime
11 Sad prime
12 Sad non-prime
13 Happy prime

Culmination

Oi, you're going to make me install Mathematica just so I can verify this! (Though I am fully convinced it is correct.) – Choler

Why do you use 10^2 instead of 100 or 99, both of which are shorter? – Kilmarnock

@Kilmarnock yep, tnx ... I started using 10^99 for allowing greater input values and forgot to change that when came down to 85 digits ... Anyway the stack crashed with very big numbers :( – Kleinstein

belisarius: I'd bump it all the way down to 9 which puts you at 99 characters total. It still managed to evaluate 2535301200456458802993406410753 in no time on my system. – Choler

@Choler The required depth is not a monotone function. For example, calculating h[30] requires 11 iterations. The max depth value for numbers under 1000 is 15. So, 9 is way toooo low. – Kleinstein

@Choler Here is the compiled version for running in the free Mathematica Player, and the instructions. Just in case. code.google.com/p/magicnumbers/wiki/… – Kleinstein

belisarius: Thanks for doing that, but as a student I have access to Mathematica. However, I'm sure others do not, so it's still useful! – Choler

9 iterations should be sufficient up to 10^974 digits. – Devitt

@Devitt I guess you've a proof there :). I just followed a few experiments and the Wikipedia statement. Without a proof I didn't want to assume that, and reaching the SAD cycle may take as much as 16 iterations for very small numbers (15999, for example) – Kleinstein

@Culmination Your test condition is >1, i.e. happy numbers will have reached 1, not unhappy numbers some arbitrary value. The smallest values for n steps: s(5) = 7, s(6) = 356, s(7) = 78999. log₁₀(s(8)) ~ 975 (=78999/81). log₁₀(log₁₀(s(9))) ~ 974. log₁₀(log₁₀(log₁₀(s(10)))) ~ 974. 9 iterations is therefore sufficient for everything up to 10^10^10^974, or 10^10^974 digits. I suppose I missed a step earlier. – Devitt

@Devitt Yep. Found the reference: research.att.com/~njas/sequences/… .. Thanks a lot – Kleinstein

Answer 4

Python - 127 chars

Beating both perl answers at this moment!

l=n=input()
while l>4:l=sum(int(i)**2for i in`l`)
print['sad','happy'][l<2],'non-prime'[4*all(n%i for i in range(2,n))*(n>1):]

I also ported this answer to GolfScript and it is just over 1/2 the size!

Whittle

That is pretty sweet. I wouldn't expect a python answer to be shorter than perl. – Choler

You can use the backticks instead of str, as I explain in my answer above. – Canasta

I never knew a space isn't required before that first for. – Humorist

+1 Never thought I'd see python beat perl in a terse contest. Incidentally, not gzip, bzip2, or even xz can not compress this smaller than it already is. – Clercq

@Seth: Most compression algorithms have some small amount of overhead and thus are miserable at compressing tiny files. – Hydroxylamine

Answer 5

C#, 380 378 374 372 364 363 315 280 275 274 Chars

By replacing the recursive function with nested loops, I was able to bring the stroke count to a respectable 280 (100 less than the original).

class P{static void Main(string[]a){var s=new System.Collections.Generic.HashSet<int>();int n=int.Parse(a[0]),p=n>1?4:0,c,d=1;for(;++d<n;)if(n%d<1)p=0;for(;n>1&s.Add(n);n=c)for(c=0;n>0;c+=d*d,n/=10)d=n%10;System.Console.Write((n>1?"sad":"happy")+" non-prime".Remove(1,p));}}

Here it is with whitespace:

class P
{
    static void Main(string[] a)
    {
        var s = new System.Collections.Generic.HashSet<int>();
        int n = int.Parse(a[0]),
            p = n > 1 ? 4 : 0,
            c,
            d = 1;
        // find out if the number is prime
        while (++d < n)
            if (n % d < 1)
                p = 0;
        // figure out happiness
        for (; n > 1 & s.Add(n); n = c)
            for (c = 0; n > 0; c += d * d, n /= 10)
                d = n % 10;

        System.Console.Write(
            (n > 1 ? "sad" : "happy")
            + " non-prime".Remove(1,p)
            );
    }
}

Haematoid

I was able to cut out quite a bit by doing some obvious golfing, not changing the algorithm at all. – Mme

Added some syntactical updates to save 5 more characters, have a bigger update which removes the need for p altogether, will post later. – Audrieaudris

shaved a character by rewriting the output – Audrieaudris

Answer 6

Python 2.6: 194 180 chars, 4 lines

import re
s=lambda n,l:0if n==1 else n in l or s(sum(int(a)**2for a in str(n)),l+[n])
n=input()
print['happy','sad'][s(n,[])],'non-'*bool(re.match(r'1?$|(11+?)\1+$','1'*n))+'prime'

The lexer being able to split 0if and 2for into two tokens each was a nice surprise to me :) (it doesn't work with else though)

Function s (sad) is recursive, and receives the list of previous numbers in the cycle as its second parameter. Primality is tested inline using the regexp trick.

