# Hailstone sequence: Difference between revisions

Hailstone sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The Hailstone sequence of numbers can be generated from a starting positive integer, n by:

• If n is 1 then the sequence ends.
• If n is even then the next n of the sequence = n/2
• If n is odd then the next n of the sequence = (3 * n) + 1

The (unproven), Collatz conjecture is that the hailstone sequence for any starting number always terminates.

1. Create a routine to generate the hailstone sequence for a number.
2. Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1
3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.
(But don't show the actual sequence!)

## ACL2

<lang Lisp>(defun hailstone (len)

   (loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))

Must be tail recursive

(defun max-hailstone-start (limit mx curr)

  (declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))</lang>


Output:

> (take 4 (hailstone 27))
(27 82 41 124)
> (nthcdr 108 (hailstone 27))
(8 4 2 1)
> (len (hailstone 27))
112
> (max-hailstone-start 100000 0 0)
(351 77031)

Similar to C method: <lang Ada>with Ada.Text_IO; use Ada.Text_IO; procedure hailstone is type int_arr is array(Positive range <>) of Integer; type int_arr_pt is access all int_arr;

function hailstones(num:Integer; pt:int_arr_pt) return Integer is stones : Integer := 1; n : Integer := num; begin if pt /= null then pt(1) := num; end if; while (n/=1) loop stones := stones + 1; if n mod 2 = 0 then n := n/2; else n := (3*n)+1; end if; if pt /= null then pt(stones) := n; end if; end loop; return stones; end hailstones;

nmax,stonemax,stones : Integer := 0; list : int_arr_pt; begin stones := hailstones(27,null); list := new int_arr(1..stones); stones := hailstones(27,list); put(" 27: "&Integer'Image(stones)); new_line; for n in 1..4 loop put(Integer'Image(list(n))); end loop; put(" .... "); for n in stones-3..stones loop put(Integer'Image(list(n))); end loop; new_line; for n in 1..100000 loop stones := hailstones(n,null); if stones>stonemax then nmax := n; stonemax := stones; end if; end loop; put_line(Integer'Image(nmax)&" max @ n= "&Integer'Image(stonemax)); end hailstone;</lang> Output:

 27:  112
27 82 41 124 ....  8 4 2 1
77031 max @ n=  351


### Alternative method

A method without pointers or dynamic memory allocation, but slower for simply counting. This is also used for the "executable library" task Executable library#Ada.

  type Integer_Sequence is array(Positive range <>) of Integer;
function Create_Sequence (N : Positive) return Integer_Sequence;


  function Create_Sequence (N : Positive) return Integer_Sequence is
begin
if N = 1 then
-- terminate
return (1 => N);
elsif N mod 2 = 0 then
-- even
return (1 => N) & Create_Sequence (N / 2);
else
-- odd
return (1 => N) & Create_Sequence (3 * N + 1);
end if;
end Create_Sequence;


procedure Main is

  package Integer_IO is new Ada.Text_IO.Integer_IO (Integer);

  procedure Print_Sequence (X : Hailstones.Integer_Sequence) is
begin
for I in X'Range loop
Integer_IO.Put (Item => X (I), Width => 0);
if I < X'Last then
end if;
end loop;
end Print_Sequence;

  Hailstone_27 : constant Hailstones.Integer_Sequence :=
Hailstones.Create_Sequence (N => 27);


begin

  Ada.Text_IO.Put_Line ("Length of 27:" & Integer'Image (Hailstone_27'Length));
Print_Sequence (Hailstone_27 (Hailstone_27'First .. Hailstone_27'First + 3));
Print_Sequence (Hailstone_27 (Hailstone_27'Last - 3 .. Hailstone_27'Last));

  declare
Longest_Length : Natural := 0;
Longest_N      : Positive;
Length         : Natural;
begin
for I in 1 .. 99_999 loop
Length := Hailstones.Create_Sequence (N => I)'Length;
if Length > Longest_Length then
Longest_Length := Length;
Longest_N := I;
end if;
end loop;
Ada.Text_IO.Put_Line ("Longest length is" & Integer'Image (Longest_Length));
Ada.Text_IO.Put_Line ("with N =" & Integer'Image (Longest_N));
end;


end Main;</lang> output:

Length of 27: 112
First four: 27, 82, 41, 124
Last four: 8, 4, 2, 1
Longest length is 351
with N = 77031

## ALGOL 68

Translation of: C

- note: This specimen retains the original C coding style.

Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - using the print routine rather than printf

<lang algol68>MODE LINT = # LONG ... # INT;

PROC hailstone = (INT in n, REF[]LINT array)INT: (

   INT hs := 1;
INT index := 0;
LINT n := in n;

WHILE n /= 1 DO
hs +:= 1;
IF array ISNT REF[]LINT(NIL) THEN array[index +:= 1] := n FI;
n := IF ODD n THEN 3*n+1 ELSE n OVER 2 FI
OD;
IF array ISNT REF[]LINT(NIL) THEN array[index +:= 1] := n FI;
hs


);

main: (

   INT j, hmax := 0;
INT jatmax, n;
INT border = 4;

FOR j TO 100000-1 DO
n := hailstone(j, NIL);
IF hmax < n THEN
hmax := n;
jatmax := j
FI
OD;

[2]INT test := (27, jatmax);
FOR key TO UPB test DO
INT val = test[key];
n := hailstone(val, NIL);
[n]LINT array;
n := hailstone(val, array);

printf(($"[ "n(border)(g(0)", ")" ..."n(border)(", "g(0))"] len="g(0)l$,
array[:border], array[n-border+1:], n))
#;free(array) #
OD;
printf(($"Max "g(0)" at j="g(0)l$, hmax, jatmax))

1. ELLA Algol68RS:
   print(("Max",hmax," at j=",jatmax, new line))


)</lang> Output:

[ 27, 82, 41, 124,  ..., 8, 4, 2, 1] len=112
[ 77031, 231094, 115547, 346642,  ..., 8, 4, 2, 1] len=351
Max 351 at j=77031


## APL

Works with: Dyalog APL

<lang APL>seq←hailstone n;next ⍝ Returns the hailstone sequence for a given number

seq←n ⍝ Init the sequence

While n≠1
   next←(n÷2) (1+3×n)  ⍝ Compute both possibilities
n←next[1+2|n]       ⍝ Pick the appropriate next step
seq,←n              ⍝ Append that to the sequence

EndWhile</lang>

Output: <lang APL> 5↑hailstone 27 27 82 41 124 62

¯5↑hailstone 27


16 8 4 2 1

⍴hailstone 27


112

1↑{⍵[⍒↑(⍴∘hailstone)¨⍵]}⍳100000


77031</lang>

## AutoHotkey

<lang autohotkey>; Submitted by MasterFocus --- http://tiny.cc/iTunis

[1] Generate the Hailstone Seq. for a number

List := varNum := 7 ; starting number is 7, not counting elements While ( varNum > 1 )

 List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )


MsgBox % List

[2] Seq. for starting number 27 has 112 elements

Count := 1, List := varNum := 27 ; starting number is 27, counting elements While ( varNum > 1 )

 Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )


MsgBox % "Sequence:n" List "nnCount: " Count

[3] Find number<100000 with longest seq. and show both values

MaxNum := Max := 0 ; reset the Maximum variables TimesToLoop := 100000 ; limit number here is 100000 Offset := 70000 ; offset - use 0 to process from 0 to 100000 Loop, %TimesToLoop% {

 If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...n-------------------n"
text .= "Current starting number: " Index "n"
text .= "Current sequence count: " Count
text .= "n-------------------n"
text .= "Maximum starting number: " MaxNum "n"
text .= "Maximum sequence count: " Max " <<" ; text split to avoid long code lines
ToolTip, %text%
Count := 1 ; going to count the elements, but no "List" required
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index ; set the new maximum values, if necessary


} ToolTip MsgBox % "Number: " MaxNum "nCount: " Max</lang>

## AutoIt

<lang autoit> $Hail = Hailstone(27) ConsoleWrite("Sequence-Lenght: "&$Hail&@CRLF) $Big = -1$Sequenzlenght = -1 For $I = 1 To 100000$Hail = Hailstone($i, False) If Number($Hail) > $Sequenzlenght Then$Sequenzlenght = Number($Hail)$Big = $i EndIf Next ConsoleWrite("Longest Sequence : "&$Sequenzlenght&" from number "&$Big&@CRLF) Func Hailstone($int, $sequence = True)$Counter = 0 While True $Counter += 1 If$sequence = True Then ConsoleWrite($int & ",") If$int = 1 Then ExitLoop If Not Mod($int, 2) Then$int = $int / 2 Else$int = 3 * $int + 1 EndIf If Not Mod($Counter, 25) AND $sequence = True Then ConsoleWrite(@CRLF) WEnd If$sequence = True Then ConsoleWrite(@CRLF) Return $Counter EndFunc ;==>Hailstone </lang> Output: 27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,412,206,103, 310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132, 566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051, 6154,3077,9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106, 53,160,80,40,20,10,5,16,8,4,2,1, Sequence-Lenght: 112 Longest Sequence : 351 from number 77031  ## AWK <lang awk> 1. !/usr/bin/awk -f function hailstone(v, verbose) { n = 1; u = v; while (1) { if (verbose) printf " "u; if (u==1) return(n); n++; if (u%2 > 0 ) u = 3*u+1; else u = u/2; } } BEGIN { i = 27; printf("hailstone(%i) has %i elements\n",i,hailstone(i,1)); ix=0; m=0; for (i=1; i<100000; i++) { n = hailstone(i,0); if (m<n) { m=n; ix=i; } } printf("longest hailstone sequence is %i and has %i elements\n",ix,m); } </lang> Output: 27 82 41 124 ....... 8 4 2 1 hailstone(27) has 112 elements longest hailstone sequence is 77031 and has 351 elements  ## BASIC ### Applesoft BASIC <lang ApplesoftBasic>10 HOME 100 N = 27 110 GOSUB 400"HAILSTONE 120 DEF FN L(I) = E(I + 4 * (I < 0)) 130IFL=112AND(S(0)=27ANDS(1)=82ANDS(2)=41ANDS(3)=124)AND(FNL(M-3)=8ANDFNL(M-2)=4ANDFNL(M-1)=2ANDFNL(M)=1)THENPRINT"THE HAILSTONE SEQUENCE FOR THE NUMBER 27 HAS 112 ELEMENTS STARTING WITH 27, 82, 41, 124 AND ENDING WITH 8, 4, 2, 1" 140 PRINT 150 V = PEEK(37) + 1 200 N = 1 210 GOSUB 400"HAILSTONE 220 MN = 1 230 ML = L 240 FOR I = 2 TO 99999 250 N = I 260 GOSUB 400"HAILSTONE 270 IFL>MLTHENMN=I:ML=L:VTABV:HTAB1:PRINT "THE NUMBER " MN " HAS A HAILSTONE SEQUENCE LENGTH OF "L" WHICH IS THE LONGEST HAILSTONE SEQUENCE OF NUMBERS LESS THAN ";:Y=PEEK(37)+1:X=PEEK(36)+1 280 IF Y THEN VTAB Y : HTAB X : PRINTI+1; 290 NEXT I 300 END 400 M = 0 410 FOR L = 1 TO 1E38 420 IF L < 5 THEN S(L-1) = N 430 M = (M + 1) * (M < 3) 440 E(M) = N 450 IF N = 1 THEN RETURN 460 EVEN = INT(N/2)=N/2 470 IF EVEN THEN N=N/2 480 IF NOT EVEN THEN N = (3 * N) + 1 490 NEXT L : STOP</lang> ### BBC BASIC <lang bbcbasic> seqlen% = FNhailstone(27, TRUE)  PRINT '"Sequence length = "; seqlen% maxlen% = 0 FOR number% = 2 TO 100000 seqlen% = FNhailstone(number%, FALSE) IF seqlen% > maxlen% THEN maxlen% = seqlen% maxnum% = number% ENDIF NEXT PRINT "The number with the longest hailstone sequence is " ; maxnum% PRINT "Its sequence length is " ; maxlen% END DEF FNhailstone(N%, S%) LOCAL L% IF S% THEN PRINT N%; WHILE N% <> 1 IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2 IF S% THEN PRINT N%; L% += 1 ENDWHILE = L% + 1</lang>  Output:  27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Sequence length = 112 The number with the longest hailstone sequence is 77031 Its sequence length is 351  ### Liberty BASIC <lang lb>print "Part 1: Create a routine to generate the hailstone sequence for a number." print "" while hailstone < 1 or hailstone <> int(hailstone)  input "Please enter a positive integer: "; hailstone  wend print "" print "The following is the 'Hailstone Sequence' for your number..." print "" print hailstone while hailstone <> 1  if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1 print hailstone  wend print "" input "Hit 'Enter' to continue to part 2...";dummy$ cls print "Part 2: Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1." print "" print "No. in Seq.","Hailstone Sequence Number for 27" print "" c = 1: hailstone = 27 print c, hailstone while hailstone <> 1

   c = c + 1
if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1
print c, hailstone


