Happy numbers

From Rosetta Code
Jump to: navigation, search
Task
Happy numbers
You are encouraged to solve this task according to the task description, using any language you may know.

From Wikipedia, the free encyclopedia:

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. Display an example of your output here.

Task: Find and print the first 8 happy numbers.

See also:


Contents

[edit] ACL2

(include-book "arithmetic-3/top" :dir :system)
 
(defun sum-of-digit-squares (n)
(if (zp n)
0
(+ (expt (mod n 10) 2)
(sum-of-digit-squares (floor n 10)))))
 
(defun is-happy-r (n seen)
(let ((next (sum-of-digit-squares n)))
(cond ((= next 1) t)
((member next seen) nil)
(t (is-happy-r next (cons next seen))))))
 
(defun is-happy (n)
(is-happy-r n nil))
 
(defun first-happy-nums-r (n i)
(cond ((zp n) nil)
((is-happy i)
(cons i (first-happy-nums-r (1- n) (1+ i))))
(t (first-happy-nums-r n (1+ i)))))
 
(defun first-happy-nums (n)
(first-happy-nums-r n 1))

Output:

(1 7 10 13 19 23 28 31)

[edit] ActionScript

function sumOfSquares(n:uint)
{
var sum:uint = 0;
while(n != 0)
{
sum += (n%10)*(n%10);
n /= 10;
}
return sum;
}
function isInArray(n:uint, array:Array)
{
for(var k = 0; k < array.length; k++)
if(n == array[k]) return true;
return false;
}
function isHappy(n)
{
var sequence:Array = new Array();
while(n != 1)
{
sequence.push(n);
n = sumOfSquares(n);
if(isInArray(n,sequence))return false;
}
return true;
}
function printHappy()
{
var numbersLeft:uint = 8;
var numberToTest:uint = 1;
while(numbersLeft != 0)
{
if(isHappy(numberToTest))
{
trace(numberToTest);
numbersLeft--;
}
numberToTest++;
}
}
printHappy();

Sample output:

1
7
10
13
19
23
28
31

[edit] Ada

with Ada.Text_IO;  use Ada.Text_IO;
with Ada.Containers.Ordered_Sets;
 
procedure Test_Happy_Digits is
function Is_Happy (N : Positive) return Boolean is
package Sets_Of_Positive is new Ada.Containers.Ordered_Sets (Positive);
use Sets_Of_Positive;
function Next (N : Positive) return Natural is
Sum  : Natural := 0;
Accum : Natural := N;
begin
while Accum > 0 loop
Sum  := Sum + (Accum mod 10) ** 2;
Accum := Accum / 10;
end loop;
return Sum;
end Next;
Current : Positive := N;
Visited : Set;
begin
loop
if Current = 1 then
return True;
elsif Visited.Contains (Current) then
return False;
else
Visited.Insert (Current);
Current := Next (Current);
end if;
end loop;
end Is_Happy;
Found : Natural := 0;
begin
for N in Positive'Range loop
if Is_Happy (N) then
Put (Integer'Image (N));
Found := Found + 1;
exit when Found = 8;
end if;
end loop;
end Test_Happy_Digits;

Sample output:

 1 7 10 13 19 23 28 31

[edit] ALGOL 68

Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
INT base10 = 10, num happy = 8;
 
PROC next = (INT in n)INT: (
INT n := in n;
INT out := 0;
WHILE n NE 0 DO
out +:= ( n MOD base10 ) ** 2;
n := n OVER base10
OD;
out
);
 
PROC is happy = (INT in n)BOOL: (
INT n := in n;
FOR i WHILE n NE 1 AND n NE 4 DO n := next(n) OD;
n=1
);
 
INT count := 0;
FOR i WHILE count NE num happy DO
IF is happy(i) THEN
count +:= 1;
print((i, new line))
FI
OD

Output:

         +1
         +7
        +10
        +13
        +19
        +23
        +28
        +31

[edit] APL

     ∇ HappyNumbers arg;⎕IO;∆roof;∆first;bin;iroof
[1] ⍝0: Happy number
[2] ⍝1: http://rosettacode.org/wiki/Happy_numbers
[3] ⎕IO←1 ⍝ Index origin
[4] ∆roof ∆first←2↑arg,10 ⍝
[5]
[6] bin←{
[7] ⍺←⍬ ⍝ Default left arg
[8] ⍵=1:1 ⍝ Always happy!
[9]
[10] numbers←⍎¨1⊂⍕⍵ ⍝ Split numbers into parts
[11] next←+/{⍵*2}¨numbers ⍝ Sum and square of numbers
[12]
[13] next∊⍺:0 ⍝ Return 0, if already exists
[14] (⍺,next)∇ next ⍝ Check next number (recursive)
[15]
[16] }¨iroof←⍳∆roof ⍝ Does all numbers upto ∆root smiles?
[17]
[18] ⎕←~∘0¨∆first↑bin/iroof ⍝ Show ∆first numbers, but not 0
      HappyNumbers 100 8
 1  7  10  13  19  23  28  31 

[edit] AutoHotkey

Loop {
If isHappy(A_Index) {
out .= (out="" ? "" : ",") . A_Index
i ++
If (i = 8) {
MsgBox, The first 8 happy numbers are: %out%
ExitApp
}
}
}
 
isHappy(num, list="") {
list .= (list="" ? "" : ",") . num
Loop, Parse, num
sum += A_LoopField ** 2
If (sum = 1)
Return true
Else If sum in %list%
Return false
Else Return isHappy(sum, list)
}
The first 8 happy numbers are: 1,7,10,13,19,23,28,31

[edit] Alternative version

while h < 8
if (Happy(A_Index)) {
Out .= A_Index A_Space
h++
}
MsgBox, % Out
 
Happy(n) {
Loop, {
Loop, Parse, n
t += A_LoopField ** 2
if (t = 89)
return, 0
if (t = 1)
return, 1
n := t, t := 0
}
}
1 7 10 13 19 23 28 31

[edit] AutoIt

 
$c = 0
$k = 0
While $c < 8
$k += 1
$n = $k
While $n <> 1
$s = StringSplit($n, "")
$t = 0
For $i = 1 To $s[0]
$t += $s[$i] ^ 2
Next
$n = $t
Switch $n
Case 4,16,37,58,89,145,42,20
ExitLoop
EndSwitch
WEnd
If $n = 1 Then
ConsoleWrite($k & " is Happy" & @CRLF)
$c += 1
EndIf
WEnd
 
Use a set of numbers (4,16,37,58,89,145,42,20) to indicate a loop and exit.
Output:
1 is Happy
7 is Happy
10 is Happy
13 is Happy
19 is Happy
23 is Happy
28 is Happy
31 is Happy

[edit] Alternative version

 
$c = 0
$k = 0
While $c < 8
$a = ObjCreate("System.Collections.ArrayList")
$k += 1
$n = $k
While $n <> 1
If $a.Contains($n) Then
ExitLoop
EndIf
$a.add($n)
$s = StringSplit($n, "")
$t = 0
For $i = 1 To $s[0]
$t += $s[$i] ^ 2
Next
$n = $t
WEnd
If $n = 1 Then
ConsoleWrite($k & " is Happy" & @CRLF)
$c += 1
EndIf
$a.Clear
WEnd
 
Saves all numbers in a list, duplicate entry indicates a loop.
Output:
1 is Happy
7 is Happy
10 is Happy
13 is Happy
19 is Happy
23 is Happy
28 is Happy
31 is Happy

[edit] AWK

function is_happy(n)
{
if ( n in happy ) return 1;
if ( n in unhappy ) return 0;
cycle[""] = 0
while( (n!=1) && !(n in cycle) ) {
cycle[n] = n
new_n = 0
while(n>0) {
d = n % 10
new_n += d*d
n = int(n/10)
}
n = new_n
}
if ( n == 1 ) {
for (i_ in cycle) {
happy[cycle[i_]] = 1
delete cycle[i_]
}
return 1
} else {
for (i_ in cycle) {
unhappy[cycle[i_]] = 1
delete cycle[i_]
}
return 0
}
}
 
BEGIN {
cnt = 0
happy[""] = 0
unhappy[""] = 0
for(j=1; (cnt < 8); j++) {
if ( is_happy(j) == 1 ) {
cnt++
print j
}
}
}

Result:

1
7
10
13
19
23
28
31

[edit] Alternative version

Alternately, for legibility one might write:

BEGIN {
for (i = 1; i < 50; ++i){
if (isHappy(i)) {
print i;
}
}
exit
}
 
function isHappy(n, seen) {
delete seen;
while (1) {
n = sumSqrDig(n)
if (seen[n]) {
return n == 1
}
seen[n] = 1
}
}
 
function sumSqrDig(n, d, tot) {
while (n) {
d = n % 10
tot += d * d
n = int(n/10)
}
return tot
}

[edit] Batch File

happy.bat

@echo off
setlocal enableDelayedExpansion
::Define a list with 10 terms as a convenience for defining a loop
set "L10=0 1 2 3 4 5 6 7 8 9"
shift /1 & goto %1
exit /b
 
 
:list min count
:: This routine prints all happy numbers > min (arg1)
:: until it finds count (arg2) happy numbers.
set /a "n=%~1, cnt=%~2"
call :listInternal
exit /b
 
 
:test min [max]
:: This routine sequentially tests numbers between min (arg1) and max (arg2)
:: to see if they are happy. If max is not specified then it defaults to min.
set /a "min=%~1"
if "%~2" neq "" (set /a "max=%~2") else set max=%min%
::The FOR /L loop does not detect integer overflow, so must protect against
::an infinite loop when max=0x7FFFFFFFF
set end=%max%
if %end% equ 2147483647 set /a end-=1
for /l %%N in (%min% 1 %end%) do (
call :testInternal %%N && (echo %%N is happy :^)) || echo %%N is sad :(
)
if %end% neq %max% call :testInternal %max% && (echo %max% is happy :^)) || echo %max% is sad :(
exit /b
 
 
 :listInternal
 :: This loop sequentially tests each number >= n. The loop conditionally
 :: breaks within the body once cnt happy numbers have been found, or if
 :: the max integer value is reached. Performance is improved by using a
 :: FOR loop to perform most of the looping, with a GOTO only needed once
 :: per 100 iterations.
for %%. in (
%L10% %L10% %L10% %L10% %L10% %L10% %L10% %L10% %L10% %L10%
) do (
call :testInternal !n! && (
echo !n!
set /a cnt-=1
if !cnt! leq 0 exit /b 0
)
if !n! equ 2147483647 (
>&2 echo ERROR: Maximum integer value reached
exit /b 1
)
set /a n+=1
)
goto :listInternal
 
 
 :testInternal n
 :: This routine loops until the sum of squared digits converges on 1 (happy)
 :: or it detects a cycle (sad). It exits with errorlevel 0 for happy and 1 for sad.
 :: Performance is improved by using a FOR loop for the looping instead of a GOTO.
 :: Numbers less than 1000 never neeed more than 20 iterations, and any number
 :: with 4 or more digits shrinks by at least one digit each iteration.
 :: Since Windows batch can't handle more than 10 digits, allowance for 27
 :: iterations is enough, and 30 is more than adequate.
setlocal
set n=%1
for %%. in (%L10% %L10% %L10%) do (
if !n!==1 exit /b 0
%= Only numbers < 1000 can cycle =%
if !n! lss 1000 (
if defined t.!n! exit /b 1
set t.!n!=1
)
%= Sum the squared digits =%
%= Batch can't handle numbers greater than 10 digits so we can use =%
%= a constrained FOR loop and avoid a slow goto =%
set sum=0
for /l %%N in (1 1 10) do (
if !n! gtr 0 set /a "sum+=(n%%10)*(n%%10), n/=10"
)
set /a n=sum
)

Sample usage and output

>happy list 1 8
1
7
10
13
19
23
28
31

>happy list 1000000000 10
1000000000
1000000003
1000000009
1000000029
1000000030
1000000033
1000000039
1000000067
1000000076
1000000088

>happy test 30
30 is sad :(

>happy test 31
31 is happy :)

>happy test 1 10
1 is happy :)
2 is sad :(
3 is sad :(
4 is sad :(
5 is sad :(
6 is sad :(
7 is happy :)
8 is sad :(
9 is sad :(
10 is happy :)

