Happy numbers: Difference between revisions
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<pre>1 7 10 13 19 23 28 31</pre> |
<pre>1 7 10 13 19 23 28 31</pre> |
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=={{header|Draco}}== |
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<lang draco>proc nonrec dsumsq(byte n) byte: |
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byte r, d; |
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r := 0; |
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while n~=0 do |
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d := n % 10; |
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n := n / 10; |
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r := r + d * d |
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od; |
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r |
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corp |
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proc nonrec happy(byte n) bool: |
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[256] bool seen; |
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byte i; |
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for i from 0 upto 255 do seen[i] := false od; |
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while not seen[n] do |
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seen[n] := true; |
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n := dsumsq(n) |
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od; |
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seen[1] |
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corp |
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proc nonrec main() void: |
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byte n, seen; |
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n := 1; |
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seen := 0; |
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while seen < 8 do |
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if happy(n) then |
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writeln(n:3); |
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seen := seen + 1 |
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fi; |
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n := n + 1 |
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od |
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corp</lang> |
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{{out}} |
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<pre> 1 |
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7 |
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10 |
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13 |
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19 |
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23 |
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28 |
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31</pre> |
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=={{header|DWScript}}== |
=={{header|DWScript}}== |
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<lang delphi>function IsHappy(n : Integer) : Boolean; |
<lang delphi>function IsHappy(n : Integer) : Boolean; |
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Revision as of 23:24, 13 January 2022
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 end in 1 are happy numbers,
- while those numbers that do not end in 1 are unhappy numbers.
- Task
Find and print the first 8 happy numbers.
Display an example of your output here on this page.
- See also
- The OEIS entry: The happy numbers: A007770
- The OEIS entry: The unhappy numbers; A031177
11l
<lang 11l>F happy(=n)
Set[Int] past
L n != 1
n = sum(String(n).map(с -> Int(с)^2))
I n C past
R 0B
past.add(n)
R 1B
print((0.<500).filter(x -> happy(x))[0.<8])</lang>
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
8080 Assembly
This is not just a demonstration of 8080 assembly, but also of why it pays to look closely at the problem domain. The following program only does 8-bit unsigned integer math, which not only fits the 8080's instruction set very well, it also means the cycle detection can be done using only an array of 256 flags, and all other state fits in the registers. This makes the program a good deal simpler than it would've been otherwise.
In general, 8-bit math is not good enough for numerical problems, but in this particular case, the problem only asks for the first eight happy numbers, none of which (nor any of the unhappy numbers in between) have a cycle that ever goes above 145, so eight bits is good enough. In fact, for any input under 256, the cycle never goes above 163; this program could be trivially changed to print up to 39 happy numbers.
<lang 8080asm>flags: equ 2 ; 256-byte page in which to keep track of cycles puts: equ 9 ; CP/M print string bdos: equ 5 ; CP/M entry point org 100h lxi d,0108h ; D=current number to test, E=amount of numbers ;;; Is D happy? number: mvi a,1 ; We haven't seen any numbers yet, set flags to 1 lxi h,256*flags init: mov m,a inr l jnz init mov a,d ; Get digits step: call digits mov l,a ; L = D1 * D1 mov h,a xra a sqr1: add h dcr l jnz sqr1 mov l,a mov h,b ; L += D10 * D10 xra a sqr10: add h dcr b jnz sqr10 add l mov l,a mov h,c ; L += D100 * D100 xra a sqr100: add h dcr c jnz sqr100 add l mov l,a mvi h,flags ; Look up corresponding flag dcr m ; Will give 0 the first time and not-0 afterwards mov a,l ; If we haven't seen the number before, another step jz step dcr l ; If we _had_ seen it, then is it 1? jz happy ; If so, it is happy next: inr d ; Afterwards, try next number jmp number happy: mov a,d ; D is happy - get its digits (for output) lxi h,string+3 call digits ; Write digits into string for output call sdgt ; Ones digit, mov a,b ; Tens digit, call sdgt mov a,c ; Hundreds digit call sdgt push d ; Keep counters on stack mvi c,puts ; Print string using CP/M call xchg call bdos pop d ; Restore counters dcr e ; One fewer happy number left jnz next ; If we need more, do the next one ret ;;; Store A as ASCII digit in [HL] and go to previous digit sdgt: adi '0' dcx h mov m,a ret ;;; Get digits of 8-bit number in A. ;;; Input: A = number ;;; Output: C=100s digit, B=10s digit, A=1s digit digits: lxi b,-1 ; Set B and C to -1 (correct for extra loop cycle) d100: inr c ; Calculate hundreds digit sui 100 ; By trial subtraction of 100 jnc d100 ; Until underflow occurs adi 100 ; Loop runs one cycle too many, so add 100 back d10: inr b ; Calculate 10s digit in the same way sui 10 jnc d10 adi 10 ret ; 1s digit is left in A afterwards string: db '000',13,10,'$'</lang>
- Output:
001 007 010 013 019 023 028 031
8th
<lang 8th>
- until! "not while!" eval i;
with: w with: n
- sumsqd \ n -- n
0 swap repeat
0; 10 /mod -rot sqr + swap
again ;
- cycle \ n xt -- n
>r
dup r@ exec \ -- tortoise, hare
repeat
swap r@ exec
swap r@ exec r@ exec
2dup = until!
rdrop drop ;
- happy? ' sumsqd cycle 1 = ;
- .happy \ n --
1 repeat
dup happy? if dup . space swap 1- swap then 1+
over 0 > while!
2drop cr ;
- with
- with
</lang>
- Output:
ok> 8 .happy 1 7 10 13 19 23 28 31
ACL2
<lang Lisp>(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))</lang>
Output:
(1 7 10 13 19 23 28 31)
Action!
<lang Action!>BYTE FUNC SumOfSquares(BYTE x)
BYTE sum,d
sum=0 WHILE x#0 DO d=x MOD 10 d==*d sum==+d x==/10 OD
RETURN (sum)
BYTE FUNC Contains(BYTE ARRAY a BYTE count,x)
BYTE i
FOR i=0 TO count-1 DO IF a(i)=x THEN RETURN (1) FI OD
RETURN (0)
BYTE FUNC IsHappyNumber(BYTE x)
BYTE ARRAY cache(100) BYTE count
count=0
WHILE x#1
DO
cache(count)=x
count==+1
x=SumOfSquares(x)
IF Contains(cache,count,x) THEN
RETURN (0)
FI
OD
RETURN (1)
PROC Main()
BYTE x,count
x=1 count=0
WHILE count<8
DO
IF IsHappyNumber(x) THEN
count==+1
PrintF("%I: %I%E",count,x)
FI
x==+1
OD
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
1: 1 2: 7 3: 10 4: 13 5: 19 6: 23 7: 28 8: 31
ActionScript
<lang 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();</lang> Sample output:
1 7 10 13 19 23 28 31
Ada
<lang 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;</lang> Sample output:
1 7 10 13 19 23 28 31
ALGOL 68
<lang algol68>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</lang> Output:
+1
+7
+10
+13
+19
+23
+28
+31
ALGOL-M
<lang algolm>begin integer function mod(a,b); integer a,b; mod := a-(a/b)*b;
integer function sumdgtsq(n); integer n; sumdgtsq :=
if n = 0 then 0 else mod(n,10)*mod(n,10) + sumdgtsq(n/10);
integer function happy(n); integer n; begin
integer i;
integer array seen[0:200];
for i := 0 step 1 until 200 do seen[i] := 0;
while seen[n] = 0 do
begin
seen[n] := 1;
n := sumdgtsq(n);
end;
happy := if n = 1 then 1 else 0;
end;
integer i, n; i := n := 0; while n < 8 do begin
if happy(i) = 1 then
begin
write(i);
n := n + 1;
end;
i := i + 1;
end; end</lang>
- Output:
1
7
10
13
19
23
28
31
ALGOL W
<lang algolw>begin
% find some happy numbers %
% returns true if n is happy, false otherwise; n must be >= 0 %
logical procedure isHappy( integer value n ) ;
if n < 2 then true
else begin
% seen is used to hold the values of the cycle of the %
% digit square sums, as noted in the Batch File %
% version, we do not need a large array. The digit %
% square sum of 9 999 999 999 is 810... %
integer array seen( 0 :: 32 );
integer number, trys;
number := n;
trys := -1;
while begin
logical terminated;
integer tPos;
terminated := false;
tPos := 0;
while not terminated and tPos <= trys do begin
terminated := seen( tPos ) = number;
tPos := tPos + 1
end while_not_terminated_and_tPos_lt_trys ;
number > 1 and not terminated
end do begin
integer sum;
trys := trys + 1;
seen( trys ) := number;
sum := 0;
while number > 0 do begin
integer digit;
digit := number rem 10;
number := number div 10;
sum := sum + ( digit * digit )
end while_number_gt_0 ;
number := sum
end while_number_gt_1_and_not_terminated ;
number = 1
end isHappy ;
% print the first 8 happy numbers %
begin
integer happyCount, n;
happyCount := 0;
n := 1;
write( "first 8 happy numbers: " );
while happyCount < 8 do begin
if isHappy( n ) then begin
writeon( i_w := 1, " ", n );
happyCount := happyCount + 1
end if_isHappy_n ;
n := n + 1
end while_happyCount_lt_8
end
end.</lang>
- Output:
first 8 happy numbers: 1 7 10 13 19 23 28 31
APL
Tradfn
<lang 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
∇</lang>
HappyNumbers 100 8 1 7 10 13 19 23 28 31
Dfn
<lang APL>
HappyNumbers←{ ⍝ return the first ⍵ Happy Numbers
⍺←⍬ ⍝ initial list
⍵=+/⍺:⍸⍺ ⍝ 1's mark happy numbers
sq←×⍨ ⍝ square function (times selfie)
isHappy←{ ⍝ is ⍵ a happy number?