wend print "" input "Hit 'Enter' to continue to part 3...";dummy$cls print "Part 3: Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.(But don't show the actual sequence)!" print "" print "Calculating result... Please wait... This could take a little while..." print "" print "Percent Done", "Start Number", "Seq. Length", "Maximum Sequence So Far" print "" for cc = 1 to 99999  hailstone = cc: c = 1 while hailstone <> 1 c = c + 1 if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1 wend if c > max then max = c: largesthailstone = cc locate 1, 7 print " " locate 1, 7 print using("###.###", cc / 99999 * 100);"%", cc, c, max scan  next cc print "" print "The number less than 100,000 with the longest 'Hailstone Sequence' is "; largesthailstone;". It's sequence length is "; max;"." end</lang> ### OxygenBasic <lang oxygenbasic> function Hailstone(sys *n) '========================= if n and 1  n=n*3+1  else  n=n>>1  end if end function function HailstoneSequence(sys n) as sys '======================================= count=1 do  Hailstone n Count++ if n=1 then exit do  end do return count end function 'MAIN '==== maxc=0 maxn=0 e=100000 for n=1 to e c=HailstoneSequence n if c>maxc maxc=c maxn=n end if  next print e ", " maxn ", " maxc 'result 100000, 77031, 351 </lang> ### PureBasic <lang PureBasic>NewList Hailstones.i() ; Make a linked list to use as we do not know the numbers of elements needed for an Array Procedure.i FillHailstones(n) ; Fills the list & returns the amount of elements in the list  Shared Hailstones() ; Get access to the Hailstones-List ClearList(Hailstones()) ; Remove old data Repeat AddElement(Hailstones()) ; Add an element to the list Hailstones()=n ; Fill current value in the new list element If n=1 ProcedureReturn ListSize(Hailstones()) ElseIf n%2=0 n/2 Else n=(3*n)+1 EndIf ForEver  EndProcedure If OpenConsole()  Define i, l, maxl, maxi l=FillHailstones(27) Print("#27 has "+Str(l)+" elements and the sequence is: "+#CRLF$)
ForEach Hailstones()
If i=6
print hailstone;chr$(9); if (doHailstone mod 10) = 0 then print end if  wend END FUNCTION</lang> ### Visual Basic .NET Works with: Visual Basic .NET version 2005+ <lang vbnet>Module HailstoneSequence  Sub Main() ' Checking sequence of 27.   Dim l As List(Of Long) = HailstoneSequence(27) Console.WriteLine("27 has {0} elements in sequence:", l.Count())   For i As Integer = 0 To 3 : Console.Write("{0}, ", l(i)) : Next Console.Write("... ") For i As Integer = l.Count - 4 To l.Count - 1 : Console.Write(", {0}", l(i)) : Next   Console.WriteLine()   ' Finding longest sequence for numbers below 100000.   Dim max As Integer = 0 Dim maxCount As Integer = 0   For i = 1 To 99999 l = HailstoneSequence(i) If l.Count > maxCount Then max = i maxCount = l.Count End If Next Console.WriteLine("Max elements in sequence for number below 100k: {0} with {1} elements.", max, maxCount) Console.ReadLine() End Sub   Private Function HailstoneSequence(ByVal n As Long) As List(Of Long) Dim valList As New List(Of Long)() valList.Add(n)   Do Until n = 1 n = IIf(n Mod 2 = 0, n / 2, (3 * n) + 1) valList.Add(n) Loop   Return valList End Function  End Module</lang> Output: 27 has 112 elements in sequence: 27, 82, 41, 124, ... , 8, 4, 2, 1 Max elements in sequence for number below 100k: 77031 with 351 elements.  ## Batch File 1. Create a routine to generate the hailstone sequence for a number. <lang dos>@echo off setlocal enabledelayedexpansion if "%1" equ "" goto :eof call :hailstone %1 seq cnt echo %seq% goto :eof hailstone set num=%1 set %2=%1 loop if %num% equ 1 goto :eof call :iseven %num% res if %res% equ T goto divideby2 set /a num = (3 * num) + 1 set %2=!%2! %num% goto loop divideby2 set /a num = num / 2 set %2=!%2! %num% goto loop iseven set /a tmp = %1 %% 2 if %tmp% equ 1 ( set %2=F ) else ( set %2=T ) goto :eof</lang> Demonstration <lang dos>>hailstone.cmd 20 20 10 5 16 8 4 2 1</lang> ## Befunge  This example may be incorrect. Calculates the Hailstone sequence but might not complete everything from task description. Please verify it and remove this message. If the example does not match the requirements or does not work, replace this message with Template:incorrect or fix the code yourself. <lang befunge>&>:.:1-|  >3*^ @ |%2: < v>2/>+</lang>  ## Bracmat <lang bracmat>(  ( hailstone = L len . !arg:?L & whl ' ( !arg:~1 & (!arg*1/2:~/|3*!arg+1):?arg & !arg !L:?L ) & (!L:? [?len&!len.!L) )  & ( reverse  = L e . :?L & whl'(!arg:%?e ?arg&!e !L:?L) & !L )  & hailstone$27:(?len.?list) & reverse$!list:?first4 [4 ? [-5 ?last4 & put$"Hailstone sequence starting with " & put$!first4 & put$(str$(" has " !len " elements and ends with ")) & put$(!last4 \n) & 1:?N & 0:?max:?Nmax & whl

 ' ( !N+1:<100000:?N
&   hailstone$!N : ( >!max:?max&!N:?Nmax | ? . ? ) )  & out $ ( str
$( "The number <100000 with the longest hailstone sequence is " !Nmax " with " !max " elements." ) )  );</lang> ## Brainf***  This example is incomplete. Please ensure that it meets all task requirements and remove this message. Prints the number of terms required to map the input to 1. Does not count the first term of the sequence. <lang Brainf***>>,[  [ ----------[ >>>[>>>>]+[[-]+<[->>>>++>>>>+[>>>>]++[->+<<<<<]]<<<] ++++++[>------<-]>--[>>[->>>>]+>+[<<<<]>-],< ]> ]>>>++>+>>[ <<[>>>>[-]+++++++++<[>-<-]+++++++++>[-[<->-]+[<<<<]]<[>+<-]>] >[>[>>>>]+[[-]<[+[->>>>]>+<]>[<+>[<<<<]]+<<<<]>>>[->>>>]+>+[<<<<]] >[[>+>>[<<<<+>>>>-]>]<<<<[-]>[-<<<<]]>>>>>>> ]>>+[[-]++++++>>>>]<<<<[[<++++++++>-]<.[-]<[-]<[-]<]<,  ]</lang> <lang Brainf***>27 111</lang> ## Brat <lang brat>hailstone = { num |  sequence = [num] while { num != 1 } { true? num % 2 == 0 { num = num / 2 } { num = num * 3 + 1 } sequence << num }   sequence  } 1. Check sequence for 27 seq = hailstone 27 true? (seq[0,3] == [27 82 41 124] && seq[-1, -4] == [8 4 2 1])  { p "Sequence for 27 is correct" } { p "Sequence for 27 is not correct!" }  1. Find longest sequence for numbers < 100,000 longest = [number: 0 length: 0] 1.to 99999 { index |  seq = hailstone index true? seq.length > longest[:length] { longest[:length] = seq.length longest[:number] = index p "Longest so far: #{index} @ #{longest[:length]} elements" }   index = index + 1 }  p "Longest was starting from #{longest[:number]} and was of length #{longest[:length]}"</lang> Output: Sequence for 27 is correct Longest so far: 1 @ 1 elements Longest so far: 2 @ 2 elements Longest so far: 3 @ 8 elements ... Longest so far: 52527 @ 340 elements Longest so far: 77031 @ 351 elements Longest was starting from 77031 and was of length 351 ## Burlesque <lang burlesque> blsq ) 27{^^^^2.%{3.*1.+}\/{2./}\/ie}{1!=}w!bx{\/+]}{\/isn!}w!L[ 112 </lang> ## C <lang C>#include <stdio.h> 1. include <stdlib.h> int hailstone(int n, int *arry) {  int hs = 1;   while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs;  } int main() {  int j, hmax = 0; int jatmax, n; int *arry;   for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry);   printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry);   return 0;  }</lang> Output [ 27, 82, 41, 124, ...., 8, 4, 2, 1] len= 112 Max 351 at j= 77031 ### With caching Much faster if you want to go over a million or so. <lang c>#include <stdio.h> 1. define N 10000000 2. define CS N /* cache size */ typedef unsigned long ulong; ulong cache[CS] = {0}; ulong hailstone(ulong n) { int x; if (n == 1) return 1; if (n < CS && cache[n]) return cache[n]; x = 1 + hailstone((n & 1) ? 3 * n + 1 : n / 2); if (n < CS) cache[n] = x; return x; } int main() { int i, l, max = 0, mi; for (i = 1; i < N; i++) { if ((l = hailstone(i)) > max) { max = l; mi = i; } } printf("max below %d: %d, length %d\n", N, mi, max); return 0; }</lang> ## C# <lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Hailstone {  class Program { public static List<int> hs(int n,List<int> seq) { List<int> sequence = seq; sequence.Add(n); if (n == 1) { return sequence; }else{ int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1; return hs(newn, sequence); } }   static void Main(string[] args) { int n = 27; List<int> sequence = hs(n,new List<int>()); Console.WriteLine(sequence.Count + " Elements"); List<int> start = sequence.GetRange(0, 4); List<int> end = sequence.GetRange(sequence.Count - 4, 4); Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end)); int number = 0, longest = 0; for (int i = 1; i < 100000; i++) { int count = (hs(i, new List<int>())).Count; if (count > longest) { longest = count; number = i; } } Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest); } }  }</lang> 112 Elements Starting with : 27,82,41,124 and ending with : 8,4,2,1 Number < 100000 with longest Hailstone seq.: 77031 with length of 351  ### With caching As with the C example, much faster if you want to go over a million or so. <lang csharp>using System; using System.Collections.Generic; namespace ConsoleApplication1 {  class Program { public static void Main() { int longestChain = 0, longestNumber = 0;   var recursiveLengths = new Dictionary<int, int>();   const int maxNumber = 100000;   for (var i = 1; i <= maxNumber; i++) { var chainLength = Hailstone(i, recursiveLengths); if (longestChain >= chainLength) continue;   longestChain = chainLength; longestNumber = i; } Console.WriteLine("max below {0}: {1} ({2} steps)", maxNumber, longestNumber, longestChain); }   private static int Hailstone(int num, Dictionary<int, int> lengths) { if (num == 1) return 1;   while (true) { if (lengths.ContainsKey(num)) return lengths[num];   lengths[num] = 1 + ((num%2 == 0) ? Hailstone(num/2, lengths) : Hailstone((3*num) + 1, lengths)); } } }  }</lang> max below 100000: 77031 (351 steps)  ## C++ <lang cpp>#include <iostream> 1. include <vector> 2. include <utility> std::vector<int> hailstone(int i) {  std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v;  } std::pair<int,int> find_longest_hailstone_seq(int n) {  std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq;  } int main () { // Use the routine to show that the hailstone sequence for the number 27  std::vector<int> h27; h27 = hailstone(27);  // has 112 elements  int l = h27.size(); std::cout << "length of hailstone(27) is " << l;  // starting with 27, 82, 41, 124 and  std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl;  // ending with 8, 4, 2, 1  std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl;   std::pair<int,int> m = find_longest_hailstone_seq(100000);   std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl;   return 0;  }</lang> output: length of hailstone(27) is 112 first four elements of hailstone(27) are 27 82 41 124 last four elements of hailstone(27) are 8 4 2 1 the longest hailstone sequence under 100,000 is 77031 with 351 elements.  ## CLIPS <lang clips>(deftemplate longest  (slot bound) ; upper bound for the range of values to check (slot next (default 2)) ; next value that needs to be checked (slot start (default 1)) ; starting value of longest sequence (slot len (default 1)) ; length of longest sequence  ) (deffacts startup  (query 27) (longest (bound 100000))  ) (deffunction hailstone-next  (?n) (if (evenp ?n) then (div ?n 2) else (+ (* 3 ?n) 1) )  ) (defrule extend-sequence  ?hail <- (hailstone$?sequence ?tail&:(> ?tail 1))
=>
(retract ?hail)
(assert (hailstone ?sequence ?tail (hailstone-next ?tail)))


)

(defrule start-query

 (query ?num)
=>
(assert (hailstone ?num))


)

(defrule result-query

 (query ?num)
(hailstone ?num $?sequence 1) => (bind ?sequence (create$ ?num ?sequence 1))
(printout t "Hailstone sequence starting with " ?num ":" crlf)
(bind ?len (length ?sequence))
(printout t "  Length: " ?len crlf)
(printout t "  First four: " (implode$(subseq$ ?sequence 1 4)) crlf)
(printout t "  Last four: " (implode$(subseq$ ?sequence (- ?len 3) ?len)) crlf)
(printout t crlf)


)

(defrule longest-create-next-hailstone

 (longest (bound ?bound) (next ?next))
(test (<= ?next ?bound))
(not (hailstone ?next $?)) => (assert (hailstone ?next))  ) (defrule longest-check-next-hailstone  ?longest <- (longest (bound ?bound) (next ?next) (start ?start) (len ?len)) (test (<= ?next ?bound)) ?hailstone <- (hailstone ?next$?sequence 1)
=>
(retract ?hailstone)
(bind ?thislen (+ 2 (length ?sequence)))
(if (> ?thislen ?len) then
(modify ?longest (start ?next) (len ?thislen) (next (+ ?next 1)))
else
(modify ?longest (next (+ ?next 1)))
)


)

(defrule longest-finished

 (longest (bound ?bound) (next ?next) (start ?start) (len ?len))
(test (> ?next ?bound))
=>
(printout t "The number less than " ?bound " that has the largest hailstone" crlf)
(printout t "sequence is " ?start " with a length of " ?len "." crlf)
(printout t crlf)


)</lang>

Output:

The number less than 100000 that has the largest hailstone
sequence is 77031 with a length of 351.

Hailstone sequence starting with 27:
Length: 112
First four: 27 82 41 124
Last four: 8 4 2 1

## Clojure

<lang clojure>(defn hailstone-seq [n]

 (:pre [(pos? n)])
(lazy-seq
(cond (= n 1)   '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else     (cons n (hailstone-seq (+ (* n 3) 1))))))


(def hseq27 (hailstone-seq 27)) (assert (= (count hseq27) 112)) (assert (= (take 4 hseq27) [27 82 41 124])) (assert (= (drop 108 hseq27) [8 4 2 1]))

(let [{max-i :num, max-len :len}

     (reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))</lang>


## CoffeeScript

Recursive version: <lang coffeescript>hailstone = (n) ->

 if n is 1
[n]

else if n % 2 is 0
[n].concat hailstone n/2

else
[n].concat hailstone (3*n) + 1



h27 = hailstone 27 console.log "hailstone(27) = #{h27[0..3]} ... #{h27[-4..]} (length: #{h27.length})"

maxlength = 0 maxnums = []

for i in [1..100000]

 seq = hailstone i

if seq.length is maxlength
maxnums.push i
else if seq.length > maxlength
maxlength = seq.length
maxnums = [i]



console.log "Max length: #{maxlength}; numbers generating sequences of this length: #{maxnums}"</lang>

hailstone(27) = 27,82,41,124 ... 8,4,2,1 (length: 112)
Max length: 351; numbers generating sequences of this length: 77031

## Common Lisp

<lang lisp>(defun hailstone (n)

 (cond ((= n 1) '(1))


((evenp n) (cons n (hailstone (/ n 2)))) (t (cons n (hailstone (+ (* 3 n) 1))))))

(defun longest (n)

 (let ((k 0) (l 0))
(loop for i from 1 below n do


(let ((len (length (hailstone i)))) (when (> len l) (setq l len k i))) finally (format t "Longest hailstone sequence under ~A for ~A, having length ~A." n k l))))</lang> Sample session:

ROSETTA> (length (hailstone 27))
112
ROSETTA> (subseq (hailstone 27) 0 4)
(27 82 41 124)
ROSETTA> (last (hailstone 27) 4)
(8 4 2 1)
ROSETTA> (longest-hailstone 100000)
Longest hailstone sequence under 100000 for 77031, having length 351.
NIL

## D

### Basic Version

<lang d>import std.stdio, std.algorithm, std.range, std.typecons;

auto hailstone(uint n) pure nothrow {

 auto result = [n];
while (n != 1) {
n = n & 1 ? n*3 + 1 : n/2;
result ~= n;
}
return result;


}

void main() {

 enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$- 4 ..$]);
writeln("Length hailstone(", M, ")= ", h.length);

 enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);


}</lang>

Output:
hailstone(27)= [27, 82, 41, 124] ... [8, 4, 2, 1]
Length hailstone(27)= 112
Longest sequence in [1,100000]= 77031 with len 351

### Faster Lazy Version

Same output. <lang d>import std.stdio, std.algorithm, std.range, std.typecons;

struct Hailstone {

 uint n;
bool empty() const pure nothrow { return n == 0; }
uint front() const pure nothrow { return n; }
void popFront() pure nothrow {
n = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
}


