Hailstone sequence

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Task
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.

Task Description:

  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!)

See Also:

Contents

[edit] ACL2

(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)))))

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)

[edit] Ada

Similar to C method:

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;

Output:

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

[edit] 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.

hailstones.ads:

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

hailstones.adb:

package body Hailstones is
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;
end Hailstones;

example main.adb:

with Ada.Text_IO;
with Hailstones;
 
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
Ada.Text_IO.Put (", ");
end if;
end loop;
Ada.Text_IO.New_Line;
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));
Ada.Text_IO.Put ("First four: ");
Print_Sequence (Hailstone_27 (Hailstone_27'First .. Hailstone_27'First + 3));
Ada.Text_IO.Put ("Last four: ");
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;

output:

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

[edit] 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
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))
# ELLA Algol68RS:
print(("Max",hmax," at j=",jatmax, new line))
#

)

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

[edit] APL

Works with: Dyalog 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

Output:

 5↑hailstone 27
27 82 41 124 62
¯5↑hailstone 27
16 8 4 2 1
⍴hailstone 27
112
1↑{⍵[⍒↑(⍴∘hailstone)¨⍵]}⍳100000
77031

[edit] 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 "`n`nCount: " 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

[edit] 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
 

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

[edit] AWK

 
#!/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);
}
 

Output:

27 82 41 124 ....... 8 4 2 1
hailstone(27) has 112 elements
longest hailstone sequence is 77031 and has 351 elements

[edit] BASIC

[edit] Applesoft BASIC

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

[edit] BBC BASIC

      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

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

[edit] Liberty BASIC

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

[edit] 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
 

[edit] 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$)
ForEach Hailstones()
If i=6
Print(#CRLF$)
i=0
EndIf
i+1
Print(RSet(Str(Hailstones()),5))
If Hailstones()<>1
Print(", ")
EndIf
Next
 
i=1
Repeat
l=FillHailstones(i)
If l>maxl
maxl=l
maxi=i
EndIf
i+1
Until i>=100000
Print(#CRLF$+#CRLF$+"The longest sequence below 100000 is #"+Str(maxi)+", and it has "+Str(maxl)+" elements.")
 
Print(#CRLF$+#CRLF$+"Press ENTER to exit."): Input()
CloseConsole()
EndIf

Output

#27 has 112 elements and the sequence 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 longest sequence found up to 100000 is #77031 which has 351 elements.

Press ENTER to exit.

[edit] Run BASIC

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
count = doHailstone(hailstone,"Y")
 
print: 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."
count = doHailstone(27,"Y")
 
print: 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 "Calculating result... Please wait... This could take a little while..."
print "Stone Percent Count"
for i = 1 to 99999
count = doHailstone(i,"N")
if count > maxCount then
theBigStone = i
maxCount = count
print using("#####",i);" ";using("###.#", i / 99999 * 100);"% ";using("####",count)
end if
next i
end
 
'---------------------------------------------
' pass number and print (Y/N)
FUNCTION doHailstone(hailstone,prnt$)
if prnt$ = "Y" then
print
print "The following is the 'Hailstone Sequence' for number:";hailstone
end if
while hailstone <> 1
if (hailstone and 1) then hailstone = (hailstone * 3) + 1 else hailstone = hailstone / 2
doHailstone = doHailstone + 1
if prnt$ = "Y" then
print hailstone;chr$(9);
if (doHailstone mod 10) = 0 then print
end if
wend
END FUNCTION

[edit] Visual Basic .NET

Works with: Visual Basic .NET version 2005+
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

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.

[edit] Batch File

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

@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

Demonstration

>hailstone.cmd 20
20 10 5 16 8 4 2 1

[edit] 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.
&>:.:1-|
>3*^ @
|%2: <
v>2/>+

[edit] 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."
)
)
);

[edit] 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.