⍺←⍬ ⍝ previous sums
⍵=1:1 ⍝ if we get to 1, it's happy
n←+/sq∘⍎¨⍕⍵ ⍝ sum of the square of the digits
n∊⍺:0 ⍝ if we hit this sum before, it's not happy
(⍺,n)∇ n} ⍝ recurse until it's happy or not
(⍺,isHappy 1+≢⍺)∇ ⍵ ⍝ recurse until we have ⍵ happy numbers
}
HappyNumbers 8
1 7 10 13 19 23 28 31 </lang>
AppleScript
Iteration
<lang AppleScript>on run
set howManyHappyNumbers to 8
set happyNumberList to {}
set globalCounter to 1
repeat howManyHappyNumbers times
repeat while not isHappy(globalCounter)
set globalCounter to globalCounter + 1
end repeat
set end of happyNumberList to globalCounter
set globalCounter to globalCounter + 1
end repeat
log happyNumberList
end run
on isHappy(numberToCheck)
set localCycle to {}
repeat while (numberToCheck ≠ 1)
if localCycle contains numberToCheck then
exit repeat
end if
set end of localCycle to numberToCheck
set tempNumber to 0
repeat while (numberToCheck > 0)
set digitOfNumber to numberToCheck mod 10
set tempNumber to tempNumber + (digitOfNumber ^ 2)
set numberToCheck to (numberToCheck - digitOfNumber) / 10
end repeat
set numberToCheck to tempNumber
end repeat
return (numberToCheck = 1)
end isHappy</lang>
Result: (*1, 7, 10, 13, 19, 23, 28, 31*)
Functional composition
<lang AppleScript>---------------------- HAPPY NUMBERS -----------------------
-- isHappy :: Int -> Bool on isHappy(n)
-- endsInOne :: [Int] -> Int -> Bool
script endsInOne
-- sumOfSquaredDigits :: Int -> Int
script sumOfSquaredDigits
-- digitSquared :: Int -> Int -> Int
script digitSquared
on |λ|(a, x)
(a + (x as integer) ^ 2) as integer
end |λ|
end script
on |λ|(n)
foldl(digitSquared, 0, splitOn("", n as string))
end |λ|
end script
-- [Int] -> Int -> Bool
on |λ|(s, n)
if n = 1 then
true
else
if s contains n then
false
else
|λ|(s & n, |λ|(n) of sumOfSquaredDigits)
end if
end if
end |λ|
end script
endsInOne's |λ|({}, n)
end isHappy
TEST ---------------------------
on run
-- seriesLength :: {n:Int, xs:[Int]} -> Bool
script seriesLength
property target : 8
on |λ|(rec)
length of xs of rec = target of seriesLength
end |λ|
end script
-- succTest :: {n:Int, xs:[Int]} -> {n:Int, xs:[Int]}
script succTest
on |λ|(rec)
tell rec to set {xs, n} to {its xs, its n}
script testResult
on |λ|(x)
if isHappy(x) then
xs & x
else
xs
end if
end |λ|
end script
{n:n + 1, xs:testResult's |λ|(n)}
end |λ|
end script
xs of |until|(seriesLength, succTest, {n:1, xs:{}})
--> {1, 7, 10, 13, 19, 23, 28, 31}
end run
GENERIC FUNCTIONS ---------------------
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- splitOn :: String -> String -> [String]
on splitOn(pat, src)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, pat}
set xs to text items of src
set my text item delimiters to dlm
return xs
end splitOn
-- until :: (a -> Bool) -> (a -> a) -> a -> a
on |until|(p, f, x)
set mp to mReturn(p)
set v to x
tell mReturn(f)
repeat until mp's |λ|(v)
set v to |λ|(v)
end repeat
end tell
return v
end |until|</lang>
- Output:
<lang AppleScript>{1, 7, 10, 13, 19, 23, 28, 31}</lang>
Arturo
<lang rebol>ord0: to :integer `0` happy?: function [x][
n: x past: new []
while [n <> 1][
s: to :string n
n: 0
loop s 'c [
i: (to :integer c) - ord0
n: n + i * i
]
if contains? past n -> return false
'past ++ n
]
return true
]
loop 0..31 'x [
if happy? x -> print x
]</lang>
- Output:
1 7 10 13 19 23 28 31
AutoHotkey
<lang 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)
}</lang>
The first 8 happy numbers are: 1,7,10,13,19,23,28,31
Alternative version
<lang AutoHotkey>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
}
}</lang>
1 7 10 13 19 23 28 31
AutoIt
<lang 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 </lang>
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
Alternative version
<lang autoit> $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 </lang>
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
AWK
<lang 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
}
}
}</lang> Result:
1 7 10 13 19 23 28 31
Alternative version
Alternately, for legibility one might write:
<lang awk>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
}</lang>
Batch File
happy.bat <lang dos>@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
)</lang>
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
BBC BASIC
<lang bbcbasic> 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)</lang>
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
BCPL
<lang bcpl>get "libhdr"
let sumdigitsq(n) =
n=0 -> 0, (n rem 10)*(n rem 10)+sumdigitsq(n/10)
let happy(n) = valof $( let seen = vec 255
for i = 0 to 255 do i!seen := false
$( n!seen := true
n := sumdigitsq(n)
$) repeatuntil n!seen
resultis 1!seen
$)
let start() be $( let n, i = 0, 0
while n < 8 do
$( if happy(i) do
$( n := n + 1
writef("%N ",i)
$)
i := i + 1
$)
wrch('*N')
$)</lang>
- Output:
1 7 10 13 19 23 28 31
Bori
<lang 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);
}</lang> Output:
First 8 happy numbers : [1, 7, 10, 13, 19, 23, 28, 31]
BQN
<lang bqn>SumSqDgt ← +´2⋆˜ •Fmt-'0'˙ Happy ← ⟨⟩{𝕨((⊑∊˜ )◶⟨∾𝕊(SumSqDgt⊢),1=⊢⟩)𝕩}⊢ 8↑Happy¨⊸/↕50</lang>
- Output:
⟨ 1 7 10 13 19 23 28 31 ⟩
Brat
<lang 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}"</lang> 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]
C
Recursively look up if digit square sum is happy. <lang c>#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; }</lang>
output
1 7 10 13 19 23 28 31 The 1000000th happy number: 7105849
Without caching, using cycle detection: <lang c>#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; }</lang> Output is same as above, but much slower.
C#
<lang csharp>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));
}
}
}</lang>
First 8 happy numbers : 1,7,10,13,19,23,28,31
Alternate (cacheless)
Instead of caching and checking for being stuck in a loop, one can terminate on the "unhappy" endpoint of 89. One might be temped to try caching the so-far-found happy and unhappy numbers and checking the cache to speed things up. However, I have found that the cache implementation overhead reduces performance compared to this cacheless version.
Reaching 10 million, the <34 second computation time was from Tio.run. It takes under 5 seconds on a somewhat modern CPU. If you edit it to max out at 100 million, it takes about 50 seconds (on the somewhat modern CPU).<lang csharp>using System;
using System.Collections.Generic;
class Program
{
static int[] sq = { 1, 4, 9, 16, 25, 36, 49, 64, 81 };
static bool isOne(int x)
{
while (true)
{
if (x == 89) return false;
int s = 0, t;
do if ((t = (x % 10) - 1) >= 0) s += sq[t]; while ((x /= 10) > 0);
if (s == 1) return true;
x = s;
}
}
static void Main(string[] args)
{
const int Max = 10_000_000; DateTime st = DateTime.Now;
Console.Write("---Happy Numbers---\nThe first 8:");
int c = 0, i; for (i = 1; c < 8; i++)
if (isOne(i)) Console.Write("{0} {1}", c == 0 ? "" : ",", i, ++c);
for (int m = 10; m <= Max; m *= 10)
{
Console.Write("\nThe {0:n0}th: ", m);
for (; c < m; i++) if (isOne(i)) c++;
Console.Write("{0:n0}", i - 1);
}
Console.WriteLine("\nComputation time {0} seconds.", (DateTime.Now - st).TotalSeconds);
}
}</lang>
- Output:
---Happy Numbers--- The first 8: 1, 7, 10, 13, 19, 23, 28, 31 The 10th: 44 The 100th: 694 The 1,000th: 6,899 The 10,000th: 67,169 The 100,000th: 692,961 The 1,000,000th: 7,105,849 The 10,000,000th: 71,313,350 Computation time 33.264518 seconds.
C++
<lang cpp>#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;
}</lang> Output:
1 7 10 13 19 23 28 31 32 44 49
Alternative version without caching: <lang cpp>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;
}</lang> Output:
1 7 10 13 19 23 28 31
Cycle detection in is_happy() above is done using Floyd's cycle-finding algorithm.
Clojure
<lang 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))</lang>
Output:
(1 7 10 13 19 23 28 31)
Alternate Version (with caching)
<lang clojure>(require '[clojure.set :refer [union]])
(def ^{:private true} cache {:happy (atom #{}) :sad (atom #{})})
(defn break-apart [n]
(->> (str n)
(map str)
(map #(Long/parseLong %))))
(defn next-number [n]
(->> (break-apart n)
(map #(* % %))
(apply +)))
(defn happy-or-sad? [prev n]
(cond (or (= n 1) ((deref (:happy cache)) n)) :happy
(or ((deref (:sad cache)) n) (some #(= % n) prev)) :sad :else :unknown))
(defn happy-algo [n]
(let [get-next (fn prev n [(conj prev n) (next-number n)])
my-happy-or-sad? (fn prev n [(happy-or-sad? prev n) (conj prev n)]) unknown? (fn res nums (= res :unknown)) [res nums] (->> [#{} n] (iterate get-next) (map my-happy-or-sad?) (drop-while unknown?) first) _ (swap! (res cache) union nums)]
res))
(def happy-numbers (->> (iterate inc 1)
(filter #(= :happy (happy-algo %)))))
(println (take 8 happy-numbers))</lang> Same output.
CLU
<lang clu>sum_dig_sq = proc (n: int) returns (int)
sum_sq: int := 0
while n > 0 do
sum_sq := sum_sq + (n // 10) ** 2
n := n / 10
end
return (sum_sq)
end sum_dig_sq
is_happy = proc (n: int) returns (bool)
nn: int := sum_dig_sq(n)
while nn ~= n cand nn ~= 1 do
n := sum_dig_sq(n)
nn := sum_dig_sq(sum_dig_sq(nn))
end
return (nn = 1)
end is_happy
happy_numbers = iter (start, num: int) yields (int)
n: int := start
while num > 0 do
if is_happy(n) then
yield (n)
num := num-1
end
n := n+1
end
end happy_numbers
start_up = proc ()
po: stream := stream$primary_output()
for i: int in happy_numbers(1, 8) do
stream$putl(po, int$unparse(i))
end
end start_up </lang>
- Output:
1 7 10 13 19 23 28 31
CoffeeScript
<lang 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</lang>
output
> coffee happy.coffee 1 7 10 13 19 23 28 31
Common Lisp
<lang 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)) </lang>
Output:
(1 7 10 13 19 23 28 31)
Cowgol
<lang cowgol>include "cowgol.coh";
sub sumDigitSquare(n: uint8): (s: uint8) is
s := 0;
while n != 0 loop
var d := n % 10;
s := s + d * d;
n := n / 10;
end loop;
end sub;
sub isHappy(n: uint8): (h: uint8) is
var seen: uint8[256]; MemZero(&seen[0], @bytesof seen);
while seen[n] == 0 loop
seen[n] := 1;
n := sumDigitSquare(n);
end loop;
if n == 1 then
h := 1;
else
h := 0;
end if;
end sub;
var n: uint8 := 1; var seen: uint8 := 0;
while seen < 8 loop
if isHappy(n) != 0 then
print_i8(n);
print_nl();
seen := seen + 1;
end if;
n := n + 1;
end loop;</lang>
- Output:
1 7 10 13 19 23 28 31
Crystal
<lang ruby>def happy?(n)
past = [] of Int32 | Int64 until n == 1 sum = 0; while n > 0; sum += (n % 10) ** 2; n //= 10 end return false if past.includes? (n = sum) past << n end true
end
i = count = 0 until count == 8; (puts i; count += 1) if happy?(i += 1) end puts (99999999999900..99999999999999).each { |i| puts i if happy?(i) }</lang>
- Output:
1 7 10 13 19 23 28 31 99999999999901 99999999999910 99999999999914 99999999999915 99999999999916 99999999999937 99999999999941 99999999999951 99999999999956 99999999999961 99999999999965 99999999999973
D
<lang 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;
}</lang>
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
Alternative Version
<lang d>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;
}</lang> Same output.