}

void main() {

 enum M = 27;
immutable h = M.Hailstone.array;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$- 4 ..$]);
writeln("Length hailstone(", M, ")= ", h.length);

 enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.Hailstone.walkLength, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);


}</lang>

### Lazy Version With Caching

Faster, same output. <lang d>import std.stdio, std.algorithm, std.range, std.typecons;

struct Hailstone(size_t cacheSize = 500_000) {

 size_t n;
__gshared static size_t[cacheSize] cache;

 bool empty() const pure nothrow { return n == 0; }
size_t front() const pure nothrow { return n; }

 void popFront() nothrow {
if (n >= cacheSize) {
n = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
} else if (cache[n]) {
n = cache[n];
} else {
immutable n2 = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
n = cache[n] = n2;
}
}


}

void main() {

 enum M = 27;
const h = M.Hailstone!().array;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$- 4 ..$]);
writeln("Length hailstone(", M, ")= ", h.length);

 enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.Hailstone!().walkLength, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);


}</lang>

## Déjà Vu

<lang dejavu>local hailstone: swap [ over ] while < 1 dup: if % over 2: #odd ++ * 3 else: #even / swap 2 swap push-through rot dup drop

if = (name) :(main): local :h27 hailstone 27 !. = 112 len h27 !. = 27 h27! 0 !. = 82 h27! 1 !. = 41 h27! 2 !. = 124 h27! 3 !. = 8 h27! 108 !. = 4 h27! 109 !. = 2 h27! 110 !. = 1 h27! 111

local :max 0 local :maxlen 0 for i range 1 99999: dup len hailstone i if < maxlen: set :maxlen set :max i else: drop !print( "number: " to-str max ", length: " to-str maxlen ) else: @hailstone</lang>

Output:
true
true
true
true
true
true
true
true
true
number: 77031, length: 351

## Dart

<lang dart>List<int> hailstone(int n) {

 if(n<=0) {
throw new IllegalArgumentException("start value must be >=1)");
}
Queue<int> seq=new Queue<int>();
while(n!=1) {
n=n%2==0?(n/2).toInt():3*n+1;
}
return new List<int>.from(seq);


}

// apparently List is missing toString() String iterableToString(Iterable seq) {

 String str="[";
Iterator i=seq.iterator();
while(i.hasNext()) {
str+=i.next();
if(i.hasNext()) {
str+=",";
}
}
return str+"]";


}

main() {

 for(int i=1;i<=10;i++) {
print("h($i)="+iterableToString(hailstone(i))); } List<int> h27=hailstone(27); List<int> first4=h27.getRange(0,4); print("first 4 elements of h(27): "+iterableToString(first4)); Expect.listEquals([27,82,41,124],first4);   List<int> last4=h27.getRange(h27.length-4,4); print("last 4 elements of h(27): "+iterableToString(last4)); Expect.listEquals([8,4,2,1],last4);   print("length of sequence h(27): "+h27.length); Expect.equals(112,h27.length);   int seq,max=0; for(int i=1;i<=100000;i++) { List<int> h=hailstone(i); if(h.length>max) { max=h.length; seq=i; } } print("up to 100000 the sequence h($seq) has the largest length ($max)");  }</lang> Output h(1)=[1] h(2)=[2,1] h(3)=[3,10,5,16,8,4,2,1] h(4)=[4,2,1] h(5)=[5,16,8,4,2,1] h(6)=[6,3,10,5,16,8,4,2,1] h(7)=[7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1] h(8)=[8,4,2,1] h(9)=[9,28,14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1] h(10)=[10,5,16,8,4,2,1] first 4 elements of h(27): [27,82,41,124] last 4 elements of h(27): [8,4,2,1] length of sequence h(27): 112 up to 100000 the sequence h(77031) has the largest length (351) ## Dc Firstly, this code takes the value from the stack, computes and prints the corresponding Hailstone sequence, and the length of the sequence. The q procedure is for counting the length of the sequence. The e and o procedure is for even and odd number respectively. The x procedure is for overall control. <lang Dc>27 [[--: ]nzpq]sq [d 2/ p]se [d 3*1+ p]so [d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx</lang> Output 82 41 124 62 (omitted) 8 4 2 1 --: 112  Then we could wrap the procedure x with a new procedure s, and call it with l which is loops the value of t from 1 to 100000, and cleaning up the stack after each time we finish up with a number. Register L for the length of the longest sequence and T for the corresponding number. Also, procedure q is slightly modified for storing L and T if needed, and all printouts in procedure e and o are muted. <lang Dc>0dsLsT1st [dsLltsT]sM [[zdlL<M q]sq [d 2/]se [d 3*1+ ]so [d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx]ss [lt1+dstlsxc lt100000>l]dslx lTn[:]nlLp </lang> Output (Takes quite some time on a decent machine) 77031:351 ## Delphi <lang Delphi>program ShowHailstoneSequence; {$APPTYPE CONSOLE}

uses SysUtils, Generics.Collections;

procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>); var

 n: Integer;


begin

 aHailstoneList.Clear;
n := aStartingNumber;

 while n <> 1 do
begin
if Odd(n) then
n := (3 * n) + 1
else
n := n div 2;
end;


end;

var

 i: Integer;
lList: TList<Integer>;
lMaxSequence: Integer;
lMaxLength: Integer;


begin

 lList := TList<Integer>.Create;
try
GetHailstoneSequence(27, lList);
Writeln(Format('27: %d elements', [lList.Count]));
Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
[lList[0], lList[1], lList[2], lList[3],
lList[lList.Count - 4], lList[lList.Count - 3], lList[lList.Count - 2], lList[lList.Count - 1]]));
Writeln;

   lMaxSequence := 0;
lMaxLength := 0;
for i := 1 to 100000 do
begin
GetHailstoneSequence(i, lList);
if lList.Count > lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lList.Count;
end;
end;
Writeln(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength]));
finally
lList.Free;
end;

 Readln;


end.</lang> Output:

27: 112 elements
[27 82 41 124 ... 8 4 2 1]

Longest sequence under 100,000: 77031 with 351 elements

## Elixir

<lang elixir>defmodule Hailstone do

 def step(1), do: 0
def step(n) when Integer.even?(n), do: div(n,2)
def step(n) when Integer.odd?(n), do: n*3 + 1
def sequence(n) do
Enum.to_list(Stream.take_while(Stream.iterate(n, &step/1), &(&1 > 0)))
end

 def run do
seq27 = Hailstone.sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4), "]",
String.replace(inspect(Enum.drop(seq27,len27-4)), "[", ", "))
IO.puts("Hailstone(27) has #{len27} elements: #{repr}")

   {start, len}  = Enum.max_by( Enum.map(1..100_000, fn(n) -> {n, length(Hailstone.sequence(n))} end),
fn({_,len}) -> len end )
IO.puts("Longest sequence starting under 100000 begins with #{start} and has #{len} elements.")
end


end

Hailstone.run</lang>

Output:
Hailstone(27) has 112 elements: [27, 82, 41, 124, ..., 8, 4, 2, 1]
Longest sequence starting under 100000 begins with 77031 and has 351 elements.


## Erlang

<lang erlang>-module(hailstone). -import(io). -export([main/0]).

hailstone(1) -> [1]; hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)]; hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].

max_length(Start, Stop) ->

   F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).


main() ->

   io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).</lang>


Output:

Eshell V5.8.4  (abort with ^G)
1> c(hailstone).
{ok,hailstone}
2> hailstone:main().
hailstone(4): [4,2,1]
hailstone(27) length: 112
hailstone(27) first 4: [27,82,41,124]
hailstone(27) last 4: [8,4,2,1]
finding maximum hailstone(N) length for 1 <= N <= 100000... done.
hailstone(77031) length: 351
ok

## Euler Math Toolbox

<lang Euler Math Toolbox> >function hailstone (n) ... $v=[n];$ repeat $if mod(n,2) then n=3*n+1;$ else n=n/2; $endif;$ v=v|n; $until n==1;$ end; $return v;$ endfunction >hailstone(27), length(%)

[ 27  82  41  124  62  31  94  47  142  71  214  107  322  161  484  242
121  364  182  91  274  137  412  206  103  310  155  466  233  700
350  175  526  263  790  395  1186  593  1780  890  445  1336  668
334  167  502  251  754  377  1132  566  283  850  425  1276  638  319
958  479  1438  719  2158  1079  3238  1619  4858  2429  7288  3644
1822  911  2734  1367  4102  2051  6154  3077  9232  4616  2308  1154
577  1732  866  433  1300  650  325  976  488  244  122  61  184  92
46  23  70  35  106  53  160  80  40  20  10  5  16  8  4  2  1 ]
112


>function hailstonelength (n) ... $v=zeros(1,n);$ v[1]=4; v[2]=2; $loop 3 to n;$ count=1; $n=#;$ repeat $if mod(n,2) then n=3*n+1;$ else n=n/2; $endif;$ if n<=cols(v) and v[n] then $v[#]=v[n]+count;$ break; $endif;$ count=count+1; $end;$ end; $return v;$ endfunction >h=hailstonelength(100000); >ex=extrema(h); ex[3], ex[4]

351
77031


</lang>

## Euphoria

<lang euphoria>function hailstone(atom n)

   sequence s
s = {n}
while n != 1 do
if remainder(n,2)=0 then
n /= 2
else
n = 3*n + 1
end if
s &= n
end while
return s


end function

function hailstone_count(atom n)

   integer count
count = 1
while n != 1 do
if remainder(n,2)=0 then
n /= 2
else
n = 3*n + 1
end if
count += 1
end while
return count


end function

sequence s s = hailstone(27) puts(1,"hailstone(27) =\n") ? s printf(1,"len = %d\n\n",length(s))

integer max,imax,count max = 0 for i = 2 to 1e5-1 do

   count = hailstone_count(i)
if count > max then
max = count
imax = i
end if


end for

printf(1,"The longest hailstone sequence under 100,000 is %d with %d elements.\n",

   {imax,max})</lang>


Output:

hailstone(27) =
{27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,
91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,
1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,
850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,
7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,
577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,
106,53,160,80,40,20,10,5,16,8,4,2,1}
len = 112

The longest hailstone sequence under 100,000 is 77031 with 351 elements.


## Excel

 This example may be incorrect. Calculates the Hailstone sequence but might not complete everything from task description. Please verify it and remove this message. If the example does not match the requirements or does not work, replace this message with Template:incorrect or fix the code yourself.
   In cell A1, place the starting number.
In cell A2 enter this formula =IF(MOD(A1,2)=0,A1/2,A1*3+1)
Drag and copy the formula down until 4, 2, 1


## Ezhil

Ezhil is a Tamil programming language, see | Wikipedia entry.

<lang src="Python"> நிரல்பாகம் hailstone ( எண் )

          பதிப்பி "=> ",எண் #hailstone seq


@( எண் == 1 ) ஆனால் பின்கொடு எண் முடி

@( (எண்%2) == 1 ) ஆனால் hailstone( 3*எண் + 1)

             இல்லை


hailstone( எண்/2 )

             முடி


முடி

எண்கள் = [5,17,19,23,37] @(எண்கள் இல் இவ்வெண்) ஒவ்வொன்றாக

  பதிப்பி "****** calculating hailstone seq for ",இவ்வெண்," *********"
hailstone( இவ்வெண் )
பதிப்பி "**********************************************"


முடி </lang>

## Factor

<lang factor>! rosetta/hailstone/hailstone.factor USING: arrays io kernel math math.ranges prettyprint sequences vectors ; IN: rosetta.hailstone

hailstone ( n -- seq )
   [ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;


<PRIVATE

main ( -- )
   27 hailstone dup dup
"The hailstone sequence from 27:" print
"  has length " write length .
"  starts with " write 4 head [ unparse ] map ", " join print
"  ends with " write 4 tail* [ unparse ] map ", " join print

   ! Maps n => { length n }, and reduces to longest Hailstone sequence.
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;


PRIVATE>

MAIN: main</lang>

Output:

## Forth

<lang forth>: hail-next ( n -- n )

 dup 1 and if 3 * 1+ else 2/ then ;

.hail ( n -- )
 begin dup . dup 1 > while hail-next repeat drop ;

hail-len ( n -- n )
 1 begin over 1 > while swap hail-next swap 1+ repeat nip ;


27 hail-len . cr 27 .hail cr

longest-hail ( max -- )
 0 0 rot 1+ 1 do    ( n length )
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;


100000 longest-hail</lang>

## Fortran

Works with: Fortran version 95 and later

<lang fortran>program Hailstone

 implicit none

 integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)

 call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ",  &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)

do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"

 deallocate(seq)



contains

subroutine hs(number, length, seqArray)

 integer, intent(in)  :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n

 n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do


end subroutine

end program</lang> Output:

Hailstone sequence for 27 has 112 elements
Sequence = 27, 82, 41, 124, ...., 8, 4, 2, 1