>,[
[
----------[
>>>[>>>>]+[[-]+<[->>>>++>>>>+[>>>>]++[->+<<<<<]]<<<]
++++++[>------<-]>--[>>[->>>>]+>+[<<<<]>-],<
]>
]>>>++>+>>[
<<[>>>>[-]+++++++++<[>-<-]+++++++++>[-[<->-]+[<<<<]]<[>+<-]>]
>[>[>>>>]+[[-]<[+[->>>>]>+<]>[<+>[<<<<]]+<<<<]>>>[->>>>]+>+[<<<<]]
>[[>+>>[<<<<+>>>>-]>]<<<<[-]>[-<<<<]]>>>>>>>
]>>+[[-]++++++>>>>]<<<<[[<++++++++>-]<.[-]<[-]<[-]<]<,
]
27
111

[edit] Brat

hailstone = { num |
sequence = [num]
while { num != 1 }
{ true? num % 2 == 0
{ num = num / 2 }
{ num = num * 3 + 1 }
sequence << num
}
 
sequence
}
 
#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!" }
 
#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]}"

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

[edit] Burlesque

 
blsq ) 27{^^^^2.%{3.*1.+}\/{2./}\/ie}{1!=}w!bx{\/+]}{\/isn!}w!L[
112
 

[edit] C

#include <stdio.h>
#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;
}

Output

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

[edit] With caching

Much faster if you want to go over a million or so.

#include <stdio.h>
 
#define N 10000000
#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;
}

[edit] C#

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);
}
}
}
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

[edit] With caching

As with the C example, much faster if you want to go over a million or so.

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));
}
}
}
}
max below 100000: 77031 (351 steps)

[edit] C++

#include <iostream>
#include <vector>
#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;
}

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.

[edit] 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)
)

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

[edit] 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))))))
 
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
 
(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 ")."))

[edit] CoffeeScript

Recursive version:

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

[edit] Common 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))))

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

[edit] D

[edit] Basic Version

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

[edit] Lazy Version

Same output.

import std.stdio, std.algorithm, std.typecons, std.range;
 
auto hailstone(uint m) pure nothrow {
return m
.recurrence!q{ a[n-1] & 1 ? a[n-1] * 3 + 1 : a[n-1] / 2}
.until!q{ a == 1 }(OpenRight.no);
}
 
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]);
}

[edit] Faster Lazy Version

Same output.

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]);
}

[edit] Lazy Version With Caching

Faster, same output.

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]);
}

[edit] Déjà Vu

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

[edit] Dart

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

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)

[edit] 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.

27
[[--: ]nzpq]sq
[d 2/ p]se
[d 3*1+ p]so
[d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx

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.

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
 

Output (Takes quite some time on a decent machine)

77031:351

[edit] Delphi

program ShowHailstoneSequence;
 
{$APPTYPE CONSOLE}
 
uses SysUtils, Generics.Collections;
 
procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>);
var
n: Integer;
begin
aHailstoneList.Clear;
aHailstoneList.Add(aStartingNumber);
n := aStartingNumber;
 
while n <> 1 do
begin
if Odd(n) then
n := (3 * n) + 1
else
n := n div 2;
aHailstoneList.Add(n);
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.

Output:

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

Longest sequence under 100,000: 77031 with 351 elements

[edit] 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
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.

[edit] 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]).
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

[edit] 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
 

[edit] 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})

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.

[edit] 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

[edit] Ezhil

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

 
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
 

[edit] 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
Output:
$ ./factor -run=rosetta.hailstone
Loading resource:work/rosetta/hailstone/hailstone.factor
The hailstone sequence from 27:
  has length 112
  starts with 27, 82, 41, 124
  ends with 8, 4, 2, 1
The hailstone sequence from 77031 has length 351.

[edit] FALSE

[$1&$[%3*1+0~]?~[2/]?]n:
[[$." "$1>][n;!]#%]s:
[1\[$1>][\1+\n;!]#%]c:
27s;! 27c;!."
"
0m:0f:
1[$100000\>][$c;!$m;>[m:$f:0]?%1+]#%
f;." has hailstone sequence length "m;.

[edit] 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

[edit] Fortran

Works with: Fortran version 95 and later
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

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

[edit] Frege

Translation of: Haskell
Works with: Frege version 3.20.113
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)
Output:
112
hailstone 27: [27, 82, 41, 124] ... [8, 4, 2, 1]
(351, 77031)
runtime 4.374 wallclock seconds.

[edit] F#

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

Output:

Maximum length 351 was found for hailstone(77031)

[edit] 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);
# [ 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 ]
CollatzLength(27);
# 112
 
CollatzMax(1, 100);
# [ 97, 119 ]
CollatzMax(1, 1000);
# [ 871, 179 ]
CollatzMax(1, 10000);
# [ 6171, 262 ]
CollatzMax(1, 100000);
# [ 77031, 351 ]
CollatzMax(1, 1000000);
# [ 837799, 525 ]

[edit] 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)
}

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.