By using the deprecated `n` syntax instead of str(n), one can further reduce the character count by 4 characters, but I choose not to use it.

Canasta

That looks awesome :P. And if you replace your int(sys.argv[1]) with input() it'll bring your char count down to 180. – Humorist

Thanks! I didn't notice stdin was allowed. – Canasta

Nice! You beat my solution by 60 chars and 15 lines! – Zug

@Yassin: Thanks for mentioning that's what it did! I'm just learning programming and I didn't even touch regex yet so I didn't know what it did :-). Now that I look, a lot of the answers use that regex. – Humorist

@Yassin @vlad003 I just added a link to an explanation of the trick in another SO question. – Canasta

I just submitted a python answer that uses 169 chars, but it's totally different than this one, so I didn't put it here (i really like the regex trick) – Guitar

Answer 7

C, 188 187 185 184 180 172 171 165

h(c,C,r,p){for(;C>1&&C%++p;);for(;c;c/=10)r+=c%10*(c%10);r&~5?h(r,C,0,1):printf(
"%s %sprime",r-1?"sad":"happy",p>=C&C>1?"":"non-");}main(c){h(c,c,0,scanf("%d",&c));}

$ ./a.out
139
happy prime

$ ./a.out
2
sad prime

$ ./a.out
440
happy non-prime

$ ./a.out
78
sad non-prime

This is one recursive function that never issues a return but either calls itself or prints output when it's done. The recursive function sums squared digits and determines prime-ness in two for loops. The scanf returns 1 which is put as an argument to h(), saving one ; and one 1 (and at the cost of having to use prefix ++p instead of postfix p++ which would make p>C possible instead of p>=C)

r&~5 is 0 for 1 4 5, of which 1 indicates happiness and the others sadness.

Next attempt: drop h() and make main() recursive.

Fadein

Hmmm, the golfed version seems to give me nothing but segfaults on two machines, the non-golf version works except 1 gives a floating point exception. Ideas? (compiled with gcc -o hp happyprime.c) – Choler

Golfed version does not compile at all for me (clang 1.0). It tells me that the second argument to main must be char**. – Turntable

fixed. Although with people specifying language versions (or custom golfing-languages) I would rather keep it at 178 and just specify platform and compiler: 32 bit linux using gcc. – Fadein

As long as I know that it requires a 32-bit compiler that is fine, are there any other gcc args I need? I still get an exception with the 1 case from the prime detection routine (initial call is p(1,0) leading to a modulo operation with 0 in the divisor). – Choler

No other args that I use (although if possible I would opt for a lot of -D.=... args). I'll have a look if I can blend in the non-primeness of 1. – Fadein

Answer 8

Perl, 140 chars

sub h{$_==1&& happy||$s{$_}++&& sad
||do{$m=0;$m+=$_**2for split//;$_=$m;&h}}$n=$_=pop;
die h,$",(1x$n)=~/^1?$|^(11+?)\1+$/&&"non-","prime\n"

Linebreaks are optional.

Appling

Ah, perl my first language. Interesting use of die for output, made me fix up my checker script. – Choler

Sorry, can be changed if necessary, but it's common practice here if nobody says "absolutely must go to stdout" :) – Appling

if you have perl 5.10+, you can also -E and say. also, you can drop \n and just use newline. – Nootka

Answer 9

MATLAB 7.8.0 (R2009a) - 120 characters

Whitespace, newlines, and comments added for readability

n=input('');
s=n;
c={'happy ','sad ','non-'};
while s>6,
  s=int2str(s)-48;
  s=s*s';                    %'# Comment to fix code highlighting
end;
disp([c{[s<2 s>1 ~isprime(n)]} 'prime'])

Intercrop

I'm curious about the last line ("disp") ... Could you explain it, please? – Kleinstein

@belisarius: c is a cell array of strings, and its contents are being indexed by a 1-by-3 logical array with elements s<2, s>1, and ~isprime(n). For every true entry in the index array, the contents of the corresponding cell are dumped into a set of square brackets, which concatenates them into one string along with the word 'prime'. It is then displayed using disp. – Intercrop