>happy test "50 + 10 * 5"
100 is happy :)

>happy test 0x7fffffff
2147483647 is sad :(

>happy test 0x7ffffffd
2147483645 is happy :)

>happy list 0x7ffffff0 10
2147483632
2147483645
ERROR: Maximum integer value reached

[edit] BBC BASIC

      number% = 0
total% = 0
REPEAT
number% += 1
IF FNhappy(number%) THEN
PRINT number% " is a happy number"
total% += 1
ENDIF
UNTIL total% = 8
END
 
DEF FNhappy(num%)
LOCAL digit&()
DIM digit&(10)
REPEAT
digit&() = 0
$$^digit&(0) = STR$(num%)
digit&() AND= 15
num% = MOD(digit&())^2 + 0.5
UNTIL num% = 1 OR num% = 4
= (num% = 1)

Output:

         1 is a happy number
         7 is a happy number
        10 is a happy number
        13 is a happy number
        19 is a happy number
        23 is a happy number
        28 is a happy number
        31 is a happy number

[edit] Bori

bool isHappy (int n)
{
ints cache;
 
while (n != 1)
{
int sum = 0;
 
if (cache.contains(n))
return false;
 
cache.add(n);
while (n != 0)
{
int digit = n % 10;
sum += (digit * digit);
n = (int)(n / 10);
}
n = sum;
}
return true;
}
 
void test ()
{
int num = 1;
ints happynums;
 
while (happynums.count() < 8)
{
if (isHappy(num))
happynums.add(num);
num++;
}
puts("First 8 happy numbers : " + str.newline + happynums);
}

Output:

First 8 happy numbers : 
[1, 7, 10, 13, 19, 23, 28, 31]

[edit] Brat

include :set
 
happiness = set.new 1
sadness = set.new
 
sum_of_squares_of_digits = { num |
num.to_s.dice.reduce 0 { sum, n | sum = sum + n.to_i ^ 2 }
}
 
happy? = { n, seen = set.new |
when {true? happiness.include? n } { happiness.merge seen << n; true }
{ true? sadness.include? n } { sadness.merge seen; false }
{ true? seen.include? n } { sadness.merge seen; false }
{ true } { seen << n; happy? sum_of_squares_of_digits(n), seen }
}
 
num = 1
happies = []
 
while { happies.length < 8 } {
true? happy?(num)
{ happies << num }
 
num = num + 1
}
 
p "First eight happy numbers: #{happies}"
p "Happy numbers found: #{happiness.to_array.sort}"
p "Sad numbers found: #{sadness.to_array.sort}"

Output:

First eight happy numbers: [1, 7, 10, 13, 19, 23, 28, 31]
Happy numbers found: [1, 7, 10, 13, 19, 23, 28, 31, 49, 68, 82, 97, 100, 130]
Sad numbers found: [2, 3, 4, 5, 6, 8, 9, 11, 12, 14, 15, 16, 17, 18, 20, 21, 22, 24, 25, 26, 27, 29, 30, 34, 36, 37, 40, 41, 42, 45, 50, 52, 53, 58, 61, 64, 65, 81, 85, 89, 145]

[edit] C

Recursively look up if digit square sum is happy.

#include <stdio.h>
 
#define CACHE 256
enum { h_unknown = 0, h_yes, h_no };
unsigned char buf[CACHE] = {0, h_yes, 0};
 
int happy(int n)
{
int sum = 0, x, nn;
if (n < CACHE) {
if (buf[n]) return 2 - buf[n];
buf[n] = h_no;
}
 
for (nn = n; nn; nn /= 10) x = nn % 10, sum += x * x;
 
x = happy(sum);
if (n < CACHE) buf[n] = 2 - x;
return x;
}
 
int main()
{
int i, cnt = 8;
for (i = 1; cnt || !printf("\n"); i++)
if (happy(i)) --cnt, printf("%d ", i);
 
printf("The %dth happy number: ", cnt = 1000000);
for (i = 1; cnt; i++)
if (happy(i)) --cnt || printf("%d\n", i);
 
return 0;
}
output
1 7 10 13 19 23 28 31 
The 1000000th happy number: 7105849

Without caching, using cycle detection:

#include <stdio.h>
 
int dsum(int n)
{
int sum, x;
for (sum = 0; n; n /= 10) x = n % 10, sum += x * x;
return sum;
}
 
int happy(int n)
{
int nn;
while (n > 999) n = dsum(n); /* 4 digit numbers can't cycle */
nn = dsum(n);
while (nn != n && nn != 1)
n = dsum(n), nn = dsum(dsum(nn));
return n == 1;
}
 
int main()
{
int i, cnt = 8;
for (i = 1; cnt || !printf("\n"); i++)
if (happy(i)) --cnt, printf("%d ", i);
 
printf("The %dth happy number: ", cnt = 1000000);
for (i = 1; cnt; i++)
if (happy(i)) --cnt || printf("%d\n", i);
 
return 0;
}
Output is same as above, but much slower.

[edit] C++

Translation of: Python
#include <map>
#include <set>
 
bool happy(int number) {
static std::map<int, bool> cache;
 
std::set<int> cycle;
while (number != 1 && !cycle.count(number)) {
if (cache.count(number)) {
number = cache[number] ? 1 : 0;
break;
}
cycle.insert(number);
int newnumber = 0;
while (number > 0) {
int digit = number % 10;
newnumber += digit * digit;
number /= 10;
}
number = newnumber;
}
bool happiness = number == 1;
for (std::set<int>::const_iterator it = cycle.begin();
it != cycle.end(); it++)
cache[*it] = happiness;
return happiness;
}
 
#include <iostream>
 
int main() {
for (int i = 1; i < 50; i++)
if (happy(i))
std::cout << i << std::endl;
return 0;
}

Output:

1
7
10
13
19
23
28
31
32
44
49

Alternative version without caching:

unsigned int happy_iteration(unsigned int n)
{
unsigned int result = 0;
while (n > 0)
{
unsigned int lastdig = n % 10;
result += lastdig*lastdig;
n /= 10;
}
return result;
}
 
bool is_happy(unsigned int n)
{
unsigned int n2 = happy_iteration(n);
while (n != n2)
{
n = happy_iteration(n);
n2 = happy_iteration(happy_iteration(n2));
}
return n == 1;
}
 
#include <iostream>
 
int main()
{
unsigned int current_number = 1;
 
unsigned int happy_count = 0;
while (happy_count != 8)
{
if (is_happy(current_number))
{
std::cout << current_number << " ";
++happy_count;
}
++current_number;
}
std::cout << std::endl;
}

Output:

1 7 10 13 19 23 28 31 

Cycle detection in is_happy() above is done using Floyd's cycle-finding algorithm.

[edit] C#

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
 
namespace HappyNums
{
class Program
{
public static bool ishappy(int n)
{
List<int> cache = new List<int>();
int sum = 0;
while (n != 1)
{
if (cache.Contains(n))
{
return false;
}
cache.Add(n);
while (n != 0)
{
int digit = n % 10;
sum += digit * digit;
n /= 10;
}
n = sum;
sum = 0;
}
return true;
}
 
static void Main(string[] args)
{
int num = 1;
List<int> happynums = new List<int>();
 
while (happynums.Count < 8)
{
if (ishappy(num))
{
happynums.Add(num);
}
num++;
}
Console.WriteLine("First 8 happy numbers : " + string.Join(",", happynums));
}
}
}
First 8 happy numbers : 1,7,10,13,19,23,28,31

[edit] Clojure

(defn happy? [n]
(loop [n n, seen #{}]
(cond
(= n 1) true
(seen n) false
 :else
(recur (->> (str n)
(map #(Character/digit % 10))
(map #(* % %))
(reduce +))
(conj seen n)))))
 
(def happy-numbers (filter happy? (iterate inc 1)))
 
(println (take 8 happy-numbers))
Output:
(1 7 10 13 19 23 28 31)

[edit] CoffeeScript

happy = (n) ->
seen = {}
while true
n = sum_digit_squares(n)
return true if n == 1
return false if seen[n]
seen[n] = true
 
sum_digit_squares = (n) ->
sum = 0
for c in n.toString()
d = parseInt(c)
sum += d*d
sum
 
i = 1
cnt = 0
while cnt < 8
if happy(i)
console.log i
cnt += 1
i += 1

output

> coffee happy.coffee 
1
7
10
13
19
23
28
31

[edit] Common Lisp

(defun sqr (n)
(* n n))
 
(defun sum-of-sqr-dgts (n)
(loop for i = n then (floor i 10)
while (plusp i)
sum (sqr (mod i 10))))
 
(defun happy-p (n &optional cache)
(or (= n 1)
(unless (find n cache)
(happy-p (sum-of-sqr-dgts n)
(cons n cache)))))
 
(defun happys (&aux (happys 0))
(loop for i from 1
while (< happys 8)
when (happy-p i)
collect i and do (incf happys)))
 
(print (happys))
 
Output:
(1 7 10 13 19 23 28 31)

[edit] D

bool isHappy(int n) pure nothrow {
int[int] past;
 
while (true) {
int total = 0;
while (n > 0) {
total += (n % 10) ^^ 2;
n /= 10;
}
if (total == 1)
return true;
if (total in past)
return false;
n = total;
past[total] = 0;
}
}
 
void main() {
import std.stdio, std.algorithm, std.range;
 
int.max.iota.filter!isHappy.take(8).writeln;
}
Output:
[1, 7, 10, 13, 19, 23, 28, 31]

[edit] Alternative Version

import std.stdio, std.algorithm, std.range, std.conv, std.string;
 
bool isHappy(int n) pure nothrow {
int[int] seen;
 
while (true) {
immutable t = n.text.representation.map!q{(a - '0') ^^ 2}.sum;
if (t == 1)
return true;
if (t in seen)
return false;
n = t;
seen[t] = 0;
}
}
 
void main() {
int.max.iota.filter!isHappy.take(8).writeln;
}

Same output.

[edit] Dart

main() {
HashMap<int,bool> happy=new HashMap<int,bool>();
happy[1]=true;
 
int count=0;
int i=0;
 
while(count<8) {
if(happy[i]==null) {
int j=i;
Set<int> sequence=new Set<int>();
while(happy[j]==null && !sequence.contains(j)) {
sequence.add(j);
int sum=0;
int val=j;
while(val>0) {
int digit=val%10;
sum+=digit*digit;
val=(val/10).toInt();
}
j=sum;
}
bool sequenceHappy=happy[j];
Iterator<int> it=sequence.iterator();
while(it.hasNext()) {
happy[it.next()]=sequenceHappy;
}
}
if(happy[i]) {
print(i);
count++;
}
i++;
}
}

[edit] dc

[lcI~rscd*+lc0<H]sH
[0rsclHxd4<h]sh
[lIp]s_
0sI[lI1+dsIlhx2>_z8>s]dssx

Output:

1
7
10
13
19
23
28
31

[edit] Déjà Vu

next-num:
0
while over:
over
* dup % swap 10
+
swap floor / swap 10 swap
drop swap
 
is-happy happies n:
if has happies n:
return happies! n
local :seq set{ n }
n
while /= 1 dup:
next-num
if has seq dup:
drop
set-to happies n false
return false
if has happies dup:
set-to happies n dup happies!
return
set-to seq over true
drop
set-to happies n true
true
 
local :h {}
1 0
while > 8 over:
if is-happy h dup:
!print( "A happy number: " over )
swap ++ swap
++
drop
drop
Output:
A happy number: 1
A happy number: 7
A happy number: 10
A happy number: 13
A happy number: 19
A happy number: 23
A happy number: 28
A happy number: 31

[edit] DWScript

function IsHappy(n : Integer) : Boolean;
var
cache : array of Integer;
sum : Integer;
begin
while True do begin
sum := 0;
while n>0 do begin
sum += Sqr(n mod 10);
n := n div 10;
end;
if sum = 1 then
Exit(True);
if sum in cache then
Exit(False);
n := sum;
cache.Add(sum);
end;
end;
 
var n := 8;
var i : Integer;
 
while n>0 do begin
Inc(i);
if IsHappy(i) then begin
PrintLn(i);
Dec(n);
end;
end;

Output:

1
7
10
13
19
23
28
31

[edit] E

This example does not show the output mentioned in the task description on this page (or a page linked to from here). Please ensure that it meets all task requirements and remove this message.
Note that phrases in task descriptions such as "print and display" and "print and show" for example, indicate that (reasonable length) output be a part of a language's solution.


def isHappyNumber(var x :int) {
var seen := [].asSet()
while (!seen.contains(x)) {
seen with= x
var sum := 0
while (x > 0) {
sum += (x % 10) ** 2
x //= 10
}
x := sum
if (x == 1) { return true }
}
return false
}
 
var count := 0
for x ? (isHappyNumber(x)) in (int >= 1) {
println(x)
if ((count += 1) >= 8) { break }
}

[edit] Erlang

-module(tasks).
-export([main/0]).
-import(lists, [map/2, member/2, sort/1, sum/1]).
 
is_happy(X, XS) ->
if
X == 1 ->
true;
X < 1 ->
false;
true ->
case member(X, XS) of
true -> false;
false ->
is_happy(sum(map(fun(Z) -> Z*Z end,
[Y - 48 || Y <- integer_to_list(X)])),
[X|XS])
end
end.
 
main(X, XS) ->
if
length(XS) == 8 ->
io:format("8 Happy Numbers: ~w~n", [sort(XS)]);
true ->
case is_happy(X, []) of
true -> main(X + 1, [X|XS]);
false -> main(X + 1, XS)
end
end.
main() ->
main(0, []).
 