Dart
<lang 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++;
}
}</lang>
dc
<lang dc>[lcI~rscd*+lc0<H]sH [0rsclHxd4<h]sh [lIp]s_ 0sI[lI1+dsIlhx2>_z8>s]dssx</lang> Output:
1 7 10 13 19 23 28 31
DCL
<lang DCL>$ happy_1 = 1 $ found = 0 $ i = 1 $ loop1: $ n = i $ seen_list = "," $ loop2: $ if f$type( happy_'n ) .nes. "" then $ goto happy $ if f$type( unhappy_'n ) .nes. "" then $ goto unhappy $ if f$locate( "," + n + ",", seen_list ) .eq. f$length( seen_list ) $ then $ seen_list = seen_list + f$string( n ) + "," $ else $ goto unhappy $ endif $ ns = f$string( n ) $ nl = f$length( ns ) $ j = 0 $ sumsq = 0 $ loop3: $ digit = f$integer( f$extract( j, 1, ns )) $ sumsq = sumsq + digit * digit $ j = j + 1 $ if j .lt. nl then $ goto loop3 $ n = sumsq $ goto loop2 $ unhappy: $ j = 1 $ loop4: $ x = f$element( j, ",", seen_list ) $ if x .eqs. "" then $ goto continue $ unhappy_'x = 1 $ j = j + 1 $ goto loop4 $ happy: $ found = found + 1 $ found_'found = i $ if found .eq. 8 then $ goto done $ j = 1 $ loop5: $ x = f$element( j, ",", seen_list ) $ if x .eqs. "" then $ goto continue $ happy_'x = 1 $ j = j + 1 $ goto loop5 $ continue: $ i = i + 1 $ goto loop1 $ done: $ show symbol found*</lang>
- Output:
FOUND = 8 Hex = 00000008 Octal = 00000000010 FOUND_1 = 1 Hex = 00000001 Octal = 00000000001 FOUND_2 = 7 Hex = 00000007 Octal = 00000000007 FOUND_3 = 10 Hex = 0000000A Octal = 00000000012 FOUND_4 = 13 Hex = 0000000D Octal = 00000000015 FOUND_5 = 19 Hex = 00000013 Octal = 00000000023 FOUND_6 = 23 Hex = 00000017 Octal = 00000000027 FOUND_7 = 28 Hex = 0000001C Octal = 00000000034 FOUND_8 = 31 Hex = 0000001F Octal = 00000000037
Delphi
Adaptation of #Pascal. The lib Boost.Int can be found here [1] <lang Delphi> program Happy_numbers;
{$APPTYPE CONSOLE}
uses
System.SysUtils, Boost.Int;
type
TIntegerDynArray = TArray<Integer>;
TIntHelper = record helper for Integer function IsHappy: Boolean; procedure Next; end;
{ TIntHelper }
function TIntHelper.IsHappy: Boolean; var
cache: TIntegerDynArray; sum, n: integer;
begin
n := self;
repeat
sum := 0;
while n > 0 do
begin
sum := sum + (n mod 10) * (n mod 10);
n := n div 10;
end;
if sum = 1 then
exit(True);
if cache.Has(sum) then
exit(False);
n := sum;
cache.Add(sum);
until false;
end;
procedure TIntHelper.Next; begin
inc(self);
end;
var
count, n: integer;
begin
n := 1;
count := 0;
while count < 8 do
begin
if n.IsHappy then
begin
count.Next;
write(n, ' ');
end;
n.Next;
end;
writeln;
readln;
end.</lang>
- Output:
1 7 10 13 19 23 28 31
Draco
<lang draco>proc nonrec dsumsq(byte n) byte:
byte r, d;
r := 0;
while n~=0 do
d := n % 10;
n := n / 10;
r := r + d * d
od;
r
corp
proc nonrec happy(byte n) bool:
[256] bool seen;
byte i;
for i from 0 upto 255 do seen[i] := false od;
while not seen[n] do
seen[n] := true;
n := dsumsq(n)
od;
seen[1]
corp
proc nonrec main() void:
byte n, seen;
n := 1;
seen := 0;
while seen < 8 do
if happy(n) then
writeln(n:3);
seen := seen + 1
fi;
n := n + 1
od
corp</lang>
- Output:
1 7 10 13 19 23 28 31
DWScript
<lang delphi>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;</lang> Output:
1 7 10 13 19 23 28 31
Dyalect
<lang dyalect>func happy(n) {
var m = []
while n > 1 {
m.add(n)
var x = n
n = 0
while x > 0 {
var d = x % 10
n += d * d
x /= 10
}
if m.indexOf(n) != -1 {
return false
}
}
return true
}
var (n, found) = (1, 0) while found < 8 {
if happy(n) {
print("\(n) ", terminator: "")
found += 1
}
n += 1
} print()</lang>
- Output:
1 7 10 13 19 23 28 31
Déjà Vu
<lang dejavu>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</lang>
- 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
E
<lang 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 }
}</lang>
Eiffel
<lang Eiffel> class APPLICATION
create make
feature {NONE} -- Initialization
make -- Run application. local l_val: INTEGER do from l_val := 1 until l_val > 100 loop if is_happy_number (l_val) then print (l_val.out) print ("%N") end l_val := l_val + 1 end end
feature -- Happy number
is_happy_number (a_number: INTEGER): BOOLEAN -- Is `a_number' a happy number? require positive_number: a_number > 0 local l_number: INTEGER l_set: ARRAYED_SET [INTEGER] do from l_number := a_number create l_set.make (10) until l_number = 1 or l_set.has (l_number) loop l_set.put (l_number) l_number := square_sum_of_digits (l_number) end
Result := (l_number = 1) end
feature{NONE} -- Implementation
square_sum_of_digits (a_number: INTEGER): INTEGER -- Sum of the sqares of digits of `a_number'. require positive_number: a_number > 0 local l_number, l_digit: INTEGER do from l_number := a_number until l_number = 0 loop l_digit := l_number \\ 10 Result := Result + l_digit * l_digit l_number := l_number // 10 end end
end
</lang>
Elena
ELENA 4.x : <lang elena>import extensions; import system'collections; import system'routines;
isHappy(int n) {
auto cache := new List<int>(5);
int sum := 0;
int num := n;
while (num != 1)
{
if (cache.indexOfElement:num != -1)
{
^ false
};
cache.append(num);
while (num != 0)
{
int digit := num.mod:10;
sum += (digit*digit);
num /= 10
};
num := sum;
sum := 0
};
^ true
}
public program() {
auto happynums := new List<int>(8);
int num := 1;
while (happynums.Length < 8)
{
if (isHappy(num))
{
happynums.append(num)
};
num += 1
};
console.printLine("First 8 happy numbers: ", happynums.asEnumerable())
}</lang>
- Output:
First 8 happy numbers: 1,7,10,13,19,23,28,31
Elixir
<lang elixir>defmodule Happy do
def task(num) do
Process.put({:happy, 1}, true)
Stream.iterate(1, &(&1+1))
|> Stream.filter(fn n -> happy?(n) end)
|> Enum.take(num)
end
defp happy?(n) do
sum = square_sum(n, 0)
val = Process.get({:happy, sum})
if val == nil do
Process.put({:happy, sum}, false)
val = happy?(sum)
Process.put({:happy, sum}, val)
end
val
end
defp square_sum(0, sum), do: sum
defp square_sum(n, sum) do
r = rem(n, 10)
square_sum(div(n, 10), sum + r*r)
end
end
IO.inspect Happy.task(8)</lang>
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
Erlang
<lang 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, []).
</lang> Command: <lang Bash>erl -run tasks main -run init stop -noshell</lang> Output: <lang Bash>8 Happy Numbers: [1,7,10,13,19,23,28,31]</lang>
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): <lang Erlang>-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.</lang> Output:
[1,7,10,13,19,23,28,31]
Euphoria
<lang 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</lang> Output:
1 7 10 13 19 23 28 31
F#
This requires the F# power pack to be referenced and the 2010 beta of F# <lang fsharp>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
</lang> Output:
1 7 10 13 19 23 28 31
Factor
<lang 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 ;</lang>
- Output:
<lang factor>8 happy-numbers ! { 1 7 10 13 19 23 28 31 }</lang>
FALSE
<lang 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+]# %%</lang>
- Output:
Happy numbers: 1 7 10 13 19 23 28 31
Fantom
<lang 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
}
}
} </lang> Output:
1 7 10 13 19 23 28 31
FOCAL
<lang FOCAL>01.10 S J=0;S N=1;T %2 01.20 D 3;I (K-2)1.5 01.30 S N=N+1 01.40 I (J-8)1.2;Q 01.50 T N,! 01.60 S J=J+1 01.70 G 1.3
02.10 S A=K;S R=0 02.20 S B=FITR(A/10) 02.30 S R=R+(A-10*B)^2 02.40 S A=B 02.50 I (-A)2.2
03.10 F X=0,162;S S(X)=-1 03.20 S K=N 03.30 S S(K)=0 03.40 D 2;S K=R 03.50 I (S(K))3.3</lang>
- Output:
= 1 = 7 = 10 = 13 = 19 = 23 = 28 = 31
Forth
<lang 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</lang>
Lookup Table
Every sequence either ends in 1, or contains a 4 as part of a cycle. Extending the table through 9 is a (modest) optimization/memoization. This executes '500000 happy-numbers' about 5 times faster than the above solution. <lang forth>CREATE HAPPINESS 0 C, 1 C, 0 C, 0 C, 0 C, 0 C, 0 C, 1 C, 0 C, 0 C,
- next ( n -- n')
0 swap BEGIN dup WHILE 10 /mod >r dup * + r> REPEAT drop ;
- happy? ( n -- t|f)
BEGIN dup 10 >= WHILE next REPEAT chars HAPPINESS + C@ 0<> ;
- happy-numbers ( n --) >r 0
BEGIN r@ WHILE
BEGIN 1+ dup happy? UNTIL dup . r> 1- >r
REPEAT r> drop drop ;
8 happy-numbers</lang>
- Output:
1 7 10 13 19 23 28 31
Produces the 1 millionth happy number with: <lang forth>: happy-number ( n -- n') \ produce the nth happy number
>r 0 BEGIN r@ WHILE
BEGIN 1+ dup happy? UNTIL r> 1- >r
REPEAT r> drop ;
1000000 happy-number . \ 7105849</lang> in about 9 seconds.
Fortran
<lang 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</lang> Output:
1 7 10 13 19 23 28 31
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
Function isHappy(n As Integer) As Boolean
If n < 0 Then Return False
' Declare a dynamic array to store previous sums.
' If a previous sum is duplicated before a sum of 1 is reached
' then the number can't be "happy" as the cycle will just repeat
Dim prevSums() As Integer
Dim As Integer digit, ub, sum = 0
Do
While n > 0
digit = n Mod 10
sum += digit * digit
n \= 10
Wend
If sum = 1 Then Return True
ub = UBound(prevSums)
If ub > -1 Then
For i As Integer = 0 To ub
If sum = prevSums(i) Then Return False
Next
End If
ub += 1
Redim Preserve prevSums(0 To ub)
prevSums(ub) = sum
n = sum
sum = 0
Loop
End Function
Dim As Integer n = 1, count = 0
Print "The first 8 happy numbers are : " Print While count < 8
If isHappy(n) Then count += 1 Print count;" =>"; n End If n += 1
Wend Print Print "Press any key to quit" Sleep</lang>
- Output:
1 => 1 2 => 7 3 => 10 4 => 13 5 => 19 6 => 23 7 => 28 8 => 31
Frege
<lang frege>module Happy where
import Prelude.Math -- ugh, since Frege doesn't have Set, use Map instead import Data.Map (member, insertMin, empty emptyMap)
digitToInteger :: Char -> Integer digitToInteger c = fromInt $ (ord c) - (ord '0')
isHappy :: Integer -> Bool isHappy = p emptyMap
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 _ = putStrLn $ unwords $ map show $ take 8 $ filter isHappy $ iterate (+ 1n) 1n</lang>
- Output:
1 7 10 13 19 23 28 31 runtime 0.614 wallclock seconds.
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.
In this page you can see the program(s) related to this task and their results.
Go
<lang 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() }</lang>
- Output:
1 7 10 13 19 23 28 31
Groovy
<lang groovy>Number.metaClass.isHappy = {
def number = delegate as Long
def cycle = new HashSet<Long>()
while (number != 1 && !cycle.contains(number)) {
cycle << number
number = (number as String).collect { d = (it as Long); d * d }.sum()
}
number == 1
}
def matches = [] for (int i = 0; matches.size() < 8; i++) {
if (i.happy) { matches << i }
} println matches</lang>
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
Harbour
<lang xbase>PROCEDURE Main()
LOCAL i := 8, nH := 0
? hb_StrFormat( "The first %d happy numbers are:", i ) ?
WHILE i > 0
IF IsHappy( ++nH )
?? hb_NtoS( nH ) + " " --i
ENDIF END
RETURN
STATIC FUNCTION IsHappy( nNumber )
STATIC aUnhappy := {}
LOCAL nDigit, nSum := 0, cNumber := hb_NtoS( nNumber )
FOR EACH nDigit IN cNumber
nSum += Val( nDigit ) ^ 2
NEXT
IF nSum == 1
aUnhappy := {}
RETURN .T.
ELSEIF AScan( aUnhappy, nSum ) > 0
RETURN .F.