Any reason for the while s>9, line? It seems to cause invalid output when the input is 7. Would while s>6, work? – Choler

@Choler @Intercrop That's why I asked. I thought that the "disp" line made the magic to fix that. Tnx for the explanation! – Kleinstein

Answer 10

Javascript 244 250

function h(n){p=n=n<2?10:n;a=",";w="";s=[];while((y=a+s.join(a)+a).indexOf(a+n+a)<0){s.push(n);k=""+n;n=0;for(i=0;i<k.length;)c=k.charAt(i++),n+=c*c}w+=y.indexOf(",1,")<0?"sad ":"happy ";for(i=2;i<p;)p=p%i++?p:0;w+=p?"":"non-";return w+"prime"}

The above code should work in browsers without additional fancy functions and features (such as Array.prototype.indexOf and [] notation for strings), but I haven't tested it outside of Firefox.

Be aware that all but n are global variables (I'm just being cheap).

Usage

h(139) // returns "happy prime"

Manamanacle

Nice, I had to add a } at the end, but that is probably because I was using Firefox's Web Console. I'm impressed that Javascript and your code was able to handle h(2535301200456458802993406410753)! – Choler

Leaving the } out was a cut and paste mistake. Thanks. – Manamanacle

Answer 11

Haskell 172

h s n|n`notElem`s=h(n:s)$sum[read[k]^2|k<-show n]|1`elem`s="happy "|0<1="sad "
c n|n<2||any((0==).mod n)[2..n-1]="non-"|0<1=[]
y n=h[]n++c n++"prime"
main=readLn>>=putStr.y

Manamanacle

You must assume the input is greater than or equal to 1. (1 is a happy non-prime) – Mme

Yeah, wasn't sure at first. Fixed now. – Manamanacle

The newline is not necessary after the output. so feel free to get take two characters. – Choler

Answer 12

Ruby 1.9

169 168 146 chars

 h={1=>'happy'};s=->x{y=0;(y+=(x%10)**2;x/=10)while x>0;h[y]||(h[y]='sad';s[y])}
 $><<s[n=$*[0].to_i]+" #{'non-'if '1'*n=~/^1?$|^(11+?)\1+$/}prime"

---

If we use p instead of $><<, the code is shortened by 2 characters

Usage:

$ ruby happyprime.rb 139 happy prime $ ruby happyprime.rb 2 sad prime

---

Non golfed:

hash = {1->'happy'}
is_happy = lambda do |number|
  #sum = number.scan(/\d/).reduce(0){|acum, digit| acum + digit.to_i ** 2 }
  sum=0; 
  while (number > 0)
      sum+= (number%10)**2
      number/=10
  end
  return hash[sum] if hash[sum] # If 1, or if cycled and hash contains the number already
  h[sum] = 'sad'
  return is_happy.call(sum)
end
number = ARGV[0].to_i
string = ""
string += is_happy.call(number) # either 'happy' or 'sad'
string += is_prime(number) ? " non-prime" : "prime"
puts string

Where the is_prime method is left as an exercise to the reader ;)

Chiffonier

Nice use of regular expressions as primality check! – Mme

You can save 1 character by getting rid of the () for that if statement (leaving a single space behind) as * is already higher precedence than =~. – Runesmith

Answer 13

J: 113 characters

h=.1=$:@([:+/[:*:@"."0":)`]@.(e.&1 4)
1!:2&2;(({&('sad ';'happy '))@h,({&('non-prime';'prime'))@(1&p:))".(1!:1]3)

 

$ echo -n 7 | jc happy.ijs
happy prime
$ echo -n 139 | jc happy.ijs
happy prime
$ echo -n 2 | jc happy.ijs
sad prime
$ echo -n 440 | jc happy.ijs
happy non-prime
$ echo -n 78 | jc happy.ijs
sad non-prime

Taynatayra

Answer 14

Java: 294 286 285 282 277 262 260 chars

---

• Update 1: replaced BigInteger#isProbablePrime() by regex. Saved 8 chars.

• Update 2: replaced && by & (oops). Saved 1 char.

• Update 3: refactored s a bit. Saved 3 chars.

• Update 4: the test on n!=1 was superfluous. Saved 5 chars.

• Update 5: replaced regex by for loop and refactored happy for loops little bits. Saved 15 chars.

• Update 6: replaced int/Integer by long/Long. Saved 2 chars.