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])
}

[edit] Groovy

def hailstone = { long start ->
def sequence = []
while (start != 1) {
sequence << start
start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
}
sequence << start
}

Test Code

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

Output:

[n:77031, size:351]

[edit] 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]

Output:

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

[edit] 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

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;

[edit] Icon and Unicon

A simple solution that generates (in the Icon sense) the sequence is:

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

and a test program for this solution is:

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

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.

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

A test program is:

procedure main(args)
n := integer(!args) | 27
task2(n)
write()
task3()
end
 
procedure task2(n)
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
 
procedure task3()
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

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
->

[edit] Io

Here is a simple, brute-force approach:

 
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.")
 

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.

[edit] 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.
collatz = method(n,
n println
unless(n <= 1,
if(n even?, collatz(n / 2), collatz(n * 3 + 1)))
)

[edit] Inform 7

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

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.

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.

[edit] J

Solution:

hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0

Usage:

   # 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

See also the Collatz Conjecture essay on the J wiki.

[edit] Java

Works with: Java version 1.5+
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>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
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) {
currentList.add(n);
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;
}
}

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

[edit] 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);

outputs

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

[edit] Julia

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

[edit] 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

[edit] Kotlin

import java.util.ArrayDeque
 
fun hailstone(n : Int) : ArrayDeque<Int> {
val hails = when {
n == 1 -> ArrayDeque<Int>()
n % 2 == 0 -> hailstone(n / 2)
else -> hailstone(3 * n + 1)
}
hails addFirst(n)
return hails
}
 
fun main(args : Array<String>) {
val hail27 = hailstone(27)
fun showSeq(s : List<Int>) = s map {it.toString()} reduce {a, b -> a + ", " + b}
System.out.println(
"Hailstone sequence for 27 is " +
showSeq(hail27 take(3)) + " ... " + showSeq(hail27 drop(hail27.size - 3)) +
" with length ${hail27.size}."
)
 
var longestHail = hailstone(1)
for (x in 1 .. 99999)
longestHail = array(hailstone(x), longestHail) maxBy {it.size} ?: longestHail
System.out.println(
"${longestHail.getFirst()} is the number less than 100000 with " +
"the longest sequence, having length ${longestHail.size}."
)
}
Output:
Hailstone sequence for 27 is 27, 82, 41 ... 4, 2, 1 with length 112.
77031 is the number less than 100000 with the longest sequence, having length 351.

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

[edit]

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

[edit] Limbo

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

[edit] 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.

Testing:

| ?- 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

[edit] 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.

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

[edit] 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 ) )

[edit] Maple

Define the procedure:

 
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;
 

Run the command and show the appropriate portion of the result;

 
> 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]
 

Compute the first 100000 sequences:

 
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);
 

Output:

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

[edit] Mathematica

Here are three ways to generate the sequence.

[edit] Fixed-Point formulation

HailstoneFP[n_Integer] := Most[FixedPointList[Which[# == 1, 1, EvenQ[#] , #/2, OddQ[#], (3*# + 1)] &, n]]

[edit] Recursive formulation using piece-wise function definitions

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

[edit] Procedural implementation

hailstone[n_Integer] := Block[{sequence = {}, c = n},
While[c > 1, c = If[EvenQ[c], c/2, 3 c + 1];
AppendTo[sequence, c]];
sequence]

[edit] Nested function-call formulation

I use this version to do the validation:

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];

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

[edit] MATLAB / Octave

   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;

Show sequence of hailstone(27) and number of elements

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

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
  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)

Output:

maxLength =  351
n =  77031

[edit] 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] */

[edit] Mercury

The actual calculation (including module ceremony) providing both a function and a predicate implementation:

:- module hailstone.
 
:- interface.
 
:- import_module int, list.
 
:- func hailstone(int) = list(int).
:- pred hailstone(int::in, list(int)::out) is det.
 
:- implementation.
 
hailstone(N) = S :- hailstone(N, S).
 
hailstone(N, [N|S]) :-
( N = 1 -> S = []
 ; N mod 2 = 0 -> hailstone(N/2, S)
 ; hailstone(3 * N + 1, S) ).
 
:- end_module hailstone.

The mainline test driver (making use of unification for more succinct tests):

:- module test_hailstone.
 
:- interface.
 
