Happy numbers
From Rosetta Code
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: The happy numbers on OEIS
[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
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] 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] 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>output
#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;
}
1 7 10 13 19 23 28 31 The 1000000th happy number: 7105849
Without caching, using cycle detection:
#include <stdio.h>Output is same as above, but much slower.
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;
}
[edit] C++
#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- digit-to-num [d] (Character/digit d 10))Output:
(defn- square [n] (* n n))
(defn happy? [n]
(loop [n n, seen #{}]
(cond (= n 1) true
(seen n) false
:else
(recur (reduce + (map (comp square digit-to-num) (str n)))
(conj seen n)))))
(def happy-numbers (filter happy? (iterate inc 1)))
(println (take 8 happy-numbers))
(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 sum-of-sqr-dgts (n)Output:
(let ((digit (mod n 10)))
(if (zerop n)
0
(+ (* digit digit) (sum-of-sqr-dgts (floor (/ n 10)))))))
(defun happy-p (n &optional cache)
(or (= n 1)
(when (not (find n cache))
(happy-p (sum-of-sqr-dgts n) (cons n cache)))))
(defun print-happys()
(let ((i 1) (happys 0))
(loop while (< happys 8) do
(when (happy-p i)
(print i)
(incf happys))
(incf i))))
(print-happys)
1 7 10 13 19 23 28 31
[edit] D
import std.stdio, std.algorithm, std.range;
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() {
int.max.iota().filter!isHappy().take(8).writeln();
}
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
[edit] Alternative version
Same output:
import std.stdio, std.algorithm, std.range, std.conv;
bool isHappy(int n) /*pure nothrow*/ {
int[int] seen;
while (true) {
const t = n.text().map!q{(a - '0') ^^ 2}().reduce!q{a + b}();
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();
}
[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] 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
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).Command:
-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, []).
erl -run tasks main -run init stop -noshellOutput:
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
[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] 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] Go
package main
import (
"fmt"
"strconv"
)
func happy(n int) bool {
m := make(map[int]int)
for n > 1 {
m[n] = 0
s := strconv.Itoa(n)
n = 0
for _, d := range s {
x := int(d) - '0'
n += x * x
}
if _, ok := m[n]; ok {
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
[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
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 </lang>
=={{header|K}}==
<lang 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</lang>
=={{header|Liberty BASIC}}==
<lang lb> 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</lang>
Output:-
<pre>1 1
2 7
3 10
4 13
5 19
6 23
7 28
8 31
[edit] Logo
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] 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] 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] MUMPS
ISHAPPY(N)Output:
;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=+$$ISHAPPY(I) WRITE:Q !,I SET:Q C=C-1
KILL I
QUIT
USER>D HAPPY^ROSETTA(8) The first 8 happy numbers are: 1 7 10 13 19 23 28 31 USER>W:+$$ISHAPPY^ROSETTA(320) "Happy Number" Happy Number USER>W:+$$ISHAPPY^ROSETTA(321) "Happy Number" USER>
[edit] NetRexx
/*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] 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] Oz
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
- 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)toselect(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:
perl coins.pl 1 7 10 13 19 23 28 31
[edit] Perl 6
sub happy (Int $n is copy --> Bool) {
my %seen;
loop {
($n = [+] map * ** 2, $n.comb) == 1 and return True;
%seen{$n}++ and return False;
}
}
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:
1 7 10 13 19 23 28 31
There's more than one way to do it...
[edit] PHP
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
[edit] Traditional
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
$i
}
$hist[$sum]=$sum
$sum2=0
foreach($j in $sum.ToString().ToCharArray()) {
$k=([int]$j)-0x30
$sum2+=$k*$k
}
$sum=$sum2
}
}
}
[edit] With Pipeline
function happy([int] $n) {
1..$n | ForEach-Object {
$sum=$_
$hist=@{}
while( $hist[$sum] -eq $null ) {
if($sum -eq 1) {
$_
}
$hist[$sum]=$sum
$sum2=0
foreach($j in $sum.ToString().ToCharArray()) {
$k=([int]$j)-0x30
$sum2+=$k*$k
}
$sum=$sum2
}
}
}
[edit] 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 True:
total = sum(int(i)**2 for i in str(n))
if total == 1:
return True
if total in past:
return False
n = total
past.add(total)
>>> [x for x in range(1,500) 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] REXX
/*REXX program to display the 1st 8 (or specified arg) happy numbers*/
arg limit . /*get argument for LIMIT. */
if limit=='' then limit=8 /*if not specified, set LIMIT to 8*/
haps=0 /*count of happy numbers so far. */
do n=1 while haps<limit /*search integers starting at one.*/
q=n /*Q may or may not be "happy". */
a.=0
do 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. */
do j=1 for length(q) /*add the squares of the numerals.*/
sum=sum+substr(q,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
Sample 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] Ruby
Use of cache from Python
require 'set'
def happy?(n)
seen = Set[]
state = while n>1 and not seen.include?(n)
if $happy_cache[:happy].include?(n): break :happy
elsif $happy_cache[:sad].include?(n): break :sad
end
seen << n
n = sum_of_squares_of_digits(n)
end
state.nil? and state = n == 1 ? :happy : :sad
$happy_cache[state] += seen
state == :happy
end
def sum_of_squares_of_digits(n)
n.to_s.each_char.inject(0) {|sum,c| sum += (c.to_i)**2}
end
$happy_cache = Hash.new(Set[])
happy_numbers = []
i = 1
while happy_numbers.length < 8
happy_numbers << i if happy? i
i += 1
end
p happy_numbers
p $happy_cache
[1, 7, 10, 13, 19, 23, 28, 31]
{:happy=>[7, 49, 97, 130, 10, 13, 19, 82, 68, 100, 23, 28, 31],
:sad=>[2, 4, 16, 37, 58, 89, 145, 42, 20, 3, 9, 81, 65, 61, 5, 25, 29, 85, 6, 36, 45, 41, 17, 50, 8, 64, 52, 11, 12, 14, 15, 26, 40, 18, 21, 22, 24, 27, 53, 34, 30]}
[edit] Run BASIC
for j = 1 to 100 ' test 1 to 100 for happy number
if isHappy(j) = 1 then print j;" is happy"
next j
function isHappy(n)
for i = 1 to 100
isHappy = 0
while n <> 0 ' Form the sum of squares of the digits.
isHappy = isHappy + (n mod 10)^2
n = int(n/10)
wend
if isHappy = 1 then goto [endIt]
n = isHappy ' Replace n with the new number formed from digits.
next i
[endIt]
end function
1 is happy 7 is happy 10 is happy 13 is happy 19 is happy 23 is happy
[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] 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
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
[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
#!/bin/bashOutput:
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
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
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
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