---

import java.util.*;class H{public static void main(String[]a){long n=new Long(a[0]),p=n>1?1:0,s,d=1;while(++d<n)if(n%d<1)p=0;for(Set c=new HashSet();c.add(n);n=s)for(s=0;n>0;s+=d*d,n/=10)d=n%10;System.out.printf("%s %sprime",n>1?"sad":"happy",p>0?"":"non-");}}

---

With newlines:

import java.util.*;
class H{
 public static void main(String[]a){
  long n=new Long(a[0]),p=n>1?1:0,s,d=1;
  while(++d<n)if(n%d<1)p=0;
  for(Set c=new HashSet();c.add(n);n=s)for(s=0;n>0;s+=d*d,n/=10)d=n%10;
  System.out.printf("%s %sprime",n>1?"sad":"happy",p>0?"":"non-");
 }
}

Marlysmarmaduke

Shouldn't that be new Char[p]? Anyway, good job. You beat me – Coulee

@BalusC, Character Count = 286. :-/ – Alkyd

@BalusC, I can save 3 chars. :) I can't seem to Save my Edit some proxy error on my side, so i'll post it as a comment. Please Check it and if it's any good update the answer. – Alkyd

import java.util.*;class H{public static void main(String[]a){int n=new Integer(a[0]),p=n,s,d;Set c=new HashSet();for(;n!=1&c.add(n);n=s)for(s=0;n>0;d=n%10,s+=d*d,n/=10);System.out.printf("%s %sprime",n==1?"happy":"sad",new String(new char[p]).matches(".?|(..+?)\\1+")?"non-":"");}} – Alkyd

@st0le: the trailing newline shouldn't count. It's really 285. As per your suggestion, yes that's better. – Marlysmarmaduke

@BalusC, 1 more byte. Move the Set c=new HashSet() into the first part of the for(;c.add(n);) loop. that saves a semicolon. :) for(Set c=new HashSet();c.add(n);n=s) – Alkyd

@BalusC,Haha, Ok, 7 bytes more. Instead of new String(new char[p]) use (""+new char[p]). :D – Alkyd

@st0le: I saved after all 15 chars. Thanks you for the hints :) – Marlysmarmaduke

@BalusC, I'm not done yet, if(n%d<1)p=0 change it to p*=(n%d)...;) or &= (same thing) – Alkyd

@st0le: 2 more chars off, those parentheses are superfluous. Regardless, that gives a false negative for 139. – Marlysmarmaduke

also p=n>1?1:0 can be changed to p=n, i think. Since Behavior would be undefined for input values less than 1 So it ought to do. :D – Alkyd

@st0le: 1 is valid input. It should become 0 for the case that while loop doesn't enter (so that p>0 in the print is still valid for 1). Changing int to long is a good one, thought of it as well in previous stages, but it didn't work well with new char[n] when the regex was still there, so I didn't use it. – Marlysmarmaduke

Answer 15

Python 2.6

happy.py: 280 314 333 chars, 14 lines.

import re
def q(z):
 while z!=1:z=sum((int(a)**2 for a in `z`));yield z
def h(g):
 l=[]
 while 1:
  try:z=g.next()
  except:return 'happy '
  if z in l:return 'sad '
  l.append(z)
p=lambda n:not re.match(r'^1$|^(11+?)\1+$','1'*n)
n=int(input())
print h(q(n))+'non-prime'[4*p(n):]

Usage:

$ echo 139 | python happy.py
happy prime
$ echo 2 | python happy.py
sad prime
$ echo 440 | python happy.py
happy non-prime
$ echo 1234567 | python happy.py
sad non-prime

--

Readable version:

import re, sys

def happy_generator(z):
    while z != 1:
        z = sum((int(a)**2 for a in str(z)))
        yield z

def is_happy(number):
    last = []
    hg = happy_generator(number)
    while True:
        try:
            z = hg.next()
        except StopIteration:
            return True

        if z in last:
            return False
        last.append(z)

def is_prime(number):
    """Prime test using regular expressions :)"""
    return re.match(r'^1?$|^(11+?)\1+$', '1'*number) is None

n = int(sys.argv[1])

print "%s %sprime" % (('sad','happy')[is_happy(n)], ('non-','')[is_prime(n)])

Clercq

Couple of minor things I noticed at a brief glance: 1: while True == while 1. 2: Don't need a space after the final print statement. 3: Rename the ge function to a single char. – Graeae

If you use the match function (instead of search), you don't need the ^ anchor in the regexp. – Canasta

This is a shorter p: p=lambda n:not re.match(r'1?$|(11+?)\1+$','1'*n) – Canasta

Since h is used only for indexing, it could return 0 and 1 instead of bools. – Beech

Answer 16

MATLAB - 166 chars

function happyprime(a)
h={'non-prime','prime'};
h=h{isprime(str2num(a))+1};
for i=1:99
   a=num2str(sum(str2num((a)').^2));
end
s={'Sad ','Happy '};
[s{(str2num(a)==1)+1},h]

Usage

happyprime 139
ans =
Happy prime

Promising

Answer 17

VBA 245 characters

Good starter, will trim if time allows. Its only my 2nd go at code golf!