Longest sequence under 100000 is for 77031 with 351 elements

## Frege

Works with: Frege version 3.20.113

<lang frege>module Hailstone where

import Data.List (maximumBy)

hailstone :: Int -> [Int] hailstone 1 = [1] hailstone n | even n = n : hailstone (n div 2)

           | otherwise = n : hailstone (n * 3 + 1)


withResult :: (t -> t1) -> t -> (t1, t) withResult f x = (f x, x)

main _ = do

let h27 = hailstone 27
printStrLn $show$ length h27
let h4 = show $take 4 h27 let t4 = show$ drop (length h27 - 4) h27
printStrLn ("hailstone 27: " ++ h4 ++ " ... " ++ t4)
printStrLn $show$ maximumBy (comparing fst) $map (withResult (length . hailstone)) (1..100000)</lang>  Output: 112 hailstone 27: [27, 82, 41, 124] ... [8, 4, 2, 1] (351, 77031) runtime 4.374 wallclock seconds.  ## F# <lang fsharp>let rec hailstone n = seq {  match n with | 1 -> yield 1 | n when n % 2 = 0 -> yield n; yield! hailstone (n / 2) | n -> yield n; yield! hailstone (n * 3 + 1)  } let hailstone27 = hailstone 27 |> Array.ofSeq assert (Array.length hailstone27 = 112) assert (hailstone27.[..3] = [|27;82;41;124|]) assert (hailstone27.[108..] = [|8;4;2;1|]) let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i} printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI</lang> Output: Maximum length 351 was found for hailstone(77031) ## GAP <lang gap>CollatzSequence := function(n)  local v; v := [ n ]; while n > 1 do if IsEvenInt(n) then n := QuoInt(n, 2); else n := 3*n + 1; fi; Add(v, n); od; return v;  end; CollatzLength := function(n)  local m; m := 1; while n > 1 do if IsEvenInt(n) then n := QuoInt(n, 2); else n := 3*n + 1; fi; m := m + 1; od; return m;  end; CollatzMax := function(a, b)  local n, len, nmax, lmax; lmax := 0; for n in [a .. b] do len := CollatzLength(n); if len > lmax then nmax := n; lmax := len; fi; od; return [ nmax, lmax ];  end; CollatzSequence(27); 1. [ 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 2. 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 3. 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 4. 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 5. 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 ] CollatzLength(27); 1. 112 CollatzMax(1, 100); 1. [ 97, 119 ] CollatzMax(1, 1000); 1. [ 871, 179 ] CollatzMax(1, 10000); 1. [ 6171, 262 ] CollatzMax(1, 100000); 1. [ 77031, 351 ] CollatzMax(1, 1000000); 1. [ 837799, 525 ]</lang> ## Go <lang go>package main import "fmt" // 1st arg is the number to generate the sequence for. // 2nd arg is a slice to recycle, to reduce garbage. func hs(n int, recycle []int) []int {  s := append(recycle[:0], n) for n > 1 { if n&1 == 0 { n = n / 2 } else { n = 3*n + 1 } s = append(s, n) } return s  } func main() {  seq := hs(27, nil) fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n", len(seq), seq[0], seq[1], seq[2], seq[3], seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])   var maxN, maxLen int for n := 1; n < 100000; n++ { seq = hs(n, seq) if len(seq) > maxLen { maxN = n maxLen = len(seq) } } fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)  }</lang> Output: hs(27): 112 elements: [27 82 41 124 ... 8 4 2 1] hs(77031): 351 elements  Alternate solution (inspired both by recent news of a new proof submitted for publication and by recent chat on #rosettacode about generators.) This solution interprets the wording of the task differently, and takes the word "generate" to mean use a generator. This has the advantage of not storing the whole sequence in memory at once. Elements are generated one at a time, counted and discarded. A time optimization added for task 3 is to store the sequence lengths computed so far. Output is the same as version above. <lang go>package main import "fmt" // Task 1 implemented with a generator. Calling newHg will "create // a routine to generate the hailstone sequence for a number." func newHg(n int) func() int {  return func() (n0 int) { n0 = n if n&1 == 0 { n = n / 2 } else { n = 3*n + 1 } return }  } func main() {  // make generator for sequence starting at 27 hg := newHg(27) // save first four elements for printing later s1, s2, s3, s4 := hg(), hg(), hg(), hg() // load next four elements in variables to use as shift register. e4, e3, e2, e1 := hg(), hg(), hg(), hg() // 4+4= 8 that we've generated so far ec := 8 // until we get to 1, generate another value, shift, and increment. // note that intermediate elements--those shifted off--are not saved. for e1 > 1 { e4, e3, e2, e1 = e3, e2, e1, hg() ec++ } // Complete task 2: fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n", ec, s1, s2, s3, s4, e4, e3, e2, e1)   // Task 3: strategy is to not store sequences, but just the length // of each sequence. as soon as the sequence we're currently working on // dips into the range that we've already computed, we short-circuit // to the end by adding the that known length to whatever length // we've accumulated so far.   var nMaxLen int // variable holds n with max length encounted so far // slice holds sequence length for each n as it is computed var computedLen [1e5]int computedLen[1] = 1 for n := 2; n < 1e5; n++ { var ele, lSum int for hg := newHg(n); ; lSum++ { ele = hg() // as soon as we get an element in the range we have already // computed, we're done... if ele < n { break } } // just add the sequence length already computed from this point. lSum += computedLen[ele] // save the sequence length for this n computedLen[n] = lSum // and note if it's the maximum so far if lSum > computedLen[nMaxLen] { nMaxLen = n } } fmt.Printf("hs(%d): %d elements\n", nMaxLen, computedLen[nMaxLen])  }</lang> ## Groovy <lang groovy>def hailstone = { long start ->  def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start  }</lang> Test Code <lang groovy>def sequence = hailstone(27) assert sequence.size() == 112 assert sequence[0..3] == [27, 82, 41, 124] assert sequence[-4..-1] == [8, 4, 2, 1] def results = (1..100000).collect { [n:it, size:hailstone(it).size()] }.max { it.size } println results</lang> Output: [n:77031, size:351] ## Haskell <lang haskell>import Data.List (maximumBy) import Data.Ord (comparing) hailstone :: Int -> [Int] hailstone 1 = [1] hailstone n | even n = n : hailstone (n div 2)  | otherwise = n : hailstone (n * 3 + 1)  withResult :: (t -> t1) -> t -> (t1, t) withResult f x = (f x, x) main :: IO () main = do let h27 = hailstone 27 print$ length h27
let h4 = show $take 4 h27 let t4 = show$ drop (length h27 - 4) h27
putStrLn ("hailstone 27: " ++ h4 ++ " ... " ++ t4)
print $maximumBy (comparing fst)$ map (withResult (length . hailstone)) [1..100000]</lang>


Output:

112
hailstone 27: [27,82,41,124] ... [8,4,2,1]
(351,77031)

## HicEst

<lang HicEst>DIMENSION stones(1000)

H27 = hailstone(27) ALIAS(stones,1, first4,4) ALIAS(stones,H27-3, last4,4) WRITE(ClipBoard, Name) H27, first4, "...", last4

longest_sequence = 0 DO try = 1, 1E5

 elements = hailstone(try)
IF(elements >= longest_sequence) THEN
number = try
longest_sequence = elements
WRITE(StatusBar, Name) number, longest_sequence
ENDIF


ENDDO WRITE(ClipBoard, Name) number, longest_sequence END

FUNCTION hailstone( n )

  USE : stones

  stones(1) = n
DO i = 1, LEN(stones)
IF(stones(i) == 1) THEN
hailstone = i
RETURN
ELSEIF( MOD(stones(i),2) ) THEN
stones(i+1) = 3*stones(i) + 1
ELSE
stones(i+1) = stones(i) / 2
ENDIF
ENDDO


END</lang> H27=112; first4(1)=27; first4(2)=82; first4(3)=41; first4(4)=124; ...; last4(1)=8; last4(2)=4; last4(3)=2; last4(4)=1;
number=77031; longest_sequence=351;

## Icon and Unicon

A simple solution that generates (in the Icon sense) the sequence is: <lang icon>procedure hailstone(n)

   while n > 1 do {
suspend n
n := if n%2 = 0 then n/2 else 3*n+1
}
suspend 1


end</lang> and a test program for this solution is: <lang icon>procedure main(args)

   n := integer(!args) | 27
every writes(" ",hailstone(n))


end</lang> but this solution is computationally expensive when run repeatedly (task 3).

The following solution uses caching to improve performance on task 3 at the expense of space. <lang icon>procedure hailstone(n)

   static cache
initial {
cache := table()
cache[1] := [1]
}
/cache[n] := [n] ||| hailstone(if n%2 = 0 then n/2 else 3*n+1)
return cache[n]


end</lang>

A test program is: <lang icon>procedure main(args)

   n := integer(!args) | 27
write()


end

   count := 0
every writes(" ",right(!(sequence := hailstone(n)),5)) do
if (count +:= 1) % 15 = 0 then write()
write()
write(*sequence," value",(*sequence=1,"")|"s"," in the sequence.")


end

   maxHS := 0
every n := 1 to 100000 do {
count := *hailstone(n)
if maxHS <:= count then maxN := n
}
write(maxN," has a sequence of ",maxHS," values")


end</lang> A sample run is:

->hs
27    82    41   124    62    31    94    47   142    71   214   107   322   161   484
242   121   364   182    91   274   137   412   206   103   310   155   466   233   700
350   175   526   263   790   395  1186   593  1780   890   445  1336   668   334   167
502   251   754   377  1132   566   283   850   425  1276   638   319   958   479  1438
719  2158  1079  3238  1619  4858  2429  7288  3644  1822   911  2734  1367  4102  2051
6154  3077  9232  4616  2308  1154   577  1732   866   433  1300   650   325   976   488
244   122    61   184    92    46    23    70    35   106    53   160    80    40    20
10     5    16     8     4     2     1
112 values in the sequence.

77031 has a sequence of 351 values
->


## Io

Here is a simple, brute-force approach: <lang io> makeItHail := method(n,

 stones := list(n)
while (n != 1,
if(n isEven,
n = n / 2,
n = 3 * n + 1
)
stones append(n)
)


)

out := makeItHail(27) writeln("For the sequence beginning at 27, the number of elements generated is ", out size, ".") write("The first four elements generated are ") for(i, 0, 3,

 write(out at(i), " ")


) writeln(".")

write("The last four elements generated are ") for(i, out size - 4, out size - 1,

 write(out at(i), " ")


) writeln(".")

numOfElems := 0 nn := 3 for(x, 3, 100000,

 out = makeItHail(x)
if(out size > numOfElems,
numOfElems = out size
nn = x
)


)

writeln("For numbers less than or equal to 100,000, ", nn, " has the longest sequence of ", numOfElems, " elements.") </lang>

Output:

For the sequence beginning at 27, the number of elements generated is 112.
The first four elements generated are 27 82 41 124 .
The last four elements generated are 8 4 2 1 .
For numbers less than or equal to 100,000, 77031 has the longest sequence of 351 elements.


## Ioke

 This example may be incorrect. Calculates the Hailstone sequence but might not complete everything from task description. Please verify it and remove this message. If the example does not match the requirements or does not work, replace this message with Template:incorrect or fix the code yourself.

<lang ioke>collatz = method(n,

 n println
unless(n <= 1,
if(n even?, collatz(n / 2), collatz(n * 3 + 1)))


)</lang>

## Inform 7

This solution uses a cache to speed up the length calculation for larger numbers.

<lang inform7>Home is a room.

To decide which list of numbers is the hailstone sequence for (N - number): let result be a list of numbers; add N to result; while N is not 1: if N is even, let N be N / 2; otherwise let N be (3 * N) + 1; add N to result; decide on result.

Hailstone length cache relates various numbers to one number.

To decide which number is the hailstone sequence length for (N - number): let ON be N; let length so far be 0; while N is not 1: if N relates to a number by the hailstone length cache relation: let result be length so far plus the number to which N relates by the hailstone length cache relation; now the hailstone length cache relation relates ON to result; decide on result; if N is even, let N be N / 2; otherwise let N be (3 * N) + 1; increment length so far; let result be length so far plus 1; now the hailstone length cache relation relates ON to result; decide on result.

To say first and last (N - number) entry/entries in (L - list of values of kind K): let length be the number of entries in L; if length <= N * 2: say L; else: repeat with M running from 1 to N: if M > 1, say ", "; say entry M in L; say " ... "; repeat with M running from length - N + 1 to length: say entry M in L; if M < length, say ", ".

When play begins: let H27 be the hailstone sequence for 27; say "Hailstone sequence for 27 has [number of entries in H27] element[s]: [first and last 4 entries in H27]."; let best length be 0; let best number be 0; repeat with N running from 1 to 99999: let L be the hailstone sequence length for N; if L > best length: let best length be L; let best number be N; say "The number under 100,000 with the longest hailstone sequence is [best number] with [best length] element[s]."; end the story.</lang>

Output:

Hailstone sequence for 27 has 112 elements: 27, 82, 41, 124 ... 8, 4, 2, 1.
The number under 100,000 with the longest hailstone sequence is 77031 with 351 elements.

## J

Solution: <lang j>hailseq=: -:(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0</lang> Usage: <lang j> # hailseq 27 NB. sequence length 112

  4 _4 {."0 1 hailseq 27       NB. first & last 4 numbers in sequence


27 82 41 124

8  4  2   1
(>:@(i. >./) , >./) #@hailseq }.i. 1e5  NB. number < 100000 with max seq length & its seq length


77031 351</lang> See also the Collatz Conjecture essay on the J wiki.

## Java

Works with: Java version 1.5+

<lang java5>import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map;

class Hailstone {

 public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
}
return list;
}

public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);

long MAX = 100000;
// Simple way
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}

// More memory efficient way
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}

// Efficient for analyzing all sequences
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));

List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}


}</lang> Output:

Sequence for 27 has 112 elements: [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Method 1, number 77031 has the longest sequence, with a length of 351
Method 2, number 77031 has the longest sequence, with a length of 351
Method 3, number 77031 has the longest sequence, with a length of 351


## JavaScript

<lang javascript>function hailstone (n) {

   var seq = [n];
while (n > 1) {
n = n % 2 ? 3 * n + 1 : n / 2;
seq.push(n);
}
return seq;


}

// task 2: verify the sequence for n = 27 var h = hailstone(27), hLen = h.length; print("sequence 27 is (" + h.slice(0, 4).join(", ") + " ... "

   + h.slice(hLen - 4, hLen).join(", ") + "). length: " + hLen);


// task 3: find the longest sequence for n < 100000 for (var n, max = 0, i = 100000; --i;) {

   var seq = hailstone(i), sLen = seq.length;
if (sLen > max) {
n = i;
max = sLen;
}


} print("longest sequence: " + max + " numbers for starting point " + n);</lang> outputs

sequence 27 is (27, 82, 41, 124 ... 8, 4, 2, 1). length: 112
longest sequence: 351 numbers for starting point 77031

## Julia

<lang julia>function hailstone(n) seq = [n] while n>1 n = n % 2 == 0 ? n >> 1 : 3n + 1 push!(seq,n) end return seq end</lang>

julia> h = hailstone(27);

julia> @assert length(h) == 112

julia> @assert h[1:4] == [27,82,41,124]

julia> @assert h[end-3:end] == [8,4,2,1]

julia> maximum([(length(hailstone(i)),i) for i in 1:100000])
(351,77031)

## K

<lang k> hail: (1<){:[x!2;1+3*x;_ x%2]}\

 seqn: hail 27

 #seqn


112

 4#seqn


27 82 41 124

 -4#seqn


8 4 2 1

 {m,x@s?m:|/s:{#hail x}'x}{x@&x!2}!:1e5


351 77031</lang>

## Limbo

<lang>implement Hailstone;

include "sys.m"; sys: Sys; include "draw.m";

Hailstone: module { init: fn(ctxt: ref Draw->Context, args: list of string); };

init(nil: ref Draw->Context, nil: list of string) { sys = load Sys Sys->PATH;

seq := hailstone(big 27); l := len seq;

sys->print("hailstone(27): "); for(i := 0; i < 4; i++) { sys->print("%bd, ", hd seq); seq = tl seq; } sys->print("⋯");

while(len seq > 4) seq = tl seq;

while(seq != nil) { sys->print(", %bd", hd seq); seq = tl seq; } sys->print(" (length %d)\n", l);

max := 1; maxn := big 1; for(n := big 2; n < big 100000; n++) { cur := len hailstone(n); if(cur > max) { max = cur; maxn = n; } } sys->print("hailstone(%bd) has length %d\n", maxn, max); }

hailstone(i: big): list of big { if(i == big 1) return big 1 :: nil; if(i % big 2 == big 0) return i :: hailstone(i / big 2); return i :: hailstone((big 3 * i) + big 1); } </lang>

Output:
hailstone(27):  27, 82, 41, 124, ⋯, 8, 4, 2, 1 (length 112)
hailstone(77031) has length 351


## Lasso

<lang Lasso>[ define_tag("hailstone", -required="n", -type="integer", -copy); local("sequence") = array(#n); while(#n != 1); ((#n % 2) == 0) ? #n = (#n / 2) | #n = (#n * 3 + 1); #sequence->insert(#n); /while; return(#sequence); /define_tag;

local("result"); #result = hailstone(27); while(#result->size > 8); #result->remove(5); /while; #result->insert("...",5);

"Hailstone sequence for n = 27 -> { " + #result->join(", ") + " }";

local("longest_sequence") = 0; local("longest_index") = 0; loop(-from=1, -to=100000); local("length") = hailstone(loop_count)->size; if(#length > #longest_sequence); #longest_index = loop_count; #longest_sequence = #length; /if; /loop;

"
"; "Number with the longest sequence under 100,000: " #longest_index + ", with " + #longest_sequence + " elements."; ]</lang>

## Logo

<lang logo>to hail.next :n

 output ifelse equal? 0 modulo :n 2 [:n/2] [3*:n + 1]


end

to hail.seq :n

 if :n = 1 [output [1]]
output fput :n hail.seq hail.next :n


end

show hail.seq 27 show count hail.seq 27

to max.hail :n

 localmake "max.n 0
localmake "max.length 0
repeat :n [if greater? count hail.seq repcount  :max.length [
make "max.n repcount
make "max.length count hail.seq repcount
] ]
(print :max.n [has hailstone sequence length] :max.length)


end

max.hail 100000</lang>

## Logtalk

<lang logtalk>:- object(hailstone).