Command:
erl -run tasks main -run init stop -noshell
Output:
8 Happy Numbers: [1,7,10,13,19,23,28,31]

In a more functional style (assumes integer_to_list/1 will convert to the ASCII value of a number, which then has to be converted to the integer value by subtracting 48):

-module(tasks).
 
-export([main/0]).
 
main() -> io:format("~w ~n", [happy_list(1, 8, [])]).
 
happy_list(_, N, L) when length(L) =:= N -> lists:reverse(L);
happy_list(X, N, L) ->
Happy = is_happy(X),
if Happy -> happy_list(X + 1, N, [X|L]);
true -> happy_list(X + 1, N, L) end.
 
is_happy(1) -> true;
is_happy(4) -> false;
is_happy(N) when N > 0 ->
N_As_Digits = [Y - 48 || Y <- integer_to_list(N)],
is_happy(lists:foldl(fun(X, Sum) -> (X * X) + Sum end, 0, N_As_Digits));
is_happy(_) -> false.

Output:

[1,7,10,13,19,23,28,31]

[edit] Euphoria

function is_happy(integer n)
sequence seen
integer k
seen = {}
while n > 1 do
seen &= n
k = 0
while n > 0 do
k += power(remainder(n,10),2)
n = floor(n/10)
end while
n = k
if find(n,seen) then
return 0
end if
end while
return 1
end function
 
integer n,count
n = 1
count = 0
while count < 8 do
if is_happy(n) then
? n
count += 1
end if
n += 1
end while

Output:

1
7
10
13
19
23
28
31

[edit] F#

This requires the F# power pack to be referenced and the 2010 beta of F#

open System.Collections.Generic
open Microsoft.FSharp.Collections
 
let answer =
let sqr x = x*x // Classic square definition
let rec AddDigitSquare n =
match n with
| 0 -> 0 // Sum of squares for 0 is 0
| _ -> sqr(n % 10) + (AddDigitSquare (n / 10)) // otherwise add square of bottom digit to recursive call
let dict = new Dictionary<int, bool>() // Dictionary to memoize values
let IsHappy n =
if dict.ContainsKey(n) then // If we've already discovered it
dict.[n] // Return previously discovered value
else
let cycle = new HashSet<_>(HashIdentity.Structural) // Set to keep cycle values in
let rec isHappyLoop n =
if cycle.Contains n then n = 1 // If there's a loop, return true if it's 1
else
cycle.Add n |> ignore // else add this value to the cycle
isHappyLoop (AddDigitSquare n) // and check the next number in the cycle
let f = isHappyLoop n // Keep track of whether we're happy or not
cycle |> Seq.iter (fun i -> dict.[i] <- f) // and apply it to all the values in the cycle
f // Return the boolean
 
1 // Starting with 1,
|> Seq.unfold (fun i -> Some (i, i + 1)) // make an infinite sequence of consecutive integers
|> Seq.filter IsHappy // Keep only the happy ones
|> Seq.truncate 8 // Stop when we've found 8
|> Seq.iter (Printf.printf "%d\n") // Print results
 

Output:

1
7
10
13
19
23
28
31

[edit] Factor

USING: combinators kernel make math sequences ;
 
: squares ( n -- s )
0 [ over 0 > ] [ [ 10 /mod sq ] dip + ] while nip ;
 
: (happy?) ( n1 n2 -- ? )
[ squares ] [ squares squares ] bi* {
{ [ dup 1 = ] [ 2drop t ] }
{ [ 2dup = ] [ 2drop f ] }
[ (happy?) ]
} cond ;
 
: happy? ( n -- ? )
dup (happy?) ;
 
: happy-numbers ( n -- seq )
[
0 [ over 0 > ] [
dup happy? [ dup , [ 1 - ] dip ] when 1 +
] while 2drop
] { } make ;
Output:
8 happy-numbers ! { 1 7 10 13 19 23 28 31 }

[edit] FALSE

[$10/$10*@\-$*\]m:             {modulo squared and division}
[$m;![$9>][m;!@@+\]#$*+]s: {sum of squares}
[$0[1ø1>][1ø3+ø3ø=|\1-\]#\%]f: {look for duplicates}
 
{check happy number}
[
$1[f;!~2ø1=~&][1+\s;!@]# {loop over sequence until 1 or duplicate}
1ø1= {return value}
\[$0=~][@%1-]#% {drop sequence and counter}
]h:
 
0 1
"Happy numbers:"
[1ø8=~][h;![" "$.\1+\]?1+]#
%%
Output:
Happy numbers: 1 7 10 13 19 23 28 31

[edit] Fantom

class Main
{
static Bool isHappy (Int n)
{
Int[] record := [,]
while (n != 1 && !record.contains(n))
{
record.add (n)
// find sum of squares of digits
newn := 0
while (n > 0)
{
newn += (n.mod(10) * n.mod(10))
n = n.div(10)
}
n = newn
}
return (n == 1)
}
 
public static Void main ()
{
i := 1
count := 0
while (count < 8)
{
if (isHappy (i))
{
echo (i)
count += 1
}
i += 1
}
}
}
 

Output:

1
7
10
13
19
23
28
31

[edit] Forth

: next ( n -- n )
0 swap begin 10 /mod >r dup * + r> ?dup 0= until ;
 
: cycle? ( n -- ? )
here dup @ cells +
begin dup here >
while 2dup @ = if 2drop true exit then
1 cells -
repeat
1 over +! dup @ cells + ! false ;
 
: happy? ( n -- ? )
0 here ! begin next dup cycle? until 1 = ;
 
: happy-numbers ( n -- )
0 swap 0 do
begin 1+ dup happy? until dup .
loop drop ;
 
8 happy-numbers \ 1 7 10 13 19 23 28 31

[edit] Fortran

program happy
 
implicit none
integer, parameter :: find = 8
integer :: found
integer :: number
 
found = 0
number = 1
do
if (found == find) then
exit
end if
if (is_happy (number)) then
found = found + 1
write (*, '(i0)') number
end if
number = number + 1
end do
 
contains
 
function sum_digits_squared (number) result (result)
 
implicit none
integer, intent (in) :: number
integer :: result
integer :: digit
integer :: rest
integer :: work
 
result = 0
work = number
do
if (work == 0) then
exit
end if
rest = work / 10
digit = work - 10 * rest
result = result + digit * digit
work = rest
end do
 
end function sum_digits_squared
 
function is_happy (number) result (result)
 
implicit none
integer, intent (in) :: number
logical :: result
integer :: turtoise
integer :: hare
 
turtoise = number
hare = number
do
turtoise = sum_digits_squared (turtoise)
hare = sum_digits_squared (sum_digits_squared (hare))
if (turtoise == hare) then
exit
end if
end do
result = turtoise == 1
 
end function is_happy
 
end program happy

Output:

1
7
10
13
19
23
28
31

[edit] Frege

Translation of: Haskell
Works with: Frege version 3.20.113
module Happy where
 
import Prelude.Math
-- ugh, since Frege doesn't have Set, use Map instead
import Data.Map (member, insertMin, empty)
 
digitToInteger :: Char -> Integer
digitToInteger c = fromInt $ (ord c) - (ord '0')
 
isHappy :: Integer -> Bool
isHappy = p empty
where p _ 1n = true
p s n | n `member` s = false
| otherwise = p (insertMin n () s) (f n)
f = sum . map (sqr . digitToInteger) . unpacked . show
 
main _ = printStrLn $ unwords $ map show $ take 8 $ filter isHappy $ iterate (+ 1n) 1n
Output:
1 7 10 13 19 23 28 31
runtime 0.251 wallclock seconds.

[edit] Go

package main
 
import "fmt"
 
func happy(n int) bool {
m := make(map[int]bool)
for n > 1 {
m[n] = true
var x int
for x, n = n, 0; x > 0; x /= 10 {
d := x % 10
n += d * d
}
if m[n] {
return false
}
}
return true
}
 
func main() {
for found, n := 0, 1; found < 8; n++ {
if happy(n) {
fmt.Print(n, " ")
found++
}
}
fmt.Println()
}
Output:
1 7 10 13 19 23 28 31

[edit] Haskell

import Data.Char (digitToInt)
import Data.Set (member, insert, empty)
 
isHappy :: Integer -> Bool
isHappy = p empty
where p _ 1 = True
p s n | n `member` s = False
| otherwise = p (insert n s) (f n)
f = sum . map ((^2) . toInteger . digitToInt) . show
 
main = mapM_ print $ take 8 $ filter isHappy [1..]

Output:

1
7
10
13
19
23
28
31

We can create a cache for small numbers to greatly speed up the process:

import Data.Array
 
happy x = if xx <= 150 then seen!xx else happy xx where
xx = dsum x
seen :: Array Int Bool
seen = listArray (1,150) $ True:False:False:False:(map happy [5..150])
dsum n | n < 10 = n * n
| otherwise = let (q,r) = n `divMod` 10 in r*r + dsum q
 
main = print $ sum $ take 10000 $ filter happy [1..]

[edit] Icon and Unicon

procedure main(arglist)
local n
n := arglist[1] | 8 # limiting number of happy numbers to generate, default=8
writes("The first ",n," happy numbers are:")
every writes(" ", happy(seq()) \ n )
write()
end
 
procedure happy(i) #: returns i if i is happy
local n
 
if 4 ~= (0 <= i) then { # unhappy if negative, 0, or 4
if i = 1 then return i
every (n := 0) +:= !i ^ 2
if happy(n) then return i
}
end

Usage and Output:

| happynum.exe 

The first 8 happy numbers are: 1 7 10 13 19 23 28 31

[edit] J

   8{. (#~1=+/@(*:@(,.&.":))^:(1&~:*.4&~:)^:_ "0) 1+i.100
1 7 10 13 19 23 28 31

This is a repeat while construction

 f ^: cond ^: _   input

that produces an array of 1's and 4's, which is converted to 1's and 0's forming a binary array having a 1 for a happy number. Finally the happy numbers are extracted by a binary selector.