Public Sub G(N)
Dim Z, D, X, O
X = N
Z = N
Do Until Z = 1 Or X > N Or X = 0
    X = 0
    For D = 1 To Len(CStr(Z))
        X = X + CLng(Mid(CStr(Z), D, 1) ^ 2)
    Next D
    Z = X
Loop
If Z = 1 Then O = "Happy" Else O = "Sad"
D = 2
Do
    If N / D = Int(N / D) Then O = O & " Not Prime": Debug.Print O: Exit Sub
    D = D + 1
Loop While D < N
O = O & " Prime"
Debug.Print O
End Sub

Sedation

Also, VisualBasic is not a great language for golfing. (English keywords, not free form, etc.) – Tailing

I do golf challenges without regard to char count just to practice :P – Querida

Answer 18

C++, 258 231 230 227 chars

#include<iostream>
#define w while
int m,n,i,j,t=10;int main(){w(std::cin>>n){j=0,m=n;w(n>1){i=0;do i+=n%t*(n%t);w(n/=t);n=n*n+i;n=++j&0xFFFF?n:0;}i=1;w(m%++i&&j>1);std::cout<<(n?"happy":"sad")<<(i-m?" non-":" ")<<"prime\n";}}

not the best golfing language, gave it a good shot anyway. Most of this is straight C so would probably be shorter in C as well.

EDIT

Generally tidied it up, think it's pretty much at the limit now without a complete rerwrite.

Also forgot to add that this assumes that there are no numbers with a sequence with over 0xFFFF numbers which a pretty sensible assumption.

EDIT 2

fixed a bug. rearranged to remove the excessive calls to std::cout.

Latitudinarian

A few C golfing tricks which should shave off some: You could try to use the ternary conditional cond?true:false which also has useful low precedence properties, comparison for (non-)equality is best done using a-b rather than a!=b, and looping is shorter with for - although while() takes as much room as for(;;), the latter gives space to put statements in without the cost of an extra ;. And there are a few more chars to win still. – Fadein

I know this will cost you a few character, but I get stuck in the last while loop when the input is 1. – Choler

@mjschultz: thanks fixed it now. thanks for doing my debugging for me as well. ;) – Longing

Answer 19

Perl, 135C

sub h{my$s;$s+=$_**2for split//,pop;($s-4)?($s-1)?&h($s):1:0}$a=pop;
print h($a)?happy:sad,$",(1x$a)=~/^1?$|^(11+?)\1+$/&&"non-",prime

Combined C and Perl

Aeneus

It looks like you have "non-prime" and "prime" flip-flopped. – Choler

Whoops! Fixed and still 135C. – Xerography

Answer 20

Perl, 113 109 105 chars

Beating all Python answers at this moment! SCNR.

$n=$s=<>;$s=0,s/\d/$s+=$&*$&/ge while($_=$s)>4;die$s>1?sad:happy,$","non-"x(1x$n)=~/^1$|(^11+)\1+$/,prime

Intermingle

About time! Now see if you can beat the GolfScript :) – Whittle

I'll pass on beating your Golfscript. Instead, I'll beat your python-fu with my sed-fu :P – Intermingle

Answer 21

Javascript, 192 190 185 182 165 158 chars

The prime checking runs from 2 to square root of N. I wasted few chars there...

In one line:

for(x=2,y=m=n=prompt();x*x<y&&n%x++;);for(s={};!s[m];m=p)for(s[m]=1,p=0;m;m=(m-=k=m%10)/10,p+=k*k);alert((m-1?'sad':'happy')+(n-1&&x*x>y?' ':' non-')+'prime')

Formatted:

// Getting the number from the input and checking for primeness
// (ie. if after the loop x>y => n is prime)
for (x=2, y=m=n=prompt(); x*x<y && n%x++;)

// Checking for happiness
// the loop is broken out of if m is already encountered
// the m==1 after the loop indicates happy number
for(s={}; !s[m]; m=p)
    for (s[m]=1, p=0; m; m=(m -= k=m%10)/10, p+=k * k);

alert((m-1 ? 'sad' : 'happy') + (n-1 && x*x>y ? ' ' : ' non-') + 'prime')