:- public(generate_sequence/2). :- mode(generate_sequence(+natural, -list(natural)), zero_or_one). :- info(generate_sequence/2, [ comment is 'Generates the Hailstone sequence that starts with its first argument. Fails if the argument is not a natural number.', argnames is ['Start', 'Sequence'] ]).

:- public(write_sequence/1). :- mode(write_sequence(+natural), zero_or_one). :- info(write_sequence/1, [ comment is 'Writes to the standard output the Hailstone sequence that starts with its argument. Fails if the argument is not a natural number.', argnames is ['Start'] ]).

:- public(sequence_length/2). :- mode(sequence_length(+natural, -natural), zero_or_one). :- info(sequence_length/2, [ comment is 'Calculates the length of the Hailstone sequence that starts with its first argument. Fails if the argument is not a natural number.', argnames is ['Start', 'Length'] ]).

:- public(longest_sequence/4). :- mode(longest_sequence(+natural, +natural, -natural, -natural), zero_or_one). :- info(longest_sequence/4, [ comment is 'Calculates the longest Hailstone sequence in the interval [Start, End]. Fails if the interval is not valid.', argnames is ['Start', 'End', 'N', 'Length'] ]).

generate_sequence(Start, Sequence) :- integer(Start), Start >= 1, sequence(Start, Sequence).

sequence(1, [1]) :- !. sequence(N, [N| Sequence]) :- ( N mod 2 =:= 0 -> M is N // 2 ; M is (3 * N) + 1 ), sequence(M, Sequence).

write_sequence(Start) :- integer(Start), Start >= 1, sequence(Start).

sequence(1) :- !, write(1), nl. sequence(N) :- write(N), write(' '), ( N mod 2 =:= 0 -> M is N // 2 ; M is (3 * N) + 1 ), sequence(M).

sequence_length(Start, Length) :- integer(Start), Start >= 1, sequence_length(Start, 1, Length).

sequence_length(1, Length, Length) :- !. sequence_length(N, Length0, Length) :- Length1 is Length0 + 1, ( N mod 2 =:= 0 -> M is N // 2 ; M is (3 * N) + 1 ), sequence_length(M, Length1, Length).

longest_sequence(Start, End, N, Length) :- integer(Start), integer(End), Start >= 1, Start =< End, longest_sequence(Start, End, 1, N, 1, Length).

longest_sequence(Current, End, N, N, Length, Length) :- Current > End, !. longest_sequence(Current, End, N0, N, Length0, Length) :- sequence_length(Current, 1, CurrentLength), Next is Current + 1, ( CurrentLength > Length0 -> longest_sequence(Next, End, Current, N, CurrentLength, Length) ; longest_sequence(Next, End, N0, N, Length0, Length) ).

- end_object.</lang>

Testing: <lang logtalk>| ?- hailstone::write_sequence(27). 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 true

| ?- hailstone::sequence_length(27, Length). Length = 112 true

| ?- hailstone::longest_sequence(1, 100000, N, Length). N = 77031, Length = 351 true</lang>

## LOLCODE

There is presently no way to query a BUKKIT for the existence of a given key, thus making memoization infeasible. This solution takes advantage of prior knowledge to run in reasonable time. <lang LOLCODE>HAI 1.3

HOW IZ I hailin YR stone

   I HAS A sequence ITZ A BUKKIT
sequence HAS A length ITZ 1
sequence HAS A SRS 0 ITZ stone

   IM IN YR stoner
BOTH SAEM stone AN 1, O RLY?
YA RLY, FOUND YR sequence
OIC

       MOD OF stone AN 2, O RLY?
YA RLY, stone R SUM OF PRODUKT OF stone AN 3 AN 1
NO WAI, stone R QUOSHUNT OF stone AN 2
OIC

       sequence HAS A SRS sequence'Z length ITZ stone
sequence'Z length R SUM OF sequence'Z length AN 1
IM OUTTA YR stoner


IF U SAY SO

I HAS A hail27 ITZ I IZ hailin YR 27 MKAY VISIBLE "hail(27) = "!

IM IN YR first4 UPPIN YR i TIL BOTH SAEM i AN 4

   VISIBLE hail27'Z SRS i " "!


IM OUTTA YR first4 VISIBLE "..."!

IM IN YR last4 UPPIN YR i TIL BOTH SAEM i AN 4

   VISIBLE " " hail27'Z SRS SUM OF 108 AN i!


IM OUTTA YR last4 VISIBLE ", length = " hail27'Z length

I HAS A max, I HAS A len ITZ 0

BTW, DIS IZ RLY NOT FAST SO WE ONLY CHEK N IN [75000, 80000) IM IN YR maxer UPPIN YR n TIL BOTH SAEM n AN 5000

   I HAS A n ITZ SUM OF n AN 75000
I HAS A seq ITZ I IZ hailin YR n MKAY
BOTH SAEM len AN SMALLR OF len AN seq'Z length, O RLY?
YA RLY, max R n, len R seq'Z length
OIC


IM OUTTA YR maxer

VISIBLE "len(hail(" max ")) = " len

KTHXBYE</lang>

Output:
hail(27) = 27 82 41 124 ... 8 4 2 1, length = 112
len(hail(77031)) = 351

## Lua

<lang lua>function hailstone( n, print_numbers )

   local n_iter = 1

   while n ~= 1 do
if print_numbers then print( n ) end
if n % 2 == 0 then
n = n / 2
else
n = 3 * n + 1
end

n_iter = n_iter + 1
end
if print_numbers then print( n ) end

return n_iter;


end

hailstone( 27, true )

max_i, max_iter = 0, 0 for i = 1, 100000 do

   num = hailstone( i, false )
if num >= max_iter then
max_i = i
max_iter = num
end


end

print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )</lang>

## Maple

Define the procedure: <lang Maple> hailstone := proc( N )

   local n := N, HS := Array([n]);
while n > 1 do
if type(n,even) then
n := n/2;
else
n := 3*n+1;
end if;
HS(numelems(HS)+1) := n;
end do;
HS;


end proc; </lang> Run the command and show the appropriate portion of the result; <lang Maple> > r := hailstone(27):

                             [ 1..112 1-D Array     ]
r := [ Data Type: anything  ]
[ Storage: rectangular ]
[ Order: Fortran_order ]


> r(1..4) ... r(-4..);

                      [27, 82, 41, 124] .. [8, 4, 2, 1]


</lang> Compute the first 100000 sequences: <lang Maple> longest := 0; n := 0; for i from 1 to 100000 do

   len := numelems(hailstone(i));
if len > longest then
longest := len;
n := i;
end if;


od: printf("The longest Hailstone sequence in the first 100k is n=%d, with %d terms\n",n,longest); </lang> Output:

The longest Hailstone sequence in the first 100k is n=77031, with 351 terms


## Mathematica

Here are three ways to generate the sequence.

### Fixed-Point formulation

<lang Mathematica>HailstoneFP[n_] := Drop[FixedPointList[If[# != 1, Which[Mod[#, 2] == 0, #/2, True, ( 3*# + 1) ], 1] &, n], -1]</lang>

### Recursive formulation using piece-wise function definitions

<lang Mathematica>HailstoneR[1] := {1} HailstoneR[n_Integer] := Prepend[HailstoneR[3 n + 1], n] /; OddQ[n] && n > 0 HailstoneR[n_Integer] := Prepend[HailstoneR[n/2], n] /; EvenQ[n] && n > 0 </lang>

### Nested function-call formulation

I use this version to do the validation: <lang Mathematica>Hailstone[n_] :=

NestWhileList[Which[Mod[#, 2] == 0, #/2, True, ( 3*# + 1) ] &, n, # != 1 &];


c27 = Hailstone@27; Print["Hailstone sequence for n = 27: [", c27;; 4, "...", c27-4 ;;, "]"] Print["Length Hailstone[27] = ", Length@c27]

longest = -1; comp = 0; Do[temp = Length@Hailstone@i;

If[comp < temp, comp = temp; longest = i],
{i, 100000}
]


Print["Longest Hailstone sequence at n = ", longest, "\nwith length = ", comp]; </lang> Output:

Hailstone sequence for n = 27: [{27,82,41,124}...{8,4,2,1}]
Length Hailstone[27] = 112
Longest Hailstone sequence at n = 77031
with length = 351


I think the fixed-point and the recursive piece-wise function formulations are more idiomatic for Mathematica

## MATLAB / Octave

<lang Matlab> function x = hailstone(n)

     % iterative definition
global VERBOSE;
x = 1;
while (1)
if VERBOSE,
printf('%i ',n);   % print element
end;

if n==1,
return;
elseif mod(n,2),
n = 3*n+1;
else
n = n/2;
end;
x = x + 1;
end;
end;</lang>


Show sequence of hailstone(27) and number of elements <lang Matlab> global VERBOSE;

 VERBOSE = 1;    % display of sequence elements turned on
N = hailstone(27);   %display sequence
printf('\n\n%i\n',N);  % </lang>


Output:

>> global VERBOSE; VERBOSE=1; hailstone(27)
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1

112


<lang Matlab> global VERBOSE;

 VERBOSE = 0;    % display of sequence elements turned off
N = 100000;
M = zeros(N,1);
for k=1:N,
M(k) = hailstone(k);   %display sequence
end;
[maxLength, n] = max(M)</lang>


Output:

maxLength =  351
n =  77031


## Maxima

<lang maxima>collatz(n) := block([L], L: [n], while n > 1 do (n: if evenp(n) then n/2 else 3*n + 1, L: endcons(n, L)), L)$collatz_length(n) := block([m], m: 1, while n > 1 do (n: if evenp(n) then n/2 else 3*n + 1, m: m + 1), m)$

collatz_max(n) := block([j, m, p], m: 0, for i from 1 thru n do

  (p: collatz_length(i), if p > m then (m: p, j: i)),


[j, m])$collatz(27); /* [27, 82, 41, ..., 4, 2, 1] */ length(%); /* 112 */ collatz_length(27); /* 112 */ collatz_max(100000); /* [77031, 351] */</lang> ## Modula-2 <lang modula2>MODULE hailst; IMPORT InOut; CONST maxCard = MAX (CARDINAL) DIV 3; TYPE action = (List, Count, Max); VAR a : CARDINAL; PROCEDURE HailStone (start : CARDINAL; type : action) : CARDINAL; VAR n, max, count : CARDINAL; BEGIN  count := 1; n := start; max := n; LOOP IF type = List THEN InOut.WriteCard (n, 12); IF count MOD 6 = 0 THEN InOut.WriteLn END END; IF n = 1 THEN EXIT END; IF ODD (n) THEN IF n < maxCard THEN n := 3 * n + 1; IF n > max THEN max := n END ELSE InOut.WriteString ("Exceeding max value for type CARDINAL at count = "); InOut.WriteCard (count, 10); InOut.WriteString (" for intermediate value "); InOut.WriteCard (n, 10); InOut.WriteString (". Aborting."); HALT END ELSE n := n DIV 2 END; INC (count) END; IF type = Max THEN RETURN max ELSE RETURN count END  END HailStone; PROCEDURE FindMax (num : CARDINAL); VAR val, maxCount, maxVal, cnt : CARDINAL; BEGIN  maxCount := 0; maxVal := 0; FOR val := 2 TO num DO cnt := HailStone (val, Count); IF cnt > maxCount THEN maxVal := val; maxCount := cnt END END; InOut.WriteString ("Longest sequence below "); InOut.WriteCard (num, 1); InOut.WriteString (" is "); InOut.WriteCard (HailStone (maxVal, Count), 1); InOut.WriteString (" for n = "); InOut.WriteCard (maxVal, 1); InOut.WriteString (" with an intermediate maximum of "); InOut.WriteCard (HailStone (maxVal, Max), 1); InOut.WriteLn  END FindMax; BEGIN  a := HailStone (27, List); InOut.WriteLn; InOut.WriteString ("Iterations total = "); InOut.WriteCard (HailStone (27, Count), 12); InOut.WriteString (" max value = "); InOut.WriteCard (HailStone (27, Max) , 12); InOut.WriteLn; FindMax (100000); InOut.WriteString ("Done."); InOut.WriteLn  END hailst.</lang> Producing: jan@Beryllium:~/modula/rosetta$ hailst
27          82          41         124          62          31
94          47         142          71         214         107
322         161         484         242         121         364
182          91         274         137         412         206
103         310         155         466         233         700
350         175         526         263         790         395
1186         593        1780         890         445        1336
668         334         167         502         251         754
377        1132         566         283         850         425
1276         638         319         958         479        1438
719        2158        1079        3238        1619        4858
2429        7288        3644        1822         911        2734
1367        4102        2051        6154        3077        9232
4616        2308        1154         577        1732         866
433        1300         650         325         976         488
244         122          61         184          92          46
23          70          35         106          53         160
80          40          20          10           5          16
8           4           2           1
Iterations total =          112 max value =         9232
Longest sequence below 100000 is 351 for n = 77031 with an intermediate maximum of 21933016
Done.

When trying the same for all values below 1 million:

Exceeding max value for type CARDINAL at n = 159487 , count = 60 and intermediate value 1699000271. Aborting.

