# Hailstone sequence

(Redirected from Collatz conjecture)
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.

This sequence was named by Lothar Collatz in 1937   (or possibly in 1939),   and is also known as (the):

•   hailstone sequence,   hailstone numbers
•   3x + 2 mapping,   3n + 1 problem
•   Collatz sequence
•   Hasse's algorithm
•   Kakutani's problem
•   Syracuse algorithm,   Syracuse problem
•   Thwaites conjecture
•   Ulam's problem

The hailstone sequence is also known as   hailstone numbers   (because the values are usually subject to multiple descents and ascents like hailstones in a cloud).

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!)

## 11l

Translation of: Python
F hailstone(=n)
V seq = [n]
L n > 1
n = I n % 2 != 0 {3 * n + 1} E n I/ 2
seq.append(n)
R seq

V h = hailstone(27)
assert(h.len == 112 & h[0.<4] == [27, 82, 41, 124] & h[(len)-4 ..] == [8, 4, 2, 1])

V m = max((1..99999).map(i -> (hailstone(i).len, i)))
print(‘Maximum length #. was found for hailstone(#.) for numbers <100,000’.format(m, m))
Output:
Maximum length 351 was found for hailstone(77031) for numbers <100,000


## 360 Assembly

*        Hailstone sequence        16/08/2015
HAILSTON CSECT
USING  HAILSTON,R12
LR     R12,R15
ST     R14,SAVER14
BEGIN    L      R11,=F'100000'     nmax
LA     R8,27              n=27
LR     R1,R8
MVI    FTAB,X'01'         ftab=true
BAL    R14,COLLATZ
LR     R10,R1             p
XDECO  R8,XDEC            n
MVC    BUF1+10(6),XDEC+6
XDECO  R10,XDEC           p
MVC    BUF1+18(5),XDEC+7
LA     R5,6
LA     R3,0               i
LA     R4,BUF1+25
LOOPED   L      R2,TAB(R3)         tab(i)
XDECO  R2,XDEC
MVC    0(7,R4),XDEC+5
LA     R3,4(R3)           i=i+1
LA     R4,7(R4)
C      R5,=F'4'
BNE    BCT
LA     R4,7(R4)
BCT      BCT    R5,LOOPED
XPRNT  BUF1,80            print hailstone(n)=p,tab(*)
MVC    LONGEST,=F'0'      longest=0
MVI    FTAB,X'00'         ftab=true
LA     R8,1               i
LOOPI    CR     R8,R11             do i=1 to nmax
BH     ELOOPI
LR     R1,R8              n
BAL    R14,COLLATZ
LR     R10,R1             p
L      R4,LONGEST
CR     R4,R10             if longest<p
BNL    NOTSUP
ST     R8,IVAL            ival=i
ST     R10,LONGEST        longest=p
NOTSUP   LA     R8,1(R8)           i=i+1
B      LOOPI
ELOOPI   EQU    *                  end i
XDECO  R11,XDEC           maxn
MVC    BUF2+9(6),XDEC+6
L      R1,IVAL            ival
XDECO  R1,XDEC
MVC    BUF2+28(6),XDEC+6
L      R1,LONGEST         longest
XDECO  R1,XDEC
MVC    BUF2+36(5),XDEC+7
XPRNT  BUF2,80            print maxn,hailstone(ival)=longest
B      RETURN
*        *      *                  r1=collatz(r1)
COLLATZ  LR     R7,R1              m=n  (R7)
LA     R6,1               p=1  (R6)
LOOPP    C      R7,=F'1'           do p=1 by 1 while(m>1)
BNH    ELOOPP
CLI    FTAB,X'01'         if ftab
BNE    NONOK
C      R6,=F'1'           if p>=1
BL     NONOK
C      R6,=F'3'           & p<=3
BH     NONOK
LR     R1,R6              then
BCTR   R1,0
SLA    R1,2
ST     R7,TAB(R1)         tab(p)=m
NONOK    LR     R4,R7              m
N      R4,=F'1'           m&1
LTR    R4,R4              if m//2=0  (if not(m&1))
BNZ    ODD
EVEN     SRA    R7,1               m=m/2
B      EIFM
ODD      LA     R3,3
MR     R2,R7              *m
LA     R7,1(R3)           m=m*3+1
EIFM     CLI    FTAB,X'01'         if ftab
BNE    NEXTP
MVC    TAB+12,TAB+16      tab(4)=tab(5)
MVC    TAB+16,TAB+20      tab(5)=tab(6)
ST     R7,TAB+20          tab(6)=m
NEXTP    LA     R6,1(R6)           p=p+1
B      LOOPP
ELOOPP   LR     R1,R6              end p; return(p)
BR     R14                end collatz
*
RETURN   L      R14,SAVER14        restore caller address
XR     R15,R15            set return code
SAVER14  DS     F
IVAL     DS     F
LONGEST  DS     F
N        DS     F
TAB      DS     6F
FTAB     DS     X
BUF1     DC     CL80'hailstone(nnnnnn)=nnnnn : nnnnnn nnnnnn nnnnnn ...*
... nnnnnn nnnnnn nnnnnn'
BUF2     DC     CL80'longest <nnnnnn : hailstone(nnnnnn)=nnnnn'
XDEC     DS     CL12
YREGS
END    HAILSTON
Output:
hailstone(    27)=  112 :     27     82     41 ......      4      2      1
longest <100000 : hailstone( 77031)=  351


## ABAP

CLASS lcl_hailstone DEFINITION.
PUBLIC SECTION.
TYPES: tty_sequence TYPE STANDARD TABLE OF i
WITH NON-UNIQUE EMPTY KEY,
BEGIN OF ty_seq_len,
start TYPE i,
len   TYPE i,
END OF ty_seq_len,
tty_seq_len TYPE HASHED TABLE OF ty_seq_len
WITH UNIQUE KEY start.

CLASS-METHODS:
get_next
IMPORTING
n                           TYPE i
RETURNING
VALUE(r_next_hailstone_num) TYPE i,

get_sequence
IMPORTING
start             TYPE i
RETURNING
VALUE(r_sequence) TYPE tty_sequence,

get_longest_sequence_upto
IMPORTING
limit                     TYPE i
RETURNING
VALUE(r_longest_sequence) TYPE ty_seq_len.

PRIVATE SECTION.
TYPES: BEGIN OF ty_seq,
start TYPE i,
seq   TYPE tty_sequence,
END OF ty_seq.
CLASS-DATA: sequence_buffer TYPE HASHED TABLE OF ty_seq
WITH UNIQUE KEY start.
ENDCLASS.

CLASS lcl_hailstone IMPLEMENTATION.
METHOD get_next.
r_next_hailstone_num = COND #( WHEN n MOD 2 = 0 THEN n / 2
ELSE ( 3 * n ) + 1 ).
ENDMETHOD.

METHOD get_sequence.
INSERT start INTO TABLE r_sequence.
IF start = 1.
RETURN.
ENDIF.

WITH TABLE KEY start = start.
IF sy-subrc = 0.
INSERT LINES OF <buff>-seq INTO TABLE r_sequence.
ELSE.
DATA(seq) = get_sequence( get_next( start ) ).
INSERT LINES OF seq INTO TABLE r_sequence.
INSERT VALUE ty_seq( start = start
seq   = seq ) INTO TABLE sequence_buffer.
ENDIF.
ENDMETHOD.

METHOD get_longest_sequence_upto.
DATA: max_seq TYPE ty_seq_len,
act_seq TYPE ty_seq_len.

DO limit TIMES.
act_seq-len = lines( get_sequence( sy-index ) ).

IF act_seq-len > max_seq-len.
max_seq-len   = act_seq-len.
max_seq-start = sy-index.
ENDIF.
ENDDO.

r_longest_sequence = max_seq.
ENDMETHOD.
ENDCLASS.

START-OF-SELECTION.
cl_demo_output=>begin_section( |Hailstone sequence of 27 is: | ).
cl_demo_output=>write( REDUCE string( INIT result = 
FOR item IN lcl_hailstone=>get_sequence( 27 )
NEXT result = |{ result } { item }| ) ).
cl_demo_output=>write( |With length: { lines( lcl_hailstone=>get_sequence( 27 ) ) }| ).
cl_demo_output=>begin_section( |Longest hailstone sequence upto 100k| ).
cl_demo_output=>write( lcl_hailstone=>get_longest_sequence_upto( 100000 ) ).
cl_demo_output=>display( ).

Output:
Hailstone sequence of 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

With length: 112

Longest hailstone sequence upto 100k

Structure
START LEN
77031 351



## 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)

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


### 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.

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


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;


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
end if;
end loop;
end Print_Sequence;

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

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

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

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

## Aime

void
print_hailstone(integer h)
{
list l;

while (h ^ 1) {
lb_p_integer(l, h);
h = h & 1 ? 3 * h + 1 : h / 2;
}

o_form("hailstone sequence for ~ is ~1 ~ ~ ~ .. ~ ~ ~ ~, it is ~ long\n",
l, l, l, l, l[-3], l[-2], l[-1], 1, ~l + 1);
}

void
max_hailstone(integer x)
{
integer e, i, m;
index r;

m = 0;
i = 1;
while (i < x) {
integer h, k, l;

h = i;
l = 1;
while (h ^ 1) {
if (i_j_integer(k, r, h)) {
l += k;
break;
} else {
l += 1;
h = h & 1 ? 3 * h + 1 : h / 2;
}
}

r[i] = l - 1;

if (m < l) {
m = l;
e = i;
}

i += 1;
}

o_form("hailstone sequence length for ~ is ~\n", e, m);
}

integer
main(void)
{
print_hailstone(27);
max_hailstone(100000);

return 0;
}
Output:
hailstone sequence for 27 is 27 82 41 124 .. 8 4 2 1, it is 112 long
hailstone sequence length for 77031 is 351

## ALGOL 60

Works with: A60
begin
comment Hailstone sequence - Algol 60;
integer array collatz[1:400]; integer icollatz;

integer procedure mod(i,j); value i,j; integer i,j;
mod:=i-(i div j)*j;

integer procedure hailstone(num);
value num; integer num;
begin
integer i,n;
icollatz:=1; n:=num; i:=0;
collatz[icollatz]:=n;
for i:=i+1 while n notequal 1 do begin
if mod(n,2)=0 then n:=n div 2
else n:=(3*n)+1;
icollatz:=icollatz+1;
collatz[icollatz]:=n
end;
hailstone:=icollatz
end hailstone;

integer i,nn,ncollatz,count,nlongest,nel,nelcur,nnn;
nn:=27;
ncollatz:=hailstone(nn);
outstring(1,"sequence for"); outinteger(1,nn); outstring(1," :\n");
for i:=1 step 1 until ncollatz do outinteger(1,collatz[i]);
outstring(1,"\n");
outstring(1,"number of elements:"); outinteger(1,ncollatz);
outstring(1,"\n\n");
nlongest:=0; nel:=0; nnn:=100000;
for count:=1, count+1 while count<nnn do begin
nelcur:=hailstone(count);
if nelcur>nel then begin
nel:=nelcur;
nlongest:=count
end
end;
outstring(1,"number <"); outinteger(1,nnn);
outstring(1,"with the longest sequence:"); outinteger(1,nlongest);
outstring(1,", with"); outinteger(1,nel); outstring(1,"elements.");
outstring(1,"\n")
end
Output:
sequence for 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
number of elements: 112

number < 100000 with the longest sequence: 77031 , with 351 elements.


## 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;

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


## ALGOL-M

The limitations of ALGOL-M's 15-bit integer data type will not allow the required search up to 100000 for the longest sequence, so we stick with what is possible.