ENDIF
AAdd( aUnhappy, nSum )
RETURN IsHappy( nSum )</lang>
Output:
The first 8 happy numbers are: 1 7 10 13 19 23 28 31
Haskell
<lang 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 . fmap ((^ 2) . toInteger . digitToInt) . show
main :: IO () main = mapM_ print $ take 8 $ filter isHappy [1 ..]</lang>
- Output:
1 7 10 13 19 23 28 31
We can create a cache for small numbers to greatly speed up the process: <lang haskell>import Data.Array (Array, (!), listArray)
happy :: Int -> Bool happy x
| xx <= 150 = seen ! xx
| otherwise = happy xx
where
xx = dsum x
seen :: Array Int Bool
seen =
listArray (1, 150) $ True : False : False : False : (happy <$> [5 .. 150])
dsum n
| n < 10 = n * n
| otherwise =
let (q, r) = n `divMod` 10
in r * r + dsum q
main :: IO () main = print $ sum $ take 10000 $ filter happy [1 ..]</lang>
- Output:
327604323
Icon and Unicon
<lang Icon>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</lang> Usage and Output:
| happynum.exe The first 8 happy numbers are: 1 7 10 13 19 23 28 31
J
<lang j> 8{. (#~1=+/@(*:@(,.&.":))^:(1&~:*.4&~:)^:_ "0) 1+i.100 1 7 10 13 19 23 28 31</lang> This is a repeat while construction <lang j> f ^: cond ^: _ input</lang> 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. <lang j> (binary array) # 1..100</lang> So for easier reading the solution could be expressed as: <lang j> 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</lang>
Java
<lang java5>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++;
}
}
}
}</lang> Output:
1 7 10 13 19 23 28 31
Java 1.8
<lang java>
import java.util.Arrays; import java.util.HashSet; import java.util.List;
public class HappyNumbers {
public static void main(String[] args) {
for (int current = 1, total = 0; total < 8; current++)
if (isHappy(current)) {
System.out.println(current);
total++;
}
}
public static boolean isHappy(int number) {
HashSet<Integer> cycle = new HashSet<>();
while (number != 1 && cycle.add(number)) {
List<String> numStrList = Arrays.asList(String.valueOf(number).split(""));
number = numStrList.stream().map(i -> Math.pow(Integer.parseInt(i), 2)).mapToInt(i -> i.intValue()).sum();
}
return number == 1;
}
}</lang> Output:
1 7 10 13 19 23 28 31
JavaScript
ES5
Iteration
<lang 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++ ;
}</lang> Output:
1 7 10 13 19 23 28 31
ES6
Functional composition
<lang JavaScript>(() => {
// isHappy :: Int -> Bool
const isHappy = n => {
const f = n =>
foldl(
(a, x) => a + raise(read(x), 2), // ^2
0,
splitOn(, show(n))
),
p = (s, n) => n === 1 ? (
true
) : member(n, s) ? (
false
) : p(
insert(n, s), f(n)
);
return p(new Set(), n);
};
// GENERIC FUNCTIONS ------------------------------------------------------
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
// filter :: (a -> Bool) -> [a] -> [a] const filter = (f, xs) => xs.filter(f);
// foldl :: (b -> a -> b) -> b -> [a] -> b const foldl = (f, a, xs) => xs.reduce(f, a);
// insert :: Ord a => a -> Set a -> Set a const insert = (e, s) => s.add(e);
// member :: Ord a => a -> Set a -> Bool const member = (e, s) => s.has(e);
// read :: Read a => String -> a const read = JSON.parse;
// show :: a -> String const show = x => JSON.stringify(x);
// splitOn :: String -> String -> [String] const splitOn = (cs, xs) => xs.split(cs);
// raise :: Num -> Int -> Num const raise = (n, e) => Math.pow(n, e);
// take :: Int -> [a] -> [a] const take = (n, xs) => xs.slice(0, n);
// TEST -------------------------------------------------------------------
return show(
take(8, filter(isHappy, enumFromTo(1, 50)))
);
})()</lang>
- Output:
<lang JavaScript>[1, 7, 10, 13, 19, 23, 28, 31]</lang>
Or, to stop immediately at the 8th member of the series, we can preserve functional composition while using an iteratively implemented until() function: <lang JavaScript>(() => {
// isHappy :: Int -> Bool
const isHappy = n => {
const f = n =>
foldl(
(a, x) => a + raise(read(x), 2), // ^2
0,
splitOn(, show(n))
),
p = (s, n) => n === 1 ? (
true
) : member(n, s) ? (
false
) : p(
insert(n, s), f(n)
);
return p(new Set(), n);
};
// GENERIC FUNCTIONS ------------------------------------------------------
// filter :: (a -> Bool) -> [a] -> [a] const filter = (f, xs) => xs.filter(f);
// foldl :: (b -> a -> b) -> b -> [a] -> b const foldl = (f, a, xs) => xs.reduce(f, a);
// insert :: Ord a => a -> Set a -> Set a const insert = (e, s) => s.add(e);
// member :: Ord a => a -> Set a -> Bool const member = (e, s) => s.has(e);
// read :: Read a => String -> a const read = JSON.parse;
// show :: a -> String const show = x => JSON.stringify(x);
// splitOn :: String -> String -> [String] const splitOn = (cs, xs) => xs.split(cs);
// raise :: Num -> Int -> Num const raise = (n, e) => Math.pow(n, e);
// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
};
// TEST -------------------------------------------------------------------
return show(
until(
m => m.xs.length === 8,
m => {
const n = m.n;
return {
n: n + 1,
xs: isHappy(n) ? m.xs.concat(n) : m.xs
};
}, {
n: 1,
xs: []
}
)
.xs
);
})();</lang>
- Output:
<lang JavaScript>[1, 7, 10, 13, 19, 23, 28, 31]</lang>
jq
<lang jq>def is_happy_number:
def next: tostring | explode | map( (. - 48) | .*.) | add;
def last(g): reduce g as $i (null; $i);
# state: either 1 or [i, o]
# where o is an an object with the previously encountered numbers as keys
def loop:
recurse( if . == 1 then empty # all done
elif .[0] == 1 then 1 # emit 1
else (.[0]| next) as $n
| if $n == 1 then 1
elif .[1]|has($n|tostring) then empty
else [$n, (.[1] + {($n|tostring):true}) ]
end
end );
1 == last( [.,{}] | loop );</lang>
Emit a stream of the first n happy numbers:
<lang jq># Set n to -1 to continue indefinitely:
def happy(n):
def subtask: # state: [i, found]
if .[1] == n then empty
else .[0] as $n
| if ($n | is_happy_number) then $n, ([ $n+1, .[1]+1 ] | subtask)
else (.[0] += 1) | subtask
end
end;
[0,0] | subtask;
happy($n|tonumber)</lang>
- Output:
<lang sh>$ jq --arg n 8 -n -f happy.jq 1 7 10 13 19 23 28 31 </lang>
Julia
<lang 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</lang> Output
julia> happy(8) 8-element Int32 Array: 1 7 10 13 19 23 28 31
A recursive version: <lang julia>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]] </lang>
- Output:
julia> show(happy(8)) [1,7,10,13,19,23,28,31,32]
Alternate, Translation of C
Faster with use of cache
<lang julia>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</lang>
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>
Another implementation which is easy to follow is given below: <lang K> / happynum.k
/ sum of squares of digits of an integer dgtsmsqr: {d::(); (0<){d::d,x!10; x%:10}/x; +/d*d} / Test if an integer is a Happy number isHappy: {s::(); while[1<x;a:(dgtsmsqr x); :[(a _in s); :0; s::s,a]; x:a];:1} / Returns 1 if Happy / Generate first x Happy numbers and display the list hnum: {[x]; h::();i:1;while[(#h)<x; :[(isHappy i); h::(h,i)]; i+:1]; `0: ,"List of ", ($x), " Happy Numbers"; h}
</lang>
The output of a session with this implementation is given below:
- Output:
K Console - Enter \ for help \l happynum hnum 8 List of 8 Happy Numbers 1 7 10 13 19 23 28 31
Kotlin
<lang scala>// version 1.0.5-2
fun isHappy(n: Int): Boolean {
val cache = mutableListOf<Int>()
var sum = 0
var nn = n
var digit: Int
while (nn != 1) {
if (nn in cache) return false
cache.add(nn)
while (nn != 0) {
digit = nn % 10
sum += digit * digit
nn /= 10
}
nn = sum
sum = 0
}
return true
}
fun main(args: Array<String>) {
var num = 1
val happyNums = mutableListOf<Int>()
while (happyNums.size < 8) {
if (isHappy(num)) happyNums.add(num)
num++
}
println("First 8 happy numbers : " + happyNums.joinToString(", "))
}</lang>
- Output:
First 8 happy numbers : 1, 7, 10, 13, 19, 23, 28, 31
Lasso
<lang lasso>#!/usr/bin/lasso9
define isHappy(n::integer) => {
local(past = set)
while(#n != 1) => {
#n = with i in string(#n)->values sum math_pow(integer(#i), 2)
#past->contains(#n) ? return false | #past->insert(#n)
}
return true
}
with x in generateSeries(1, 500)
where isHappy(#x) take 8
select #x</lang> Output: <lang lasso>1, 7, 10, 13, 19, 23, 28, 31</lang>
Liberty BASIC
<lang lb> ct = 0
n = 0
DO
n = n + 1
IF HappyN(n, sqrInt$) = 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:-
1 1 2 7 3 10 4 13 5 19 6 23 7 28 8 31
Locomotive Basic
<lang locobasic>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</lang>
Logo
<lang 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</lang>
Output:
1 7 10 13 19 23 28 31
LOLCODE
<lang lolcode>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</lang >
Output:
1 7 10 13 19 23 28 31
Lua
<lang 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</lang> Output:
1 7 10 13 19 23 28 31
M2000 Interpreter
Lambda Function PrintHappy has a closure another lambda function IsHappy which has a closure of another lambda function the sumOfSquares.
<lang M2000 Interpreter>
Function FactoryHappy {
sumOfSquares= lambda (n) ->{
k$=str$(abs(n),"")
Sum=0
For i=1 to len(k$)
sum+=val(mid$(k$,i,1))**2
Next i
=sum
}
IsHappy=Lambda sumOfSquares (n) ->{
Inventory sequence
While n<>1 {
Append sequence, n
n=sumOfSquares(n)
if exist(sequence, n) then =false : Break
}
=True
}
=Lambda IsHappy ->{
numleft=8
numToTest=1
While numleft {
if ishappy(numToTest) Then {
Print numToTest
numleft--
}
numToTest++
}
}
} PrintHappy=factoryHappy() Call PrintHappy() </lang>
- Output:
1 7 10 13 19 23 28 31
MAD
<lang MAD> NORMAL MODE IS INTEGER
BOOLEAN CYCLE
DIMENSION CYCLE(200)
VECTOR VALUES OUTFMT = $I2*$
SEEN = 0
I = 0
NEXNUM THROUGH ZERO, FOR K=0, 1, K.G.200 ZERO CYCLE(K) = 0B
I = I + 1
SUMSQR = I
CHKLP N = SUMSQR
SUMSQR = 0
SUMLP DIG = N-N/10*10
SUMSQR = SUMSQR + DIG*DIG
N = N/10
WHENEVER N.NE.0, TRANSFER TO SUMLP
WHENEVER SUMSQR.E.1, TRANSFER TO HAPPY
WHENEVER CYCLE(SUMSQR), TRANSFER TO NEXNUM
CYCLE(SUMSQR) = 1B
TRANSFER TO CHKLP
HAPPY PRINT FORMAT OUTFMT,I
SEEN = SEEN+1
WHENEVER SEEN.L.8, TRANSFER TO NEXNUM
END OF PROGRAM
</lang>
- Output:
1 7 10 13 19 23 28 31
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.) <lang Maple>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:</lang> (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, <lang Maple> > SumSqDigits( 1234567890987654321 );
570
</lang> We can check this by computing it another way (more directly). <lang Maple> > n := 1234567890987654321: > `+`( op( map( parse, StringTools:-Explode( convert( n, 'string' ) ) )^~2) );
570
</lang> 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). <lang Maple>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:</lang> We can use this to determine the number of happy (H) and sad (S) numbers up to one million as follows. <lang Maple> > H, S := selectremove( Happy?, [seq]( 1 .. N ) ): > nops( H ), nops( S );
143071, 856929
</lang> 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. <lang Maple>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:</lang> With input equal to 8, we get <lang Maple> > FindHappiness( 8 );
{1, 7, 10, 13, 19, 23, 28, 31}
</lang> For completeness, here is an implementation of the cycle detection algorithm for recognizing happy numbers. It is much slower, however. <lang Maple>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:</lang>
Mathematica / Wolfram Language
Custom function HappyQ: <lang Mathematica>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</lang> Examples for a specific number: <lang Mathematica>HappyQ[1337] HappyQ[137]</lang> gives back: <lang Mathematica>True False</lang> Example finding the first 8: <lang Mathematica>m = 8; n = 1; i = 0; happynumbers = {}; While[n <= m,
i++; If[HappyQ[i], n++; AppendTo[happynumbers, i] ] ]
happynumbers</lang> gives back: <lang Mathematica>{1, 7, 10, 13, 19, 23, 28, 31}</lang>
MATLAB
Recursive version: <lang MATLAB>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</lang>
- Output:
1 7 10 13 19 23 28 31
MAXScript
<lang MAXScript> fn isHappyNumber n = ( local pastNumbers = #() while n != 1 do ( n = n as string local newNumber = 0 for i = 1 to n.count do ( local digit = n[i] as integer newNumber += pow digit 2 ) n = newNumber if (finditem pastNumbers n) != 0 do return false append pastNumbers newNumber ) n == 1 ) printed = 0 for i in (for h in 1 to 500 where isHappyNumber h collect h) do ( if printed == 8 do exit print i as string printed += 1
) </lang> Output: <lang MAXScript> 1 7 10 13 19 23 28 31 </lang>
Mercury
<lang Mercury>:- module happy.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is det.