 (binary array) # 1..100

So for easier reading the solution could be expressed as:

   cond=: 1&~: *. 4&~:     NB. not equal to 1 and not equal to 4
sumSqrDigits=: +/@(*:@(,.&.":))
 
sumSqrDigits 123 NB. test sum of squared digits
14
8{. (#~ 1 = sumSqrDigits ^: cond ^:_ "0) 1 + i.100
1 7 10 13 19 23 28 31

[edit] Java

Works with: Java version 1.5+
Translation of: JavaScript
import java.util.HashSet;
public class Happy{
public static boolean happy(long number){
long m = 0;
int digit = 0;
HashSet<Long> cycle = new HashSet<Long>();
while(number != 1 && cycle.add(number)){
m = 0;
while(number > 0){
digit = (int)(number % 10);
m += digit*digit;
number /= 10;
}
number = m;
}
return number == 1;
}
 
public static void main(String[] args){
for(long num = 1,count = 0;count<8;num++){
if(happy(num)){
System.out.println(num);
count++;
}
}
}
}

Output:

1
7
10
13
19
23
28
31

[edit] JavaScript

function happy(number) {
var m, digit ;
var cycle = [] ;
 
while(number != 1 && cycle[number] !== true) {
cycle[number] = true ;
m = 0 ;
while (number > 0) {
digit = number % 10 ;
m += digit * digit ;
number = (number - digit) / 10 ;
}
number = m ;
}
return (number == 1) ;
}
 
var cnt = 8 ;
var number = 1 ;
 
while(cnt-- > 0) {
while(!happy(number))
number++ ;
document.write(number + " ") ;
number++ ;
}

Output:

1 7 10 13 19 23 28 31 

[edit] Julia

 
function happy(x)
happy_ints = ref(Int)
int_try = 1
while length(happy_ints) < x
n = int_try
past = ref(Int)
while n != 1
n = sum([y^2 for y in digits(n)])
contains(past,n) ? break : push!(past,n)
end
n == 1 && push!(happy_ints,int_try)
int_try += 1
end
return happy_ints
end

Output

 julia> happy(8)
8-element Int32 Array:
  1
  7
 10
 13
 19
 23
 28
 31

A recursive version:

sumhappy(n) = sum(x->x^2, digits(n))
 
function ishappy(x, mem = [])
x == 1? true :
x in mem? false :
ishappy(sumhappy(x),[mem, x])
end
 
nexthappy (x) = ishappy(x+1) ? x+1 : nexthappy(x+1)
 
happy(n) = [z = 1, [z = nexthappy(z) for i = 1:n-1]]
 
Output:
julia> show(happy(8))
[1,7,10,13,19,23,28,31,32]

Alternate, Translation of C
Faster with use of cache

Translation of: C
const CACHE = 256
buf = zeros(Int,CACHE)
buf[1] = 1
#happy(n) returns 1 if happy, 0 if not
function happy(n)
if n < CACHE
buf[n] > 0 && return 2-buf[n]
buf[n] = 2
end
sum = 0
nn = n
while nn != 0
x = nn%10
sum += x*x
nn = int8(nn/10)
end
x = happy(sum)
n < CACHE && (buf[n] = 2-x)
return x
end
function main()
i = 1; counter = 1000000
while counter > 0
if happy(i) == 1
counter -= 1
end
i += 1
end
return i-1
end

[edit] K

  hpy: {x@&1={~|/x=1 4}{_+/_sqr 0$'$x}//:x}
 
hpy 1+!100
1 7 10 13 19 23 28 31 32 44 49 68 70 79 82 86 91 94 97 100
 
8#hpy 1+!100
1 7 10 13 19 23 28 31

[edit] Lasso

#!/usr/bin/lasso9
 
define isHappy(n::integer) => {
local(past = set)
while(#n != 1) => {
#n = with i in string(#n)->values sum math_pow(integer(#i), 2)
#past->contains(#n) ? return false | #past->insert(#n)
}
return true
}
 
with x in generateSeries(1, 500)
where isHappy(#x)
take 8
select #x

Output:

1, 7, 10, 13, 19, 23, 28, 31

[edit] Liberty BASIC

    ct = 0
n = 0
DO
n = n + 1
IF HappyN(n, sqrInt$) = 1 THEN
ct = ct + 1
PRINT ct, n
END IF
LOOP UNTIL ct = 8
END
 
FUNCTION HappyN(n, sqrInts$)
n$ = Str$(n)
sqrInts = 0
FOR i = 1 TO Len(n$)
sqrInts = sqrInts + Val(Mid$(n$, i, 1)) ^ 2
NEXT i
IF sqrInts = 1 THEN
HappyN = 1
EXIT FUNCTION
END IF
IF Instr(sqrInts$, ":";Str$(sqrInts);":") > 0 THEN
HappyN = 0
EXIT FUNCTION
END IF
sqrInts$ = sqrInts$ + Str$(sqrInts) + ":"
HappyN = HappyN(sqrInts, sqrInts$)
END FUNCTION

Output:-

1             1
2             7
3             10
4             13
5             19
6             23
7             28
8             31

[edit] Locomotive Basic

10 mode 1:defint a-z
20 for i=1 to 100
30 i2=i
40 for l=1 to 20
50 a$=str$(i2)
60 i2=0
70 for j=1 to len(a$)
80 d=val(mid$(a$,j,1))
90 i2=i2+d*d
100 next j
110 if i2=1 then print i;"is a happy number":n=n+1:goto 150
120 if i2=4 then 150 ' cycle found
130 next l
140 ' check if we have reached 8 numbers yet
150 if n=8 then end
160 next i

Happy Numbers, Locomotive BASIC.png

[edit]

to sum_of_square_digits :number
output (apply "sum (map [[d] d*d] ` :number))
end
 
to is_happy? :number [:seen []]
output cond [
[ [:number = 1] "true ]
[ [member? :number :seen] "false ]
[ else (is_happy? (sum_of_square_digits :number) (lput :number :seen))]
]
end
 
to n_happy :count [:start 1] [:result []]
output cond [
[ [:count <= 0] :result ]
[ [is_happy? :start]
(n_happy (:count-1) (:start+1) (lput :start :result)) ]
[ else
(n_happy :count (:start+1) :result) ]
]
end
 
print n_happy 8
bye

Output:

1 7 10 13 19 23 28 31

[edit] LOLCODE

Works with: lci 0.10.3
OBTW 
Happy Numbers Rosetta Code task in LOLCODE
Requires 1.3 for BUKKIT availability
TLDR

HAI 1.3
CAN HAS STDIO?
 
BTW Simple list implementation.
BTW Used for the list of numbers already seen in IZHAPPY
 
BTW Create a list
HOW IZ I MAEKLIST
I HAS A LIST ITZ A BUKKIT
LIST HAS A LENGTH ITZ 0
FOUND YR LIST
IF U SAY SO
 
BTW Append an item to list
HOW IZ I PUTIN YR LIST AN YR ITEM
LIST HAS A SRS LIST'Z LENGTH ITZ ITEM
LIST'Z LENGTH R SUM OF LIST'Z LENGTH AN 1
IF U SAY SO
 
BTW Check for presence of an item in the list
HOW IZ I DUZLISTHAS YR HAYSTACK AN YR NEEDLE
IM IN YR BARN UPPIN YR INDEX WILE DIFFRINT INDEX AN HAYSTACK'Z LENGTH
I HAS A ITEM ITZ HAYSTACK'Z SRS INDEX
BOTH SAEM ITEM AN NEEDLE
O RLY?
YA RLY
FOUND YR WIN
OIC
IM OUTTA YR BARN
FOUND YR FAIL
IF U SAY SO
 
BTW Calculate the next number using the happy formula
HOW IZ I HAPPYSTEP YR NUM
I HAS A NEXT ITZ 0
IM IN YR LOOP
BOTH SAEM NUM AN 0
O RLY?
YA RLY
GTFO
OIC
I HAS A DIGIT ITZ MOD OF NUM AN 10
NUM R QUOSHUNT OF NUM AN 10
I HAS A SQUARE ITZ PRODUKT OF DIGIT AN DIGIT
NEXT R SUM OF NEXT AN SQUARE
IM OUTTA YR LOOP
FOUND YR NEXT
IF U SAY SO
 
BTW Check to see if a number is happy
HOW IZ I IZHAPPY YR NUM
I HAS A SEENIT ITZ I IZ MAEKLIST MKAY
IM IN YR LOOP
BOTH SAEM NUM AN 1
O RLY?
YA RLY
FOUND YR WIN
OIC
I IZ DUZLISTHAS YR SEENIT AN YR NUM MKAY
O RLY?
YA RLY
FOUND YR FAIL
OIC
I IZ PUTIN YR SEENIT AN YR NUM MKAY
NUM R I IZ HAPPYSTEP YR NUM MKAY
IM OUTTA YR LOOP
IF U SAY SO
 
BTW Print out the first 8 happy numbers
I HAS A KOUNT ITZ 0
IM IN YR LOOP UPPIN YR NUM WILE DIFFRINT KOUNT AN 8
I IZ IZHAPPY YR NUM MKAY
O RLY?
YA RLY
KOUNT R SUM OF KOUNT AN 1
VISIBLE NUM
OIC
IM OUTTA YR LOOP
KTHXBYE
Output:
1
7
10
13
19
23
28
31

[edit] Lua

function digits(n)
if n > 0 then return n % 10, digits(math.floor(n/10)) end
end
function sumsq(a, ...)
return a and a ^ 2 + sumsq(...) or 0
end
local happy = setmetatable({true, false, false, false}, {
__index = function(self, n)
self[n] = self[sumsq(digits(n))]
return self[n]
end } )
i, j = 0, 1
repeat
i, j = happy[j] and (print(j) or i+1) or i, j + 1
until i == 8

Output:

1
7
10
13
19
23
28
31

[edit] Maple

To begin, here is a procedure to compute the sum of the squares of the digits of a positive integer. It uses the built-in procedure irem, which computes the integer remainder and, if passed a name as the optional third argument, assigns it the corresponding quotient. (In other words, it performs integer division with remainder. There is also a dual, companion procedure iquo, which returns the integer quotient and assigns the remainder to the (optional) third argument.)

SumSqDigits := proc( n :: posint )
local s := 0;
local m := n;
while m <> 0 do
s := s + irem( m, 10, 'm' )^2
end do;
s
end proc:

(Note that the unevaluation quotes on the third argument to irem are essential here, as that argument must be a name and, if m were passed without quotes, it would evaluate to a number.)

For example,

 
> SumSqDigits( 1234567890987654321 );
570
 

We can check this by computing it another way (more directly).

 
> n := 1234567890987654321:
> `+`( op( map( parse, StringTools:-Explode( convert( n, 'string' ) ) )^~2) );
570
 

The most straight-forward way to check whether a number is happy or sad seems also to be the fastest (that I could think of).

Happy? := proc( n )
if n = 1 then
true
elif n = 4 then
false
else
local s := SumSqDigits( n );
while not ( s in { 1, 4 } ) do
s := SumSqDigits( s )
end do;
evalb( s = 1 )
end if
end proc:

We can use this to determine the number of happy (H) and sad (S) numbers up to one million as follows.

 
> H, S := selectremove( Happy?, [seq]( 1 .. N ) ):
> nops( H ), nops( S );
143071, 856929
 

Finally, to solve the stated problem, here is a completely straight-forward routine to locate the first N happy numbers, returning them in a set.

FindHappiness := proc( N )
local count := 0;
local T := table();
for local i while count < N do
if Happy?( i ) then
count := 1 + count;
T[ count ] := i
end if
end do;
{seq}( T[ i ], i = 1 .. count )
end proc:

With input equal to 8, we get

 
> FindHappiness( 8 );
{1, 7, 10, 13, 19, 23, 28, 31}
 

For completeness, here is an implementation of the cycle detection algorithm for recognizing happy numbers. It is much slower, however.