BEGIN

INTEGER N, LEN, YES, NO, LIMIT, LONGEST, NLONG;

% RETURN P MOD Q %
INTEGER FUNCTION MOD(P, Q);
INTEGER P, Q;
BEGIN
MOD := P - Q * (P / Q);
END;

% COMPUTE AND OPTIONALLY DISPLAY HAILSTONE SEQUENCE FOR N. %
% RETURN LENGTH OF SEQUENCE OR ZERO ON OVERFLOW. %
INTEGER FUNCTION HAILSTONE(N, DISPLAY);
INTEGER N, DISPLAY;
BEGIN
INTEGER LEN;
LEN := 1;
IF DISPLAY = 1 THEN WRITE("");
WHILE (N <> 1) AND (N > 0) DO
BEGIN
IF DISPLAY = 1 THEN WRITEON(N,"  ");
IF MOD(N,2) = 0 THEN
N := N / 2
ELSE
N := (N * 3) + 1;
LEN := LEN + 1;
END;
IF DISPLAY = 1 THEN WRITEON(N);
HAILSTONE := (IF N < 0 THEN 0 ELSE LEN);
END;

% EXERCISE THE FUNCTION %
YES := 1; NO := 0;
WRITE("DISPLAY HAILSTONE SEQUENCE FOR WHAT NUMBER?");
LEN := HAILSTONE(N, YES);
WRITE("SEQUENCE LENGTH =", LEN);

% FIND LONGEST SEQUENCE BEFORE OVERFLOW %
N := 2;
LONGEST := 1;
LEN := 2;
NLONG := 2;
LIMIT := 1000;
WRITE("SEARCHING FOR LONGEST SEQUENCE UP TO N =",LIMIT," ...");
WHILE (N < LIMIT) AND (LEN <> 0) DO
BEGIN
LEN := HAILSTONE(N, NO);
IF LEN > LONGEST THEN
BEGIN
LONGEST := LEN;
NLONG := N;
END;
N := N + 1;
END;
IF LEN = 0 THEN WRITE("SEARCH TERMINATED WITH OVERFLOW AT N =",N-1);
WRITE("MAXIMUM SEQUENCE LENGTH =", LONGEST, " FOR N =", NLONG);

END
Output:
DISPLAY HAILSTONE SEQUENCE FOR WHAT NUMBER?
-> 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
SEQUENCE LENGTH =   112
SEARCHING FOR LONGEST SEQUENCE UP TO N = 10000 ...
SEARCH TERMINATED WITH OVERFLOW AT N =   447
MAXIMUM SEQUENCE LENGTH =   144 FOR N = 327

## ALGOL W

begin
% show some Hailstone Sequence related information                       %
% calculates the length of the sequence generated by n,                  %
% if showFirstAndLast is true, the first and last 4 elements of the      %
% sequence are stored in first and last                                  %
% hs holds a cache of the upbHs previously calculated sequence lengths   %
% if showFirstAndLast is false, the cache will be used                   %
procedure hailstone ( integer value  n
; integer array  first, last ( * )
; integer result length
; integer array  hs          ( * )
; integer value  upbHs
; logical value  showFirstAndLast
) ;
if not showFirstAndLast and n <= upbHs and hs( n ) not = 0 then begin
% no need to store the start and end of the sequence and we already  %
% know the length of the sequence for n                              %
length := hs( n )
end
else begin
% must calculate the sequence length                                 %
integer sv;
for i := 1 until 4 do first( i ) := last( i ) := 0;
length := 0;
sv     := n;
if sv > 0 then begin
while begin
length := length + 1;
if showFirstAndLast then begin
if length <= 4 then first( length ) := sv;
for lPos := 1 until 3 do last( lPos ) := last( lPos + 1 );
last( 4 ) := sv
end
else if sv <= upbHs and hs( sv ) not = 0 then begin
% have a known value                                 %
length := ( length + hs( sv ) ) - 1;
sv     := 1
end ;
sv not = 1
end do begin
sv := if odd( sv ) then ( 3 * sv ) + 1 else sv div 2
end while_sv_ne_1 ;
if n < upbHs then hs( n ) := length
end if_sv_gt_0
end hailstone ;
begin
% test the hailstone procedure                                       %
integer HS_CACHE_SIZE;
HS_CACHE_SIZE := 100000;
begin
integer array first, last ( 1 :: 4 );
integer       length, maxLength, maxNumber;
integer array hs          ( 1 :: HS_CACHE_SIZE );
for i := 1 until HS_CACHE_SIZE do hs( i ) := 0;
hailstone( 27, first, last, length, hs, HS_CACHE_SIZE, true );
write( i_w := 1, s_w := 0
, "27: length ", length, ", first: ["
, first( 1 ), " ", first( 2 ), " ", first( 3 ), " ", first( 4 )
, "] last: ["
, last( 1 ), " ", last( 2 ), " ", last( 3 ), " ", last( 4 )
, "]"
);
maxNumber := 0;
maxLength := 0;
for n := 1 until 100000 do begin
hailstone( n, first, last, length, hs, HS_CACHE_SIZE, false );
if length > maxLength then begin
maxNumber := n;
maxLength := length
end if_length_gt_maxLength
end for_n ;
write( i_w := 1, s_w := 1, "Maximum sequence length: ", maxLength, " for: ", maxNumber )
end
end
end.
Output:
27: length 112, first: [27 82 41 124] last: [8 4 2 1]
Maximum sequence length: 351  for: 77031


## Amazing Hopper

#include <basico.h>

#proto Hailstone(_X_,_SW_)

algoritmo

valor=27, máxima secuencia=0, vtemp=0
imprimir ( _Hailstone(27,1) ---copiar en 'máxima secuencia'--- , NL )

i=28
iterar
_Hailstone(i,0), copiar en 'vtemp'
cuando( sea mayor que 'máxima secuencia' ) {
máxima secuencia = vtemp
valor=i
}
++i
hasta que ' #(i==100000) '
imprimir ( #(utf8("Máxima longitud ")),máxima secuencia,\
#(utf8(") para números <100,000")), NL )

terminar

subrutinas

Hailstone(n, sw)
largo_de_secuencia = 0
v={}, n, mete(v)
iterar
tomar si ( es par(n), #(n/2), \
tomar si ( #(n<>1), #(3*n+1), 1) )
---copiar en 'n'--- mete(v)
hasta que ' #(n==1) '
#(length(v)), mover a 'largo_de_secuencia'
cuando (sw){
decimales '0'
#( v[1:4] ), ",...,",
#( v[largo_de_secuencia-4 : largo_de_secuencia] )
NL, #(utf8("Tamaño de la secuencia: "))
imprime esto; decimales normales
}
retornar ' largo_de_secuencia '

Output:
27,82,41,124,...,16,8,4,2,1
Tamaño de la secuencia: 112
Máxima longitud 351 fue encontrada para Hailstone(77031) para números <100,000


## APL

Works with: Dyalog APL
⍝ recursive dfn:
dfnHailstone←{
c←⊃⌽⍵ ⍝ last element
1=c:1 ⍝ if it is 1, stop.
⍵,∇(1+2|c)⊃(c÷2)(1+3×c) ⍝ otherwise pick the next step, and append the result of the recursive call
}

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


## AppleScript

on hailstoneSequence(n)
script o
property sequence : {n}
end script

repeat until (n = 1)
if (n mod 2 is 0) then
set n to n div 2
else
set n to 3 * n + 1
end if
set end of o's sequence to n
end repeat

return o's sequence
end hailstoneSequence

set n to 27
tell hailstoneSequence(n)
return {n:n, |length of sequence|:(its length), |first 4 numbers|:items 1 thru 4, |last 4 numbers|:items -4 thru -1}
end tell

Output:
{|length of sequence|:112, |first 4 numbers|:{27, 82, 41, 124}, |last 4 numbers|:{8, 4, 2, 1}}
-- Number(s) below 100,000 giving the longest sequence length, using the hailstoneSequence(n) handler above.
set nums to {}
set longestLength to 1
repeat with n from 2 to 99999
set thisLength to (count hailstoneSequence(n))
if (thisLength < longestLength) then
else if (thisLength > longestLength) then
set nums to {n}
set longestLength to thisLength
else
set end of nums to n
end if
end repeat
return {|number(s) giving longest sequence length|:nums, |length of sequence|:longestLength}

Output:
{|number(s) giving longest sequence length|:{77031}, |length of sequence|:351}

## ARM Assembly

Output is in hexadecimal but is otherwise correct. Because of the Game Boy Advance's limited screen size, only the first 4 and last 4 entries are printed to the screen. The emulator's memory dump can show the rest. In addition, the task was split into two separate programs.

### Hailstone Sequence of N equal to 27

        .org 0x08000000

b ProgramStart

;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
; Program Start

.equ ramarea, 0x02000000
.equ CursorX,ramarea
.equ CursorY,ramarea+1
.equ hailstoneram,0x02000004

ProgramStart:
mov sp,#0x03000000			;Init Stack Pointer

mov r4,#0x04000000  		        ;DISPCNT - LCD Video Controller
mov r2,#0x403    			;4= Layer 2 on / 3= ScreenMode 3
str	r2,[r4]         	        ;now the user can see the screen

bl ResetTextCursors			;set text cursors to top left of screen. This routine, as well as the other I/O
;     routines, were omitted to keep this entry short.
mov r0,#27
bl PrintString
bl NewLine
bl ShowHex32
bl NewLine
bl NewLine

bl Hailstone

;function is complete, return the output
bl PrintString
bl NewLine
ldr r1,HailStoneRam_Mirror		;mov r2,0x02000004

ldr r0,[r1],#4
bl ShowHex32
bl NewLine

ldr r0,[r1],#4
bl ShowHex32
bl NewLine

ldr r0,[r1],#4
bl ShowHex32
bl NewLine

ldr r0,[r1],#4
bl ShowHex32
bl NewLine
bl NewLine

bl PrintString
bl NewLine

ldr r0,[r2],#4
bl ShowHex32
bl NewLine

ldr r0,[r2],#4
bl ShowHex32
bl NewLine

ldr r0,[r2],#4
bl ShowHex32
bl NewLine

ldr r0,[r2],#4
bl ShowHex32
bl NewLine
bl NewLine

bl PrintString
bl NewLine
mov r0,r3
bl ShowHex32

forever:
b forever              ;we're done, so trap the program counter.

Hailstone:
;input: R0 = n.
;out: 	r2 = pointer to last 4 entries
;	r3 = length of sequence

;reg usage:
;R3 = counter for entries in the sequence.
;R5 = pointer to output ram
stmfd sp!,{r4-r12,lr}

mov r5,#0x02000000
str r0,[r5],#4				;store in hailstone ram and post-inc by 4
mov r3,#0
loop_hailstone:
add r3,r3,#1				;represents number of entries in the sequence
cmp r0,#1
beq hailstone_end
tst r0,#1
;;;; executes only if r0 was even
moveq r0,r0,lsr #1			;divide

;;;; executes only if r0 was odd
movne r1,r0
movne r0,r0,lsl #1

str r0,[r5],#4			;store in hailstone ram, post inc by 4

b loop_hailstone

hailstone_end:
sub r5,r5,#16			;subtract 16 to get pointer to last 4 entries.
mov r2,r5				;output ptr to last 4 entries to r2.
;pointer to first 4 entries is 0x02000004
ldmfd sp!,{r4-r12,pc}

HailStoneRam_Mirror:
.long 0x02000004
HailstoneMessage_Init:
.align 4
HailstoneMessage_0:
.byte "First 4 numbers are:",255
.align 4
HailstoneMessage_1:
.byte "Last 4 numbers are:",255
.align 4
HailstoneMessage_2:
.byte "Sequence length is:",255
.align 4

;;;;;;;;;;; EVERYTHING PAST THIS POINT IS JUST I/O ROUTINES FOR PRINTING NUMBERS AND WORDS TO THE GAME BOY ADVANCE'S SCREEN
;;;;;;;;;;; I ASSURE YOU THAT ALL OF IT WORKS BUT CHANCES ARE YOU DIDN'T COME HERE TO SEE THAT.
;;;;;;;;;;; THANKS TO KEITH OF CHIBIAKUMAS.COM FOR WRITING THEM!
Output:
Your input was:
0000001B

First 4 numbers are:
0000001B
00000052
00000029
0000007C

Last 4 numbers are:
00000008
00000004
00000002
00000001

Sequence length is:
00000070


### Biggest Sequence Between 2 and 100,000

To keep this short, I'm only including the part that changed, and the output. This goes after the call to ResetTextCursors but before the infinite loop:

	mov r0,#1
bl Hailstone
mov r6,r3

mov r0,#2
mov r8,#100000

loop_getBiggestHailstone:
mov r10,r0
bl Hailstone
mov r0,r10

cmp r3,r6
movgt r6,r3				;if greater than, store in r6
movgt r7,r0				;if greater than, store in r7
cmp r0,r8
blt loop_getBiggestHailstone

bl PrintString
bl NewLine
bl PrintString
bl NewLine

mov r0,r7
bl ShowHex32
bl NewLine

bl PrintString
bl NewLine

mov r0,r6
bl ShowHex32
bl NewLine
Output:
The number that makes the
biggest sequence is:
00012CE7
And that sequence has a length
of:
0000015F


## Arturo

hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2

'ret ++ n
]
ret
]

print "Hailstone sequence for 27:"
print hailstone 27

maxHailstoneLength: 0
maxHailstone: 0

loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]

print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]