- - implementation.
- - import_module int, list, set_tree234.
main(!IO) :-
print_line(get_n_happy_numbers(8, 1), !IO).
- - func get_n_happy_numbers(int, int) = list(int).
get_n_happy_numbers(NumToFind, N) =
( if NumToFind > 0 then
( if is_happy(N, init)
then [N | get_n_happy_numbers(NumToFind - 1, N + 1)]
else get_n_happy_numbers(NumToFind, N + 1)
)
else
[]
).
- - pred is_happy(int::in, set_tree234(int)::in) is semidet.
is_happy(1, _). is_happy(N, !.Seen) :-
not member(N, !.Seen), insert(N, !Seen), is_happy(sum_sqr_digits(N), !.Seen).
- - func sum_sqr_digits(int) = int.
sum_sqr_digits(N) =
( if N < 10 then sqr(N) else sqr(N mod 10) + sum_sqr_digits(N div 10) ).
- - func sqr(int) = int.
sqr(X) = X * X.</lang>
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
MiniScript
This solution uses the observation that any infinite cycle of this algorithm hits the number 89, and so that can be used to know when we've found an unhappy number. <lang MiniScript>isHappy = function(x)
while true
if x == 89 then return false
sum = 0
while x > 0
sum = sum + (x % 10)^2
x = floor(x / 10)
end while
if sum == 1 then return true
x = sum
end while
end function
found = [] i = 1 while found.len < 8
if isHappy(i) then found.push i i = i + 1
end while print "First 8 happy numbers: " + found</lang>
- Output:
First 8 happy numbers: [1, 7, 10, 13, 19, 23, 28, 31]
ML
mLite
<lang ocaml>(* 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.
- )
local fun get_digits (d, s) where (d = 0) = s | (d, s) = get_digits( d div 10, (d mod 10) :: s) | n = get_digits( n div 10, [n mod 10] ) ; fun mem (x, []) = false | (x, a :: as) where (x = a) = true | (x, _ :: as) = mem (x, as) in fun happy 1 = "happy" | n = let val this = (fold (+,0) ` map (fn n = n ^ 2) ` get_digits n); val sads = [2, 4, 16, 37, 58, 89, 145, 42, 20] in if (mem (n,sads)) then "unhappy" else happy this end end
foreach (fn n = (print n; print " is "; println ` happy n)) ` iota 10; </lang> Output:
1 is happy 2 is unhappy 3 is unhappy 4 is unhappy 5 is unhappy 6 is unhappy 7 is happy 8 is unhappy 9 is unhappy 10 is happy
Modula-2
<lang modula2>MODULE HappyNumbers; FROM InOut IMPORT WriteCard, WriteLn;
CONST Amount = 8; VAR seen, num: CARDINAL;
PROCEDURE SumDigitSquares(n: CARDINAL): CARDINAL; VAR sum, digit: CARDINAL; BEGIN
sum := 0;
WHILE n>0 DO
digit := n MOD 10;
n := n DIV 10;
sum := sum + digit * digit;
END;
RETURN sum;
END SumDigitSquares;
PROCEDURE Happy(n: CARDINAL): BOOLEAN; VAR i: CARDINAL;
seen: ARRAY [0..255] OF BOOLEAN;
BEGIN
FOR i := 0 TO 255 DO
seen[i] := FALSE;
END;
REPEAT
seen[n] := TRUE;
n := SumDigitSquares(n);
UNTIL seen[n];
RETURN seen[1];
END Happy;
BEGIN
seen := 0;
num := 0;
WHILE seen < Amount DO
IF Happy(num) THEN
INC(seen);
WriteCard(num,2);
WriteLn();
END;
INC(num);
END;
END HappyNumbers.</lang>
- Output:
1 7 10 13 19 23 28 31
MUMPS
<lang 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</lang>
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>
NetRexx
<lang 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</lang>
- 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
Nim
<lang nim>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</lang>
Output:
1 7 10 13 19 23 28 31
Objeck
<lang 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("");
}
}
}</lang> output:
First 8 happy numbers: 1,7,10,13,19,23,28,31,
OCaml
Using Floyd's cycle-finding algorithm. <lang ocaml>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)) ;;</lang> Output:
$ ocaml nums.cma happy_numbers.ml 1 7 10 13 19 23 28 31
Oforth
<lang Oforth>: isHappy(n) | cycle |
ListBuffer new ->cycle
while(n 1 <>) [
cycle include(n) ifTrue: [ false return ]
cycle add(n)
0 n asString apply(#[ asDigit sq + ]) ->n
]
true ;
- happyNum(N)
| numbers |
ListBuffer new ->numbers 1 while(numbers size N <>) [ dup isHappy ifTrue: [ dup numbers add ] 1+ ] numbers println ;</lang>
Output:
>happyNum(8) [1, 7, 10, 13, 19, 23, 28, 31]
ooRexx
<lang 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
</lang>
First 8 happy numbers are: 1 7 10 13 19 23 28 31
Oz
<lang oz>functor import
System
define
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
{System.show {List.take HappyNumbers 8}}
end</lang> Output:
[1 7 10 13 19 23 28 31]
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. <lang parigp>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))</lang> Output:
%1 = [1, 7, 10, 13, 19, 23, 28, 31]
Pascal
<lang 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.</lang> Output:
:> ./HappyNumbers 1 7 10 13 19 23 28 31
alternative for counting fast
The Cache is limited to maximum value of the sum of squared digits and filled up in a blink of an eye.Even for cDigit2=1e9 takes 0.7s.Calculation of sum of squared digits is improved.Saving this SqrdSumCache speeds up tremendous. So i am able to check if the 1'000'000 th happy number is 7105849 as stated in C language.This seems to be true. Extended to 10e18 Tested with Free Pascal 3.0.4 <lang pascal>Program HappyNumbers (output); {$IFDEF FPC}
{$MODE DELPHI}
{$OPTIMIZATION ON,All}
{$ELSE}
{$APPLICATION CONSOLE}
{$ENDIF} //{$DEFINE Use1E9} uses
sysutils,//Timing strutils;//Numb2USA
const
base = 10; HighCache = 20*(sqr(base-1));//sum of sqr digit of Uint64
{$IFDEF Use1E9}
cDigit1 = sqr(base)*sqr(base);//must be power of base cDigit2 = Base*sqr(cDigit1);// 1e9 cMaxPot = 18;
{$ELSE}
cDigit1 = base*sqr(base);//must be power of base cDigit2 = sqr(cDigit1);// 1e6 cMaxPot = 14;
{$ENDIF}
type
tSumSqrDgts = array[0..cDigit2] of word; tCache = array[0..2*HighCache] of word; tSqrdSumCache = array[0..2*HighCache] of Uint32;
var
SumSqrDgts :tSumSqrDgts; Cache : tCache;
SqrdSumCache1, SqrdSumCache2 :tSqrdSumCache;
T1,T0 : TDateTime; MAX2,Max1 : NativeInt;
procedure InitSumSqrDgts; //calc all sum of squared digits 0..cDigits2 //using already calculated values var
i,j,n,sq,Base1: NativeInt;
begin
For i := 0 to Base-1 do
SumSqrDgts[i] := i*i;
Base1 := Base;
n := Base;
repeat
For i := 1 to base-1 do
Begin
sq := SumSqrDgts[i];
For j := 0 to base1-1 do
Begin
SumSqrDgts[n] := sq+SumSqrDgts[j];
inc(n);
end;
end;
Base1 := Base1*base;
until Base1 >= cDigit2;
SumSqrDgts[n] := 1;
end;
function SumSqrdDgt(n: Uint64):NativeUint;inline; var
r: Uint64;
begin
result := 0; while n>cDigit2 do Begin r := n; n := n div cDigit2; r := r-n*cDigit2; inc(result,SumSqrDgts[r]); end; inc(result,SumSqrDgts[n]);
end;
procedure CalcSqrdSumCache1; var
Count : tSqrdSumCache; i,sq,result : NativeInt;
begin
For i :=High(Count) downto 0 do
Count[i] := 0;
//count the manifold
For i := cDigit1-1 downto 0 do
inc(count[SumSqrDgts[i]]);
For i := High(Count) downto 0 do
if count[i] <> 0 then
Begin
Max1 := i;
BREAK;
end;
For sq := 0 to (20-3)*81 do
Begin
result := 0;
For i := Max1 downto 0 do
inc(result,Count[i]*Cache[sq+i]);
SqrdSumCache1[sq] := result;
end;
end;
procedure CalcSqrdSumCache2; var
Count : tSqrdSumCache; i,sq,result : NativeInt;
begin
For i :=High(Count) downto 0 do
Count[i] := 0;
For i := cDigit2-1 downto 0 do
inc(count[SumSqrDgts[i]]);
For i := High(Count) downto 0 do
if count[i] <> 0 then
Begin
Max2 := i;
BREAK;
end;
For sq := 0 to (20-6)*81 do
Begin
result := 0;
For i := Max2 downto 0 do
inc(result,Count[i]*Cache[sq+i]);
SqrdSumCache2[sq] := result;
end;
end;
procedure Inithappy; var
n,s,p : NativeUint;
Begin
fillchar(SqrdSumCache1,SizeOf(SqrdSumCache1),#0); fillchar(SqrdSumCache2,SizeOf(SqrdSumCache2),#0); InitSumSqrDgts; fillChar(Cache,SizeOf(Cache),#0);
Cache[1] := 1;
For n := 1 to High(Cache) do
Begin
If Cache[n] = 0 then
Begin
//start a linked list
Cache[n] := n;
p := n;
s := SumSqrdDgt(p);
while Cache[s] = 0 do
Begin
Cache[s] := p;
p := s;
s := SumSqrdDgt(p);
end;
//mark linked list backwards as happy number
IF Cache[s] = 1 then
Begin
repeat
s := Cache[p];
Cache[p] := 1;
p := s;
until s = n;
Cache[n] := 1;
end;
end;
end;
//mark all unhappy numbers with 0
For n := 1 to High(Cache) do
If Cache[n] <> 1 then
Cache[n] := 0;
CalcSqrdSumCache1;
CalcSqrdSumCache2;
end;
function is_happy(n: NativeUint): boolean;inline; begin
is_happy := Boolean(Cache[SumSqrdDgt(n)])
end;