Happy? := proc( n :: posint )
local a, b;
a, b := n, SumSqDigits( n );
while a <> b do
a := SumSqDigits( a );
b := (SumSqDigits@@2)( b )
end do;
evalb( a = 1 )
end proc:

[edit] Mathematica

Custom function HappyQ:

AddSumSquare[input_]:=Append[input,Total[IntegerDigits[Last[input]]^2]]
NestUntilRepeat[a_,f_]:=NestWhile[f,{a},!MemberQ[Most[Last[{##}]],Last[Last[{##}]]]&,All]
HappyQ[a_]:=Last[NestUntilRepeat[a,AddSumSquare]]==1

Examples for a specific number:

HappyQ[1337]
HappyQ[137]

gives back:

True
False

Example finding the first 8:

m = 8;
n = 1;
i = 0;
happynumbers = {};
While[n <= m,
i++;
If[HappyQ[i],
n++;
AppendTo[happynumbers, i]
]
]
happynumbers

gives back:

{1, 7, 10, 13, 19, 23, 28, 31}

[edit] MATLAB

Recursive version:

function findHappyNumbers
nHappy = 0;
k = 1;
while nHappy < 8
if isHappyNumber(k, [])
fprintf('%d ', k)
nHappy = nHappy+1;
end
k = k+1;
end
fprintf('\n')
end
 
function hap = isHappyNumber(k, prev)
if k == 1
hap = true;
elseif ismember(k, prev)
hap = false;
else
hap = isHappyNumber(sum((sprintf('%d', k)-'0').^2), [prev k]);
end
end
Output:
1 7 10 13 19 23 28 31 

[edit] MUMPS

ISHAPPY(N)
 ;Determines if a number N is a happy number
 ;Note that the returned strings do not have a leading digit unless it is a happy number
IF (N'=N\1)!(N<0) QUIT "Not a positive integer"
NEW SUM,I
 ;SUM is the sum of the square of each digit
 ;I is a loop variable
 ;SEQ is the sequence of previously checked SUMs from the original N
 ;If it isn't set already, initialize it to an empty string
IF $DATA(SEQ)=0 NEW SEQ SET SEQ=""
SET SUM=0
FOR I=1:1:$LENGTH(N) DO
.SET SUM=SUM+($EXTRACT(N,I)*$EXTRACT(N,I))
QUIT:(SUM=1) SUM
QUIT:$FIND(SEQ,SUM)>1 "Part of a sequence not containing 1"
SET SEQ=SEQ_","_SUM
QUIT $$ISHAPPY(SUM)
HAPPY(C) ;Finds the first C happy numbers
NEW I
 ;I is a counter for what integer we're looking at
WRITE !,"The first "_C_" happy numbers are:"
FOR I=1:1 QUIT:C<1 SET Q=+$$ISHAPPY(I) WRITE:Q !,I SET:Q C=C-1
KILL I
QUIT
Output:
USER>D HAPPY^ROSETTA(8)
 
The first 8 happy numbers are:
1
7
10
13
19
23
28
31
USER>W:+$$ISHAPPY^ROSETTA(320) "Happy Number"
Happy Number
USER>W:+$$ISHAPPY^ROSETTA(321) "Happy Number"
 
USER>

[edit] NetRexx

Translation of: REXX
/*NetRexx program to display the 1st 8 (or specified arg) happy numbers*/
limit = arg[0] /*get argument for LIMIT. */
say limit
if limit = null, limit ='' then limit=8 /*if not specified, set LIMIT to 8*/
haps = 0 /*count of happy numbers so far. */
 
loop n=1 while haps < limit /*search integers starting at one.*/
q=n /*Q may or may not be "happy". */
a=0
 
loop forever /*see if Q is a happy number. */
if q==1 then do /*if Q is unity, then it's happy*/
haps = haps + 1 /*bump the count of happy numbers.*/
say n /*display the number. */
iterate n /*and then keep looking for more. */
end
 
sum=0 /*initialize sum to zero. */
 
loop j=1 for q.length /*add the squares of the numerals.*/
sum = sum + q.substr(j,1) ** 2
end
 
if a[sum] then iterate n /*if already summed, Q is unhappy.*/
a[sum]=1 /*mark the sum as being found. */
q=sum /*now, lets try the Q sum. */
end
end
Output
1
7
10
13
19
23
28
31

Sample output when 100 is specified as the program's argument.

1
7
10
13
19
23
28
31
32
44
49
68
70
79
82
86
91
94
97
100
103
109
129
130
133
139
167
176
188
190
192
193
203
208
219
226
230
236
239
262
263
280
291
293
301
302
310
313
319
320
326
329
331
338
356
362
365
367
368
376
379
383
386
391
392
397
404
409
440
446
464
469
478
487
490
496
536
556
563
565
566
608
617
622
623
632
635
637
638
644
649
653
655
656
665
671
673
680
683
694

[edit] Nimrod

Translation of: Python
import intsets
 
proc happy(n: int): bool =
var
n = n
past = initIntSet()
while n != 1:
let s = $n
n = 0
for c in s:
let i = ord(c) - ord('0')
n += i * i
if n in past:
return false
past.incl(n)
return true
 
for x in 0..31:
if happy(x):
echo x

Output:

1
7
10
13
19
23
28
31

[edit] Objeck

use IO;
use Structure;
 
bundle Default {
class HappyNumbers {
function : native : IsHappy(n : Int) ~ Bool {
cache := IntVector->New();
sum := 0;
while(n <> 1) {
if(cache->Has(n)) {
return false;
};
 
cache->AddBack(n);
while(n <> 0) {
digit := n % 10;
sum += (digit * digit);
n /= 10;
};
 
n := sum;
sum := 0;
};
 
return true;
}
 
function : Main(args : String[]) ~ Nil {
num := 1;
happynums := IntVector->New();
 
while(happynums->Size() < 8) {
if(IsHappy(num)) {
happynums->AddBack(num);
};
 
num += 1;
};
 
Console->Print("First 8 happy numbers: ");
each(i : happynums) {
Console->Print(happynums->Get(i))->Print(",");
};
Console->PrintLine("");
}
}
}

output:

First 8 happy numbers: 1,7,10,13,19,23,28,31,

[edit] OCaml

Using Floyd's cycle-finding algorithm.

open Num
 
let step =
let rec aux s n =
if n =/ Int 0 then s else
let q = quo_num n (Int 10)
and r = mod_num n (Int 10)
in aux (s +/ (r */ r)) q
in aux (Int 0) ;;
 
let happy n =
let rec aux x y =
if x =/ y then x else aux (step x) (step (step y))
in (aux n (step n)) =/ Int 1 ;;
 
let first n =
let rec aux v x n =
if n = 0 then v else
if happy x
then aux (x::v) (x +/ Int 1) (n - 1)
else aux v (x +/ Int 1) n
in aux [ ] (Int 1) n ;;
 
List.iter print_endline (
List.rev_map string_of_num (first 8)) ;;

Output:

$ ocaml nums.cma happy_numbers.ml
1
7
10
13
19
23
28
31

[edit] ooRexx

 
count = 0
say "First 8 happy numbers are:"
loop i = 1 while count < 8
if happyNumber(i) then do
count += 1
say i
end
end
 
::routine happyNumber
use strict arg number
 
-- use to trace previous cycle results
previous = .set~new
loop forever
-- stop when we hit the target
if number = 1 then return .true
-- stop as soon as we start cycling
if previous[number] \== .nil then return .false
previous~put(number)
next = 0
-- loop over all of the digits
loop digit over number~makearray('')
next += digit * digit
end
-- and repeat the cycle
number = next
end
 
First 8 happy numbers are:
1
7
10
13
19
23
28
31

[edit] Oz

This example does not show the output mentioned in the task description on this page (or a page linked to from here). Please ensure that it meets all task requirements and remove this message.
Note that phrases in task descriptions such as "print and display" and "print and show" for example, indicate that (reasonable length) output be a part of a language's solution.


declare
fun {IsHappy N}
{IsHappy2 N nil}
end
 
fun {IsHappy2 N Seen}
if N == 1 then true
elseif {Member N Seen} then false
else
Next = {Sum {Map {Digits N} Square}}
in
{IsHappy2 Next N|Seen}
end
end
 
fun {Sum Xs}
{FoldL Xs Number.'+' 0}
end
 
fun {Digits N}
{Map {Int.toString N} fun {$ D} D - &0 end}
end
 
fun {Square N} N*N end
 
fun lazy {Nat I}
I|{Nat I+1}
end
 
%% List.filter is eager. But we need a lazy Filter:
fun lazy {LFilter Xs P}
case Xs of X|Xr andthen {P X} then X|{LFilter Xr P}
[] _|Xr then {LFilter Xr P}
[] nil then nil
end
end
 
HappyNumbers = {LFilter {Nat 1} IsHappy}
in
{Show {List.take HappyNumbers 8}}

[edit] PARI/GP

Works with: PARI/GP version 2.4.3 and above
This code uses the select() function, which was added in PARI version 2.4.2. The order of the arguments changed between versions; to use in 2.4.2 change select(function, vector) to select(vector, function).

If the number has more than three digits, the sum of the squares of its digits has fewer digits than the number itself. If the number has three digits, the sum of the squares of its digits is at most 3 * 9^2 = 243. A simple solution is to look up numbers up to 243 and calculate the sum of squares only for larger numbers.

H=[1,7,10,13,19,23,28,31,32,44,49,68,70,79,82,86,91,94,97,100,103,109,129,130,133,139,167,176,188,190,192,193,203,208,219,226,230,236,239];
isHappy(n)={
if(n<262,
setsearch(H,n)>0
,
n=eval(Vec(Str(n)));
isHappy(sum(i=1,#n,n[i]^2))
)
};
select(isHappy, vector(31,i,i))

Output:

%1 = [1, 7, 10, 13, 19, 23, 28, 31]

[edit] Pascal

Program HappyNumbers (output);
 
uses
Math;
 
function find(n: integer; cache: array of integer): boolean;
var
i: integer;
begin
find := false;
for i := low(cache) to high(cache) do
if cache[i] = n then
find := true;
end;
 
function is_happy(n: integer): boolean;
var
cache: array of integer;
sum: integer;
begin
setlength(cache, 1);
repeat
sum := 0;
while n > 0 do
begin
sum := sum + (n mod 10)**2;
n := n div 10;
end;
if sum = 1 then
begin
is_happy := true;
break;
end;
if find(sum, cache) then
begin
is_happy := false;
break;
end;
n := sum;
cache[high(cache)]:= sum;
setlength(cache, length(cache)+1);
until false;
end;
 
var
n, count: integer;
 
begin
n := 1;
count := 0;
while count < 8 do
begin
if is_happy(n) then
begin
inc(count);
write(n, ' ');
end;
inc(n);
end;
writeln;
end.

Output:

:> ./HappyNumbers
1 7 10 13 19 23 28 31

[edit] Perl

use List::Util qw(sum);
 
sub is_happy ($)
{for (my ($n, %seen) = shift ;; $n = sum map {$_**2} split //, $n)
{$n == 1 and return 1;
$seen{$n}++ and return 0;}}
 
for (my ($n, $happy) = (1, 0) ; $happy < 8 ; ++$n)
{is_happy $n or next;
print "$n\n";
++$happy;}
Output:
1
7
10
13
19
23
28
31
Translation of: Perl 6
use List::Util qw(sum);
sub is_happy {
my ($n) = @_;
my %seen;
while (1) {
$n = sum map { $_ ** 2 } split //, $n;
return 1 if $n == 1;
return 0 if $seen{$n}++;
}
}
 
my $n;
is_happy( ++$n ) and print "$n " or redo for 1..8;
Output:
1 7 10 13 19 23 28 31

[edit] Perl 6

sub happy (Int $n is copy --> Bool) {
my %seen;
loop {
$n = [+] $n.comb.map: { $_ ** 2 }
return True if $n == 1;
return False if %seen{$n}++;
}
}
 
say join ' ', grep(&happy, 1 .. *)[^8];
Output:
1 7 10 13 19 23 28 31

Here's another approach that uses a different set of tricks including lazy lists, gather/take, repeat-until, and the cross metaoperator X.

my @happy := gather for 1..* -> $number {
my %stopper = 1 => 1;
my $n = $number;
repeat until %stopper{$n}++ {
$n = [+] $n.comb X** 2;
}
take $number if $n == 1;
}
 
say ~@happy[^8];

Output is the same as above.

Here is a version using a subset and an anonymous recursion (we cheat a little bit by using the knowledge that 7 is the second happy number):

subset Happy of Int where sub ($n) {
$n == 1 ?? True !!
$n < 7 ?? False !!
&?ROUTINE([+] $n.comb »**» 2);
}
 
say (grep Happy, 1 .. *)[^8];

Again, output is the same as above. It is not clear whether this version returns in finite time for any integer, though.

There's more than one way to do it...