Output:
Hailstone sequence for 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
max hailstone sequence found (<100000): of length 351 for 77031


## AutoHotkey

; Submitted by MasterFocus --- http://tiny.cc/iTunis

;  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

;  Seq. for starting number 27 has 112 elements

Count := 1, List := varNum := 27 ; starting number is 27, counting elements
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:n" List "nnCount: " Count

;  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


## 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  ## 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  ## BASIC ### 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 ### 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  ### Commodore BASIC 100 PRINT : PRINT "HAILSTONE SEQUENCE FOR N = 27:" 110 N=27 : SHOW=1 120 GOSUB 1000 130 PRINT X"ELEMENTS" 140 PRINT : PRINT "FINDING N WITH THE LONGEST HAILSTONE SEQUENCE" 150 SHOW=0 160 T0 = TI 170 FOR N=2 TO 100000 180 : GOSUB 1000 190 : IF X>MAX THEN MAX=X : NMAX = N 200 : REM' PRINT N,X,MAX 210 NEXT 230 PRINT "LONGEST HAILSTONE SEQUENCE STARTS WITH "NMAX"." 240 PRINT "IT HAS"MAX"ELEMENTS" 260 END 1000 REM '*** HAILSTONE SEQUENCE SUBROUTINE *** 1010 X = 0 : S = N 1020 IF SHOW THEN PRINT S, 1030 X = X+1 1040 IF S=1 THEN RETURN 1050 IF INT(S/2)=S/2 THEN S = S/2 : GOTO 1020 1060 S = 3*S+1 1070 GOTO 1020  ### FreeBASIC ' version 17-06-2015 ' compile with: fbc -s console Function hailstone_fast(number As ULongInt) As ULongInt ' faster version ' only counts the sequence Dim As ULongInt count = 1 While number <> 1 If (number And 1) = 1 Then number += number Shr 1 + 1 ' 3*n+1 and n/2 in one count += 2 Else number Shr= 1 ' divide number by 2 count += 1 End If Wend Return count End Function Sub hailstone_print(number As ULongInt) ' print the number and sequence Dim As ULongInt count = 1 Print "sequence for number "; number Print Using "########"; number; 'starting number While number <> 1 If (number And 1) = 1 Then number = number * 3 + 1 ' n * 3 + 1 count += 1 Else number = number \ 2 ' n \ 2 count += 1 End If Print Using "########"; number; Wend Print : Print Print "sequence length = "; count Print Print String(79,"-") End Sub Function hailstone(number As ULongInt) As ULongInt ' normal version ' only counts the sequence Dim As ULongInt count = 1 While number <> 1 If (number And 1) = 1 Then number = number * 3 + 1 ' n * 3 + 1 count += 1 End If number = number \ 2 ' divide number by 2 count += 1 Wend Return count End Function ' ------=< MAIN >=------ Dim As ULongInt number Dim As UInteger x, max_x, max_seq hailstone_print(27) Print For x As UInteger = 1 To 100000 number = hailstone(x) If number > max_seq Then max_x = x max_seq = number End If Next Print "The longest sequence is for "; max_x; ", it has a sequence length of "; max_seq ' empty keyboard buffer While Inkey <> "" : Wend Print : Print : Print "hit any key to end program" Sleep End Output: sequence for number 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 sequence length = 112 ------------------------------------------------------------------------------- The longest sequence is for 77031, it has a sequence length of 351 ### GW-BASIC 10 N# = 27 20 P = 1 30 GOSUB 130 40 PRINT "That took";C;"steps." 50 P = 0 : A = 0 : B = 0 60 FOR M = 1 TO 99999! 70 N# = M 80 GOSUB 130 90 IF C > B THEN B = C: A = M 100 NEXT M 110 PRINT "The longest sequence is for n=";A;" and is ";B;" steps long." 120 END 130 C = 1 140 IF P = 1 THEN PRINT N# 150 IF N# < 2 THEN RETURN 160 IF N#/2 = INT(N#/2) THEN N# = N#/2 ELSE N# = 3*N# + 1 170 C = C + 1 180 GOTO 140 ### 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$cls print "Part 3: Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.(But don't show the actual sequence)!" print "" print "Calculating result... Please wait... This could take a little while..." print "" print "Percent Done", "Start Number", "Seq. Length", "Maximum Sequence So Far" print "" for cc = 1 to 99999 hailstone = cc: c = 1 while hailstone <> 1 c = c + 1 if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1 wend if c > max then max = c: largesthailstone = cc locate 1, 7 print " " locate 1, 7 print using("###.###", cc / 99999 * 100);"%", cc, c, max scan next cc print "" print "The number less than 100,000 with the longest 'Hailstone Sequence' is "; largesthailstone;". It's sequence length is "; max;"." end ### OxygenBasic function Hailstone(sys *n) '========================= if n and 1 n=n*3+1 else n=n>>1 end if end function function HailstoneSequence(sys n) as sys '======================================= count=1 do Hailstone n Count++ if n=1 then exit do end do return count end function 'MAIN '==== maxc=0 maxn=0 e=100000 for n=1 to e c=HailstoneSequence n if c>maxc maxc=c maxn=n end if next print e ", " maxn ", " maxc 'result 100000, 77031, 351 ### PureBasic NewList Hailstones.i() ; Make a linked list to use as we do not know the numbers of elements needed for an Array Procedure.i FillHailstones(n) ; Fills the list & returns the amount of elements in the list Shared Hailstones() ; Get access to the Hailstones-List ClearList(Hailstones()) ; Remove old data Repeat AddElement(Hailstones()) ; Add an element to the list Hailstones()=n ; Fill current value in the new list element If n=1 ProcedureReturn ListSize(Hailstones()) ElseIf n%2=0 n/2 Else n=(3*n)+1 EndIf ForEver EndProcedure If OpenConsole() Define i, l, maxl, maxi l=FillHailstones(27) Print("#27 has "+Str(l)+" elements and the sequence is: "+#CRLF$)
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$+"Press ENTER to exit."): Input() CloseConsole() EndIf Output:  #27 has 112 elements and the sequence is: 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 The longest sequence found up to 100000 is #77031 which has 351 elements. Press ENTER to exit.  ### Run BASIC print "Part 1: Create a routine to generate the hailstone sequence for a number." print "" while hailstone < 1 or hailstone <> int(hailstone) input "Please enter a positive integer: "; hailstone wend count = doHailstone(hailstone,"Y") print: print "Part 2: Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1." count = doHailstone(27,"Y") print: print "Part 3: Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.(But don't show the actual sequence)!" print "Calculating result... Please wait... This could take a little while..." print "Stone Percent Count" for i = 1 to 99999 count = doHailstone(i,"N") if count > maxCount then theBigStone = i maxCount = count print using("#####",i);" ";using("###.#", i / 99999 * 100);"% ";using("####",count) end if next i end '--------------------------------------------- ' pass number and print (Y/N) FUNCTION doHailstone(hailstone,prnt$)
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 ### Tiny BASIC Tiny BASIC is limited to signed integers from -32768 to 32767. This code combines two integers into one: number = 32766A + B, to emulate integers up to 1.07 billion. Dealing with integer truncation, carries, and avoiding overflows requires some finesse. Even so one number, namely 77671, causes an overflow because one of its steps exceeds 1.07 billion.  PRINT "Enter a positive integer" INPUT N REM unit column LET M = 0 REM 32766 column LET C = 1 REM count LET P = 1 REM print the sequence? LET L = 1 REM finite state label GOSUB 10 LET F = 1 REM current champion LET E = 0 REM 32766 part of current champ LET Y = 1 REM length of current longest sequence LET P = 0 REM no more printing LET W = 0 REM currently testing this number LET V = 0 REM 32766 column of the number PRINT "Testing for longest chain for n<100000..." 5 LET W = W + 1 REM PRINT V, " ", W LET N = W LET M = V LET C = 1 REM reset count IF W = 32766 THEN GOSUB 50 GOSUB 10 IF C > Y THEN GOSUB 60 IF V = 3 THEN IF W = 1702 THEN GOTO 8 GOTO 5 8 PRINT "The longest sequence starts at 32766x",E," + ",F PRINT "And goes for ",Y," steps." END 10 IF P = 1 THEN IF M > 0 THEN PRINT C," 32766x",M," + ",N IF P = 1 THEN IF M = 0 THEN PRINT C," ",N IF M = 0 THEN IF N = 1 THEN RETURN LET C = C + 1 IF 2*(N/2)=N THEN GOTO 20 IF M > 10922 THEN GOTO 100 IF N > 21844 THEN GOTO 30 IF N > 10922 THEN GOTO 40 LET N = 3*N + 1 LET M = 3*M GOTO 10 20 LET N = N/2 IF (M/2)*2<>M THEN LET N = N + 16383 REM account for integer truncation LET M=M/2 GOTO 10 30 LET N = N - 21844 REM two ways of accounting for overflow LET N = 3*N + 1 LET M = 3*M + 2 GOTO 10 40 LET N = N - 10922 LET N = 3*N + 1 LET M = 3*M + 1 GOTO 10 50 LET W = 0 REM addition with carry LET V = V + 1 RETURN 60 LET Y = C REM tracking current champion LET F = W LET E = V RETURN 100 PRINT "Uh oh, getting an overflow for 32766x",V," + ",W PRINT "at step number ",C PRINT "trying to triple 32766x",M," + ",N RETURN  Output: Enter a positive integer 27 1 27 2 82 3 41 .... 110 4 111 2 112 1 Testing for longest chain for n<100000... Uh oh, getting an overflow for 32766x2 + 12139 at step number 72 trying to triple 32766x15980 + 7565 The longest sequence starts at 32766x2 + 11499 And goes for 351 steps. ## Batch File 1. Create a routine to generate the hailstone sequence for a number. 2. Show that the hailstone sequence for the number 27 has 112 elements... @echo off setlocal enabledelayedexpansion echo. ::Task #1 call :hailstone 111 echo Task #1: (Start:!sav!) echo !seq! echo. echo Sequence has !cnt! elements. echo. ::Task #2 call :hailstone 27 echo Task #2: (Start:!sav!) echo !seq! echo. echo Sequence has !cnt! elements. echo. pause>nul exit /b 0 ::The Function :hailstone set num=%1 set seq=%1 set sav=%1 set cnt=0 :loop set /a cnt+=1 if !num! equ 1 goto :eof set /a isodd=%num%%%2 if !isodd! equ 0 goto divideby2 set /a num=(3*%num%)+1 set seq=!seq! %num% goto loop :divideby2 set /a num/=2 set seq=!seq! %num% goto loop Output: Task #1: (Start:111) 111 334 167 502 251 754 377 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 has 70 elements. Task #2: (Start: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 Sequence has 112 elements. The script above could only be used in smaller inputs. Thus, for the third task, a slightly different script will be used. However, this script is still slow. I tried this on a fast computer and it took about 40-45 minutes to complete. @echo off setlocal enableDelayedExpansion if "%~1"=="test" ( for /l %%. in () do ( set /a "test1=num %% 2, cnt=cnt+1" if !test1! equ 0 (set /a num/=2 & if !num! equ 1 exit !cnt!) else (set /a num=3*num+1) ) ) set max=0 set record=0 for /l %%X in (2,1,100000) do ( set num=%%X & cmd /c "%~f0" test if !errorlevel! gtr !max! (set /a "max=!errorlevel!,record=%%X") ) set /a max+=1 echo.Number less than 100000 with longest sequence: %record% echo.With length %max%. pause>nul exit /b 0 Output: Number less than 100000 with longest sequence: 77031 With length 351. ## beeswax This approach reuses the main hailstone sequence function for all three tasks. The pure hailstone sequence function, returning the sequence for any number entered in the console:  >@:N q >%"d3~@.PNp d~2~pL~1F{<T_ Returning the sequence for the starting value 27  >@:N q >%"d3~@.PNq d~2~qL~1Ff{<BF3_ {NNgA< Output of the sequence, followed by the length of the sequence: 27 82 41 124 62 31 94 47 ... 2158 1079 3238 1619 4858 2429 7288 3644 1822 ... 16 8 4 2 1 112  Number below 100,000 with the longest hailstone sequence, and the length of that sequence:  >@: q pf1_# >%"d3~@.Pqf#{g? {gpK@~BP9~5@P@q'M< d~2~pL~1Ff< < >?d >zAg?MM@1~y@~gLpz2~yg@~3~hAg?M d >?~fz1~y?yg@hhAg?Mb Output: 77031 351  ## Befunge 93*:. v > :2%v > v+1*3_2/ >" ",:.v v< <v v-1:< < +1\_$1+v^ \
v .,+94<>^>::v
>" "03pv  :* p
v67:" "<  0: 1
>p78p25  *^*p0
v!-1:  <<*^<
9$_:0\ ^-^< v v01g00:< 1 4 >g"@"*+v^ <+ v01/"@":_$ ^,
>p"@"%00p\$:^. vg01g00 ,+49< >"@"*+.@  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 1 112 77031 351 ## BQN Works in: CBQN Collatz ← ⥊⊸{ 𝕨𝕊1: 𝕨; (𝕨⊸∾ 𝕊 ⊢) (2|𝕩)⊑⟨𝕩÷2⋄1+3×𝕩⟩ } Collatz1 ← ⌽∘{ 1: ⟨1⟩; 𝕩∾˜𝕊(2|𝕩)⊑⟨𝕩÷2⋄1+3×𝕩⟩ } •Show Collatz1 5 •Show (⊑∾≠){𝕩⊑˜⊑⍒≠¨𝕩}Collatz1¨1+↕99999  ⟨ 5 16 8 4 2 1 ⟩ ⟨ 77031 351 ⟩  ## 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." ) ) ); ## Brainf***  This example is incomplete. Please ensure that it meets all task requirements and remove this message. >>>>>>,>,>,<< [ .[-<+>] ] > [ .[-<+>] ] > [ .[-<+>] ] <<<< >------------------------------------------------[<<+>>-]> [ <<< [<+>-]< [>++++++++++<-]> >>> ------------------------------------------------ [<<<+>>>-]> [ <<<< [<+>-]< [>++++++++++<-]> >>>> ------------------------------------------------ [<<<<+>>>>-] ] < <<<[>+<<<+>>-]>[-<+>]>>>>>>>>>++++[>+++++++++++<-]++++[>>++++++++<<-]<<<<<<<<<< [ >>>>>>>>>>+>.>.<<<<<<<<<<<< >>+>+<<< [-[->]<]+ >>>[>] <[-<]<[-]< [>+>+<<-]>[<+>-]+ >[ <<<[->>>>+>+>+<<<<<<]>>>>>> [-<<<<<<+>>>>>>]<[-<<<<<+>>>>>]<[-<<<<+>>>>] <<<<+>> - >[-]] <<[-]>[ <<[-<+>[-<->>>>>+>]<<<<<]>>>>[-<<<<+>>>>]<< -] <<[->+>+<<]>[-<+>]> [>>+>+<<<-]>>>[<<<+>>>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-] ++++++++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[->>++++++++[<++++++>-]]<[.[-]<]< -[+>]< ] [This program never terminates! ] [This program isn't complete, (it only prints the hailstone ] [sequence of a number until 1) but it may help other people ] [to make complete versions. ] [ ] [This program only takes in up to 3 digit numbers as input ] [If you want to input 1 digit integers, add a 0 before. e.g ] [04. ] [ ] [Summary: ] [This program takes 16 memory cells of space. Their data is ] [presented below: ] [ ] [Cell 0: Temp cell. ] [Cell 1: Displays the current number. This changes based on ] [Collatz' Conjecture. ] [Cell 14: Displays length of the hailstone sequence. ] [Cell 15: ASCII code for ",". ] [Cell 16: ASCII code for " " (Space). ] [Rest of the cells: Temp cells. ] ## 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 ## Burlesque blsq ) 27{^^^^2.%{3.*1.+}\/{2./}\/ie}{1!=}w!bx{\/+]}{\/isn!}w!L[ 112 ## 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,arry,arry,arry, 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 ### 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; }  ## 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  ### 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)  ## 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 << " " << h27 << " " << h27 << " " << h27 << 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.  ### Library: Qt Uses: Qt Templated solution works for all of Qt's sequential container classes (QLinkedList, QList, QVector). #include <QDebug> #include <QVector> template <class T> T hailstone(typename T::value_type n) { T seq; for (seq << n; n != 1; seq << n) { n = (n&1) ? (3*n)+1 : n/2; } return seq; } template <class T> T longest_hailstone_seq(typename T::value_type n) { T maxSeq; for (; n > 0; --n) { const auto seq = hailstone<T>(n); if (seq.size() > maxSeq.size()) { maxSeq = seq; } } return maxSeq; } int main(int, char *[]) { const auto seq = hailstone<QVector<uint_fast16_t>>(27); qInfo() << "hailstone(27):"; qInfo() << " length:" << seq.size() << "elements"; qInfo() << " first 4 elements:" << seq.mid(0,4); qInfo() << " last 4 elements:" << seq.mid(seq.size()-4); const auto max = longest_hailstone_seq<QVector<uint_fast32_t>>(100000); qInfo() << "longest sequence with starting element under 100000:"; qInfo() << " length:" << max.size() << "elements"; qInfo() << " starting element:" << max.first(); }  Output: hailstone(27): length: 112 elements first 4 elements: QVector(27, 82, 41, 124) last 4 elements: QVector(8, 4, 2, 1) longest sequence with starting element under 100000: length: 351 elements starting element: 77031  ## Ceylon shared void run() { {Integer*} hailstone(variable Integer n) { variable [Integer*] stones = [n]; while(n != 1) { n = if(n.even) then n / 2 else 3 * n + 1; stones = stones.append([n]); } return stones; } value hs27 = hailstone(27); print("hailstone sequence for 27 is hs27.take(3)...hs27.skip(hs27.size - 3).take(3) with length hs27.size"); variable value longest = hailstone(1); for(i in 2..100k - 1) { value current = hailstone(i); if(current.size > longest.size) { longest = current; } } print("the longest sequence under 100,000 starts with longest.first else "what?" and has length longest.size"); }  ## 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