Check: http://jsfiddle.net/TwxAW/6/

Discover

This fails in the case where n=1 where it should say "happy non-prime" instead of "happy prime". – Choler

Answer 22

PHP 217 Chars

$t=$argv[1];for($z=$t-1,$p=1;$z&&++$p<$t;)$z=$t%$p;$f=array(1);while(!in_array($t,$f,1)){$f[]=$t;$t=array_reduce(str_split($t),function($v,$c){return $v+=$c*$c;});}print($t<2?"happy ":"sad ").(!$z?"non-":"")."prime";

Usage:

$ php -r '$t=$argv[1];for($z=$t-1,$p=1;$z&&++$p<$t;)$z=$t%$p;$f=array(1);while(!in_array($t,$f,1)){$f[]=$t;$t=array_reduce(str_split($t),function($v,$c){return $v+=$c*$c;});}print($t<2?"happy ":"sad ").(!$z?"non-":"")."prime";' 139
happy prime

Haustorium

This does extremely well for everything except for the 1 case. – Choler

Oops! Fixed, added 3 characters (total 217 now) to handle the proper prime check for 1. – Haustorium

Answer 23

Scala, 253 247 246

object H{def main(a:Array[String]){var s=Set(0)
val n=a(0)toInt
def r(m:Int):String={val k=""+m map(c=>c*(c-96)+2304)sum;if(k<2)"happy"else if(s(k))"sad"else{s+=k;r(k)}}
printf("%s %sprime",r(n),if(n<2|(2 to n-1 exists(n%_==0)))"non-"else"")}}

Probably there is some room for improvements. The damn test for 1 as non-prime costs 6 chars :-(

Lenora

Answer 24

F#, 249 chars

let n=stdin.ReadLine()|>int
let rec s x=seq{yield x;yield!string x|>Seq.sumBy(fun c->(int c-48)*(int c-48))|>s}
printfn"%s %sprime"(if s n|>Seq.take 99|>Seq.exists((=)1)then"happy"else"sad")(if[2..n/2]|>Seq.exists(fun d->n%d=0)then"non-"else"")

Hydroxylamine

Answer 25

Clojure, 353 318 298 261 230 chars

(defn h[x m](cond(= x 1)"happy "(m x)"sad ":else(recur(reduce +(for[n(map #(-(int %)48)(str x))](* n n)))(assoc m x 1))))(println(let [x (read)](str(h x{})(if(re-matches #"^1$|^(11+)?\1+"(apply str(repeat x\1)))"non-""")"prime")))

ptimac:clojure pti$ clj  happy.clj 139
CP=/Users/pti/playpen/clojure:/Users/pti/Library/Clojure/lib/clojure.jar:/Users/pti/Library/Clojure/lib/jline.jar:/Users/pti/Library/Clojure/lib/clojure-contrib.jar
happy prime
ptimac:clojure pti$ clj  happy.clj 440
CP=/Users/pti/playpen/clojure:/Users/pti/Library/Clojure/lib/clojure.jar:/Users/pti/Library/Clojure/lib/jline.jar:/Users/pti/Library/Clojure/lib/clojure-contrib.jar
happy non-prime
ptimac:clojure pti$ clj  happy.clj 2
CP=/Users/pti/playpen/clojure:/Users/pti/Library/Clojure/lib/clojure.jar:/Users/pti/Library/Clojure/lib/jline.jar:/Users/pti/Library/Clojure/lib/clojure-contrib.jar
sad prime
ptimac:clojure pti$ clj  happy.clj 78
CP=/Users/pti/playpen/clojure:/Users/pti/Library/Clojure/lib/clojure.jar:/Users/pti/Library/Clojure/lib/jline.jar:/Users/pti/Library/Clojure/lib/clojure-contrib.jar
sad non-prime

I am leaning on the clojure contrib for the primes sequence. I wonder if using for loops would be shorter than recursion?

I read up on the regex prime number check. It's awesome and removes 30 chars and my dependency on clojure.contrib. I also refactored the command line parsing somwhat and inlined a function.