function nthHappy(Limit: Uint64):Uint64; var
d,e,sE: NativeUint;
begin
result := 0;
d := 0;
e := 0;
sE := SumSqrDgts[e];
//big steps
while Limit >= cDigit2 do
begin
dec(Limit,SqrdSumCache2[SumSqrDgts[d]+sE]);
inc(result,cDigit2);
inc(d);
IF d >=cDigit2 then
Begin
inc(e);
sE := SumSqrdDgt(e);//SumSqrDgts[e];
d :=0;
end;
end;
//small steps
while Limit >= cDigit1 do
Begin
dec(Limit,SqrdSumCache1[SumSqrdDgt(result)]);
inc(result,cDigit1);
end;
//ONE BY ONE
while Limit > 0 do
begin
dec(Limit,Cache[SumSqrdDgt(result)]);
inc(result);
end;
result -= 1;
end;
var
n, count :Uint64; Limit: NativeUint;
begin
write('cDigit1 = ',Numb2USA(IntToStr(cDigit1)));
writeln(' cDigit2 = ',Numb2USA(IntToStr(cDigit2)));
T0 := now;
Inithappy;
writeln('Init takes ',FormatDateTime(' HH:NN:SS.ZZZ',now-T0));
n := 1;
count := 0;
while count < 10 do
begin
if is_happy(n) then
begin
inc(count);
write(n, ' ');
end;
inc(n);
end;
writeln;
T0 := now;
T1 := T0;
n := 1;
Limit := 10;
repeat
writeln('1E',n:2,' n.th happy number ',Numb2USA(IntToStr(nthHappy(Limit))):26,
FormatDateTime(' HH:NN:SS.ZZZ',now-T1));
T1 := now;
inc(n);
Limit := limit*10;
until n> cMaxPot;
writeln('Total time counting ',FormatDateTime('HH:NN:SS.ZZZ',now-T0));
end. </lang>
- output
cDigit1 = 1,000 cDigit2 = 1,000,000 Init takes 00:00:00.004 1 7 10 13 19 23 28 31 32 44 1E 1 n.th happy number 44 00:00:00.000 1E 2 n.th happy number 694 00:00:00.000 1E 3 n.th happy number 6,899 00:00:00.000 1E 4 n.th happy number 67,169 00:00:00.000 1E 5 n.th happy number 692,961 00:00:00.000 1E 6 n.th happy number 7,105,849 00:00:00.000 1E 7 n.th happy number 71,313,350 00:00:00.000 1E 8 n.th happy number 698,739,425 00:00:00.000 1E 9 n.th happy number 6,788,052,776 00:00:00.000 1E10 n.th happy number 66,305,148,869 00:00:00.000 1E11 n.th happy number 660,861,957,662 00:00:00.001 1E12 n.th happy number 6,745,877,698,967 00:00:00.008 1E13 n.th happy number 70,538,879,028,725 00:00:00.059 1E14 n.th happy number 744,083,563,164,178 00:00:00.612 Total time counting 00:00:00.680 real 0m0,685s cDigit1 = 10,000 cDigit2 = 1,000,000,000 Init takes 00:00:02.848 1 7 10 13 19 23 28 31 32 44 1E 1 n.th happy number 44 00:00:00.000 1E 2 n.th happy number 694 00:00:00.000 1E 3 n.th happy number 6,899 00:00:00.000 1E 4 n.th happy number 67,169 00:00:00.000 1E 5 n.th happy number 692,961 00:00:00.000 1E 6 n.th happy number 7,105,849 00:00:00.000 1E 7 n.th happy number 71,313,350 00:00:00.000 1E 8 n.th happy number 698,739,425 00:00:00.001 1E 9 n.th happy number 6,788,052,776 00:00:00.008 1E10 n.th happy number 66,305,148,869 00:00:00.010 1E11 n.th happy number 660,861,957,662 00:00:00.009 1E12 n.th happy number 6,745,877,698,967 00:00:00.008 1E13 n.th happy number 70,538,879,028,725 00:00:00.008 1E14 n.th happy number 744,083,563,164,178 00:00:00.011 1E15 n.th happy number 7,888,334,045,397,315 00:00:00.019 1E16 n.th happy number 82,440,929,809,838,249 00:00:00.079 1E17 n.th happy number 845,099,936,580,193,833 00:00:00.698 1E18 n.th happy number 8,489,964,903,498,345,213 00:00:06.920 Total time counting 00:00:07.771 real 0m10,627s
Perl
Since all recurrences end with 1 or repeat (37,58,89,145,42,20,4,16), we can do this test very quickly without having to make hashes of seen numbers. <lang perl>use List::Util qw(sum);
sub ishappy {
my $s = shift;
while ($s > 6 && $s != 89) {
$s = sum(map { $_*$_ } split(//,$s));
}
$s == 1;
}
my $n = 0; print join(" ", map { 1 until ishappy(++$n); $n; } 1..8), "\n";</lang>
- Output:
1 7 10 13 19 23 28 31
Or we can solve using only the rudimentary task knowledge as below. Note the slightly different ways of doing the digit sum and finding the first 8 numbers where ishappy(n) is true -- this shows there's more than one way to do even these small sub-tasks.
<lang perl>use List::Util qw(sum); sub is_happy {
my ($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 "$n " or redo for 1..8;</lang>
- Output:
1 7 10 13 19 23 28 31
Phix
Copy of Euphoria tweaked to give a one-line output
function is_happy(integer n) sequence seen = {} while n>1 do seen &= n integer k = 0 while n>0 do k += power(remainder(n,10),2) n = floor(n/10) end while n = k if find(n,seen) then return false end if end while return true end function integer n = 1 sequence s = {} while length(s)<8 do if is_happy(n) then s &= n end if n += 1 end while ?s
- Output:
{1,7,10,13,19,23,28,31}
PHP
<lang php>function isHappy($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++;
}
$i++;
}</lang>
1 7 10 13 19 23 28 31
PicoLisp
<lang PicoLisp>(de happy? (N)
(let Seen NIL
(loop
(T (= N 1) T)
(T (member N Seen))
(setq N
(sum '((C) (** (format C) 2))
(chop (push 'Seen N)) ) ) ) ) )
(let H 0
(do 8
(until (happy? (inc 'H)))
(printsp H) ) )</lang>
Output:
1 7 10 13 19 23 28 31
PILOT
<lang pilot>C :max=8
:n=0 :i=0
- test
U :*happy T (a=1):#n C (a=1):i=i+1 C :n=n+1 J (i<max):*test E :
- happy
C :a=n
:x=n
U :*sumsq C :b=s
- loop
C :x=a U :*sumsq C :a=s C :x=b U :*sumsq C :x=s U :*sumsq C :b=s J (a<>b):*loop E :
- sumsq
C :s=0
- digit
C :y=x/10
:z=x-y*10 :s=s+z*#z :x=y
J (x):*digit E :</lang>
- Output:
1 7 10 13 19 23 28 31
PL/I
<lang PL/I>test: proc options (main); /* 19 November 2011 */
declare (i, j, n, m, nh initial (0) ) fixed binary (31);
main_loop:
do j = 1 to 100;
n = j;
do i = 1 to 100;
m = 0;
/* Form the sum of squares of the digits. */
do until (n = 0);
m = m + mod(n, 10)**2;
n = n/10;
end;
if m = 1 then
do;
put skip list (j || ' is a happy number');
nh = nh + 1;
if nh = 8 then return;
iterate main_loop;
end;
n = m; /* Replace n with the new number formed from digits. */
end;
end;
end test; </lang> OUTPUT:
1 is a happy number
7 is a happy number
10 is a happy number
13 is a happy number
19 is a happy number
23 is a happy number
28 is a happy number
31 is a happy number
PL/M
<lang plm>100H:
/* FIND SUM OF SQUARE OF DIGITS OF NUMBER */ DIGIT$SQUARE: PROCEDURE (N) BYTE;
DECLARE (N, T, D) BYTE;
T = 0;
DO WHILE N > 0;
D = N MOD 10;
T = T + D * D;
N = N / 10;
END;
RETURN T;
END DIGIT$SQUARE;
/* CHECK IF NUMBER IS HAPPY */ HAPPY: PROCEDURE (N) BYTE;
DECLARE (N, I) BYTE;
DECLARE FLAG (256) BYTE;
DO I=0 TO 255;
FLAG(I) = 0;
END;
DO WHILE NOT FLAG(N);
FLAG(N) = 1;
N = DIGIT$SQUARE(N);
END;
RETURN N = 1;
END HAPPY;
/* CP/M BDOS CALL */ BDOS: PROCEDURE (FN, ARG);
DECLARE FN BYTE, ARG ADDRESS; GO TO 5;
END BDOS;
/* PRINT STRING */ PRINT: PROCEDURE (STR);
DECLARE STR ADDRESS; CALL BDOS(9, STR);
END PRINT;
/* PRINT NUMBER */ PRINT$NUMBER: PROCEDURE (N);
DECLARE S (6) BYTE INITIAL ('...',13,10,'$');
DECLARE P ADDRESS;
DECLARE (N, C BASED P) BYTE;
P = .S(3);
DIGIT:
P = P - 1; C = (N MOD 10) + '0'; N = N / 10; IF N > 0 THEN GO TO DIGIT; CALL PRINT(P);
END PRINT$NUMBER;
/* FIND FIRST 8 HAPPY NUMBERS */ DECLARE SEEN BYTE INITIAL (0); DECLARE N BYTE INITIAL (1);
DO WHILE SEEN < 8;
IF HAPPY(N) THEN DO;
CALL PRINT$NUMBER(N);
SEEN = SEEN + 1;
END;
N = N + 1;
END;
CALL BDOS(0,0); EOF</lang>
- Output:
1 7 10 13 19 23 28 31
Potion
<lang potion>sqr = (n): n * n.
isHappy = (n) :
loop :
if (n == 1): return true.
if (n == 4): return false.
sum = 0
n = n string
n length times (i): sum = sum + sqr(n(i) number integer).
n = sum
.
.
firstEight = () i = 0 while (firstEight length < 8) :
i++ if (isHappy(i)): firstEight append(i).
. firstEight string print</lang>
PowerShell
<lang 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
}
}
$a -join ','
}</lang> Output : <lang PowerShell>happy(8) 7,10,13,19,23,28,31,32</lang>
Prolog
<lang Prolog>happy_numbers(L, Nb) :-
% creation of the list length(L, Nb), % Process of this list get_happy_number(L, 1).
% the game is over
get_happy_number([], _).
% querying the newt happy_number get_happy_number([H | T], N) :-
N1 is N+1,
(is_happy_number(N) ->
H = N,
get_happy_number(T, N1);
get_happy_number([H | T], N1)).
% we must memorized the numbers reached is_happy_number(N) :-
is_happy_number(N, [N]).
% a number is happy when we get 1 is_happy_number(N, _L) :-
get_next_number(N, 1), !.
% or when this number is not already reached ! is_happy_number(N, L) :-
get_next_number(N, NN), \+member(NN, L), is_happy_number(NN, [NN | L]).