## 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 ")."))


## CLU

% Generate the hailstone sequence for a number
hailstone = iter (n: int) yields (int)
while true do
yield(n)
if n=1 then break end
if n//2 = 0 then
n := n/2
else
n := 3*n + 1
end
end
end hailstone

% Make an array from an iterator
iter_array = proc [T,U: type] (i: itertype (U) yields (T), s: U) returns (array[T])
arr: array[T] := array[T]$[] for item: T in i(s) do array[T]$addh(arr, item) end
return(arr)
end iter_array

start_up = proc ()
po: stream := stream$primary_output() % Generate the hailstone sequence for 27 h27: array[int] := iter_array[int,int](hailstone, 27) lo27: int := array[int]$low(h27)
hi27: int := array[int]$high(h27) stream$putl(po, "The hailstone sequence for 27 has "
|| int$unparse(array[int]$size(h27)) || " elements.")
stream$puts(po, "The first 4 elements are:") for i: int in int$from_to(lo27, lo27+3) do
stream$puts(po, " " || int$unparse(h27[i]))
end
stream$puts(po, ", and the last 4 elements are:") for i: int in int$from_to(hi27-3, hi27) do
stream$puts(po, " " || int$unparse(h27[i]))
end
stream$putl(po, "") % Find whichever sequence < 100 000 has the longest sequence maxnum: int := 0 maxlen: int := 0 for i: int in int$from_to(1, 99999) do
len: int := array[int]$size(iter_array[int,int](hailstone, i)) if len > maxlen then maxnum, maxlen := i, len end end stream$putl(po, int$unparse(maxnum) || " has the longest hailstone sequence < 100000: " || int$unparse(maxlen))
end start_up
Output:
The hailstone sequence for 27 has 112 elements.
The first 4 elements are: 27 82 41 124, and the last 4 elements are: 8 4 2 1
77031 has the longest hailstone sequence < 100000: 351

## COBOL

Testing with GnuCOBOL

       identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.

data division.
working-storage section.
01 most                 constant as 1000000.
01 coverage             constant as 100000.
01 stones               usage binary-long.
01 n                    usage binary-long.
01 storm                usage binary-long.

01 show-arg             pic 9(6).
01 show-default         pic 99 value 27.
01 show-sequence        usage binary-long.
01 longest              usage binary-long occurs 2 times.

01 filler.
05 hail              usage binary-long
occurs 0 to most depending on stones.
01 show                 pic z(10).
01 low-range            usage binary-long.
01 high-range           usage binary-long.
01 range                usage binary-long.

01 remain               usage binary-long.
01 unused               usage binary-long.

procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence

move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.

*> **************************************************************
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if

move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
move n to hail(stones)

if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if

if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.

end program hailstones.

Output:
prompt$cobc -x hailstones.cob prompt$ ./hailstones
+0000000027: 27, 82, 41, 124, ..., 8, 4, 2, 1: +0000000112 elements
Longest at: +0000077031 with +0000000351 elements
prompt$./hailstones 42 +0000000042: 42, 21, 64, 32, ..., 8, 4, 2, 1: +0000000009 elements Longest at: +0000077031 with +0000000351 elements  ## 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 ## 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 ## Cowgol include "cowgol.coh"; # Generate the hailstone sequence for the given N and return the length. # If a non-NULL pointer to a buffer is given, then store the sequence there. sub hailstone(n: uint32, buf: [uint32]): (len: uint32) is len := 0; loop if buf != 0 as [uint32] then [buf] := n; buf := @next buf; end if; len := len + 1; if n == 1 then break; elseif n & 1 == 0 then n := n / 2; else n := 3*n + 1; end if; end loop; end sub; # Generate hailstone sequence for 27 var h27: uint32; var h27len := hailstone(27, &h27); # Print information about it print("The hailstone sequence for 27 has "); print_i32(h27len); print(" elements.\nThe first 4 elements are:"); var n: @indexof h27 := 0; while n < 4 loop print_char(' '); print_i32(h27[n]); n := n + 1; end loop; print(", and the last 4 elements are:"); n := h27len as @indexof h27 - 4; while n as uint32 < h27len loop print_char(' '); print_i32(h27[n]); n := n + 1; end loop print(".\n"); # Find longest hailstone sequence < 100,000 var i: uint32 := 1; var max_i := i; var len: uint32 := 0; var max_len := len; while i < 100000 loop len := hailstone(i, 0 as [uint32]); if len > max_len then max_i := i; max_len := len; end if; i := i + 1; end loop; print_i32(max_i); print(" has the longest hailstone sequence < 100000: "); print_i32(max_len); print_nl(); Output: The hailstone sequence for 27 has 112 elements. The first 4 elements are: 27 82 41 124, and the last 4 elements are: 8 4 2 1. 77031 has the longest hailstone sequence < 100000: 351 ## Crystal def hailstone(n) seq = [n] until n == 1 n = n.even? ? n // 2 : n * 3 + 1 seq << n end seq end max_len = (1...100_000).max_by{|n| hailstone(n).size } max = hailstone(max_len) puts ([max_len, max.size, max.max, max.first(4), max.last(4)]) # => [77031, 351, 21933016, [77031, 231094, 115547, 346642], [8, 4, 2, 1]] twenty_seven = hailstone(27) puts ([twenty_seven.size, twenty_seven.first(4), max.last(4)]) # => [112, [27, 82, 41, 124], [8, 4, 2, 1]]  ## D ### 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 .. $]); writeln("Length hailstone(", M, ")= ", h.length); enum N = 100_000; immutable p = iota(1, N) .map!(i => tuple(i.hailstone.length, i)) .reduce!max; writeln("Longest sequence in [1,", N, "]= ",p," with len ",p); }  Output: hailstone(27)= [27, 82, 41, 124] ... [8, 4, 2, 1] Length hailstone(27)= 112 Longest sequence in [1,100000]= 77031 with len 351 ### Lazy Version Same output. import std.stdio, std.algorithm, std.typecons, std.range; auto hailstone(uint m) pure nothrow @nogc { return m .recurrence!q{ a[n - 1] & 1 ? a[n - 1] * 3 + 1 : a[n - 1]/2} .until!q{ a == 1 }(OpenRight.no); } void main() { enum M = 27; immutable h = M.hailstone.array; writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]); writeln("Length hailstone(", M, ")= ", h.length); enum N = 100_000; immutable p = iota(1, N) .map!(i => tuple(i.hailstone.walkLength, i)) .reduce!max; writeln("Longest sequence in [1,", N, "]= ",p," with len ",p); }  ### Faster Lazy Version Same output. struct Hailstone { uint n; bool empty() const pure nothrow @nogc { return n == 0; } uint front() const pure nothrow @nogc { return n; } void popFront() pure nothrow @nogc { n = n == 1 ? 0 : (n & 1 ? (n * 3 + 1) : n / 2); } } void main() { import std.stdio, std.algorithm, std.range, std.typecons; enum M = 27; immutable h = M.Hailstone.array; writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]); writeln("Length hailstone(", M, ")= ", h.length); enum N = 100_000; immutable p = iota(1, N) .map!(i => tuple(i.Hailstone.walkLength, i)) .reduce!max; writeln("Longest sequence in [1,", N, "]= ",p," with len ",p); }  ### Lazy Version With Caching Faster, same output. import std.stdio, std.algorithm, std.range, std.typecons; struct Hailstone(size_t cacheSize = 500_000) { size_t n; __gshared static size_t[cacheSize] cache; bool empty() const pure nothrow @nogc { return n == 0; } size_t front() const pure nothrow @nogc { return n; } void popFront() nothrow { if (n >= cacheSize) { n = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2); } else if (cache[n]) { n = cache[n]; } else { immutable n2 = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2); n = cache[n] = n2; } } } void main() { enum M = 27; const h = M.Hailstone!().array; writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]); writeln("Length hailstone(", M, ")= ", h.length); enum N = 100_000; immutable p = iota(1, N) .map!(i => tuple(i.Hailstone!().walkLength, i)) .reduce!max; writeln("Longest sequence in [1,", N, "]= ",p," with len ",p); }  ### Generator Range Version import std.stdio, std.algorithm, std.range, std.typecons, std.concurrency; auto hailstone(size_t n) { return new Generator!size_t({ yield(n); while (n > 1) { n = (n & 1) ? (3 * n + 1) : (n / 2); yield(n); } }); } void main() { enum M = 27; const h = M.hailstone.array; writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]); writeln("Length hailstone(", M, ")= ", h.length); enum N = 100_000; immutable p = iota(1, N) .map!(i => tuple(i.hailstone.walkLength, i)) .reduce!max; writeln("Longest sequence in [1,", N, "]= ",p," with len ",p); }  ## Dart import 'package:collection/collection.dart'; import 'dart:collection'; List<int> hailstone(int n) { if(n <= 0) { throw ArgumentError("start value must be >=1)"); } var seq = Queue<int>(); seq.add(n); while(n != 1) { n = n%2 == 0 ? n ~/ 2 : 3 * n + 1; seq.add(n); } return seq.toList(); } main() { for(int i = 1; i <= 10; i++) { print("h($i) = ${hailstone(i)}"); } var h27 = hailstone(27); var first4 = h27.take(4).toList(); print("first 4 elements of h(27):$first4");
assert(ListEquality().equals([27, 82, 41, 124], first4));