<lang MUMPS>hailstone(n) ; If n=1 Quit n If n#2 Quit n_" "_$$hailstone(3*n+1) Quit n_" "_$$hailstone(n\2) Set x=hailstone(27) Write !,$Length(x," ")," terms in ",x,! 112 terms in 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1</lang> ## NetRexx <lang NetRexx>/* NetRexx */ options replace format comments java crossref savelog symbols binary do  start = 27 hs = hailstone(start) hsCount = hs.words say 'The number' start 'has a hailstone sequence comprising' hsCount 'elements' say ' its first four elements are:' hs.subword(1, 4) say ' and last four elements are:' hs.subword(hsCount - 3)   hsMax = 0 hsCountMax = 0 llimit = 100000 loop x_ = 1 to llimit - 1 hs = hailstone(x_) hsCount = hs.words if hsCount > hsCountMax then do hsMax = x_ hsCountMax = hsCount end end x_   say 'The number' hsMax 'has the longest hailstone sequence in the range 1 to' llimit - 1 'with a sequence length of' hsCountMax  catch ex = Exception  ex.printStackTrace  end return method hailstone(hn = long) public static returns Rexx signals IllegalArgumentException  hs = Rexx() if hn <= 0 then signal IllegalArgumentException('Invalid start point. Must be a positive integer greater than 0')   loop label n_ while hn > 1 hs = hs' 'hn if hn // 2 \= 0 then hn = hn * 3 + 1 else hn = hn % 2 end n_ hs = hs' 'hn   return hs.strip</lang>  Output The number 27 has a hailstone sequence comprising 112 elements its first four elements are: 27 82 41 124 and last four elements are: 8 4 2 1 The number 77031 has the longest hailstone sequence in the range 1 to 99999 with a sequence length of 351  ## Oberon-2 <lang oberon2>MODULE hailst; IMPORT Out; CONST maxCard = MAX (INTEGER) DIV 3;  List = 1; Count = 2; Max = 3;  VAR a : INTEGER; PROCEDURE HailStone (start, type : INTEGER) : INTEGER; VAR n, max, count : INTEGER; BEGIN  count := 1; n := start; max := n; LOOP IF type = List THEN Out.Int (n, 12); IF count MOD 6 = 0 THEN Out.Ln END END; IF n = 1 THEN EXIT END; IF ODD (n) THEN IF n < maxCard THEN n := 3 * n + 1; IF n > max THEN max := n END ELSE Out.String ("Exceeding max value for type INTEGER at: "); Out.String (" n = "); Out.Int (start, 12); Out.String (" , count = "); Out.Int (count, 12); Out.String (" and intermediate value "); Out.Int (n, 1); Out.String (". Aborting."); Out.Ln; HALT (2) END ELSE n := n DIV 2 END; INC (count) END; IF type = Max THEN RETURN max ELSE RETURN count END  END HailStone; PROCEDURE FindMax (num : INTEGER); VAR val, maxCount, maxVal, cnt : INTEGER; BEGIN  maxCount := 0; maxVal := 0; FOR val := 2 TO num DO cnt := HailStone (val, Count); IF cnt > maxCount THEN maxVal := val; maxCount := cnt END END; Out.String ("Longest sequence below "); Out.Int (num, 1); Out.String (" is "); Out.Int (HailStone (maxVal, Count), 1); Out.String (" for n = "); Out.Int (maxVal, 1); Out.String (" with an intermediate maximum of "); Out.Int (HailStone (maxVal, Max), 1); Out.Ln  END FindMax; BEGIN  a := HailStone (27, List); Out.Ln; Out.String ("Iterations total = "); Out.Int (HailStone (27, Count), 12); Out.String (" max value = "); Out.Int (HailStone (27, Max) , 12); Out.Ln; FindMax (1000000); Out.String ("Done."); Out.Ln  END hailst.</lang> Producing  27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Iterations total = 112 max value = 9232 Exceeding max value for type INTEGER at: n = 113383 , count = 120 and intermediate value 827370449. Aborting. ## OCaml <lang ocaml>#load "nums.cma";; open Num;; (* generate Hailstone sequence *) let hailstone n =  let one = Int 1 and two = Int 2 and three = Int 3 in let rec g s x = if x =/ one then x::s else g (x::s) (if mod_num x two =/ one then three */ x +/ one else x // two) in g [] (Int n)  (* compute only sequence length *) let haillen n =  let one = Int 1 and two = Int 2 and three = Int 3 in let rec g s x = if x =/ one then s+1 else g (s+1) (if mod_num x two =/ one then three */ x +/ one else x // two) in g 0 (Int n)  (* max length for starting values in 1..n *) let hailmax =  let rec g idx len = function | 0 -> (idx, len) | i -> let a = haillen i in if a > len then g i a (i-1) else g idx len (i-1) in g 0 0  hailmax 100000 ;; (* - : int * int = (77031, 351) *) List.rev_map string_of_num (hailstone 27) ;; (* - : string list = ["27"; "82"; "41"; "124"; "62"; "31"; "94"; "47"; "142"; "71"; "214"; "107"; "322"; "161"; "484"; "242"; "121"; "364"; "182"; "91"; "274"; "137"; "412"; "206"; "103"; "310"; "155"; "466"; "233"; "700"; "350"; "175"; "526"; "263"; "790"; "395"; "1186"; "593"; "1780"; "890"; "445"; "1336"; "668"; "334"; "167"; "502"; "251"; "754"; "377"; "1132"; "566"; "283"; "850"; "425"; "1276"; "638"; "319"; "958"; "479"; "1438"; "719"; "2158"; "1079"; "3238"; "1619"; "4858"; "2429"; "7288"; "3644"; "1822"; "911"; "2734"; "1367"; "4102"; "2051"; "6154"; "3077"; "9232"; "4616"; "2308"; "1154"; "577"; "1732"; "866"; "433"; "1300"; "650"; "325"; "976"; "488"; "244"; "122"; "61"; "184"; "92"; "46"; "23"; "70"; "35"; "106"; "53"; "160"; "80"; "40"; "20"; "10"; "5"; "16"; "8"; "4"; "2"; "1"] *)</lang>  ## ooRexx <lang ooRexx> sequence = hailstone(27) say "Hailstone sequence for 27 has" sequence~items "elements and is ["sequence~toString('l', ", ")"]" highestNumber = 1 highestCount = 1 loop i = 2 to 100000  sequence = hailstone(i) count = sequence~items if count > highestCount then do highestNumber = i highestCount = count end  end say "Number" highestNumber "has the longest sequence with" highestCount "elements" -- short routine to generate a hailstone sequence routine hailstone  use arg n   sequence = .array~of(n) loop while n \= 1 if n // 2 == 0 then n = n / 2 else n = 3 * n + 1 sequence~append(n) end return sequence  </lang> Output: Hailstone sequence for 27 has 112 elements and is [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 77, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 102, 051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 0, 40, 20, 10, 5, 16, 8, 4, 2, 1] Number 77031 has the longest sequence with 351 elements  ## Order To display the length, and first and last elements, of the hailstone sequence for 27, we could do this: <lang c>#include <order/interpreter.h> 1. define ORDER_PP_DEF_8hailstone ORDER_PP_FN( \ 8fn(8N, \  8cond((8equal(8N, 1), 8seq(1)) \ (8is_0(8remainder(8N, 2)), \ 8seq_push_front(8N, 8hailstone(8quotient(8N, 2)))) \ (8else, \ 8seq_push_front(8N, 8hailstone(8inc(8times(8N, 3))))))) )  ORDER_PP(  8lets((8H, 8seq_map(8to_lit, 8hailstone(27))) (8S, 8seq_size(8H)), 8print(8(h(27) - length:) 8to_lit(8S) 8comma 8space 8(starts with:) 8seq_take(4, 8H) 8comma 8space 8(ends with:) 8seq_drop(8minus(8S, 4), 8H)) ) )</lang>  Output: <lang>h(27) - length:112, starts with:(27)(82)(41)(124), ends with:(8)(4)(2)(1)</lang> Unfortunately, the C preprocessor not really being designed with large amounts of garbage collection in mind, trying to compute the hailstone sequences up to 100000 is almost guaranteed to run out of memory (and take a very, very long time). If we wanted to try, we could add this to the program, which in most languages would use relatively little memory: <lang c>#define ORDER_PP_DEF_8h_longest ORDER_PP_FN( \ 8fn(8M, 8P, \  8if(8is_0(8M), \ 8P, \ 8let((8L, 8seq_size(8hailstone(8M))), \ 8h_longest(8dec(8M), \ 8if(8greater(8L, 8tuple_at_1(8P)), \ 8pair(8M, 8L), 8P))))) )  ORDER_PP(  8let((8P, 8h_longest(8nat(1,0,0,0,0,0), 8pair(0, 0))), 8pair(8to_lit(8tuple_at_0(8P)), 8to_lit(8tuple_at_1(8P))))  )</lang> ...or even this "more elegant" version, which will run out of memory very quickly indeed (but in practice seems to work better for smaller ranges): <lang c>ORDER_PP(  8let((8P, 8seq_head( 8seq_sort(8fn(8P, 8Q, 8greater(8tuple_at_1(8P), 8tuple_at_1(8Q))), 8seq_map(8fn(8N, 8pair(8N, 8seq_size(8hailstone(8N)))), 8seq_iota(1, 8nat(1,0,0,0,0,0)))))), 8pair(8to_lit(8tuple_at_0(8P)), 8to_lit(8tuple_at_1(8P)))) )</lang>  Notice that large numbers (>100) must be entered as digit sequences with 8nat. 8to_lit converts a digit sequence back to a readable number. ## Oz <lang oz>declare  fun {HailstoneSeq N} N > 0 = true %% assert if N == 1 then [1] elseif {IsEven N} then N|{HailstoneSeq N div 2} else N|{HailstoneSeq 3*N+1} end end   HSeq27 = {HailstoneSeq 27} {Length HSeq27} = 112 {List.take HSeq27 4} = [27 82 41 124] {List.drop HSeq27 108} = [8 4 2 1]   fun {MaxBy2nd A=A1#A2 B=B1#B2} if B2 > A2 then B else A end end   Pairs = {Map {List.number 1 99999 1} fun {$ I} I#{Length {HailstoneSeq I}} end}

 MaxI#MaxLen = {List.foldL Pairs MaxBy2nd 0#0}
{System.showInfo
"Maximum length "#MaxLen#" was found for hailstone("#MaxI#")"}</lang>


Output:

Maximum length 351 was found for hailstone(77031)


## PARI/GP

<lang parigp>show(n)={

 my(t=1);
while(n>1,
print1(n",");
n=if(n%2,
3*n+1
,
n/2
);
t++
);
print(1);
t


};

len(n)={

 my(t=1);
while(n>1,
if(n%2,
t+=2;
n+=(n>>1)+1
,
t++;
n>>=1
)
);
t


};

show(27) r=0;for(n=1,1e5,t=len(n);if(t>r,r=t;ra=n));print(ra"\t"r)</lang> Output:

27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,4
12,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,133
6,668,334,167,502,251,754,377,1132,566,283,850,425,1276,638,319,958,479,1438,719
,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,
9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,2
3,70,35,106,53,160,80,40,20,10,5,16,8,4,2,1

and

77031   351

See Delphi

## Perl

### Straightforward

<lang Perl>#!/usr/bin/perl

use warnings; use strict;

my @h = hailstone(27); print "Length of hailstone(27) = " . scalar @h . "\n"; print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";

my ($max,$n) = (0, 0); for my $x (1 .. 99_999) {  @h = hailstone($x);
if (scalar @h > $max) { ($max, $n) = (scalar @h,$x);
}


}

print "Max length $max was found for hailstone($n) for numbers < 100_000\n";

sub hailstone {

   my ($n) = @_;   my @sequence = ($n);

   while ($n > 1) { if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n =$n * 3 + 1;
}

       push @sequence, $n; }   return @sequence;  }</lang> Output: Length of hailstone(27) = 112 [27, 82, 41, 124, ..., 8, 4, 2, 1] Max length 351 was found for hailstone(77031) for numbers < 100_000  ### Compact A more compact version: <lang Perl>#!/usr/bin/perl use strict; sub hailstone {  @_ = local$_ = shift;
push @_, $_ =$_ % 2 ? 3 * $_ + 1 :$_ / 2 while $_ > 1; @_;  } my @h = hailstone($_ = 27); print "$_: @h[0 .. 3] ... @h[-4 .. -1] (".@h.")\n"; @h = (); for (1 .. 99_999) { @h = ($_, $h[2]) if ($h[2] = hailstone($_)) >$h[1] } printf "%d: (%d)\n", @h;</lang>

The same approach as in the compact version above, obfuscated: <lang Perl>sub _{my$_=$_[];push@_,$_&1?$_+=$_++<<1:($_>>=1)while$_^1;@_} @_=_($_=031^2);print "$_: @_[0..3] ... @_[-4..-1] (".@_.")\n";$_[1]<($_[2]=_($_))and@_=($_,$_[2])for 1..1e5-1;printf "%d: (%d)\n", @_;</lang>

Output in either case:

27: 27 82 41 124 ... 8 4 2 1 (112)
77031: (351)


## Perl 6

<lang perl6>sub hailstone($n) {$n, { $_ %% 2 ??$_ div 2 !! $_ * 3 + 1 } ... 1 } my @h = hailstone(27); say "Length of hailstone(27) = {+@h}"; say ~@h; my$m max= +hailstone($_) =>$_ for 1..99_999; say "Max length $m.key() was found for hailstone($m.value()) for numbers < 100_000";</lang>

## PHP

<lang php>function hailstone($n,$seq=array()){ $sequence =$seq; $sequence[] =$n; if($n == 1){ return$sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n,$sequence); } }

$result = hailstone(27); echo count($result) . ' Elements.
'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . ' ';$maxResult = array(0);

for($i=1;$i<=100000;$i++){$result = count(hailstone($i)); if($result > max($maxResult)){$maxResult = array($i=>$result); } } foreach($maxResult as$key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' .$key . ' with length of ' . $val; }</lang> 112 Elements. Starting with : 27,82,41,124 and ending with : 8,4,2,1 Number < 100000 with longest Hailstone seq.: 77031 with length of 351  ## PicoLisp <lang PicoLisp>(de hailstone (N)  (make (until (= 1 (link N)) (setq N (if (bit? 1 N) (inc (* N 3)) (/ N 2) ) ) ) ) )  (let L (hailstone 27)  (println 27 (length L) (head 4 L) '- (tail 4 L)) )  (let N (maxi '((N) (length (hailstone N))) (range 1 100000))  (println N (length (hailstone N))) )</lang>  Output: 27 112 (27 82 41 124) - (8 4 2 1) 77031 351 ## Pike <lang Pike>#!/usr/bin/env pike int next(int n) {  if (n==1) return 0; if (n%2) return 3*n+1; else return n/2;  } array(int) hailstone(int n) {  array seq = ({ n }); while (n=next(n)) seq += ({ n }); return seq;  } void main() {  array(int) two = hailstone(27); if (equal(two[0..3], ({ 27, 82, 41, 124 })) && equal(two[<3..], ({ 8,4,2,1 }))) write("sizeof(({ %{%d, %}, ... %{%d, %} }) == %d\n", two[0..3], two[<3..], sizeof(two));   mapping longest = ([ "length":0, "start":0 ]);   foreach(allocate(100000); int start; ) { int length = sizeof(hailstone(start)); if (length > longest->length) { longest->length = length; longest->start = start; } } write("longest sequence starting at %d has %d elements\n", longest->start, longest->length);  }</lang> Output: sizeof(({ 27, 82, 41, 124, , ... 8, 4, 2, 1, }) == 112 longest sequence starting at 77031 has 351 elements  ## PL/I <lang PL/I>test: proc options (main);  declare (longest, n) fixed (15); declare flag bit (1); declare (i, value) fixed (15);   /* Task 1: */ flag = '1'b; put skip list ('The sequence for 27 is'); i = hailstones(27);   /* Task 2: */ flag = '0'b; longest = 0; do i = 1 to 99999; if longest < hailstones(i) then do; longest = hailstones(i); value = i; end; end; put skip edit (value, ' has the longest sequence of ', longest) (a);  hailstones: procedure (n) returns ( fixed (15));  declare n fixed (15) nonassignable; declare (m, p) fixed (15);   m = n; p = 1; if flag then put skip list (m); do p = 1 by 1 while (m > 1); if iand(m, 1) = 0 then m = m/2; else m = 3*m + 1; if flag then put skip list (m); end; if flag then put skip list ('The hailstone sequence has length' || p); return (p);  end hailstones; end test;</lang> Output: The sequence for 27 is 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 The hailstone sequence has length 112 77031 has the longest sequence of 351  ## Powershell  This example does not show the output mentioned in the task description on this page (or a page linked to from here). Please ensure that it meets all task requirements and remove this message. Note that phrases in task descriptions such as "print and display" and "print and show" for example, indicate that (reasonable length) output be a part of a language's solution. Works with: Powershell version 3.0 <lang Powershell> function Get-HailStone ($n) {

  switch($n) { 1 {$n;return}
{$n % 2 -eq 0}{$n; return Get-Hailstone ($n =$n / 2)}
{$n % 2 -ne 0}{$n; return Get-Hailstone ($n = ($n * 3) +1)}