:- import_module io.
 
:- pred main(io.state::di, io.state::uo) is det.
 
:- implementation.
 
:- import_module int, list.
:- import_module hailstone.
 
:- pred longest(int::in, int::out, int::out) is det.
:- pred longest(int::in, int::in, int::in, int::out, int::out) is det.
 
longest(M, N, L) :- longest(M, 0, 0, N, L).
 
longest(N, CN, CL, MN, ML) :-
( N > 1 ->
L = list.length(hailstone(N)),
( L > CL -> longest(N - 1, N, L, MN, ML)
 ; longest(N - 1, CN, CL, MN, ML) )
 ; MN = CN, ML = CL ).
 
 
main(!IO) :-
S = hailstone(27),
( list.length(S) = 112,
list.append([27, 82, 41, 124], _, S),
list.remove_suffix(S, [8, 4, 2, 1], _),
longest(100000, 77031, 351) ->
io.write_string("All tests succeeded.\n", !IO)
 ; io.write_string("At least one test failed.\n", !IO) ).
 
:- end_module test_hailstone.

The output of running this program is:

All tests succeeded.

For those unused to logic programming languages it seems that nothing has been proved in terms of confirming anything, but if you look at the predicate declaration for longest/3

:- pred longest(int::in, int::out, int::out) is det.

… you see that the second and third parameters are output parameters. This by calling longest(100000, 77031, 351) you prove, through unification, that the the longest sequence is with the number 77031 and that it is 351 cycles long.

Similarly, using list.append([27, 82, 41, 124], _, S) automatically proves that the generated sequence begins with the provided sequence, etc. Thus we know that the correct sequences and values were generated without bothering to print them out.

[edit] Modula-2

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.

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.

[edit] 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

[edit] 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
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

[edit] Oberon-2

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.

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.

[edit] 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"] *)

[edit] 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
 

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

[edit] Order

To display the length, and first and last elements, of the hailstone sequence for 27, we could do this:

#include <order/interpreter.h>
 
#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))
) )
Output:
h(27) - length:112, starts with:(27)(82)(41)(124), ends with:(8)(4)(2)(1)

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:

#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))))
)

...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):

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)))) )

Notice that large numbers (>100) must be entered as digit sequences with 8nat. 8to_lit converts a digit sequence back to a readable number.

[edit] 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#")"}

Output:

Maximum length 351 was found for hailstone(77031)

[edit] PARI/GP

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)

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

[edit] Pascal

See Delphi

[edit] Perl

[edit] Straightforward

#!/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;
}

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

[edit] Compact

A more compact version:

#!/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;

The same approach as in the compact version above, obfuscated:

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", @_;

Output in either case:

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

[edit] Perl 6

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";

[edit] 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.<br>';
echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>';
 
$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;
}
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

[edit] 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))) )

Output:

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

[edit] 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);
}

Output:

sizeof(({ 27, 82, 41, 124, , ... 8, 4, 2, 1,  }) == 112
longest sequence starting at 77031 has 351 elements

[edit] 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;

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

[edit] PowerShell

Works with: PowerShell version 3.0
# Author M. McNabb
function Get-HailStoneBelowLimit {
param($UpperLimit)
begin {
function Get-HailStone {
param($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)}
}
}
$Counts = @()
}
 
process {
for ($i = 1; $i -lt $UpperLimit; $i++) {
$Object = [pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
$Counts += $Object
}
}
end {$Counts}
}

Output:

PS C:\> Get-HailStone 27
27
82
41
...
8
4
2
1

PS C:\> (Get-HailStone 27).count
112

PS C:\> Get-HailStoneBelowLimit 100000 | Sort Count -Descending | Select -first 1
Number         Count
------         -----
 77031           351

[edit] Prolog

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

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).

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.

hailstone(27,X),
length(X,112),
append([27, 82, 41, 124], _, X),
append(_, [8, 4, 2, 1], X).

3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequences length.

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).

run this query.

longestHailstoneSequence(100000, Seq, Len).

to get the following result

Seq = 77031,
Len = 351 

[edit] 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

:- 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).

Code for task one :

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)).

Output :

 ?- task1.
ok
true.

Code for task two :

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.

Output :

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

[edit] 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)));

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

[edit] 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)))

Sample Output

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

[edit] 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)))
}
 
### PART 2:
twenty.seven <- makeHailstone(27)
twenty.seven$hseq
twenty.seven$seq.length
 
### 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="")
Output:
> 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.