var last4 = h27.skip(h27.length - 4).take(4).toList();
print("last 4 elements of h(27): $last4"); assert(ListEquality().equals([8, 4, 2, 1], last4)); print("length of sequence h(27):${h27.length}");
assert(112 == h27.length);

int seq = 0, max = 0;
for(int i = 1; i <= 100000; i++) {
var 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) = 
h(2) = [2, 1]
h(3) = [3, 10, 5, 16, 8, 4, 2, 1]
h(4) = [4, 2, 1]
h(5) = [5, 16, 8, 4, 2, 1]
h(6) = [6, 3, 10, 5, 16, 8, 4, 2, 1]
h(7) = [7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
h(8) = [8, 4, 2, 1]
h(9) = [9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
h(10) = [10, 5, 16, 8, 4, 2, 1]
first 4 elements of h(27): [27, 82, 41, 124]
last 4 elements of h(27): [8, 4, 2, 1]
length of sequence h(27): 112
up to 100000 the sequence h(77031) has the largest length (351)

## Dc

Firstly, this code takes the value from the stack, computes and prints the corresponding Hailstone sequence, and the length of the sequence. The q procedure is for counting the length of the sequence. The e and o procedure is for even and odd number respectively. The x procedure is for overall control.

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

## DCL

$n = f$integer( p1 )
$i = 1$ loop:
$if p2 .nes. "QUIET" then$ s'i = n
$if n .eq. 1 then$ goto done
$i = i + 1$  if .not. n
$then$   n = n / 2
$else$   if n .gt. 715827882 then $exit ! avoid overflowing$   n = 3 * n + 1
$endif$  goto loop
$done:$ if p2 .nes. "QUIET"
$then$  penultimate_i = i - 1
$antepenultimate_i = i - 2$  preantepenultimate_i = i - 3
$write sys$output "sequence has ", i, " elements starting with ", s1, ", ", s2, ", ", s3, ", ", s4, " and ending with ", s'preantepenultimate_i, ", ", s'antepenultimate_i, ", ", s'penultimate_i, ", ", s'i
$endif$ sequence_length == i
Output:
$@hailstone 27 sequence has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1 $ limit = f$integer( p1 )$ i = 1
$max_so_far = 0$ loop:
$call hailstone 'i quiet$  if sequence_length .gt. max_so_far
$then$   max_so_far = sequence_length
$current_record_holder = i$  endif
$i = i + 1$  if i .lt. limit then $goto loop$ write sys$output current_record_holder, " is the number less than ", limit, " which has the longest hailstone sequence which is ", max_so_far, " in length"$ exit
 hailstone: subroutine
$n = f$integer( p1 )
$i = 1$ loop:
$if p2 .nes. "QUIET" then$ s'i = n
$if n .eq. 1 then$ goto done
$i = i + 1$  if .not. n
$then$   n = n / 2
$else$   if n .gt. 715827882 then $exit ! avoid overflowing$   n = 3 * n + 1
$endif$  goto loop
$done:$ if p2 .nes. "QUIET"
$then$  penultimate_i = i - 1
$antepenultimate_i = i - 2$  preantepenultimate_i = i - 3
$write sys$output "sequence has ", i, " elements starting with ", s1, ", ", s2, ", ", s3, ", ", s4, " and ending with ", s'preantepenultimate_i, ", ", s'antepenultimate_i, ", ", s'penultimate_i, ", ", s'i
$endif$ sequence_length == I
$exit$ endsubroutine
Output:
$@longest_hailstone 100000 77031 is the number less than 100000 which has the longest hailstone sequence which is 351 in length ## Delphi ### Using List<Integer> program ShowHailstoneSequence; {$APPTYPE CONSOLE}

uses SysUtils, Generics.Collections;

procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>);
var
n: Integer;
begin
aHailstoneList.Clear;
n := aStartingNumber;

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

var
i: Integer;
lList: TList<Integer>;
lMaxSequence: Integer;
lMaxLength: Integer;
begin
lList := TList<Integer>.Create;
try
GetHailstoneSequence(27, lList);
Writeln(Format('27: %d elements', [lList.Count]));
Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
[lList, lList, lList, lList,
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;

end.

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

Longest sequence under 100,000: 77031 with 351 elements

### Using Boost.Algorithm and TParallel.For

Library: Boost.Int

program ShowHailstoneSequence;