[edit] PHP

Translation of: D
function isHappy($n) {
while (1) {
$total = 0;
while ($n > 0) {
$total += pow(($n % 10), 2);
$n /= 10;
}
if ($total == 1)
return true;
if (array_key_exists($total, $past))
return false;
$n = $total;
$past[$total] = 0;
}
}
 
$i = $cnt = 0;
while ($cnt < 8) {
if (isHappy($i)) {
echo "$i ";
$cnt++;
}
$i++;
}
1 7 10 13 19 23 28 31 

[edit] PicoLisp

(de happy? (N)
(let Seen NIL
(loop
(T (= N 1) T)
(T (member N Seen))
(setq N
(sum '((C) (** (format C) 2))
(chop (push 'Seen N)) ) ) ) ) )
 
(let H 0
(do 8
(until (happy? (inc 'H)))
(printsp H) ) )

Output:

1 7 10 13 19 23 28 31

[edit] PL/I

test: proc options (main); /* 19 November 2011 */
declare (i, j, n, m, nh initial (0) ) fixed binary (31);
 
main_loop:
do j = 1 to 100;
n = j;
do i = 1 to 100;
m = 0;
/* Form the sum of squares of the digits. */
do until (n = 0);
m = m + mod(n, 10)**2;
n = n/10;
end;
if m = 1 then
do;
put skip list (j || ' is a happy number');
nh = nh + 1;
if nh = 8 then return;
iterate main_loop;
end;
n = m; /* Replace n with the new number formed from digits. */
end;
end;
end test;
 

OUTPUT:

             1 is a happy number
             7 is a happy number 
            10 is a happy number 
            13 is a happy number 
            19 is a happy number 
            23 is a happy number 
            28 is a happy number 
            31 is a happy number 

[edit] PowerShell

function happy([int] $n) {
$a=@()
for($i=2;$a.count -lt $n;$i++) {
$sum=$i
$hist=@{}
while( $hist[$sum] -eq $null ) {
if($sum -eq 1) {
$a+=$i
}
$hist[$sum]=$sum
$sum2=0
foreach($j in $sum.ToString().ToCharArray()) {
$k=([int]$j)-0x30
$sum2+=$k*$k
}
$sum=$sum2
}
}
$a -join ','
}

Output :

happy(8)
7,10,13,19,23,28,31,32

[edit] Prolog

Works with: SWI-Prolog
happy_numbers(L, Nb) :-
% creation of the list
length(L, Nb),
% Process of this list
get_happy_number(L, 1).
 
 
% the game is over
get_happy_number([], _).
 
% querying the newt happy_number
get_happy_number([H | T], N) :-
N1 is N+1,
(is_happy_number(N) ->
H = N,
get_happy_number(T, N1);
get_happy_number([H | T], N1)).
 
% we must memorized the numbers reached
is_happy_number(N) :-
is_happy_number(N, [N]).
 
% a number is happy when we get 1
is_happy_number(N, _L) :-
get_next_number(N, 1), !.
 
% or when this number is not already reached !
is_happy_number(N, L) :-
get_next_number(N, NN),
\+member(NN, L),
is_happy_number(NN, [NN | L]).
 
% Process of the next number from N
get_next_number(N, NewN) :-
get_list_digits(N, LD),
maplist(square, LD, L),
sumlist(L, NewN).
 
get_list_digits(N, LD) :-
number_chars(N, LCD),
maplist(number_chars_, LD, LCD).
 
number_chars_(D, CD) :-
number_chars(D, [CD]).
 
square(N, SN) :-
SN is N * N.

Output :

 ?- happy_numbers(L, 8).
L = [1,7,10,13,19,23,28,31].

[edit] PureBasic

#ToFind=8
#MaxTests=100
#True = 1: #False = 0
Declare is_happy(n)
 
If OpenConsole()
Define i=1,Happy
Repeat
If is_happy(i)
Happy+1
PrintN("#"+Str(Happy)+RSet(Str(i),3))
EndIf
i+1
Until Happy>=#ToFind
;
Print(#CRLF$+#CRLF$+"Press ENTER to exit"): Input()
CloseConsole()
EndIf
 
Procedure is_happy(n)
Protected i,j=n,dig,sum
Repeat
sum=0
While j
dig=j%10
j/10
sum+dig*dig
Wend
If sum=1: ProcedureReturn #True: EndIf
j=sum
i+1
Until i>#MaxTests
ProcedureReturn #False
EndProcedure

Sample output:

#1  1
#2  7
#3 10
#4 13
#5 19
#6 23
#7 28
#8 31

[edit] Python

>>> def happy(n):
past = set()
while n != 1:
n = sum(int(i)**2 for i in str(n))
if n in past:
return False
past.add(n)
return True
 
>>> [x for x in xrange(500) if happy(x)][:8]
[1, 7, 10, 13, 19, 23, 28, 31]

[edit] R

is.happy <- function(n)
{
stopifnot(is.numeric(n) && length(n)==1)
getdigits <- function(n)
{
as.integer(unlist(strsplit(as.character(n), "")))
}
digits <- getdigits(n)
previous <- c()
repeat
{
sumsq <- sum(digits^2, na.rm=TRUE)
if(sumsq==1L)
{
happy <- TRUE
break
} else if(sumsq %in% previous)
{
happy <- FALSE
attr(happy, "cycle") <- previous
break
} else
{
previous <- c(previous, sumsq)
digits <- getdigits(sumsq)
}
}
happy
}

Example usage

is.happy(2)
[1] FALSE
attr(,"cycle")
[1]   4  16  37  58  89 145  42  20
#Find happy numbers between 1 and 50
which(apply(rbind(1:50), 2, is.happy))
1  7 10 13 19 23 28 31 32 44 49
#Find the first 8 happy numbers
happies <- c()
i <- 1L
while(length(happies) < 8L)
{
if(is.happy(i)) happies <- c(happies, i)
i <- i + 1L
}
happies
1  7 10 13 19 23 28 31

[edit] Racket

#lang racket
(define (sum-of-squared-digits number (result 0))
(if (zero? number)
result
(sum-of-squared-digits (quotient number 10)
(+ result (expt (remainder number 10) 2)))))
 
(define (happy-number? number (seen null))
(define next (sum-of-squared-digits number))
(cond ((= 1 next)
#t)
((memq next seen)
#f)
(else
(happy-number? next (cons number seen)))))
 
(define (get-happys max)
(for/list ((x (in-range max))
#:when (happy-number? x))
x))
 
(display (take (get-happys 100) 8)) ;displays (1 7 10 13 19 23 28 31)

[edit] REXX

[edit] unoptimized

/*REXX program displays  eight  (or a specified limit)  happy  numbers. */
parse arg limit . /*get optional argument LIMIT. */
if limit=='' | limit==',' then limit=8 /*Not specified? Set LIMIT to 8.*/
haps=0 /*count of happy numbers so far. */
 
do n=1 while haps<limit; q=n; a.=0 /*search integers starting at 1.*/
do until q==1 /*see if Q is a happy number.*/
s=0 /*prepare to add squares of digs.*/
do j=1 for length(q) /*sum the squares of the digits. */
s=s+substr(q,j,1)**2 /*add the square of a digit. */
end /*j*/
 
if a.s then iterate n /*if already summed, Q is unhappy*/
a.s=1; q=s /*mark sum as found, try Q sum.*/
end /*until*/
say n /*display the number (N is happy)*/
haps=haps+1 /*bump the count of happy numbers*/
end /*n*/
/*stick a fork in it, we're done.*/

output (using the default input of 8)

1
7
10
13
19
23
28
31

[edit] optimized

This REXX code uses additional memorization (by keeping track of happy and unhappy numbers),
it's about 25% faster than the unoptimized version.

This REXX version also accepts a range of happy numbers to be shown, that is,
it can show the 2000th through the 2032nd (inclusive) happy numbers.

/*REXX program displays  eight (or a specified range of) happy  numbers.*/
parse arg L H . /*get optional args: low & high */
if L=='' | L==',' then L=8 /*Not specified? Set L to default*/
if H=='' | H==',' then do; H=L; L=1; end /*use a range for the showing.*/
#.0=0; #.1=1; #.2=4; #.3=9; #.4=16; #.5=25; #.6=36; #.7=49; #.8=64; #.9=81
haps=0 /*count of happy numbers so far. */
@.=0;  !.=0 /*sparse array: happy&unhappy #s.*/
 
do n=1 while haps<H; q=n; a.=0 /*search integers starting at 1.*/
if !.n then iterate /*if N is unhappy, try another.*/
 
do until q==1 /*see if Q is a happy number. */
s=0 /*prepare to add squares of digs.*/
do j=1 for length(q) /*sum the squares of the digits. */
_=substr(q,j,1) /*get a single digit (in base 10)*/
s=s+#._ /*add the square of a digit. */
end /*j*/
 
if @.s then leave /*we have found a happy number.*/
if !.s then iterate n /*Sum unhappy? Then Q is unhappy*/
if a.s then do /*If already summed? Q is unhappy*/
 !.q=1;  !.s=1 /*mark Q & S as unhappy numbers*/
iterate n /*previously summed, so Q unhappy*/
end
a.s=1; q=s /*mark sum as found, try Q sum.*/
end /*until*/
 
@.n=1 /*mark N as a happy number. */
haps=haps+1 /*bump the count of happy numbers*/
if haps<L then iterate /*don't display, N is too low. */
say n /*display the happy N number.*/
end /*n*/
/*stick a fork in it, we're done.*/

output when the input is 2000 2032

13141
13142
13148
13158
13177
13182
13184
13185
13188
13203
13212
13214
13218
13221
13228
13230
13233
13241
13247
13248
13258
13266
13274
13281
13282
13284
13285
13299
13300
13302
13303
13305
13307

[edit] optimized, horizontal list

This REXX version is identical to the optimized version, but displays the numbers in a horizontal list.

/*REXX program displays  eight (or a specified range of) happy  numbers.*/
parse arg L H . /*get optional args: low & high */
if L=='' | L==',' then L=8 /*Not specified? Set L to default*/
if H=='' | H==',' then do; H=L; L=1; end /*use a range for the showing.*/
#.0=0; #.1=1; #.2=4; #.3=9; #.4=16; #.5=25; #.6=36; #.7=49; #.8=64; #.9=81
haps=0 /*count of happy numbers so far. */
@.=0;  !.=0 /*sparse array: happy&unhappy #s.*/
out= /*the output line (of happy nums)*/
sw=linesize() /*obtain the linesize of term scr*/
 
do n=1 while haps<H; q=n; a.=0 /*search integers starting at 1.*/
if !.n then iterate /*if N is unhappy, try another.*/
 
do until q==1 /*see if Q is a happy number. */
s=0 /*prepare to add squares of digs.*/
do j=1 for length(q) /*sum the squares of the digits. */
_=substr(q,j,1) /*get a single digit (in base 10)*/
s=s+#._ /*add the square of a digit. */
end /*j*/
 
if @.s then leave /*we have found a happy number.*/
if !.s then iterate n /*Sum unhappy? Then Q is unhappy*/
if a.s then do /*If already summed? Q is unhappy*/
 !.q=1;  !.s=1 /*mark Q & S as unhappy numbers*/
iterate n /*if already summed, Q is unhappy*/
end
a.s=1; q=s /*mark sum as found, try Q sum.*/
end /*until*/
 
@.n=1 /*mark N as a happy number. */
haps=haps+1 /*bump the count of happy numbers*/
if haps<L then iterate /*don't display, N is too low. */
if length(out n)>sw then do /*maybe display the happy number.*/
say strip(out) /*line is too long, tell it*/
out= /*nullify the OUT (line).*/
end
out=out n /*append the happy number to OUT.*/
end /*n*/
 
if out\=='' then say strip(out) /*handle any residuals for OUT. */
/*stick a fork in it, we're done.*/

This REXX program makes use of   LINESIZE   REXX program (or BIF) which is used to determine the screen width (or linesize) of the terminal (console).
The   LINESIZE.REX   REXX program is included here ──► LINESIZE.REX.

output when the input being used is: 1 1500
(linesize   is   200.)