Pre golf(somewhat outdated):

(defn h[x m]
  (cond
   (= x 1) "happy "
   (m x) "sad "
   :else (recur
               (reduce +
                 (for [n (map #(- (int %) 48) (str x))] (* n n))) 
               (assoc m x 1))))

    (println
      (let [x (read)]
        (str
           (h x{})
           (if (re-matches #"^1$|^(11+)?\1+"(apply str(repeat x \1)))
             "non-"
             "")
           "prime")))

Tricornered

Nice, I've never worked with clojure before, but I had to change the (nth *command-line-args* 1) to (nth *command-line-args* 2). (1 being your script's name). – Choler

Yes that seems to have to do with the bash scripts and the change from the Script class to clojure.main for script support. – Tricornered

Answer 26

PARI/GP

184 bytes

A Bit late to participate, but here's a short one.

a(x,v=[])=s=0;while(x,s+=(x%10)^2;x\=10);for(k=1,length(v),if(v[k]==s,return("sad ")));return(if(s==1,"happy ",v=concat(v,s);a(s,v)));f(x)=print(a(x),if(!isprime(x),"non-",""),"prime")

To use it. Write f(139)

Alkyd

Answer 27

GNU sed, 146 125 chars

Run with sed -rf file. Using -r saves 5 backslashes.

Needs bc, printf, and a shell with support for here-strings.

h
s/^/printf %*s /e
s/^ $|^(  +)\1+$/non-/
s/ *$/prime/
x
:a
s/./+&*&/g
s//bc<<</e
tb
:b
s/^1$/happy/
s/^4$/sad/
Ta
G
s/\n/ /

GNU sed, 155 141 chars (needs neither printf nor here-strings)

Uses the more standard traditional yes and head instead of printf.

h
:a
s/./+&*&/g
s/.*/echo 0&|bc/e
tb
:b
s/^1$/happy/
s/^4$/sad/
Ta
x
s/^/yes|head -/e
s/\n//g
s/^y$|^(yy+)\1+$/non-/
s/y*$/prime/
x
G
s/\n/ /

GNU sed, 134 115 chars (slightly bad formatted output)

Slightly shorter version, doesn't respect output formatting (has extra spaces and a newline between happy/sad and (non-)prime).

h
:a
s/./+&*&/g
s//bc<<</e
tb
:b
s/^1$/happy/
s/^4$/sad/
Ta
p
g
s/^/printf %*s /e
s/^ $|^(  +)\1+$/non-/
s/$/prime/

Intermingle

Answer 28

Python - 142 chars

I was playing around with this idea, but it turned out too long. Perhaps someone can find a way to make it shorter. Maybe it'll turn out better in Ruby. Should be fun to understand how it works anyway :)

n=input();L=[n];print"%s non-prime"[4*([1for f in range(1,n)if L.append(sum(int(x)**2for x in`L[-1]`))or n%f<1]==[1]):]%['sad','happy'][1in L]

Whittle

Answer 29

Python, 169 168 158 157 166 164 162 chars, 4 lines

l=n=input()
while l>4:l=sum(int(i)**2for i in str(l))
print['sad','happy'][l==1and str(n)!=1],
print['non-',''][n!=1 and sum(n%i==0for i in range(1,n))<2]+"prime"

Takes a number from stdin and doesn't muck around with regexes like the other python answer, although I must admit that is pretty cool. I could also shave off 6 chars by using backticks instead of the str-function, but let's play nice.

EDIT: Fixed a bug with 1 being a prime, which bumped up the charcount by 10. I figure there must be a more concise way than mine for doing this.

EDIT 2: Apparently, python 2.6 allows print[1, 2] without a space between the two.

EDIT 3: Used another calculation for the happy numbers

Guitar

I do appreciate the non-use of regex, but yours seems to detect 1 as prime and 7 as sad. (Also, strictly speaking the last comma should be a +.) – Choler

Hmmm, it still claims that 7 is a "sad prime", instead of "happy prime". Also, you could remove the space before both the fors as well. – Choler

Answer 30

Python (285 270 269 246 241 247 240 237 chars, 21 20 21 18 19 lines)

n=input()
s='prime'
for i in range(2,n):
    if n%i==0: 
        s='non'+s
        break
f=list(str(n))
g=set()
while n!=1:
    n=sum([int(z)**2 for z in f])
    if n in g:
        s='sad '+s
        break
    else:
        f=list(str(n))
        g.add(n)
else:
    s='happy '+s
print s

EDIT: Yes, the number went up, there was a bug :-P

Zug

using n=input() will really cut down on your char count. And it seems to be accepted in the rules. And some things to fix: since you use a for loop it appends lots of non's to the s. Also, 1 is not prime so you'd need to fix that. – Humorist

Yes, I realised that as soon as I posted. Edited now. – Zug

You don't need the int(). input() already evaluates the input and makes it an int if it already is one. – Humorist

I fixed the loop problem and the unnecessary int(). You just have to fix the 1 case. – Humorist

Oh, I didn't know that. Also managed to cut down one more line by putting the if and statement inside a single line. Also, the lambda was unnecessary, so, one more line and a bunch of characters gone there... – Zug