{$APPTYPE CONSOLE} uses System.SysUtils, System.Types, System.Threading, System.SyncObjs, Boost.Algorithm, Boost.Int, System.Diagnostics; var lList: TIntegerDynArray; lMaxSequence, lMaxLength, i: Integer; StopWatch: TStopwatch; begin lList := Hailstone(27); Writeln(Format('27: %d elements', [lList.Count])); Writeln(lList.toString(4), #10); lMaxSequence := 0; lMaxLength := 0; StopWatch := TStopwatch.Create; StopWatch.Start; TParallel.for (1, 1, 100000, procedure(idx: Integer) var lList: TIntegerDynArray; begin lList := Hailstone(idx); if lList.Count > lMaxLength then begin TInterlocked.Exchange(lMaxSequence, idx); TInterlocked.Exchange(lMaxLength, lList.Count); end; end); StopWatch.Stop; Write(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength])); Writeln(Format(' in %d ms', [StopWatch.ElapsedMilliseconds])); Readln; end.  Output: 27: 112 elements [27, 82, 41, 124 ... 8, 4, 2, 1] Longest sequence under 100,000: 77031 with 351 elements in 520 ms  ## 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 ## EasyLang proc hailstone n . list[] . list[] = [ ] while n <> 1 list[] &= n if n mod 2 = 0 n = n / 2 else n = 3 * n + 1 . . list[] &= 1 . hailstone 27 l[] write "27 has length " & len l[] & " with " for i to 4 write l[i] & " " . write "... " for i = len l[] - 3 to len l[] write l[i] & " " . print "" for i = 1 to 100000 hailstone i l[] if len l[] >= max_iter max_i = i max_iter = len l[] end end print max_i & " has length " & max_iter ## EchoLisp (lib 'hash) (lib 'sequences) (lib 'compile) (define (hailstone n) (when (> n 1) (if (even? n) (/ n 2) (1+ (* n 3))))) (define H (make-hash)) ;; (iterator/f seed f) returns seed, (f seed) (f(f seed)) ... (define (hlength seed) (define collatz (iterator/f hailstone seed)) (or (hash-ref H seed) ;; known ? (hash-set H seed (for ((i (in-naturals)) (h collatz)) ;; add length of subsequence if already known #:break (hash-ref H h) => (+ i (hash-ref H h)) (1+ i))))) (define (task (nmax 100000)) (for ((n [1 .. nmax])) (hlength n)) ;; fill hash table (define hmaxlength (apply max (hash-values H))) (define hmaxseed (hash-get-key H hmaxlength)) (writeln 'maxlength= hmaxlength 'for hmaxseed))  Output: (define H27 (iterator/f hailstone 27)) (take H27 6) → (27 82 41 124 62 31) (length H27) → 112 (list-tail (take H27 112) -6) → (5 16 8 4 2 1) (task) maxlength= 351 for 77031 ;; more ... (lib 'bigint) (task 200000) maxlength= 383 for 156159 (task 300000) maxlength= 443 for 230631 (task 400000) maxlength= 443 for 230631 (task 500000) maxlength= 449 for 410011 (task 600000) maxlength= 470 for 511935 (task 700000) maxlength= 509 for 626331 (task 800000) maxlength= 509 for 626331 (task 900000) maxlength= 525 for 837799 (task 1000000) maxlength= 525 for 837799  ## EDSAC order code This program uses no optimization, and is best run on a fast simulator. Even with the storage-related code cut out, Part 2 of the task executes 182 million EDSAC orders. At 650 orders per second, the original EDSAC would have taken 78 hours.  [Hailstone (or Collatz) task for Rosetta Code. EDSAC program, Initial Orders 2.] [This program shows how subroutines can be called via the phi, H, N, ..., V parameters, so that the code doesn't have to be changed if the subroutines are moved about in store. See Wilkes, Wheeler and Gill, 1951 edition, page 18.] [Library subroutine P7, prints long strictly positive integer; 10 characters, right justified, padded left with spaces. Input: 0D = integer to be printed. Closed, even; 35 storage locations; working position 4D.] T 55 K [call subroutine via V parameter] P 56 F [address of subroutine] E 25 K T V GKA3FT26@H28#@NDYFLDT4DS27@TFH8@S8@T1FV4DAFG31@SFLDUFOFFFSFL4F T4DA1FA27@G11@XFT28#ZPFT27ZP1024FP610D@524D!FO30@SFL8FE22@ [Subroutine to print a string placed after the subroutine call. One location per character, with character in top 5 bits. Last character flagged by having bit 0 set. 17 locations, workspace 0F.] T 54 K [call subroutine via C parameter] P 91 F [address of subroutine] E 25 K T C GKH16@A2FG4@A6@A2FT6@AFTFOFCFSFE3@A6@A3FT15@EFV2047F [************ Rosetta Code task ************ Subroutine to generate and optionally store the hailstone (Collatz) sequence for the passed-in initial term n. Input: 4D = n, 35-bit positive integer 6F = start address of sequence if stored; must be even; 0 = don't store Output: 7F = number of terms in sequence, or -1 if error Workspace: 0D (general), 8D (term of sequence) Must be loaded at an even address.] T 45 K [call subroutine via H parameter] P 108 F [address of subroutine] E 25 K T H G K A 3 F T 46 @ H 54#@ [mult reg := 1 to test odd/even] A 4 D [load n passed in by caller] T 8 D [term := n] A 54 @ [load 1 (single)] T 7 F [include initial term in count] A 6 F [load address for store] S 56 @ [test for 0; allow for pre-inc] G 11 @ [skip next if storing not wanted] A 12 @ [make 'T addr D' order]  T 21 @ [plant T order, or -ve value if not storing (note that a T order is +ve as an integer)] [Loop: deal with current term in sequence First store it, if user requested that]  T D [clear acc; also serves to make 'T addr D' order] A 21 @ [load T order to store term] G 22 @ [jump if caller doesn't want store] A 56 @ [pre-inc the address] U 21 @ [update T order] S 51 @ [check not gone beyond max EDSAC address] E 47 @ [error exit if it has] T F [clear acc] A 8 D [load term]  T D [store]  T F [clear acc] A 54#@ [load 1 (double)] S 8 D [1 - term] E 46 @ [if term = 1, jump out with acc = 0] T F [clear acc] C 8 D [acc := term AND 1] S 54#@ [test whether 0 or 1] G 38 @ [jump if term is even] [Here if term is odd; acc = 0] A 8 D [load term] S 52#@ [guard against numeric overflow] E 47 @ [jump if overflow] A 52#@ [restore term after test] L D [term*2] A 8 D [term*3] A 54#@ [plus 1] E 41 @ [join common code] [Here if term is even]  T F [clear acc] A 8 D [load term] R D [term/2] [Common code, acc = new term]  T 8 D [store new term] A 7 F [load count] A 54 @ [add 1] T 7 F [update count] E 12 @ [loop back] [Here when sequence has reached 1 Assume jump here with acc = 0]  E F [return with acc = 0]  T F [here on error] S 54 F [acc := -1] T 7 F [return that as count] E 46 @ [Arrange the following to ensure even addresses for 35-bit values]  T 1024 F [for checking valid address]  H 682 DT 682 D [(2^34 - 1)/3]  P DP F   P 2 F [to change addresses by 2] [Program to demonstrate Rosetta Code subroutine] T 180 K G K [Double constants] [P 500 F P F] [maximum n = 1000"]  & 848 F PF [maximum n = 100000]  P 13 D PF [n = 27 as demo of sequence]  P D PF  [Double variables]  P F P F [n, start of Collatz sequence]  P F P F [n with maximum count] [Single constants]  P 400 F [where to store sequence]  P 2 F [to change addresses by 2]  @ F [carriage return]  & F [line feed]  K 4096 F [null char]  A D [used for maiking 'A addr D' order]  P 8 F [ used for adding 8 to address] [Single variables]  P F [maximum number of terms]  P F [temporary store]  P F [marks end of printing] [Subroutine to print 4 numbers starting at address in 6F. Prints new line (CR, LF) at end.]  A 3 F [plant link for return] T 40 @ A 6 F [load start address] A 15 @ [make 'A addr D' order] A 16 @ [inc address by 8 (4 double values)] U 19 @ [store as test for end] S 16 @ [restore 'A addr D' order for start]  U 31 @ [plant 'A addr D' order in code] S 19 @ [test for end] E 38 @ [out if so] T F [clear acc]  A D [load number] T D [to 0D for printing]  A 33 @ [call print subroutine] G V A 31 @ [load 'A addr D' order] A 11 @ [inc address to next double value] G 27 @ [loop back]  O 12 @ [here when done, print CR LF] O 13 @  E F [return] [Enter with acc = 0] [PART 1]  A 2#@ [load demo value of n] T 4 D [to 4D for subroutine] A 10 @ [address to store sequence] T 6 F [to 6F for subroutine]  A 45 @ [call subroutine to generate sequence] G H A 7 F [load length of sequence] G 198 @ [out if error] T 18 @ [Print result]  A 50 @ [print 'start' message] G C K2048F SF TF AF RF TF !F !F #D A 2#@ [load demo value of n] T D [to 0D for printing]  A 63 @ [print demo n] G V  A 65 @ [print 'length' string] G C K2048F @F &F LF EF NF GF TF HF !F #D T D [ensure 1F and sandwich bit are 0] A 18 @ [load length] T F [to 0F (effectively 0D) for printing]  A 81 @ G V  A 83 @ [print 'first and last four' string] G C K2048F @F &F FF IF RF SF TF !F AF NF DF !F LF AF SF TF !F FF OF UF RF @F &F #D A 18 @ [load length of sequence] L 1 F [times 4] A 6 F [make address of last 4] S 16 @ T 18 @ [store address of last 4]  A 115 @ [print first 4 terms] G 20 @ A 18 @ [retrieve address of last 4] T 6 F [pass as parameter]  A 119 @ [print last 4 terms] G 20 @ [PART 2] T F T 17 @ [max count := 0] T 6#@ [n := 0] [Loop: update n, start new sequence]  T F [clear acc] A 6#@ [load n] A 4#@ [add 1 (double)] U 6#@ [update n] T 4 D [n to 4D for subroutine] T 6 F [say no store]  A 130 @ [call subroutine to generate sequence] G H A 7 F [load count returned by subroutine] G 198 @ [out if error] S 17 @ [compare with max count so far] G 140 @ [skip if less] A 17 @ [restore count after test] T 17 @ [update max count] A 6#@ [load n] T 8#@ [remember n that gave max count]  T F [clear acc] A 6#@ [load n just done] S #@ [compare with max(n)] G 124 @ [loop back if n < max(n) else fall through with acc = 0] [Here whan reached maximum n. Print result.]  A 144 @ [print 'max n' message] G C K2048F MF AF XF !F NF !F !F #D A #@ [load maximum n] T D [to 0D for printing]  A 157 @ [call print subroutine] G V  A 159 @ [print 'max len' message] G C K2048F @F &F MF AF XF !F LF EF NF #D T D [clear 1F and sandwich bit] A 17 @ [load max count (single)] T F [to 0F, effectively to 0D]  A 175 @ [call print subroutine] G V  A 177 @ [print 'at n =' message] G C K2048F @F &F AF TF !F NF !F #F VF !D A 8#@ [load n for which max count occurred] T D [to 0D for printing]  A 192 @ [call print subroutine] G V  O 12 @ [print CR, LF] O 13 @ O 14 @ [print null to flush teleprinter buffer] Z F [stop] [Here if term would overflow EDSAC 35-bit value. With a maximum n of 100,000 this doesn't happen.]  A 198 @ [print 'overflow' message] G C K2048F @F &F OF VF EF RF FF LF OF WD E 194 @ [jump to exit] E 41 Z [define entry point] P F [acc = 0 on entry] Output: START 27 LENGTH 112 FIRST AND LAST FOUR 27 82 41 124 8 4 2 1 MAX N 100000 MAX LEN 351 AT N = 77031  ## Egel import "prelude.eg" namespace Hailstone ( using System using List def even = [ N -> (N%2) == 0 ] def hailstone = [ 1 -> {1} | N -> if even N then cons N (hailstone (N/2)) else cons N (hailstone (N * 3 + 1)) ] def hailpair = [ N -> (N, length (hailstone N)) ] def hailmax = [ (N, NMAX), (M, MMAX) -> if (NMAX < MMAX) then (M, MMAX) else (N, NMAX) ] def largest = [ 1 -> (1, 1) | N -> let M0 = hailpair N in let M1 = largest (N - 1) in hailmax M0 M1 ] ) using System using List using Hailstone def task0 = let H27 = hailstone 27 in length H27 def task1 = let H27 = hailstone 27 in let L = length H27 in (take 4 H27, drop (L - 4) H27) def task2 = largest 100000 def main = (task0, task1, task2) ## Eiffel class APPLICATION create make feature make local test: LINKED_LIST [INTEGER] count, number, te: INTEGER do create test.make test := hailstone_sequence (27) io.put_string ("There are " + test.count.out + " elements in the sequence for the number 27.") io.put_string ("%NThe first 4 elements are: ") across 1 |..| 4 as t loop io.put_string (test [t.item].out + "%T") end io.put_string ("%NThe last 4 elements are: ") across (test.count - 3) |..| test.count as t loop io.put_string (test [t.item].out + "%T") end across 1 |..| 99999 as c loop test := hailstone_sequence (c.item) te := test.count if te > count then count := te number := c.item end end io.put_string ("%NThe longest sequence for numbers below 100000 is " + count.out + " for the number " + number.out + ".") end hailstone_sequence (n: INTEGER): LINKED_LIST [INTEGER] -- Members of the Hailstone Sequence starting from 'n'. require n_is_positive: n > 0 local seq: INTEGER do create Result.make from seq := n until seq = 1 loop Result.extend (seq) if seq \\ 2 = 0 then seq := seq // 2 else seq := ((3 * seq) + 1) end end Result.extend (seq) ensure sequence_terminated: Result.last = 1 end end  Output: There are 112 elements in the sequence for the number 27. The first 4 elements are: 27 82 41 124 The last 4 elements are: 8 4 2 1 The longest sequence for numbers below 100000 is 351 for the number 77031.  ## Elena ELENA 4.x : import system'collections; import extensions; const int maxNumber = 100000; Hailstone(int n,Map<int,int> lengths) { if (n == 1) { ^ 1 }; while (true) { if (lengths.containsKey(n)) { ^ lengths[n] } else { if (n.isEven()) { lengths[n] := 1 + Hailstone(n/2, lengths) } else { lengths[n] := 1 + Hailstone(3*n + 1, lengths) } } } } public program() { int longestChain := 0; int longestNumber := 0; auto recursiveLengths := new Map<int,int>(4096,4096); for(int i := 1, i < maxNumber, i+=1) { var chainLength := Hailstone(i, recursiveLengths); if (longestChain < chainLength) { longestChain := chainLength; longestNumber := i } }; console.printFormatted("max below {0}: {1} ({2} steps)", maxNumber, longestNumber, longestChain) } Output: max bellow 100000: 77031 (351 steps)  ## Elixir defmodule Hailstone do require Integer def step(1) , do: 0 def step(n) when Integer.is_even(n), do: div(n,2) def step(n) , do: n*3 + 1 def sequence(n) do Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list end def run do seq27 = sequence(27) len27 = length(seq27) repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ") IO.puts "Hailstone(27) has #{len27} elements: #{repr}" {len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max 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.  ## EMal fun hailstone = List by int n List h = int[n] while n != 1 n = when((n % 2 == 0), n / 2, 3 * n + 1) h.append(n) end return h end int NUMBER = 27 int LESS_THAN = 100000 List sequence = hailstone(NUMBER) writeLine("The hailstone sequence for the number " + NUMBER + " has " + sequence.length + " elements") writeLine("starting with " + sequence.extractStart(4).join(", ") + " and ending with " + sequence.extractEnd(4).join(", ") + ".") int number = 0 sequence = int[] for int i = 1; i < LESS_THAN; ++i List current = hailstone(i) if current.length > sequence.length sequence = current number = i end end writeLine("The number less than 100000 with longest hailstone sequence is " + number + ", with length of " + sequence.length + ".") 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 less than 100000 with longest hailstone sequence is 77031, with length of 351.  ## Erlang -module(hailstone). -import(io). -export([main/0]). hailstone(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 Erlang 2 This version has one collatz function for just calculating totals (just for fun) and the second generating lists. -module(collatz). -export([main/0,collatz/1,coll/1,max_atz_under/1]). collatz(1) -> 1; collatz(N) when N rem 2 == 0 -> 1 + collatz(N div 2); collatz(N) when N rem 2 > 0 -> 1 + collatz(3 * N +1). max_atz_under(N) -> F = fun (X) -> {collatz(X), X} end, {_, Index} = lists:max(lists:map(F, lists:seq(1, N))), Index. coll(1) -> ; coll(N) when N rem 2 == 0 -> [N|coll(N div 2)]; coll(N) -> [N|coll(3 * N + 1)]. main() -> io:format("collatz(4) non-list total: ~w~n", [collatz(4)]), io:format("coll(4) with lists ~w~n", [coll(4)] ), Seq27 = coll(27), Seq1000 = coll(max_atz_under(100000)), io:format("coll(27) length: ~B~n", [length(Seq27)]), io:format("coll(27) first 4: ~w~n", [lists:sublist(Seq27, 4)]), io:format("collatz(27) last 4: ~w~n", [lists:nthtail(length(Seq27) - 4, Seq27)]), io:format("maximum N <= 100000..."), io:format("Max: ~w~n", [max_atz_under(100000)]), io:format("Total: ~w~n", [ length( Seq1000 ) ] ).  Output 64> collatz:main(). collatz(4) non-list total: 3 coll(4) with lists [4,2,1] coll(27) length: 112 coll(27) first 4: [27,82,41,124] collatz(27) last 4: [8,4,2,1] maximum N <= 100000...Max: 77031 Total: 351 ok  ## ERRE In Italy it's known also as "Ulam conjecture". PROGRAM ULAM !$DOUBLE

PROCEDURE HAILSTONE(X,PRT%->COUNT)
COUNT=1
IF PRT% THEN PRINT(X,) END IF
REPEAT
IF X/2<>INT(X/2) THEN
X=X*3+1
ELSE
X=X/2
END IF
IF PRT% THEN PRINT(X,) END IF
COUNT=COUNT+1
UNTIL X=1
IF PRT% THEN PRINT END IF
END PROCEDURE

BEGIN
HAILSTONE(27,TRUE->COUNT)
PRINT("Sequence length for 27:";COUNT)
MAX_COUNT=2
NMAX=2
FOR I=3 TO 100000 DO
HAILSTONE(I,FALSE->COUNT)
IF COUNT>MAX_COUNT THEN NMAX=I MAX_COUNT=COUNT END IF
END FOR
PRINT("Max. number is";NMAX;" with";MAX_COUNT;"elements")
END PROGRAM
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 for 27: 112
Max. number is 77031 with 351 elements

## 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=4; v=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, ex
351
77031

## 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.


## Excel

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


## Ezhil

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

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

## 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)

## 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. ## 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;.

## Fermat

Array g

Func Collatz(n, d) =
{Runs the Collatz procedure for the number n and returns the number of steps.}
{If d is nonzero, prints the terms in the sequence.}
steps := 1;
while n>1 do
if n|2=0 then n:=n/2 else n:=3n+1 fi;
if d then !!n fi;
steps := steps + 1
od;
steps.