[edit] Racket

 
#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))
 

Output:

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

[edit] REXX

[edit] non-optimized

/*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

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

[edit] optimized

This optimized version is about seven times faster than the unoptimized version.

/*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

output is the same as the non-optimized version.

[edit] Ruby

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

Works with: Ruby version 1.8.7
def hailstone n
seq = [n]
until n == 1
n = (n.even?) ? (n / 2) : (3 * n + 1)
seq << n
end
seq
end
 
# for n = 27, show sequence length and first and last 4 elements
hs27 = hailstone 27
p [hs27.length, hs27[0..3], hs27[-4..-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}"

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

[edit] 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
module Hailstone
class ListNode
include Enumerable
attr_reader :value, :size, :succ
 
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
 
# 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]]
 
# 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}"

[edit] Scala

Library: Scala
Works with: Scala version 2.10.2
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.")
}
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.

[edit] 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)

[edit] 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;

Output:

hailstone(27):
27, 82, 41, 124, ...., 8, 4, 2, 1  length=112
Maximum length 351 at number=77031

[edit] Smalltalk

Works with: GNU 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) ].
]
].
|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.

[edit] SNUSP

   /@+@@@+++# 27
   |    halve odd   /===count<<\    /recurse\    #/?\ zero
$>@/===!/===-?\==>?!/-<+++\    \!/=!\@\>?!\@/<@\.!\-/
 /+<-\!>\?-<+>/++++<\?>+++/*6+4  |    |   \=/  \=itoa=@@@+@+++++#
 \=>?/<=!=\   |                  |    !     /+ !/+ !/+ !/+   \    mod10
        |//!==/========\         |    /<+> -\!?-\!?-\!?-\!?-\!
 /=>?\<=/\<+>!\->+>+<<?/>>=print@/\ln \?!\-?!\-?!\-?!\-?!\-?/\    div10
 \+<-/!<     ----------.++++++++++/      #  +/! +/! +/! +/! +/

[edit] Tcl

The core looping structure is an example of an n-plus-one-half loop, except the loop is officially infinite here.

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"

Output:

h27 len=112
head4 = 27 82 41 124
tail4 = 8 4 2 1
max is 77031, with length 351

[edit] 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)))
$ txr -l hailstone.txr
longest sequence is 351 for n = 77031

[edit] UNIX Shell

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

Works with: Bash
#!/bin/bash
# seq is the array genereated by hailstone
# index is used for seq
declare -a seq
declare -i index
 
# 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
}
 
# 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
i=27
hailstone $i
echo "$i: ${#seq[@]}"
echo "${seq[@]:0:4} ... ${seq[@]:(-4):4}"
 
# Show the number less than 100,000 which has the longest hailstone
# sequence together with that sequences length.
# (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}"

output

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

[edit] 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
# Outputs a hailstone sequence from $1, with one element per line.
# 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."

[edit] 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.

# Outputs a hailstone sequence from !:1, with one element per line.
# 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-]"
 
# hailstone-length $i
# acts like
# @ len = `hailstone $i | wc -l | tr -d ' '`
# 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."
$ 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.

[edit] Ursala

#import std
#import nat
 
hail = @iNC ~&h~=1->x ^C\~& @h ~&h?\~&t successor+ sum@iNiCX
 
#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>

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

[edit] 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);
]

Output:

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

[edit] zkl

fcn collatz(n,z=L()){ z.append(n); if(n==1) return(z);
if(n.isEven) return(self.fcn(n/2,z)); return(self.fcn(n*3+1,z)) }

This uses tail recursion and thus is stack efficient.

Output:
var n=collatz(27)
n.len()
112
n[0,4]
L(27,82,41,124)
n[-4,*]
L(8,4,2,1)

Rather than write a function that calculates the length, just roll through all 100,000 sequences and save the largest (length,sequence start) pair. Creating all those Collatz lists isn't quick. This works by using a [mutable] list to hold state as the pump does the basic looping.

[2..0d100_000].pump(Void,  // loop n from 2 to 100,000
collatz, // generate Collatz sequence(n)
fcn(c,n){ // if new longest sequence, save length/C, return longest
if(c.len()>n[0]) n.clear(c.len(),c[0]); n}.fp1(L(0,0)))
Output:
L(351,77031)  // length, hailstone
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