% Process of the next number from N get_next_number(N, NewN) :-
get_list_digits(N, LD), maplist(square, LD, L), sumlist(L, NewN).
get_list_digits(N, LD) :- number_chars(N, LCD), maplist(number_chars_, LD, LCD).
number_chars_(D, CD) :- number_chars(D, [CD]).
square(N, SN) :- SN is N * N.</lang> Output : <lang Prolog> ?- happy_numbers(L, 8). L = [1,7,10,13,19,23,28,31].</lang>
PureBasic
<lang PureBasic>#ToFind=8
- MaxTests=100
- True = 1: #False = 0
Declare is_happy(n)
If OpenConsole()
Define i=1,Happy
Repeat
If is_happy(i)
Happy+1
PrintN("#"+Str(Happy)+RSet(Str(i),3))
EndIf
i+1
Until Happy>=#ToFind
;
Print(#CRLF$+#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</lang> Sample output:
#1 1 #2 7 #3 10 #4 13 #5 19 #6 23 #7 28 #8 31
Python
Procedural
<lang 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]</lang>
Composition of pure functions
Drawing 8 terms from a non finite stream, rather than assuming prior knowledge of the finite sample size required: <lang python>Happy numbers
from itertools import islice
- main :: IO ()
def main():
Test
print(
take(8)(
happyNumbers()
)
)
- happyNumbers :: Gen [Int]
def happyNumbers():
Generator :: non-finite stream of happy numbers.
x = 1
while True:
x = until(isHappy)(succ)(x)
yield x
x = succ(x)
- isHappy :: Int -> Bool
def isHappy(n):
Happy number sequence starting at n reaches 1 ? seen = set()
# p :: Int -> Bool
def p(x):
if 1 == x or x in seen:
return True
else:
seen.add(x)
return False
# f :: Int -> Int
def f(x):
return sum(int(d)**2 for d in str(x))
return 1 == until(p)(f)(n)
- GENERIC -------------------------------------------------
- succ :: Int -> Int
def succ(x):
The successor of an integer. return 1 + x
- take :: Int -> [a] -> [a]
- take :: Int -> String -> String
def take(n):
The prefix of xs of length n,
or xs itself if n > length xs.
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
- until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
The result of repeatedly applying f until p holds.
The initial seed value is x.
def go(f, x):
v = x
while not p(v):
v = f(v)
return v
return lambda f: lambda x: go(f, x)
if __name__ == '__main__':
main()</lang>
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
Quackery
<lang Quackery>
[ 0 swap
[ 10 /mod 2 **
rot + swap
dup 0 = until ]
drop ] is digitsquare ( n --> n )
[ [ digitsquare
dup 1 != while
dup 42 != while
again ]
1 = ] is happy ( n --> b )
[ [] 1
[ dip
[ 2dup size > ]
swap while
dup happy if
[ tuck join swap ]
1+ again ]
drop nip ] is happies ( n --> [ )
8 happies echo</lang>
- Output:
[ 1 7 10 13 19 23 28 31 ]
R
<lang 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
}</lang> Example usage <lang R>is.happy(2)</lang>
[1] FALSE attr(,"cycle") [1] 4 16 37 58 89 145 42 20
<lang R>#Find happy numbers between 1 and 50 which(apply(rbind(1:50), 2, is.happy))</lang>
1 7 10 13 19 23 28 31 32 44 49
<lang R>#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</lang>
1 7 10 13 19 23 28 31
Racket
<lang 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)</lang>
Raku
(formerly Perl 6)
<lang perl6>sub happy (Int $n is copy --> Bool) {
loop {
state %seen;
$n = [+] $n.comb.map: { $_ ** 2 }
return True if $n == 1;
return False if %seen{$n}++;
}
}
say join ' ', grep(&happy, 1 .. *)[^8];</lang>
- 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. <lang perl6>my @happy = lazy gather for 1..* -> $number {
my %stopper = 1 => 1;
my $n = $number;
repeat until %stopper{$n}++ {
$n = [+] $n.comb X** 2;
}
take $number if $n == 1;
}
say ~@happy[^8];</lang> Output is the same as above.
Here is a version using a subset and an anonymous recursion (we cheat a little bit by using the knowledge that 7 is the second happy number): <lang perl6>subset Happy of Int where sub ($n) {
$n == 1 ?? True !! $n < 7 ?? False !! &?ROUTINE([+] $n.comb »**» 2);
}
say (grep Happy, 1 .. *)[^8];</lang> 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...
Relation
<lang Relation> function happy(x) set y = x set lasty = 0 set found = " " while y != 1 and not (found regex "\s".y."\s") set found = found . y . " " set m = 0 while y > 0 set digit = y mod 10 set m = m + digit * digit set y = (y - digit) / 10 end while set y = format(m,"%1d") end while set found = found . y . " " if y = 1 set result = 1 else set result = 0 end if end function
set c = 0 set i = 1 while c < 8 and i < 100 if happy(i) echo i set c = c + 1 end if set i = i + 1 end while </lang>
1 7 10 13 19 23 28 31
REXX
unoptimized
<lang REXX>/*REXX program computes and displays a specified amount of happy numbers. */ parse arg limit . /*obtain optional argument from the CL.*/ if limit== | limit=="," then limit=8 /*Not specified? Then use the default.*/ haps=0 /*count of the happy numbers (so far).*/
do n=1 while haps<limit; @.=0; q=n /*search the integers starting at unity*/
do until q==1 /*determine if Q is a happy number.*/
s=0 /*prepare to add squares of digits. */
do j=1 for length(q) /*sum the squares of the decimal digits*/
s=s + substr(q, j, 1) **2 /*add the square of a decimal digit.*/
end /*j*/
if @.s then iterate n /*if already summed, Q is unhappy. */
@.s=1; q=s /*mark the 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 all done. */</lang>
- output when using the input of: 8
1 7 10 13 19 23 28 31
optimized, vertical list
This REXX code uses additional memorization (by keeping track of happy and unhappy numbers),
it's about 2 1/2 times 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 (as shown below).
<lang rexx>/*REXX program computes and displays a specified range of happy numbers. */
parse arg L H . /*obtain optional arguments from the CL*/
if L== | L=="," then L=8 /*Not specified? Then use the default.*/
if H== | H=="," then do; H=L; L=1; end /*use a range for the displaying of #s.*/
do i=0 to 9; #.i=i**2; end /*i*/ /*build a squared decimal digit table. */
@.=0; @.1=1; !.=@.; !.2=1; !.4=1 /*sparse array: @≡happy, !≡unhappy. */ haps=0 /*count of the happy numbers (so far).*/
do n=1 while haps<H /*search integers starting at unity. */
if !.n then iterate /*if N is unhappy, then try another. */
q=n /* [↓] Q is the number being tested*/
do until q==1; s=0 /*see if Q is a happy number. */
?=q /* [↓] ? is destructively parsed. */
do length(q) /*parse all the decimal digits of ? */
parse var ? _ +1 ? /*obtain a single decimal digit of ? */
s=s + #._ /*add the square of that decimal digit.*/
end /*length(q)*/ /* [↑] perform the DO W times. */
if !.s then do; !.n=1; iterate n; end /*is S unhappy? Then Q is also. */
if @.s then leave /*Have we found a happy number? */
q=s /*try the Q sum to see if it's happy.*/
end /*until*/
@.n=1 /*mark N as a happy number.*/
haps=haps+1 /*bump the counter of the happy numbers*/
if haps<L then iterate /*don't display if N is too low.*/
say right(n, 30) /*display right justified happy number.*/
end /*n*/
/*stick a fork in it, we're all done. */</lang>
- output when using the input of: 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
optimized, horizontal list
This REXX version is identical to the optimized version, but displays the numbers in a horizontal list. <lang rexx>/*REXX program computes and displays a specified range of happy numbers. */ sw=linesize() - 1 /*obtain the screen width (less one). */ parse arg limit . /*obtain optional argument from the CL.*/ if L== | L=="," then L=8 /*Not specified? Then use the default.*/ if H== | H=="," then do; H=L; L=1; end /*use a range for the displaying of #s.*/
do i=0 to 9; #.i=i**2; end /*i*/ /*build a squared decimal digit table. */
@.=0; @.1=1; !.=@.; !.2=1; !.4=1 /*sparse array: @≡happy, !≡unhappy. */ haps=0 /*count of the happy numbers (so far).*/ $=
do n=1 while haps<H /*search integers starting at unity. */
if !.n then iterate /*if N is unhappy, then try another. */
q=n /*(below) Q is the number tested. */
do until q==1; s=0 /*see if Q is a happy number. */
?=q /* [↓] ? is destructively PARSEd. */
do length(q) /*parse all the decimal digits of ? */
parse var ? _ +1 ? /*obtain a single decimal digit of ? */
s=s + #._ /*add the square of that decimal digit.*/
end /*length(q)*/ /* [↑] perform the DO W times. */
if !.s then do; !.n=1; iterate n; end /*is S unhappy? Then Q is also. */
if @.s then leave /*Have we found a happy number? */
q=s /*try the Q sum to see if it's happy.*/
end /*until*/
@.n=1 /*mark N as a happy number. */
haps=haps+1 /*bump the count of the happy numbers. */
if haps<L then iterate /*don't display it, N is too low. */
$=$ n /*add N to the horizontal list. */
if length($ n)>sw then do /*if the list is too long, then split */
say strip($) /* ··· and display what we've got. */
$=n /*Set the next line to overflow. */
end /* [↑] new line now contains overflow.*/
end /*n*/
if $\= then say strip($) /*display any residual happy numbers. */
/*stick a fork in it, we're all done. */</lang>
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).
Some REXXes don't have this BIF, so the linesize.rex REXX program is included here ──► LINESIZE.REX.
- output when using the input of: 1 15002
(The linesize for the terminal being used for this example was 200.)
(Shown at two-thirds size.)
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
Ring
<lang ring>n = 1 found = 0
While found < 8
If IsHappy(n)
found += 1
see string(found) + " : " + string(n) + nl
ok
n += 1
End
Func IsHappy n
cache = []
While n != 1
Add(cache,n)
t = 0
strn = string(n)
for e in strn
t += pow(number(e),2)
next
n = t
If find(cache,n) Return False ok
End
Return True
</lang>
- Output:
1 : 1 2 : 7 3 : 10 4 : 13 5 : 19 6 : 23 7 : 28 8 : 31
Ruby
<lang ruby>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</lang>
Helper method to produce output: <lang ruby>def print_happy
happy_numbers = []
1.step do |i| break if happy_numbers.length >= 8 happy_numbers << i if happy?(i) end
p happy_numbers
end
print_happy</lang>
- Output:
<lang ruby>[1, 7, 10, 13, 19, 23, 28, 31]</lang>
Alternative version
<lang ruby>@memo = [0,1] def happy(n)
sum = n.to_s.chars.map{|c| c.to_i**2}.inject(:+)
return @memo[sum] if @memo[sum]==0 or @memo[sum]==1
@memo[sum] = 0 # for the cycle check
@memo[sum] = happy(sum) # return 1:Happy number, 0:other
end
i = count = 0 while count < 8
i += 1 puts i or count+=1 if happy(i)==1
end
puts for i in 99999999999900..99999999999999
puts i if happy(i)==1
end</lang>
- Output:
1 7 10 13 19 23 28 31 99999999999901 99999999999910 99999999999914 99999999999915 99999999999916 99999999999937 99999999999941 99999999999951 99999999999956 99999999999961 99999999999965 99999999999973
Simpler Alternative
<lang ruby>def happy?(n)
past = []
until n == 1
n = n.digits.sum { |d| d * d }
return false if past.include? n
past << n
end
true
end
i = count = 0 until count == 8; puts i or count += 1 if happy?(i += 1) end puts (99999999999900..99999999999999).each { |i| puts i if happy?(i) }</lang>
- Output:
1 7 10 13 19 23 28 31 99999999999901 99999999999910 99999999999914 99999999999915 99999999999916 99999999999937 99999999999941 99999999999951 99999999999956 99999999999961 99999999999965 99999999999973
Run BASIC
<lang runbasic>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</lang>
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
Rust
In Rust, using a tortoise/hare cycle detection algorithm (generic for integer types) <lang rust>#![feature(core)]
fn sumsqd(mut n: i32) -> i32 {
let mut sq = 0;
while n > 0 {
let d = n % 10;
sq += d*d;
n /= 10
}
sq
}
use std::num::Int; fn cycle<T: Int>(a: T, f: fn(T) -> T) -> T {
let mut t = a; let mut h = f(a);
while t != h {
t = f(t);
h = f(f(h))
}
t
}
fn ishappy(n: i32) -> bool {
cycle(n, sumsqd) == 1
}
fn main() {
let happy = std::iter::count(1, 1)
.filter(|&n| ishappy(n))
.take(8)
.collect::<Vec<i32>>();
println!("{:?}", happy)
}</lang>
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
Salmon
<lang 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;
};
};</lang>
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.