Function LongestTo(n) =
{Finds the number up to n for which the Collatz algorithm takes the most number of steps.}
{The result is stored in the array [g]: g is the number, g is how many steps it takes.}
champ:=0;
record:=0;
for i = 1, n do
q:=Collatz(i, 0);
if q > record then
champ:=i; record:=q; fi;
od;
g:=champ;
g:=record;
.

## 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


## 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

## Frege

Works with: Frege version 3.21.586-g026e8d7
module Hailstone where

import Data.List (maximumBy)

hailstone :: Int -> [Int]
hailstone 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
putStrLn $show$ length h27
let h4 = show $take 4 h27 let t4 = show$ drop (length h27 - 4) h27
putStrLn ("hailstone 27: " ++ h4 ++ " ... " ++ t4)
putStrLn $show$ maximumBy (comparing fst) $map (withResult (length . hailstone)) [1..100000] Output: 112 hailstone 27: [27, 82, 41, 124] ... [8, 4, 2, 1] (351, 77031) runtime 0.969 wallclock seconds.  ## Frink hailstone[n] := { results = new array while n != 1 { results.push[n] if n mod 2 == 0 // n is even? n = n / 2 else n = (3n + 1) } results.push return results } longestLen = 0 longestN = 0 for n = 1 to 100000 { seq = hailstone[n] if length[seq] > longestLen { longestLen = length[seq] longestN = n } } println["$longestN has length $longestLen"] ## FunL def hailstone( 1 ) =  hailstone( n ) = n # hailstone( if 2|n then n/2 else n*3 + 1 ) if _name_ == '-main-' h27 = hailstone( 27 ) assert( h27.length() == 112 and h27.startsWith([27, 82, 41, 124]) and h27.endsWith([8, 4, 2, 1]) ) val (n, len) = maxBy( snd, [(i, hailstone( i ).length()) | i <- 1:100000] ) println( n, len ) Output: 77031, 351  ## Futhark fun hailstone_step(x: int): int = if (x % 2) == 0 then x/2 else (3*x) + 1 fun hailstone_seq(x: int): []int = let capacity = 100 let i = 1 let steps = replicate capacity (-1) let steps = x loop ((capacity,i,steps,x)) = while x != 1 do let (steps, capacity) = if i == capacity then (concat steps (replicate capacity (-1)), capacity * 2) else (steps, capacity) let x = hailstone_step x let steps[i] = x in (capacity, i+1, steps, x) in #1 (split i steps) fun hailstone_len(x: int): int = let i = 1 loop ((i,x)) = while x != 1 do (i+1, hailstone_step x) in i fun max (x: int) (y: int): int = if x < y then y else x fun main (x: int) (n: int): ([]int, int) = (hailstone_seq x, reduce max 0 (map hailstone_len (map (1+) (iota (n-1)))))  ## FutureBasic local fn Hailstone( n as NSInteger ) as NSInteger NSInteger count = 1 while ( n != 1 ) if ( n and 1 ) == 1 n = n * 3 + 1 count++ end if n = n / 2 count++ wend end fn = count void local fn PrintHailstone( n as NSInteger ) NSInteger count = 1, col = 1 print "Sequence for number "; n; ":" : print print using "########"; n; col = 2 while ( n != 1 ) if ( n and 1 ) == 1 n = n * 3 + 1 count++ else n = n / 2 count++ end if print using "########"; n; if col == 10 then print : col = 1 else col++ wend print : print print "Sequence length = "; count end fn window 1, @"Hailstone Sequence", ( 0, 0, 620, 400 ) NSInteger n, seq_num, x, max_x, max_seq seq_num = 27 print fn PrintHailstone( seq_num ) print for x = 1 to 100000 n = fn Hailstone( x ) if n > max_seq max_x = x max_seq = n end if next print "The longest sequence is for "; max_x; ", it has a sequence length of "; max_seq; "." HandleEvents Output: Sequence for number 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 Sequence length = 112 The longest sequence is for 77031, it has a sequence length of 351.  ## Fōrmulæ Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition. Programs in Fōrmulæ are created/edited online in its website. In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation. Part 1. Create a routine to generate the hailstone sequence for a number 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} Part 3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length The sequence for the number 77,031 is the longest one, with 351 terms Part 4. (Additional, not a requirement) Show the number less than 100,000 which has the highest value in its hailstone sequence together with that highest value The sequence for the number 77,671 has the highest term: 1,570,824,736 ## 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 ]  ## 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, seq, seq, seq, 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 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]) }  ## 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] ## Haskell import Data.List (maximumBy) import Data.Ord (comparing) -------------------- HAILSTONE SEQUENCE ------------------ collatz :: Int -> Int collatz n | even n = n div 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst$
maximumBy (comparing snd) $(,) <*> (length . hailstone) <$> [1 .. 100000]

--------------------------- TEST -------------------------
main :: IO ()
main =
mapM_
putStrLn
[ "Collatz sequence for 27: ",
(show . hailstone) 27,
"The number " <> show longestChain,
"has the longest hailstone sequence",
"for any number less then 100000. ",
"The sequence has length: "
<> (show . length . hailstone $longestChain) ]  Output: Collatz sequence for 27: [27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2] The number 77031 has the longest hailstone sequence for any number less then 100000. The sequence has length: 350 The following is an older version, which some of the language examples on this page are translated from: import Data.Ord (comparing) import Data.List (maximumBy, intercalate) hailstone :: Int -> [Int] hailstone 1 =  hailstone n | even n = n : hailstone (n div 2) | otherwise = n : hailstone (n * 3 + 1) withResult :: (Int -> Int) -> Int -> (Int, Int) withResult f x = (f x, x) h27 :: [Int] h27 = hailstone 27 main :: IO () main = mapM_ putStrLn [ (show . length) h27 , "hailstone 27: " ++ intercalate " ... " (show <$> [take 4 h27, drop (length h27 - 4) h27])
, show $maximumBy (comparing fst)$
withResult (length . hailstone) <$> [1 .. 100000] ]  Output: 112 hailstone 27: [27,82,41,124] ... [8,4,2,1] (351,77031) Or, going back to basics, we can observe that the hailstone sequence is an 'anamorphism' – it builds up a list structure from a single integer value, which makes unfoldr the obvious first thing to reach for the first main task. In turn, deriving the longest sequence for starting values below 100000 essentially involves a 'catamorphism' – it takes a list of hailstone sequences (or at least a list of their seed values and their lengths), and strips that structure down to a single (N, length) pair. This makes foldr the obvious recursion scheme to start with for the second main task. One approach to using unfoldr and then foldr might be: import Data.List (unfoldr) -------------------- HAILSTONE SEQUENCE ------------------ hailStones :: Int -> [Int] hailStones = (<> ) . unfoldr go where f x | even x = div x 2 | otherwise = 1 + 3 * x go x | 2 > x = Nothing | otherwise = Just (x, f x) mostStones :: Int -> (Int, Int) mostStones = foldr go (0, 0) . enumFromTo 1 where go x (m, ml) | l > ml = (x, l) | otherwise = (m, ml) where l = length (hailStones x) ------------------------- GENERIC ------------------------ lastN_ :: Int -> [Int] -> [Int] lastN_ = (foldr (const (drop 1)) <*>) . drop --------------------------- TEST ------------------------- h27, start27, end27 :: [Int] [h27, start27, end27] = [id, take 4, lastN_ 4] <*> [hailStones 27] maxNum, maxLen :: Int (maxNum, maxLen) = mostStones 100000 main :: IO () main = mapM_ putStrLn [ "Sequence 27 length:" , show$ length h27
, "Sequence 27 start:"
, show start27
, "Sequence 27 end:"
, show end27
, ""
, "N with longest sequence where N <= 100000"
, show maxNum
, "length of this sequence:"
, show maxLen
]

Output:
Sequence 27 length:
112
Sequence 27 start:
[27,82,41,124]
Sequence 27 end:
[8,4,2,1]

N with longest sequence where N <= 100000
77031
length of this sequence:
351

## 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;

## 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 := 
}
/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
write()
end

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

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


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
->


## 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;
while N is not 1:
if N is even, let N be N / 2;
otherwise let N be (3 * N) + 1;
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.

## 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)
)
stones
)

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.


## 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)))
)


## 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


## 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>();
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
}
return list;
}

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

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

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

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

List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
Output:
Sequence for 27 has 112 elements: [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Method 1, number 77031 has the longest sequence, with a length of 351
Method 2, number 77031 has the longest sequence, with a length of 351
Method 3, number 77031 has the longest sequence, with a length of 351


## JavaScript

### ES5

#### Imperative

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);

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

#### Functional

This simple problem turns out to be a good test of the constraints on composing (ES5) JavaScript code in a functional style.

The first sub-problem falls easily within reach of a basic recursive definition (translating one of the Haskell solutions).

(function () {

// Hailstone Sequence
// n -> [n]
function hailstone(n) {
return n === 1 ?  : (
[n].concat(
hailstone(n % 2 ? n * 3 + 1 : n / 2)
)
)
}

var lstCollatz27 = hailstone(27);

return {
length: lstCollatz27.length,
sequence: lstCollatz27
};

})();

Output:
{"length":112,"sequence":[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]}


Attempting to fold that recursive function over an array of 100,000 elements, however, (to solve the second part of the problem) soon runs out of stack space, at least on the system used here.

The stack problem can be quickly fixed, as often, by simply applying a memoized function, which reuses previously calculated paths.

(function () {

function memoizedHailstone() {
var dctMemo = {};

return function hailstone(n) {
var value = dctMemo[n];

if (typeof value === "undefined") {
dctMemo[n] = value = (n === 1) ?
 : ([n].concat(hailstone(n % 2 ? n * 3 + 1 : n / 2)));
}
return value;
}
}

// Derived a memoized version of the function,
// which can reuse previously calculated paths
var fnCollatz = memoizedHailstone();

// Iterative version of range
// [m..n]
function range(m, n) {
var a = Array(n - m + 1),
i = n + 1;
while (i--) a[i - 1] = i;
return a;
}

// Fold/reduce over an array to find the maximum length
function longestBelow(n) {
return range(1, n).reduce(
function (a, x, i) {
var lng = fnCollatz(x).length;

return lng > a.l ? {
n: i + 1,
l: lng
} : a

}, {
n: 0,
l: 0
}
)
}

return longestBelow(100000);

})();

Output:
// Number, length of sequence
{"n":77031, "l":351}


For better time (as well as space) we can continue to memoize while falling back to a function which returns the sequence length alone, and is iteratively implemented. This also proves more scaleable, and we can still use a fold/reduce pattern over a list to find the longest collatz sequences for integers below one million, or ten million and beyond, without hitting the limits of system resources.