1 7 10 13 19 23 28 31 32 44 49 68 70 79 82 86 91 94 97 100 103 109 129 130 133 139 167 176 188 190 192 193 203 208 219 226 230 236 239 262 263 280 291 293 301 302 310 313 319 320 326 329 331 338 356
362 365 367 368 376 379 383 386 391 392 397 404 409 440 446 464 469 478 487 490 496 536 556 563 565 566 608 617 622 623 632 635 637 638 644 649 653 655 656 665 671 673 680 683 694 700 709 716 736 739
748 761 763 784 790 793 802 806 818 820 833 836 847 860 863 874 881 888 899 901 904 907 910 912 913 921 923 931 932 937 940 946 964 970 973 989 998 1000 1003 1009 1029 1030 1033 1039 1067 1076 1088
1090 1092 1093 1112 1114 1115 1121 1122 1125 1128 1141 1148 1151 1152 1158 1177 1182 1184 1185 1188 1209 1211 1212 1215 1218 1221 1222 1233 1247 1251 1257 1258 1274 1275 1277 1281 1285 1288 1290 1299
1300 1303 1309 1323 1330 1332 1333 1335 1337 1339 1353 1366 1373 1390 1393 1411 1418 1427 1444 1447 1448 1457 1472 1474 1475 1478 1481 1484 1487 1511 1512 1518 1521 1527 1528 1533 1547 1557 1572 1574
1575 1578 1581 1582 1587 1599 1607 1636 1663 1666 1670 1679 1697 1706 1717 1724 1725 1727 1733 1742 1744 1745 1748 1752 1754 1755 1758 1760 1769 1771 1772 1784 1785 1796 1808 1812 1814 1815 1818 1821
1825 1828 1841 1844 1847 1851 1852 1857 1874 1875 1880 1881 1882 1888 1900 1902 1903 1920 1929 1930 1933 1959 1967 1976 1992 1995 2003 2008 2019 2026 2030 2036 2039 2062 2063 2080 2091 2093 2109 2111
2112 2115 2118 2121 2122 2133 2147 2151 2157 2158 2174 2175 2177 2181 2185 2188 2190 2199 2206 2211 2212 2221 2224 2242 2245 2254 2257 2258 2260 2275 2285 2300 2306 2309 2313 2331 2333 2338 2339 2360
2369 2383 2390 2393 2396 2417 2422 2425 2448 2452 2455 2457 2458 2471 2475 2478 2484 2485 2487 2511 2517 2518 2524 2527 2528 2542 2545 2547 2548 2554 2555 2557 2568 2571 2572 2574 2575 2581 2582 2584
2586 2602 2603 2620 2630 2639 2658 2685 2693 2714 2715 2717 2725 2741 2745 2748 2751 2752 2754 2755 2771 2784 2800 2811 2815 2818 2825 2833 2844 2845 2847 2851 2852 2854 2856 2865 2874 2881 2899 2901
2903 2910 2919 2930 2933 2936 2963 2989 2991 2998 3001 3002 3010 3013 3019 3020 3026 3029 3031 3038 3056 3062 3065 3067 3068 3076 3079 3083 3086 3091 3092 3097 3100 3103 3109 3123 3130 3132 3133 3135
3137 3139 3153 3166 3173 3190 3193 3200 3206 3209 3213 3231 3233 3238 3239 3260 3269 3283 3290 3293 3296 3301 3308 3310 3312 3313 3315 3317 3319 3321 3323 3328 3329 3331 3332 3338 3346 3351 3355 3356
3364 3365 3367 3371 3376 3380 3382 3383 3391 3392 3436 3456 3463 3465 3466 3506 3513 3531 3535 3536 3546 3553 3560 3563 3564 3602 3605 3607 3608 3616 3620 3629 3634 3635 3637 3643 3645 3646 3650 3653
3654 3661 3664 3667 3670 3673 3676 3680 3689 3692 3698 3706 3709 3713 3731 3736 3760 3763 3766 3779 3789 3790 3797 3798 3803 3806 3823 3830 3832 3833 3860 3869 3879 3896 3897 3901 3902 3907 3910 3913
3920 3923 3926 3931 3932 3962 3968 3970 3977 3978 3986 3987 4004 4009 4040 4046 4064 4069 4078 4087 4090 4096 4111 4118 4127 4144 4147 4148 4157 4172 4174 4175 4178 4181 4184 4187 4217 4222 4225 4248
4252 4255 4257 4258 4271 4275 4278 4284 4285 4287 4336 4356 4363 4365 4366 4400 4406 4414 4417 4418 4428 4441 4447 4449 4455 4460 4471 4474 4477 4481 4482 4494 4517 4522 4525 4527 4528 4536 4545 4552
4554 4555 4558 4563 4571 4572 4577 4582 4585 4599 4604 4609 4633 4635 4636 4640 4653 4663 4690 4708 4712 4714 4715 4718 4721 4725 4728 4741 4744 4747 4751 4752 4757 4774 4775 4780 4781 4782 4788 4807
4811 4814 4817 4824 4825 4827 4841 4842 4852 4855 4870 4871 4872 4878 4887 4888 4900 4906 4944 4959 4960 4995 5036 5056 5063 5065 5066 5111 5112 5118 5121 5127 5128 5133 5147 5157 5172 5174 5175 5178
5181 5182 5187 5199 5211 5217 5218 5224 5227 5228 5242 5245 5247 5248 5254 5255 5257 5268 5271 5272 5274 5275 5281 5282 5284 5286 5306 5313 5331 5335 5336 5346 5353 5360 5363 5364 5417 5422 5425 5427
5428 5436 5445 5452 5454 5455 5458 5463 5471 5472 5477 5482 5485 5499 5506 5517 5524 5525 5527 5533 5542 5544 5545 5548 5552 5554 5555 5558 5560 5569 5571 5572 5584 5585 5596 5603 5605 5606 5628 5630
5633 5634 5643 5650 5659 5660 5666 5682 5695 5712 5714 5715 5718 5721 5722 5724 5725 5741 5742 5747 5751 5752 5774 5781 5789 5798 5799 5811 5812 5817 5821 5822 5824 5826 5842 5845 5854 5855 5862 5871
5879 5897 5919 5949 5956 5965 5978 5979 5987 5991 5994 5997 6008 6017 6022 6023 6032 6035 6037 6038 6044 6049 6053 6055 6056 6065 6071 6073 6080 6083 6094 6107 6136 6163 6166 6170 6179 6197 6202 6203
6220 6230 6239 6258 6285 6293 6302 6305 6307 6308 6316 6320 6329 6334 6335 6337 6343 6345 6346 6350 6353 6354 6361 6364 6367 6370 6373 6376 6380 6389 6392 6398 6404 6409 6433 6435 6436 6440 6453 6463
6490 6503 6505 6506 6528 6530 6533 6534 6543 6550 6559 6560 6566 6582 6595 6605 6613 6616 6631 6634 6637 6643 6650 6656 6661 6665 6673 6701 6703 6710 6719 6730 6733 6736 6763 6789 6791 6798 6800 6803
6825 6830 6839 6852 6879 6893 6897 6899 6904 6917 6923 6932 6938 6940 6955 6971 6978 6983 6987 6989 6998 7000 7009 7016 7036 7039 7048 7061 7063 7084 7090 7093 7106 7117 7124 7125 7127 7133 7142 7144
7145 7148 7152 7154 7155 7158 7160 7169 7171 7172 7184 7185 7196 7214 7215 7217 7225 7241 7245 7248 7251 7252 7254 7255 7271 7284 7306 7309 7313 7331 7336 7360 7363 7366 7379 7389 7390 7397 7398 7408
7412 7414 7415 7418 7421 7425 7428 7441 7444 7447 7451 7452 7457 7474 7475 7480 7481 7482 7488 7512 7514 7515 7518 7521 7522 7524 7525 7541 7542 7547 7551 7552 7574 7581 7589 7598 7599 7601 7603 7610
7619 7630 7633 7636 7663 7689 7691 7698 7711 7712 7721 7739 7744 7745 7754 7788 7793 7804 7814 7815 7824 7839 7840 7841 7842 7848 7851 7859 7869 7878 7884 7887 7893 7895 7896 7900 7903 7916 7930 7937
7938 7958 7959 7961 7968 7973 7983 7985 7986 7995 8002 8006 8018 8020 8033 8036 8047 8060 8063 8074 8081 8088 8099 8108 8112 8114 8115 8118 8121 8125 8128 8141 8144 8147 8151 8152 8157 8174 8175 8180
8181 8182 8188 8200 8211 8215 8218 8225 8233 8244 8245 8247 8251 8252 8254 8256 8265 8274 8281 8299 8303 8306 8323 8330 8332 8333 8360 8369 8379 8396 8397 8407 8411 8414 8417 8424 8425 8427 8441 8442
8452 8455 8470 8471 8472 8478 8487 8488 8511 8512 8517 8521 8522 8524 8526 8542 8545 8554 8555 8562 8571 8579 8597 8600 8603 8625 8630 8639 8652 8679 8693 8697 8699 8704 8714 8715 8724 8739 8740 8741
8742 8748 8751 8759 8769 8778 8784 8787 8793 8795 8796 8801 8808 8810 8811 8812 8818 8821 8847 8848 8874 8877 8880 8881 8884 8909 8929 8936 8937 8957 8963 8967 8969 8973 8975 8976 8990 8992 8996 9001
9004 9007 9010 9012 9013 9021 9023 9031 9032 9037 9040 9046 9064 9070 9073 9089 9098 9100 9102 9103 9120 9129 9130 9133 9159 9167 9176 9192 9195 9201 9203 9210 9219 9230 9233 9236 9263 9289 9291 9298
9301 9302 9307 9310 9313 9320 9323 9326 9331 9332 9362 9368 9370 9377 9378 9386 9387 9400 9406 9444 9459 9460 9495 9519 9549 9556 9565 9578 9579 9587 9591 9594 9597 9604 9617 9623 9632 9638 9640 9655
9671 9678 9683 9687 9689 9698 9700 9703 9716 9730 9737 9738 9758 9759 9761 9768 9773 9783 9785 9786 9795 9809 9829 9836 9837 9857 9863 9867 9869 9873 9875 9876 9890 9892 9896 9908 9912 9915 9921 9928
9945 9951 9954 9957 9968 9975 9980 9982 9986 10000 10003 10009 10029 10030 10033 10039 10067 10076 10088 10090 10092 10093 10112 10114 10115 10121 10122 10125 10128 10141 10148 10151 10152 10158
10177 10182 10184 10185 10188 10209 10211 10212 10215 10218 10221 10222 10233 10247 10251 10257 10258 10274 10275 10277 10281 10285 10288 10290 10299 10300 10303 10309 10323 10330 10332 10333 10335
10337

[edit] Ruby

Works with: Ruby version 1.9
require 'set' # Set: Fast array lookup / Simple existence hash
 
@seen_numbers = Set.new
@happy_numbers = Set.new
 
def happy?(n)
return true if n == 1 # Base case
return @happy_numbers.include?(n) if @seen_numbers.include?(n) # Use performance cache, and stop unhappy cycles
 
@seen_numbers << n
digit_squared_sum = n.to_s.each_char.inject(0) { |sum, c| sum + c.to_i**2 } # In Rails: n.to_s.each_char.sum { c.to_i**2 }
 
if happy?(digit_squared_sum)
@happy_numbers << n
true # Return true
else
false # Return false
end
end

Helper method to produce output:

def print_happy
happy_numbers = []
 
(1..Float::INFINITY).each do |i|
break if happy_numbers.length >= 8
happy_numbers << i if happy?(i)
end
 
p happy_numbers
end

Output:

[1, 7, 10, 13, 19, 23, 28, 31]

[edit] Run BASIC

for i = 1 to 100
if happy(i) = 1 then
cnt = cnt + 1
PRINT cnt;". ";i;" is a happy number "
if cnt = 8 then end
end if
next i
 
FUNCTION happy(num)
while count < 50 and happy <> 1
num$ = str$(num)
count = count + 1
happy = 0
for i = 1 to len(num$)
happy = happy + val(mid$(num$,i,1)) ^ 2
next i
num = happy
wend
end function
1. 1 is a happy number 
2. 7 is a happy number 
3. 10 is a happy number 
4. 13 is a happy number 
5. 19 is a happy number 
6. 23 is a happy number 
7. 28 is a happy number 
8. 31 is a happy number  

[edit] Salmon

variable happy_count := 0;
outer:
iterate(x; [1...+oo])
{
variable seen := <<(* --> false)>>;
variable now := x;
while (true)
{
if (seen[now])
{
if (now == 1)
{
++happy_count;
print(x, " is happy.\n");
if (happy_count == 8)
break from outer;;
};
break;
};
seen[now] := true;
variable new := 0;
while (now != 0)
{
new += (now % 10) * (now % 10);
now /::= 10;
};
now := new;
};
};

This Salmon program produces the following output:

1 is happy.
7 is happy.
10 is happy.
13 is happy.
19 is happy.
23 is happy.
28 is happy.
31 is happy.