Scala
<lang 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 </lang>
Scheme
<lang 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))))</lang>
The output is:
happy numbers: 1 7 10 13 19 23 28 31
Scratch
Scratch is a free visual programming language. Click the link, then "See inside" to view the code.
https://scratch.mit.edu/projects/78912620/
This code will allow you to check if a positive interger (<=9999) is a happy number. It will also output a list of the first 8 happy numbers. (1 7 10 13 19 23 28 31)
Seed7
<lang 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;</lang>
Output:
1 7 10 13 19 23 28 31 32 44 49
SequenceL
<lang sequencel>import <Utilities/Math.sl>; import <Utilities/Conversion.sl>;
main(argv(2)) := findHappys(stringToInt(head(argv)));
findHappys(count) := findHappysHelper(count, 1, []);
findHappysHelper(count, n, happys(1)) :=
happys when size(happys) = count
else
findHappysHelper(count, n + 1, happys ++ [n]) when isHappy(n)
else
findHappysHelper(count, n + 1, happys);
isHappy(n) := isHappyHelper(n, []);
isHappyHelper(n, cache(1)) :=
let
digits[i] := (n / integerPower(10, i - 1)) mod 10
foreach i within 1 ... ceiling(log(10, n + 1));
newN := sum(integerPower(digits, 2));
in
false when some(n = cache)
else
true when n = 1
else
isHappyHelper(newN, cache ++ [n]);</lang>
- Output:
$>happy.exe 8 [1,7,10,13,19,23,28,31]
SETL
<lang 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;</lang> <lang SETL>happy := []; n := 1; until #happy = 8 loop
if is_happy(n) then happy with:= n; end if; n +:= 1;
end loop;
print(happy);</lang> Output:
[1 7 10 13 19 23 28 31]
Alternative version: <lang SETL>print([n : n in [1..100] | is_happy(n)](1..8));</lang> Output:
[1 7 10 13 19 23 28 31]
Sidef
<lang ruby>func happy(n) is cached {
static seen = Hash()
return true if n.is_one return false if seen.exists(n)
seen{n} = 1
happy(n.digits.sum { _*_ })
}
say happy.first(8)</lang>
- Output:
[1, 7, 10, 13, 19, 23, 28, 31]
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. <lang smalltalk>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
]
]
].</lang> <lang smalltalk>|happy| happy := HappyNumber new.
1 to: 31 do: [ :i |
(happy isHappy: i) ifTrue: [ i displayNl ]
].</lang> Output:
1 7 10 13 19 23 28 31
an alternative version is:
<lang smalltalk>|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</lang> Output: OrderedCollection(1 7 10 13 19 23 28 31)
Swift
<lang Swift>func isHappyNumber(var n:Int) -> Bool {
var cycle = [Int]()
while n != 1 && !cycle.contains(n) {
cycle.append(n)
var m = 0
while n > 0 {
let d = n % 10
m += d * d
n = (n - d) / 10
}
n = m
}
return n == 1
}
var found = 0 var count = 0 while found != 8 {
if isHappyNumber(count) {
print(count)
found++
}
count++
}</lang>
- Output:
1 7 10 13 19 23 28 31
Tcl
using code from Sum of squares#Tcl <lang 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]"</lang>
the first 8 happy numbers are: {1 7 10 13 19 23 28 31}
TUSCRIPT
<lang 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</lang> 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
uBasic/4tH
<lang> ' ************************ ' MAIN ' ************************
PROC _PRINT_HAPPY(20) END
' ************************ ' END MAIN ' ************************
' ************************ ' SUBS & FUNCTIONS ' ************************
' -------------------- _is_happy PARAM(1) ' -------------------- LOCAL (5)
f@ = 100 c@ = a@ b@ = 0
DO WHILE b@ < f@ e@ = 0
DO WHILE c@
d@ = c@ % 10
c@ = c@ / 10
e@ = e@ + (d@ * d@)
LOOP
UNTIL e@ = 1 c@ = e@ b@ = b@ + 1 LOOP
RETURN(b@ < f@)
' -------------------- _PRINT_HAPPY PARAM(1) ' -------------------- LOCAL (2)
b@ = 1 c@ = 0
DO
IF FUNC (_is_happy(b@)) THEN
c@ = c@ + 1
PRINT b@
ENDIF
b@ = b@ + 1 UNTIL c@ + 1 > a@ LOOP
RETURN
' ************************ ' END SUBS & FUNCTIONS ' ************************ </lang>
UNIX Shell
<lang bash>#!/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 "$n" let count-=1 fi done return 0
}
first_n_happy 8</lang>
Output:
1 7 10 13 19 23 28 31
Ursala
The happy function is a predicate testing whether a given number is happy, and first(p) defines a function mapping a number n to the first n positive naturals having property p. <lang Ursala>#import std
- import nat
happy = ==1+ ^== sum:-0+ product*iip+ %np*hiNCNCS+ %nP
first "p" = ~&i&& iota; ~&lrtPX/&; leql@lrPrX->lrx ^|\~& ^/successor@l ^|T\~& "p"&& ~&iNC
- cast %nL
main = (first happy) 8</lang> output:
<1,7,10,13,19,23,28,31>
Vala
<lang vala>using Gee;
/* function to sum the square of the digits */ int sum(int input){ // convert input int to string string input_str = input.to_string(); int total = 0; // read through each character in string, square them and add to total for (int x = 0; x < input_str.length; x++){ // char.digit_value converts char to the decimal value the char it represents holds int digit = input_str[x].digit_value(); total += (digit * digit); }
return total; } // end sum
/* function to decide if a number is a happy number */ bool is_happy(int total){ var past = new HashSet<int>(); while(true){ total = sum(total); if (total == 1){ return true;}
if (total in past){ return false;}
past.add(total); } // end while loop } // end happy
public static void main(){ var happynums = new ArrayList<int>(); int x = 1;
while (happynums.size < 8){ if (is_happy(x) == true) happynums.add(x); x++; }
foreach(int num in happynums) stdout.printf("%d ", num); stdout.printf("\n"); } // end main</lang> The output is:
1 7 10 13 19 23 28 31
VBA
<lang vb> Option Explicit
Sub Test_Happy() Dim i&, Cpt&
For i = 1 To 100
If Is_Happy_Number(i) Then
Debug.Print "Is Happy : " & i
Cpt = Cpt + 1
If Cpt = 8 Then Exit For
End If
Next
End Sub
Public Function Is_Happy_Number(ByVal N As Long) As Boolean Dim i&, Number$, Cpt&
Is_Happy_Number = False 'default value
Do
Cpt = Cpt + 1 'Count Loops
Number = CStr(N) 'conversion Long To String to be able to use Len() function
N = 0
For i = 1 To Len(Number)
N = N + CInt(Mid(Number, i, 1)) ^ 2
Next i
'If Not N = 1 after 50 Loop ==> Number Is Not Happy
If Cpt = 50 Then Exit Function
Loop Until N = 1
Is_Happy_Number = True
End Function </lang>
- Output:
Is Happy : 1 Is Happy : 7 Is Happy : 10 Is Happy : 13 Is Happy : 19 Is Happy : 23 Is Happy : 28 Is Happy : 31
VBScript
<lang vb> count = 0 firsteigth="" For i = 1 To 100 If IsHappy(CInt(i)) Then firsteight = firsteight & i & "," count = count + 1 End If If count = 8 Then Exit For End If Next WScript.Echo firsteight
Function IsHappy(n) IsHappy = False m = 0 Do Until m = 60 sum = 0 For j = 1 To Len(n) sum = sum + (Mid(n,j,1))^2 Next If sum = 1 Then IsHappy = True Exit Do Else n = sum m = m + 1 End If Loop End Function </lang>
- Output:
1,7,10,13,19,23,28,31,
Visual Basic .NET
This version uses Linq to carry out the calculations. <lang vbnet>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</lang> The output is:
1: 1 2: 7 3: 10 4: 13 5: 19 6: 23 7: 28 8: 31
Cacheless version
Curiously, this runs in about two thirds of the time of the cacheless C# version on Tio.run. <lang vbnet>Module Module1
Dim sq As Integer() = {1, 4, 9, 16, 25, 36, 49, 64, 81}
Function isOne(x As Integer) As Boolean
While True
If x = 89 Then Return False
Dim t As Integer, s As Integer = 0
Do
t = (x Mod 10) - 1 : If t >= 0 Then s += sq(t)
x \= 10
Loop While x > 0
If s = 1 Then Return True
x = s
End While
Return False
End Function
Sub Main(ByVal args As String())
Const Max As Integer = 10_000_000
Dim st As DateTime = DateTime.Now
Console.Write("---Happy Numbers---" & vbLf & "The first 8:")
Dim i As Integer = 1, c As Integer = 0
While c < 8
If isOne(i) Then Console.Write("{0} {1}", If(c = 0, "", ","), i, c) : c += 1
i += 1
End While
Dim m As Integer = 10
While m <= Max
Console.Write(vbLf & "The {0:n0}th: ", m)
While c < m
If isOne(i) Then c += 1
i += 1
End While
Console.Write("{0:n0}", i - 1)
m = m * 10
End While
Console.WriteLine(vbLf & "Computation time {0} seconds.", (DateTime.Now - st).TotalSeconds)
End Sub
End Module</lang>
- Output:
---Happy Numbers--- The first 8: 1, 7, 10, 13, 19, 23, 28, 31 The 10th: 44 The 100th: 694 The 1,000th: 6,899 The 10,000th: 67,169 The 100,000th: 692,961 The 1,000,000th: 7,105,849 The 10,000,000th: 71,313,350 Computation time 19.235551 seconds.
Wren
<lang ecmascript>var happy = Fn.new { |n|
var m = {}
while (n > 1) {
m[n] = true
var x = n
n = 0
while (x > 0) {
var d = x % 10
n = n + d*d
x = (x/10).floor
}
if (m[n] == true) return false // m[n] will be null if 'n' is not a key
}
return true
}
var found = 0 var n = 1 while (found < 8) {
if (happy.call(n)) {
System.write("%(n) ")
found = found + 1
}
n = n + 1
} System.print()</lang>
- Output:
1 7 10 13 19 23 28 31
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.
<lang XPL0>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 ]</lang>
Output:
1 7 10 13 19 23 28 31
Zig
<lang Zig> const std = @import("std"); const stdout = std.io.getStdOut().outStream();
pub fn main() !void {
try stdout.print("The first 8 happy numbers are: ", .{});
var n: u32 = 1;
var c: u4 = 0;
while (c < 8) {
if (isHappy(n)) {
c += 1;
try stdout.print("{} ", .{n});
}
n += 1;
}
try stdout.print("\n", .{});
}
fn isHappy(n: u32) bool {
var t = n;
var h = sumsq(n);
while (t != h) {
t = sumsq(t);
h = sumsq(sumsq(h));
}
return t == 1;
}
fn sumsq(n0: u32) u32 {
var s: u32 = 0;
var n = n0;
while (n > 0) : (n /= 10) {
const m = n % 10;
s += m * m;
}
return s;
} </lang>
- Output:
The first 8 happy numbers are: 1 7 10 13 19 23 28 31
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).
<lang zkl>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
} }
}</lang> <lang zkl>h:=Utils.Generator(happyNumbers); h.walk(8).println();</lang>
- Output:
L(1,7,10,13,19,23,28,31)
Get the one million-th happy number. Nobody would call this quick. <lang zkl>Utils.Generator(happyNumbers).drop(0d1_000_000-1).next().println();</lang>
- Output:
7105849
ZX Spectrum Basic
<lang zxbasic>10 FOR i=1 TO 100 20 GO SUB 1000 30 IF isHappy=1 THEN PRINT i;" is a happy number" 40 NEXT i 50 STOP 1000 REM Is Happy? 1010 LET isHappy=0: LET count=0: LET num=i 1020 IF count=50 OR isHappy=1 THEN RETURN 1030 LET n$=STR$ (num) 1040 LET count=count+1 1050 LET isHappy=0 1060 FOR j=1 TO LEN n$ 1070 LET isHappy=isHappy+VAL n$(j)^2 1080 NEXT j 1090 LET num=isHappy 1100 GO TO 1020</lang>
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