}


}

function Get-HailStoneBelowLimit($UpperLimit) {  begin {$Counts = @()}

   process
{
for ($i = 1;$i -lt $UpperLimit;$i++)
{
$Object = [pscustomobject]@{ 'Number' =$i
'Count' = (Get-HailStone $i).count }$Counts += $Object } }   end {$Counts}


} Get-HailStoneBelowLimit 100000 |

   Sort-Object count -descending |
Select-Object number -first 1


</lang>

## Prolog

1. Create a routine to generate the hailstone sequence for a number. <lang prolog>hailstone(1,[1]) :- !. hailstone(N,[N|S]) :- 0 is N mod 2, N1 is N / 2, hailstone(N1,S). hailstone(N,[N|S]) :- 1 is N mod 2, N1 is (3 * N) + 1, hailstone(N1, S).</lang>

2. Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1.

The following query performs the test. <lang prolog>hailstone(27,X), length(X,112), append([27, 82, 41, 124], _, X), append(_, [8, 4, 2, 1], X).</lang>

3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequences length. <lang prolog>longestHailstoneSequence(M, Seq, Len) :- longesthailstone(M, 1, 1, Seq, Len). longesthailstone(1, Cn, Cl, Mn, Ml):- Mn = Cn, Ml = Cl. longesthailstone(N, _, Cl, Mn, Ml) :- hailstone(N, X),

                                      length(X, L),
Cl < L,
N1 is N-1,
longesthailstone(N1, N, L, Mn, Ml).


longesthailstone(N, Cn, Cl, Mn, Ml) :- N1 is N-1,

                                      longesthailstone(N1, Cn, Cl, Mn, Ml).</lang>


run this query. <lang prolog>longestHailstoneSequence(100000, Seq, Len).</lang> to get the following result

Seq = 77031,
Len = 351


### Constraint Handling Rules

CHR is a programming language created by Professor Thom Frühwirth.
Works with SWI-Prolog and module chr written by Tom Schrijvers and Jan Wielemaker

<lang Prolog>:- use_module(library(chr)).

- chr_option(debug, off).
- chr_option(optimize, full).
- chr_constraint collatz/2, hailstone/1, clean/0.

% to remove all constraints hailstone/1 after computation clean @ clean \ hailstone(_) <=> true. clean @ clean <=> true.

% compute Collatz number init @ collatz(1,X) <=> X = 1 | true. collatz @ collatz(N, C) <=> (N mod 2 =:= 0 -> C is N / 2; C is 3 * N + 1).

% Hailstone loop hailstone(1) ==> true. hailstone(N) ==> N \= 1 | collatz(N, H), hailstone(H).</lang>

Code for task one : <lang Prolog>task1 :- hailstone(27), findall(X, find_chr_constraint(hailstone(X)), L), clean, % check the requirements ( (length(L, 112), append([27, 82, 41, 124 | _], [8,4,2,1], L)) -> writeln(ok); writeln(ko)).</lang> Output :

 ?- task1.
ok
true.

Code for task two : <lang Prolog>longest_sequence :- seq(2, 100000, 1-[1], Len-V), format('For ~w sequence has ~w len ! ~n', [V, Len]).

% walk through 2 to 100000 and compute the length of the sequences % memorize the longest seq(N, Max, Len-V, Len-V) :- N is Max + 1, !. seq(N, Max, CLen - CV, FLen - FV) :- len_seq(N, Len - N), ( Len > CLen -> Len1 = Len, V1 = [N] ; Len = CLen -> Len1 = Len, V1 = [N | CV] ; Len1 = CLen, V1 = CV), N1 is N+1, seq(N1, Max, Len1 - V1, FLen - FV).

% compute the len of the Hailstone sequence for a number len_seq(N, Len - N) :- hailstone(N), findall(hailstone(X), find_chr_constraint(hailstone(X)), L), length(L, Len), clean.</lang> Output :

 ?- longest_sequence.
For [77031] sequence has 351 len !
true.


## Pure

<lang pure>// 1. Create a routine to generate the hailstone sequence for a number. type odd x::int = x mod 2; type even x::int = ~odd x; odd x = typep odd x; even x = typep even x;

hailstone 1 = [1]; hailstone n::even = n:hailstone (n div 2); hailstone n::odd = n:hailstone (3*n + 1);

// 2. Use the routine to show that the hailstone sequence for the number 27 // has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1 n = 27; hs = hailstone n; l = # hs; using system;

printf

   ("the hailstone sequence for the number %d has %d elements " +
"starting with %s and ending with %s\n")
(n, l, __str__ (hs!!(0..3)), __str__ ( hs!!((l-4)..l)));


// 3. Show the number less than 100,000 which has the longest hailstone // sequence together with that sequences length. printf ("the number under 100,000 with the longest sequence is %d " +

       "with a sequence length of %d\n")
(foldr (\ (a,b) (c,d) -> if (b > d) then (a,b) else (c,d))
(0,0)
(map (\ x -> (x, # hailstone x)) (1..100000)));</lang>


Output:

the hailstone sequence for the number 27 has 112 elements starting with [27,82,41,124] and ending with [8,4,2,1]
the number under 100,000 with the longest sequence is 77031 with a sequence length of 351


## Python

<lang python>def hailstone(n):

   seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq


if __name__ == '__main__':

   h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))</lang>


Sample Output

Maximum length 351 was found for hailstone(77031) for numbers <100,000

## R

<lang r>### PART 1: makeHailstone <- function(n){

 hseq <- n
while (hseq[length(hseq)] > 1){
current.value <- hseq[length(hseq)]
if (current.value %% 2 == 0){
next.value <- current.value / 2
} else {
next.value <- (3 * current.value) + 1
}
hseq <- append(hseq, next.value)
}
return(list(hseq=hseq, seq.length=length(hseq)))


}

1. PART 2:

twenty.seven <- makeHailstone(27) twenty.seven$hseq twenty.seven$seq.length

1. PART 3:

max.length <- 0; lower.bound <- 1; upper.bound <- 100000

for (index in lower.bound:upper.bound){

 current.hseq <- makeHailstone(index)
if (current.hseq$seq.length > max.length){ max.length <- current.hseq$seq.length
max.index  <- index
}


}

cat("Between ", lower.bound, " and ", upper.bound, ", the input of ",

   max.index, " gives the longest hailstone sequence, which has length ",
max.length, ". \n", sep="")</lang>


Output: <lang r>> twenty.seven$hseq  [1] 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 [16] 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 [31] 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 [46] 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 [61] 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 [76] 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 [91] 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20  [106] 10 5 16 8 4 2 1 > twenty.seven$seq.length [1] 112

Between 1 and 1e+05, the input of 77031 gives the longest hailstone sequence, which has length 351.</lang>

## Racket

<lang Racket>

1. lang racket

(define hailstone

 (let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))


(define h27 (hailstone 27)) (printf "h(27) = ~s, ~s items\n"

       (,@(take h27 4) ... ,@(take-right h27 4))
(length h27))


(define N 100000) (define longest

 (for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))


(printf "for x<=~s, ~s has the longest sequence with ~s items\n"

       N (car longest) (cdr longest))


</lang>

Output:

h(27) = (27 82 41 124 ... 8 4 2 1), 112 items
for x<=100000, 77031 has the longest sequence with 351 items


## REXX

<lang REXX>/*REXX pgm tests a number and a range for hailstone (Collatz) sequences.*/ parse arg x y . /*get optional arguments from CL.*/ if x== | x==',' then x=27 /*Any 1st argument? Use default.*/ if y== | y==',' then y=100000-1 /*Any 2nd argument? Use default.*/ numeric digits 20; @.=0 /*handle big #s; initialize array*/ $=hailstone(x) /*═══════════════════task 1═════════════════════════*/ say x ' has a hailstone sequence of ' words($) say ' and starts with: ' subword($, 1, 4) " ∙∙∙" say ' and ends with: ∙∙∙' subword($, max(1, words($)-3)) say if y==0 then exit /*═══════════════════task 2═════════════════════════*/ w=0; do j=1 for y /*traipse through the numbers. */  call hailstone j /*compute the hailstone sequence.*/ if #hs<=w then iterate /*Not big 'nuff? Then keep going.*/ bigJ=j; w=#hs /*remember what # has biggest HS.*/ end /*j*/  say '(between 1──►'y") " bigJ ' has the longest hailstone sequence:' w exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────HAILSTONE subroutine────────────────*/ hailstone: procedure expose #hs; parse arg n 1 s /*N & S set to 1st arg*/  do #hs=1 while n\==1 /*loop while N isn't unity. */ if n//2 then n=n*3+1 /*if N is odd, calc: 3*n +1 */ else n=n%2 /* " " " even, perform fast ÷ */ s=s n /*build a sequence list (append).*/ end /*#hs*/  return s</lang> output 27 has a hailstone sequence of 112 and starts with: 27 82 41 124 ∙∙∙ and ends with: ∙∙∙ 8 4 2 1 (between 1──►99999) 77031 has the longest hailstone sequence: 351  ### optimized This optimized version is about seven times faster than the unoptimized version. <lang REXX>/*REXX pgm tests a number and a range for hailstone (Collatz) sequences.*/ parse arg x y . /*get optional arguments from CL.*/ if x== | x==',' then x=27 /*Any 1st argument? Use default.*/ if y== | y==',' then y=99999 /*Any 2nd argument? Use default.*/ numeric digits 20; @.=0 /*handle big #s; initialize array*/$=hailstone(x) /*═══════════════════task 1═════════════════════════*/ say x ' has a hailstone sequence of ' words($) say ' and starts with: ' subword($, 1, 4) " ∙∙∙" say ' and ends with: ∙∙∙' subword($, max(1, words($)-3)) say if y==0 then exit /*═══════════════════task 2═════════════════════════*/ w=0; do j=1 for y /*loop through all numbers <100k.*/

            $=hailstone(j) /*compute the hailstone sequence.*/ #hs=words($)              /*find the length of the sequence*/
if #hs<=w  then iterate   /*Not big 'nuff? Then keep going.*/
bigJ=j;    w=#hs          /*remember what # has biggest HS.*/
end   /*j*/


say '(between 1──►'y") " bigJ ' has the longest hailstone sequence:' w exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────HAILSTONE subroutine────────────────*/ hailstone: procedure expose @.; parse arg n 1 s 1 o /*N,S,O = 1st arg.*/ @.1= /*special case for unity. */

          do  forever                 /*loop while  N  isn't  unity.   */
if @.n\==0  then do; s=s @.n; leave; end  /*been here before?*/
if n//2  then n=n*3+1       /*if  N  is odd,  calc:   3*n +1 */
else n=n%2         /* "  "   " even, perform fast ÷ */
s=s n                       /*build a sequence list (append).*/
end   /*forever*/


@.o=subword(s,2) /*memoization for this hailstone.*/ return s</lang> output is the same as the non-optimized version.

## Ruby

This program uses new methods (Integer#even? and Enumerable#max_by) from Ruby 1.8.7.

Works with: Ruby version 1.8.7

<lang ruby>def hailstone n

 seq = [n]
until n == 1
n = (n.even?) ? (n / 2) : (3 * n + 1)
seq << n
end
seq


end

1. for n = 27, show sequence length and first and last 4 elements

hs27 = hailstone 27 p [hs27.length, hs27[0..3], hs27[-4..-1]]

1. find the longest sequence among n less than 100,000

n, len = (1 ... 100_000) .collect {|n|

 [n, hailstone(n).length]} .max_by {|n, len| len}


puts "#{n} has a hailstone sequence length of #{len}" puts "the largest number in that sequence is #{hailstone(n).max}"</lang> Output:

[112, [27, 82, 41, 124], [8, 4, 2, 1]]
77031 has a hailstone sequence length of 351
the largest number in that sequence is 21933016

### With shared structure

This version builds some linked lists with shared structure. Hailstone::ListNode is an adaptation of ListNode from Singly-linked list/Element definition#Ruby. When two sequences contain the same value, those two lists share a tail. This avoids recomputing the end of the sequence.

Works with: Ruby version 1.8.7

<lang ruby>module Hailstone

 class ListNode
include Enumerable

   def initialize(value, size, succ=nil)
@value, @size, @succ = value, size, succ
end

   def each
node = self
while node


yield node.value node = node.succ

     end
end
end

 @@sequence = {1 => ListNode.new(1, 1)}

 module_function

 def sequence(n)
unless @@sequence[n]
ary = []
m = n
until succ = @@sequence[m]
ary << m
m = (m.even?) ? (m / 2) : (3 * m + 1)
end
ary.reverse_each do |m|
@@sequence[m] = succ = ListNode.new(m, succ.size + 1, succ)
end
end
@@sequence[n]
end


end

1. for n = 27, show sequence length and first and last 4 elements

hs27 = Hailstone.sequence(27).to_a p [hs27.length, hs27[0..3], hs27[-4..-1]]

1. find the longest sequence among n less than 100,000

hs_big = (1 ... 100_000) .collect {|n|

 Hailstone.sequence n}.max_by {|hs| hs.size}


puts "#{hs_big.first} has a hailstone sequence length of #{hs_big.size}" puts "the largest number in that sequence is #{hs_big.max}"</lang>

## Scala

Library: Scala
Works with: Scala version 2.10.2

<lang Scala>object HailstoneSequence extends App {

 def hailstone(n: Int): Stream[Int] =
n #:: (if (n == 1) Stream.empty else hailstone(if (n % 2 == 0) n / 2 else n * 3 + 1))

 val nr = args.headOption.map(_.toInt).getOrElse(27)
val collatz = hailstone(nr)
println(s"Use the routine to show that the hailstone sequence for the number: $nr.") println(collatz.toList) println(s"It has${collatz.length} elements.")
println
println(
"Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length.")
val (n, len) = (1 until 100000).map(n => (n, hailstone(n).length)).maxBy(_._2)
println(s"Longest hailstone sequence length= $len occurring with number$n.")