[edit] Scala

scala> def isHappy(n: Int) = {
| new Iterator[Int] {
| val seen = scala.collection.mutable.Set[Int]()
| var curr = n
| def next = {
| val res = curr
| curr = res.toString.map(_.asDigit).map(n => n * n).sum
| seen += res
| res
| }
| def hasNext = !seen.contains(curr)
| }.toList.last == 1
| }
isHappy: (n: Int)Boolean
 
scala> Iterator from 1 filter isHappy take 8 foreach println
1
7
10
13
19
23
28
31
 

[edit] Scheme

(define (number->list num)
(do ((num num (quotient num 10))
(lst '() (cons (remainder num 10) lst)))
((zero? num) lst)))
 
(define (happy? num)
(let loop ((num num) (seen '()))
(cond ((= num 1) #t)
((memv num seen) #f)
(else (loop (apply + (map (lambda (x) (* x x)) (number->list num)))
(cons num seen))))))
 
(display "happy numbers:")
(let loop ((n 1) (more 8))
(cond ((= more 0) (newline))
((happy? n) (display " ") (display n) (loop (+ n 1) (- more 1)))
(else (loop (+ n 1) more))))

The output is:

happy numbers: 1 7 10 13 19 23 28 31

[edit] Seed7

$ include "seed7_05.s7i";
 
const type: cacheType is hash [integer] boolean;
var cacheType: cache is cacheType.value;
 
const func boolean: happy (in var integer: number) is func
result
var boolean: isHappy is FALSE;
local
var bitset: cycle is bitset.value;
var integer: newnumber is 0;
var integer: cycleNum is 0;
begin
while number > 1 and number not in cycle do
if number in cache then
number := ord(cache[number]);
else
incl(cycle, number);
newnumber := 0;
while number > 0 do
newnumber +:= (number rem 10) ** 2;
number := number div 10;
end while;
number := newnumber;
end if;
end while;
isHappy := number = 1;
for cycleNum range cycle do
cache @:= [cycleNum] isHappy;
end for;
end func;
 
const proc: main is func
local
var integer: number is 0;
begin
for number range 1 to 50 do
if happy(number) then
writeln(number);
end if;
end for;
end func;

Output:

1
7
10
13
19
23
28
31
32
44
49

[edit] SETL

proc is_happy(n);
s := [n];
while n > 1 loop
if (n := +/[val(i)**2: i in str(n)]) in s then
return false;
end if;
s with:= n;
end while;
return true;
end proc;
happy := [];
n := 1;
until #happy = 8 loop
if is_happy(n) then happy with:= n; end if;
n +:= 1;
end loop;
 
print(happy);

Output:

[1 7 10 13 19 23 28 31]

Alternative version:

print([n : n in [1..100] | is_happy(n)](1..8));

Output:

[1 7 10 13 19 23 28 31]

[edit] Smalltalk

Works with: GNU Smalltalk
Translation of: Python

In addition to the "Python's cache mechanism", the use of a Bag assures that found e.g. the happy 190, we already have in cache also the happy 910 and 109, and so on.

Object subclass: HappyNumber [
|cache negativeCache|
HappyNumber class >> new [ |me|
me := super new.
^ me init
]
init [ cache := Set new. negativeCache := Set new. ]
 
hasSad: aNum [
^ (negativeCache includes: (self recycle: aNum))
]
hasHappy: aNum [
^ (cache includes: (self recycle: aNum))
]
addHappy: aNum [
cache add: (self recycle: aNum)
]
addSad: aNum [
negativeCache add: (self recycle: aNum)
]
 
recycle: aNum [ |r n| r := Bag new.
n := aNum.
[ n > 0 ]
whileTrue: [ |d|
d := n rem: 10.
r add: d.
n := n // 10.
].
^r
]
 
isHappy: aNumber [ |cycle number newnumber|
number := aNumber.
cycle := Set new.
[ (number ~= 1) & ( (cycle includes: number) not ) ]
whileTrue: [
(self hasHappy: number)
ifTrue: [ ^true ]
ifFalse: [
(self hasSad: number) ifTrue: [ ^false ].
cycle add: number.
newnumber := 0.
[ number > 0 ]
whileTrue: [ |digit|
digit := number rem: 10.
newnumber := newnumber + (digit * digit).
number := (number - digit) // 10.
].
number := newnumber.
]
].
(number = 1)
ifTrue: [
cycle do: [ :e | self addHappy: e ].
^true
]
ifFalse: [
cycle do: [ :e | self addSad: e ].
^false
]
]
].
|happy|
happy := HappyNumber new.
 
1 to: 31 do: [ :i |
(happy isHappy: i)
ifTrue: [ i displayNl ]
].

Output:

1
7
10
13
19
23
28
31

an alternative version is:

Works with: Smalltalk/X
|next isHappy happyNumbers|
 
next :=
[:n |
(n printString collect:[:ch | ch digitValue squared] as:Array) sum
].
 
isHappy :=
[:n | | t already |
already := Set new.
t := n.
[ t == 1 or:[ (already includes:t)]] whileFalse:[
already add:t.
t := next value:t.
].
t == 1
].
 
happyNumbers := OrderedCollection new.
try := 1.
[happyNumbers size < 8] whileTrue:[
(isHappy value:try) ifTrue:[ happyNumbers add:try].
try := try + 1
].
happyNumbers printCR

Output: OrderedCollection(1 7 10 13 19 23 28 31)

[edit] Tcl

using code from Sum of squares#Tcl

proc is_happy n {
set seen [list]
while {$n > 1 && [lsearch -exact $seen $n] == -1} {
lappend seen $n
set n [sum_of_squares [split $n ""]]
}
return [expr {$n == 1}]
}
 
set happy [list]
set n -1
while {[llength $happy] < 8} {
if {[is_happy $n]} {lappend happy $n}
incr n
}
puts "the first 8 happy numbers are: [list $happy]"
the first 8 happy numbers are: {1 7 10 13 19 23 28 31}

[edit] TUSCRIPT

$$ MODE TUSCRIPT
SECTION check
IF (n!=1) THEN
n = STRINGS (n,":>/:")
LOOP/CLEAR nr=n
square=nr*nr
n=APPEND (n,square)
ENDLOOP
n=SUM(n)
r_table=QUOTES (n)
BUILD R_TABLE/word/EXACT chk=r_table
IF (seq.ma.chk) THEN
status="next"
ELSE
seq=APPEND (seq,n)
ENDIF
RELEASE r_table chk
ELSE
PRINT checkednr," is a happy number"
happynrs=APPEND (happynrs,checkednr)
status="next"
ENDIF
ENDSECTION
 
happynrs=""
 
LOOP n=1,100
sz_happynrs=SIZE(happynrs)
IF (sz_happynrs==8) EXIT
checkednr=VALUE(n)
status=seq=""
LOOP
IF (status=="next") EXIT
DO check
ENDLOOP
ENDLOOP

Output:

1 is a happy number
7 is a happy number
10 is a happy number
13 is a happy number
19 is a happy number
23 is a happy number
28 is a happy number
31 is a happy number 

[edit] UNIX Shell

Works with: Bourne Again SHell
#!/bin/bash
function sum_of_square_digits
{
local -i n="$1" sum=0
while (( n )); do
local -i d=n%10
let sum+=d*d
let n=n/10
done
echo "$sum"
}
 
function is_happy?
{
local -i n="$1"
local seen=()
while (( n != 1 )); do
if [ -n "${seen[$n]}" ]; then
return 1
fi
seen[n]=1
let n="$(sum_of_square_digits "$n")"
done
return 0
}
 
function first_n_happy
{
local -i count="$1"
local -i n
for (( n=0; count; n+=1 )); do
if is_happy? "$n"; then
echo "$n"
let count-=1
fi
done
return 0
}
 
first_n_happy 8
Output:
1
7
10
13
19
23
28
31

[edit] Ursala

The happy function is a predicate testing whether a given number is happy, and first(p) defines a function mapping a number n to the first n positive naturals having property p.

#import std
#import nat
 
happy = ==1+ ^== sum:-0+ product*iip+ %np*hiNCNCS+ %nP
 
first "p" = ~&i&& iota; ~&lrtPX/&; leql@lrPrX->lrx ^|\~& ^/successor@l ^|T\~& "p"&& ~&iNC
 
#cast %nL
 
main = (first happy) 8

output:

<1,7,10,13,19,23,28,31>

[edit] Vala

Library: Gee
using Gee;
 
/* function to sum the square of the digits */
int sum(int input){
// convert input int to string
string input_str = input.to_string();
int total = 0;
// read through each character in string, square them and add to total
for (int x = 0; x < input_str.length; x++){
// char.digit_value converts char to the decimal value the char it represents holds
int digit = input_str[x].digit_value();
total += (digit * digit);
}
 
return total;
} // end sum
 
/* function to decide if a number is a happy number */
bool is_happy(int total){
var past = new HashSet<int>();
while(true){
total = sum(total);
if (total == 1){
return true;}
 
if (total in past){
return false;}
 
past.add(total);
} // end while loop
} // end happy
 
public static void main(){
var happynums = new ArrayList<int>();
int x = 1;
 
while (happynums.size < 8){
if (is_happy(x) == true)
happynums.add(x);
x++;
}
 
foreach(int num in happynums)
stdout.printf("%d ", num);
stdout.printf("\n");
} // end main

The output is:

1 7 10 13 19 23 28 31

[edit] Visual Basic .NET

This version uses Linq to carry out the calculations.

Module HappyNumbers
Sub Main()
Dim n As Integer = 1
Dim found As Integer = 0
 
Do Until found = 8
If IsHappy(n) Then
found += 1
Console.WriteLine("{0}: {1}", found, n)
End If
n += 1
Loop
 
Console.ReadLine()
End Sub
 
Private Function IsHappy(ByVal n As Integer)
Dim cache As New List(Of Long)()
 
Do Until n = 1
cache.Add(n)
n = Aggregate c In n.ToString() _
Into Total = Sum(Int32.Parse(c) ^ 2)
If cache.Contains(n) Then Return False
Loop
 
Return True
End Function
End Module

The output is:

1: 1
2: 7
3: 10
4: 13
5: 19
6: 23
7: 28
8: 31

[edit] XPL0

The largest possible 32-bit integer is less than 9,999,999,999. The sum of the squares of these ten digits is 10*9^2 = 810. If a cycle consisted of all the values smaller than 810, an array size of 810 would still be sufficiently large to hold them. Actually, tests show that the array only needs to hold 16 numbers.

int List(810);          \list of numbers in a cycle
int Inx; \index for List
include c:\cxpl\codes;
 
 
func HadNum(N); \Return 'true' if number N is in the List
int N;
int I;
[for I:= 0 to Inx-1 do
if N = List(I) then return true;
return false;
]; \HadNum
 
 
func SqDigits(N); \Return the sum of the squares of the digits of N
int N;
int S, D;
[S:= 0;
while N do
[N:= N/10;
D:= rem(0);
S:= S + D*D;
];
return S;
]; \SqDigits
 
 
int N0, N, C;
[N0:= 0; \starting number
C:= 0; \initialize happy (starting) number counter
repeat N:= N0;
Inx:= 0; \reset List index
loop [N:= SqDigits(N);
if N = 1 then \happy number
[IntOut(0, N0); CrLf(0);
C:= C+1;
quit;
];
if HadNum(N) then quit; \if unhappy number then quit
List(Inx):= N; \if neither, add it to the List
Inx:= Inx+1; \ and continue the cycle
];
N0:= N0+1; \next starting number
until C=8; \done when 8 happy numbers have been found
]

Output:

1
7
10
13
19
23
28
31

[edit] zkl

Here is a function that generates a continuous stream of happy numbers. Given that there are lots of happy numbers, caching them doesn't seem like a good idea memory wise. Instead, a num of squared digits == 4 is used as a proxy for a cycle (see the Wikipedia article, there are several number that will work).

Translation of: Icon and Unicon
fcn happyNumbers{  // continously spew happy numbers
foreach N in ([1..]){
n:=N; while(1){
n=n.split().reduce(fcn(p,n){ p + n*n },0);
if(n==1) { vm.yield(N); break; }
if(n==4) break; // unhappy cycle
}
}
}
h:=Utils.Generator(happyNumbers);
h.walk(8).println();
Output:
L(1,7,10,13,19,23,28,31)

Get the one million-th happy number. Nobody would call this quick.

Utils.Generator(happyNumbers).drop(0d1_000_000-1).next().println();
Output:
7105849
Personal tools
Namespaces

Variants
Actions
Community
Explore
Misc
Toolbox