Did your changes get posted? I don't see them here. Oh, and the int(n) in your for loop is unnecessary because n is already an integer. EDIT: Nevermind; the changes are here now. – Humorist

Done :). I don't think it can be shortened any more now without serious mental effort. It's the shortest Python solution here by a long way. – Zug

You need that break statement in your for loop or else you get this result for 440: happy nonnonnonnonnonnonnonnonnonnonnonnonnonnonprime – Humorist

Made it even shorter using while: else:. – Zug

Change range(2,n/2+1) to range(2,n-1) to save some characters? Who cares about efficiency? Don't you need a - after non? – Manamanacle

Actually, since xrange(start, stop) is non-inclusive of the last number, just xrange(2,n) will do :) – Zug

Rather than s='non'+s;break you can use s='non-prime' and save a few chars. – Mme

@Gabe: but he'd still need the break statement in there so it doesn't do the nonononononononononononprime problem. – Humorist

vlad003: The break was only necessary because he was concatenating with the previous value. Since s='non-prime' is idempotent, there's no need for the break. – Mme

Answer 31

Python 2.6, 300 298 294 chars

Different from previous answer in that this doesn't use regex.

I'm sure there's some way of shortening my h(x) function, but I'm still learning Python so I've got no idea.

p(x) returns True if it's a non-prime. h(x) returns True if it's happy. I've done the t = True so that it shortens the character count when doing the truth check.

x=input()
def p(x):
 if x==1 or 1 in [1 for i in range(2,x) if x%i==0]: return True
def h(x):
 l=[]
 while x not in l:
  l.append(x)
  x=sum([int(i)**2 for i in str(x)])
 if 1 in l: return True
if h(x):print'happy',
elif not h(x):print'sad',
if p(x):print'non-prime'
elif not p(x):print'prime'

Cronin

Close, you'll need to fix it for the 1 case (1 being non-prime). Also you can strip the [] in the sum(). You could also probably just do if h(x): and the like. – Choler

Can someone please test this for me. I'm getting a weird problem: if I run this as a .py file from CLI, it says that 1 is prime. If I write the exact same function in an interpreter, it says 1 is not prime... – Humorist

So I think I fixed the 1 case this time. I'm still confused a bit about the whole logic of it though :P – Humorist

Answer 32

JavaScript - 171 (Closure Compiler Compressed)

Just found this fun challenge. This is 171 characters in JavaScript:

function h(a){for(b=2,c="";b<a;b++)a%b||(c="non-");c+="prime";for(b={};;){if(a==1)return"happy "+c;if(b[a])return"sad "+c;b[a]=1;for(e=0;a;){d=a%10;e+=d*d;a=(a-d)/10}a=e}}

I ran the source code through the Closure Compiler in Advanced Mode to compress it. The source:

function h(num)
{
    // Check for prime-ness
    // Save the word "prime" by attaching a "non-" in front if it has a factor
    // Notice that we are not concerned with performance here, otherwise should
    // break when a factor is found and should only go up to Math.sqrt(num)

    for (var x=2, prime=""; x < num; x++) {
        if (!(num % x)) prime="non-";
    }

    // Create "prime" or "non-prime"

    prime += "prime";

    // Keep all visited numbers in a hash

    var visited = {};

    while(true) {
        // If the number is one, then happy
        if (num == 1) return "happy " + prime;

        // Otherwise put it into the hash
        if (visited[num]) return "sad " + prime;
        visited[num] = 1;

        // Recalculate the new number by shifting through each digit
        var sum = 0;
        while (num) {   // Stop when num -> 0 (i.e. no more digits)
            var digit = num % 10;
            sum += digit * digit;
            num = (num - digit)/10; // Shift by one digit
        }

        num = sum;
    }
}

Poriferous

Answer 33

Racket, 286 characters

Only issue is that 1 is prime.

(define d
  (lambda (n)
    (cond
      ((= n 0) 0)
      ((+
       (expt (modulo n 10) 2)
       (d (floor (/ n 10)))
       ))
      )
    )
  )

(define dd
  (lambda (n l)
    (cond
      ((= n 1) "happy ")
      ((for/or ([v l]) (= n v)) "sad ")
      ((dd (d n) (cons n l)))
      )
    )
  )

(define cg
  (lambda (n)
    (display (dd n '()))
    (display
    (if
     (for/and
       ([i (in-range 2 (floor (+ (sqrt n) 1)))]) (> (modulo n i) 0))
       "prime" "non-prime"
     ))
    )
  )

Braca