}</lang>

Output:
Use the routine to show that the hailstone sequence for the number: 27.
List(27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1)
It has 112 elements.

Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length.
Longest hailstone sequence length= 351 occurring with number 77031.

## Scheme

<lang scheme>(define (collatz n) (if (= n 1) '(1) (cons n (collatz (if (even? n) (/ n 2) (+ 1 (* 3 n)))))))

(define (collatz-length n) (let aux ((n n) (r 1)) (if (= n 1) r (aux (if (even? n) (/ n 2) (+ 1 (* 3 n))) (+ r 1)))))

(define (collatz-max a b) (let aux ((i a) (j 0) (k 0)) (if (> i b) (list j k) (let ((h (collatz-length i))) (if (> h k) (aux (+ i 1) i h) (aux (+ i 1) j k))))))

(collatz 27)

(27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182
91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395
1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283
850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429
7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154
577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35
106 53 160 80 40 20 10 5 16 8 4 2 1)

(collatz-length 27)

112

(collatz-max 1 100000)

(77031 351)</lang>

<lang seed7>$include "seed7_05.s7i"; const func array integer: hailstone (in var integer: n) is func  result var array integer: hSequence is 0 times 0; begin while n <> 1 do hSequence &:= n; if odd(n) then n := 3 * n + 1; else n := n div 2; end if; end while; hSequence &:= n; end func;  const func integer: hailstoneSequenceLength (in var integer: n) is func  result var integer: sequenceLength is 1; begin while n <> 1 do incr(sequenceLength); if odd(n) then n := 3 * n + 1; else n := n div 2; end if; end while; end func;  const proc: main is func  local var integer: number is 0; var integer: length is 0; var integer: maxLength is 0; var integer: numberOfMaxLength is 0; var array integer: h27 is 0 times 0; begin for number range 1 to 99999 do length := hailstoneSequenceLength(number); if length > maxLength then maxLength := length; numberOfMaxLength := number; end if; end for; h27 := hailstone(27); writeln("hailstone(27):"); for number range 1 to 4 do write(h27[number] <& ", "); end for; write("...."); for number range length(h27) -3 to length(h27) do write(", " <& h27[number]); end for; writeln(" length=" <& length(h27)); writeln("Maximum length " <& maxLength <& " at number=" <& numberOfMaxLength); end func;</lang>  Output: hailstone(27): 27, 82, 41, 124, ...., 8, 4, 2, 1 length=112 Maximum length 351 at number=77031  ## Smalltalk Works with: GNU Smalltalk <lang smalltalk>Object subclass: Sequences [  Sequences class >> hailstone: n [ |seq| seq := OrderedCollection new. seq add: n. (n = 1) ifTrue: [ ^seq ]. (n even) ifTrue: [ seq addAll: (Sequences hailstone: (n / 2)) ] ifFalse: [ seq addAll: (Sequences hailstone: ( (3*n) + 1 ) ) ]. ^seq. ]   Sequences class >> hailstoneCount: n [ ^ (Sequences hailstoneCount: n num: 1) ]   "this 'version' avoids storing the sequence, it just counts its length - no memoization anyway" Sequences class >> hailstoneCount: n num: m [ (n = 1) ifTrue: [ ^m ]. (n even) ifTrue: [ ^ Sequences hailstoneCount: (n / 2) num: (m + 1) ] ifFalse: [ ^ Sequences hailstoneCount: ( (3*n) + 1) num: (m + 1) ]. ]  ].</lang> <lang smalltalk>|r| r := Sequences hailstone: 27. "hailstone 'from' 27" (r size) displayNl. "its length" "test 'head' ..." ( (r first: 4) = #( 27 82 41 124 ) asOrderedCollection ) displayNl. "... and 'tail'" ( ( (r last: 4 ) ) = #( 8 4 2 1 ) asOrderedCollection) displayNl. |longest| longest := OrderedCollection from: #( 1 1 ). 2 to: 100000 do: [ :c |  |l| l := Sequences hailstoneCount: c. (l > (longest at: 2) ) ifTrue: [ longest replaceFrom: 1 to: 2 with: { c . l } ].  ]. ('Sequence generator %1, sequence length %2' % { (longest at: 1) . (longest at: 2) })  displayNl.</lang>  ## SNUSP  /@+@@@+++# 27 | halve odd /===count<<\ /recurse\ #/?\ zero$>@/===!/===-?\==>?!/-<+++\    \!/=!\@\>?!\@/<@\.!\-/
/+<-\!>\?-<+>/++++<\?>+++/*6+4  |    |   \=/  \=itoa=@@@+@+++++#
\=>?/<=!=\   |                  |    !     /+ !/+ !/+ !/+   \    mod10
|//!==/========\         |    /<+> -\!?-\!?-\!?-\!?-\!
/=>?\<=/\<+>!\->+>+<<?/>>=print@/\ln \?!\-?!\-?!\-?!\-?!\-?/\    div10
\+<-/!<     ----------.++++++++++/      #  +/! +/! +/! +/! +/


## Tcl

The core looping structure is an example of an n-plus-one-half loop, except the loop is officially infinite here. <lang tcl>proc hailstone n {

   while 1 {


lappend seq $n if {$n == 1} {return $seq} set n [expr {$n & 1 ? $n*3+1 :$n/2}]

   }


}

set h27 [hailstone 27] puts "h27 len=[llength $h27]" puts "head4 = [lrange$h27 0 3]" puts "tail4 = [lrange $h27 end-3 end]" set maxlen [set max 0] for {set i 1} {$i<100000} {incr i} {

   set l [llength [hailstone $i]] if {$l>$maxlen} {set maxlen$l;set max $i}  } puts "max is$max, with length $maxlen"</lang> Output: h27 len=112 head4 = 27 82 41 124 tail4 = 8 4 2 1 max is 77031, with length 351  ## TXR <lang txr>@(do (defun hailstone (n)  (cons n (gen (not (eq n 1)) (set n (if (evenp n) (trunc n 2) (+ (* 3 n) 1)))))))  @(next :list @(mapcar* (fun tostring) (hailstone 27))) 27 82 41 124 @(skip) 8 4 2 1 @(eof) @(do (let ((max 0) maxi)  (each* ((i (range 1 99999)) (h (mapcar* (fun hailstone) i)) (len (mapcar* (fun length) h))) (if (> len max) (progn (set max len) (set maxi i)))) (format t "longest sequence is ~a for n = ~a\n" max maxi)))</lang>  $ txr -l hailstone.txr
longest sequence is 351 for n = 77031

## UNIX Shell

The best way is to use a shell with built-in arrays and arithmetic, such as Bash.

Works with: Bash

<lang bash>#!/bin/bash

1. seq is the array genereated by hailstone
2. index is used for seq

declare -a seq declare -i index

1. Create a routine to generate the hailstone sequence for a number

hailstone () {

 unset seq index
seq[$((index++))]=$((n=$1)) while [$n -ne 1 ]; do
[ $((n % 2)) -eq 1 ] && ((n=n*3+1)) || ((n=n/2)) seq[$((index++))]=$n done  } 1. Use the routine to show that the hailstone sequence for the number 27 2. has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1 i=27 hailstone$i echo "$i:${#seq[@]}" echo "${seq[@]:0:4} ...${seq[@]:(-4):4}"

1. Show the number less than 100,000 which has the longest hailstone
2. sequence together with that sequences length.
3. (But don't show the actual sequence)!

max=0 maxlen=0 for ((i=1;i<100000;i++)); do

 hailstone $i if [$((len=${#seq[@]})) -gt$maxlen ]; then
max=$i maxlen=$len
fi


done

echo "${max} has a hailstone sequence length of${maxlen}"</lang>

output

27: 112
27 82 41 124 ... 8 4 2 1
77031 has a hailstone sequence of 351

### Bourne Shell

This script follows tradition for the Bourne Shell; its hailstone() function writes the sequence to standard output, so the shell can capture or pipe this output. This script is very slow because it forks many processes. Each command substitution forks a subshell, and each expr(1) command forks a process.

• Therefore, this script only examines sequences from 1 to 1000, not 100000. A fast computer might run this script in 45 to 120 seconds, using most time to run system calls in kernel mode. If the script went to 100000, it would need several hours.
Works with: Bourne Shell

<lang bash># Outputs a hailstone sequence from $1, with one element per line. 1. Clobbers$n.

hailstone() { n=expr "$1" + 0 eval "test$? -lt 2 || return $?" #$n must be integer.

echo $n while test$n -ne 1; do if expr $n % 2 >/dev/null; then n=expr 3 \*$n + 1 else n=expr $n / 2 fi echo$n done }

set -- hailstone 27 echo "Hailstone sequence from 27 has $# elements:" first="$1, $2,$3, $4" shift expr$# - 4 echo " $first, ...,$1, $2,$3, $4" i=1 max=0 maxlen=0 while test$i -lt 1000; do len=hailstone $i | wc -l | tr -d ' ' test$len -gt $maxlen && max=$i maxlen=$len i=expr$i + 1 done echo "Hailstone sequence from $max has$maxlen elements."</lang>

### C Shell

This script is several times faster than the previous Bourne Shell script, because it uses C Shell expressions, not the expr(1) command. This script is slow, but it can reach 100000, and a fast computer might run it in less than 15 minutes.

<lang csh># Outputs a hailstone sequence from !:1, with one element per line.

1. Clobbers $n. alias hailstone eval \@ n = \!:1:q \\ echo$n \\ while ( $n != 1 ) \\ if ($n % 2 ) then \\ @ n = 3 * $n + 1 \\ else \\ @ n /= 2 \\ endif \\ echo$n \\ end \\ '\'

set sequence=(hailstone 27) echo "Hailstone sequence from 27 has $#sequence elements:" @ i =$#sequence - 3 echo " $sequence[1-4] ...$sequence[$i-]" 1. hailstone-length$i
2. acts like
3. @ len = hailstone $i | wc -l | tr -d ' ' 4. but without forking any subshells. alias hailstone-length eval \@ n = \!:1:q \\ @ len = 1 \\ while ($n != 1 ) \\ if ( $n % 2 ) then \\ @ n = 3 *$n + 1 \\ else \\ @ n /= 2 \\ endif \\ @ len += 1 \\ end \\ '\'

@ i = 1 @ max = 0 @ maxlen = 0 while ($i < 100000) # XXX - I must run hailstone-length in a subshell, because my # C Shell has a bug when it runs hailstone-length inside this # while ($i < 1000) loop: it forgets about this loop, and # reports an error <<end: Not in while/foreach.>> @ len = hailstone-length $i; echo$len if ($len >$maxlen) then @ max = $i @ maxlen =$len endif @ i += 1 end echo "Hailstone sequence from $max has$maxlen elements."</lang>

$csh -f hailstone.csh Hailstone sequence from 27 has 112 elements: 27 82 41 124 ... 8 4 2 1 Hailstone sequence from 77031 has 351 elements. ## Ursala <lang Ursala>#import std 1. import nat hail = @iNC ~&h~=1->x ^C\~& @h ~&h?\~&t successor+ sum@iNiCX 1. show+ main = <  ^T(@ixX take/$4; %nLP~~lrxPX; ^|TL/~& :/'...',' has length '--@h+ %nP+ length) hail 27,
^|TL(~&,:/' has sequence length ') %nP~~ nleq$^&r ^(~&,length+ hail)* nrange/1 100000></lang>  The hail function computes the sequence as follows. • Given a number as an argument, @iNC makes a list containing only that number before passing it to the rest of the function. The i in the expression stands for the identity function, N for the constant null function, and C for the cons operator. • The iteration combinator (->) is used with a predicate of ~&h~=l which tests the condition that the head (~&h) of its argument is not equal (~=) to 1. Iteration of the rest of the function continues while this predicate holds. • The x suffix says to return the reversal of the list after the iteration finishes. • The function being iterated builds a list using the cons operator (^C) with the identity function (~&) of the argument for the tail, and the result of the rest of the line for the head. • The @h operator says that the function following will be applied to the head of the list. • The conditional operator (?) has the head function (~&h) as its predicate, which tests whether the head of its argument is non-null. • In this case, the argument is a natural number, but naturals are represented as lists of booleans, so taking the head of a number is the same as testing the least significant bit. • If the condition is not met, the number has a 0 least significant bit, and therefore is even. In this case, the conditional predicate calls for taking its tail (~&t), effectively dividing it by 2 using a bit shift. • If the condition is met, the number is odd, so the rest of the function computes the successor of the number multiplied by three. • Rather than multiplying the hard way, the function sum@iNiCX computes the sum of the pair (X) of numbers given by the identity function (i) of the argument, and the doubling of the argument (NiC), also obtained by a bit shift, with a zero bit (N) consed (C) with the identity (i). Most of the main expression pertains to less interesting printing and formatting, but the part that searches for the longest sequence in the range is nleq$^&r ^(~&,length+ hail)* nrange/1 100000.

• The expression nrange/1 100000 evaluates to the list of the first 100000 positive integers.
• The map operator (*) causes a list to be made of the results of its operand applied to each number.
• The operand to the map operator, applied to an individual number in the list, constructs a pair (^) with the identity function (~&) of the number on the left, and the length of the hail sequence on the right.
• The maximizing operator (\$^) with respect to the natural less or equal relation (nleq) applied to the right sides (&r) of its pair of arguments extracts the number with the maximum length sequence.

output:

<27,82,41,124>...<8,4,2,1> has length 112
77031 has sequence length 351

## XPL0

<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations int Seq(1000); \more than enough for longest sequence

func Hailstone(N); \Return length of Hailstone sequence starting at N int N; \ also fills Seq array with sequence int I; [I:= 0; loop [Seq(I):= N; I:= I+1;

    if N=1 then return I;
N:= if N&1 then N*3+1 else N/2;
];


];

int N, SN, Len, MaxLen; [Len:= Hailstone(27); Text(0, "27's Hailstone length = "); IntOut(0, Len); CrLf(0);

Text(0, "Sequence = "); for N:= 0 to 3 do [IntOut(0, Seq(N)); ChOut(0, ^ )]; Text(0, "... "); for N:= Len-4 to Len-1 do [IntOut(0, Seq(N)); ChOut(0, ^ )]; CrLf(0);

MaxLen:= 0; for N:= 1 to 100_000-1 do

   [Len:= Hailstone(N);
if Len > MaxLen then [MaxLen:= Len;  SN:= N];       \save N with max length
];


IntOut(0, SN); Text(0, "'s Hailstone length = "); IntOut(0, MaxLen); ]</lang>

Output:

27's Hailstone length = 112
Sequence = 27 82 41 124 ... 8 4 2 1
77031's Hailstone length = 351