(function (n) {

var dctMemo = {};

// Length only of hailstone sequence
// n -> n
function collatzLength(n) {
var i = 1,
a = n,
lng;

while (a !== 1) {
lng = dctMemo[a];
if ('u' === (typeof lng)) {
a = (a % 2 ? 3 * a + 1 : a / 2);
i++;
} else return lng + i - 1;
}
return i;
}

// Iterative version of range
// [m..n]
function range(m, n) {
var a = Array(n - m + 1),
i = n + 1;
while (i--) a[i - 1] = i;
return a;
}

// Fold/reduce over an array to find the maximum length
function longestBelow(n) {

return range(1, n).reduce(
function (a, x) {

var lng = dctMemo[x] || (dctMemo[x] = collatzLength(x));

return lng > a.l ? {
n: x,
l: lng
} : a

}, {
n: 0,
l: 0
}
)
}

return [100000, 1000000, 10000000].map(longestBelow);

})();

Output:
[
{"n":77031, "l":351},   // 100,000
{"n":837799, "l":525},  // 1,000,000
{"n":8400511, "l":686}  // 10,000,000
]

longestBelow(100000000)
-> {"n":63728127, "l":950}


### ES6

(() => {

// hailstones :: Int -> [Int]
const hailstones = x => {
const collatz = memoized(n =>
even(n) ? div(n, 2) : (3 * n) + 1);
return reverse(until(
xs => xs === 1,
xs => cons(collatz(xs), xs), [x]
));
};

// collatzLength :: Int -> Int
const collatzLength = n =>
until(
xi => xi === 1,
([x, i]) => [(x % 2 ? 3 * x + 1 : x / 2), i + 1], //
[n, 1]
);

// GENERIC FUNCTIONS -----------------------------------------------------

// comparing :: (a -> b) -> (a -> a -> Ordering)
const comparing = f =>
(x, y) => {
const
a = f(x),
b = f(y);
return a < b ? -1 : (a > b ? 1 : 0);
};

// cons :: a -> [a] -> [a]
const cons = (x, xs) => [x].concat(xs);

// div :: Int -> Int -> Int
const div = (x, y) => Math.floor(x / y);

// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);

// even :: Int -> Bool
const even = n => n % 2 === 0;

// fst :: (a, b) -> a
const fst = pair => pair.length === 2 ? pair : undefined;

// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f);

// maximumBy :: (a -> a -> Ordering) -> [a] -> a
const maximumBy = (f, xs) =>
xs.length > 0 ? (
xs.slice(1)
.reduce((a, x) => f(x, a) > 0 ? x : a, xs)
) : undefined;

// memoized :: (a -> b) -> (a -> b)
const memoized = f => {
const dctMemo = {};
return x => {
const v = dctMemo[x];
return v !== undefined ? v : (dctMemo[x] = f(x));
};
};

// reverse :: [a] -> [a]
const reverse = xs =>
xs.slice(0)
.reverse();

// unlines :: [String] -> String
const unlines = xs => xs.join('\n');

// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
};

// MAIN ------------------------------------------------------------------
const
// ceiling :: Int
ceiling = 100000,

// (maxLen, maxNum) :: (Int, Int)
[maxLen, maxNum] =
maximumBy(
comparing(fst),
map(i => [collatzLength(i), i], enumFromTo(1, ceiling))
);
return unlines([
'Collatz sequence for 27: ',
${hailstones(27)}, '', The number${maxNum} has the longest hailstone sequence,
for any starting number under ${ceiling}., '', The length of that sequence is${maxLen}.
]);
})();

Output:

(Run in the Atom editor, through the Script package)

Collatz sequence for 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

The number 77031 has the longest hailstone sequence
for any starting number under 100000.

The length of that sequence is 351.

[Finished in 1.139s]


## jq

Works with: jq version 1.4
# Generate the hailstone sequence as a stream to save space (and time) when counting
def hailstone:
recurse( if . > 1 then
if . % 2 == 0 then ./2|floor else 3*. + 1 end
else empty
end );

def count(g): reduce g as $i (0; .+1); # return [i, length] for the first maximal-length hailstone sequence where i is in [1 .. n] def max_hailstone(n): # state: [i, length] reduce range(1; n+1) as$i
([0,0];
($i | count(hailstone)) as$l
| if $l > . then [$i, $l] else . end); Examples: [27|hailstone] as$h
| "[27|hailstone]|length is \($h|length)", "The first four numbers: \($h[0:4])",
"The last four numbers:  \($h|.[length-4:length])", "", max_hailstone(100000) as$m
| "Maximum length for n|hailstone for n in 1..100000 is \($m) (n == \($m))"
Output:
$jq -M -r -n -f hailstone.jq [27|hailstone]|length is 112 The first four numbers: [27,82,41,124] The last four numbers: [8,4,2,1] Maximum length for n|hailstone for n in 1..100000 is 351 (n == 77031)  ## Julia Works with: Julia version 0.6 and 1.0+ ### Dynamic solution function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing  Output: hailstonelength(27) = 112 findmax((hailstonelength(i) for i = 1:100000)) = (351, 77031)  ### Solution with iterator #### Julia 1.0 Works with: Julia version 1.0+ struct HailstoneSeq{T<:Integer} count::T end Base.eltype(::HailstoneSeq{T}) where T = T function Base.iterate(h::HailstoneSeq, state=h.count) if state == 1 (1, 0) elseif state < 1 nothing elseif iseven(state) (state, state ÷ 2) elseif isodd(state) (state, 3state + 1) end end function Base.length(h::HailstoneSeq) len = 0 for _ in h len += 1 end return len end function Base.show(io::IO, h::HailstoneSeq) f5 = collect(Iterators.take(h, 5)) print(io, "HailstoneSeq{", join(f5, ", "), "...}") end hs = HailstoneSeq(27) println("Collection of the Hailstone sequence from 27:$hs")
cl = collect(hs)
println("First 5 elements: ", join(cl[1:5], ", "))
println("Last 5 elements: ", join(cl[end-4:end], ", "))

Base.isless(h::HailstoneSeq, s::HailstoneSeq) = length(h) < length(s)
println("The number with the longest sequence under 100,000 is: ", maximum(HailstoneSeq.(1:100_000)))

Output:
Collection of the Hailstone sequence from 27: HailstoneSeq{27, 82, 411, 124, 62...}
First 5 elements: 27, 82, 41, 124, 62
Last 5 elements: 16, 8, 4, 2, 1
The number with the longest sequence under 100,000 is: HailstoneSeq{777031, 231094, 115547, 346642, 173321...}

#### Julia 0.6

Works with: Julia version 0.6
struct HailstoneSeq{T<:Integer}
start::T
end

Base.eltype(::HailstoneSeq{T}) where T = T

Base.start(hs::HailstoneSeq) = (-1, hs.start)
Base.done(::HailstoneSeq, state) = state == (1, 4)
function Base.next(::HailstoneSeq, state)
_, s2 = state
s1 = s2
if iseven(s2)
s2 = s2 ÷ 2
else
s2 = 3s2 + 1
end
return s1, (s1, s2)
end

function Base.length(hs::HailstoneSeq)
r = 0
for _ in hs
r += 1
end
return r
end

function Base.show(io::IO, hs::HailstoneSeq)
f5 = collect(Iterators.take(hs, 5))
print(io, "HailstoneSeq(", join(f5, ", "), "...)")
end

hs = HailstoneSeq(27)
println("Collection of the Hailstone sequence from 27: $hs") cl = collect(hs) println("First 5 elements: ", join(cl[1:5], ", ")) println("Last 5 elements: ", join(cl[end-4:end], ", ")) Base.isless(h::HailstoneSeq, s::HailstoneSeq) = length(h) < length(s) println("The number with the longest sequence under 100,000 is: ", maximum(HailstoneSeq.(1:100_000)))  Output: Collection of the Hailstone sequence from 27: HailstoneSeq(27, 82, 41, 124, 62...) First 5 elements: 27, 82, 41, 124, 62 Last 5 elements: 16, 8, 4, 2, 1 The number with the longest sequence under 100,000 is: HailstoneSeq(77031, 231094, 115547, 346642, 173321...) ## K  hail: (1<){:[x!2;1+3*x;_ x%2]}\ seqn: hail 27 #seqn 112 4#seqn 27 82 41 124 -4#seqn 8 4 2 1 {m,x@s?m:|/s:{#hail x}'x}{x@&x!2}!:1e5 351 77031  ## Kotlin import java.util.ArrayDeque fun hailstone(n: Int): ArrayDeque<Int> { val hails = when { n == 1 -> ArrayDeque<Int>() n % 2 == 0 -> hailstone(n / 2) else -> hailstone(3 * n + 1) } hails.addFirst(n) return hails } fun main(args: Array<String>) { val hail27 = hailstone(27) fun showSeq(s: List<Int>) = s.map { it.toString() }.reduce { a, b -> a + ", " + b } println("Hailstone sequence for 27 is " + showSeq(hail27.take(3)) + " ... " + showSeq(hail27.drop(hail27.size - 3)) + " with length${hail27.size}.")

var longestHail = hailstone(1)
for (x in 1..99999)
longestHail = arrayOf(hailstone(x), longestHail).maxBy { it.size } ?: longestHail
println("${longestHail.first} is the number less than 100000 with " + "the longest sequence, having length${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.

## 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.";
]


## 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)
{

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


## Lingo

on hailstone (n, sequenceList)
len = 1
repeat while n<>1
if n mod 2 = 0 then
n = n / 2
else
n = 3 * n + 1
end if
len = len + 1
end repeat
return len
end

Usage:

sequenceList = []
hailstone(27, sequenceList)
put sequenceList
-- [27, 82, 41, 124, ... , 8, 4, 2, 1]

n = 0
maxLen = 0
repeat with i = 1 to 99999
len = hailstone(i)
if len>maxLen then
n = i
maxLen = len
end if
end repeat
put n, maxLen
-- 77031 351

## Logo

to hail.next :n
output ifelse equal? 0 modulo :n 2 [:n/2] [3*:n + 1]
end

to hail.seq :n
if :n = 1 [output ]
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

## 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, ) :-
!.
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


## 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

## 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 ) )


## M2000 Interpreter

Use of two versions of Hailstone, one which return each n, and another one which return only the length of sequence.

Also we use current stack as FIFO to get the last 4 numbers

Module hailstone.Task {
hailstone=lambda  (n as long)->{
=lambda n  (&val) ->{
if n=1 then =false: exit
=true
if n mod 2=0 then n/=2 : val=n: exit
n*=3 : n++: val=n
}
}
Count=Lambda (n) ->{
m=lambda n ->{
if n=1 then =false: exit
=true :if n mod 2=0 then n/=2 :exit
n*=3 : n++
}
c=1
While m() {c++}
=c

}
k=Hailstone(27)
counter=1
x=0
Print 27,
While k(&x) {
counter++
Print x,
if counter=4 then exit
}
Print
Flush  ' empty current stack
While k(&x) {
counter++
data x   ' send to end of stack -used as FIFO
if stack.size>4 then drop
}
\\ [] return a stack object and leave empty current stack
\\ Print use automatic iterator to print all values in columns.
Print []
Print "counter:";counter
m=0
For i=2 to 99999 {
m1=max.data(count(i), m)
if m1<>m then m=m1: im=i
}
Print Format$("Number {0} has then longest hailstone sequence of length {1}", im, m) } hailstone.Task Output:  27 82 41 124 8 4 2 1 counter:112 Number 77031 has then longest hailstone sequence of length 351  ## 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  ## Mathematica / Wolfram Language Here are four ways to generate the sequence. ### Nested function call formulation HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]  This is probably the most readable and shortest implementation. ### Fixed-Point formulation HailstoneFP[n_] := Most@FixedPointList[Switch[#, 1, 1, _?OddQ , 3# + 1, _, #/2] &, n]  ### Recursive formulation HailstoneR = {1} HailstoneR[n_?OddQ] := Prepend[HailstoneR[3 n + 1], n] HailstoneR[n_] := Prepend[HailstoneR[n/2], n]  ### Procedural implementation HailstoneP[n_] := Module[{x = {n}, s = n}, While[s > 1, x = {x, s = If[OddQ@s, 3 s + 1, s/2]}]; Flatten@x]  ### Validation I use this version to do the validation: Hailstone[n_] := NestWhileList[If[Mod[#, 2] == 0, #/2, ( 3*# + 1) ] &, n, # != 1 &]; c27 = Hailstone@27; Print["Hailstone sequence for n = 27: [", c27[[;; 4]], "...", c27[[-4 ;;]], "]"] Print["Length Hailstone = ", 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 = 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 #### Sequence 27 With[{seq = HailstoneFP}, { Length[seq], Take[seq, 4], Take[seq, -4]}]  Output: {112, {27, 82, 41, 124}, {8, 4, 2, 1}} Alternatively, Short[HailstoneFP,0.45]  Output: {27, 82, 41, 124, <<104>>, 8, 4, 2, 1} #### Longest sequence length MaximalBy[Table[{i, Length[HailstoneFP[i]]}, {i, 100000}], Last]  Output: {{77031, 351}} ## MATLAB / Octave ### Hailstone Sequence For N function x = hailstone(n) x = n; while n > 1 % faster than mod(n, 2) if n ~= floor(n / 2) * 2 n = n * 3 + 1; else n = n / 2; end x(end + 1) = n; %#ok end Show sequence of hailstone(27) and number of elements: x = hailstone(27); fprintf('hailstone(27): %d %d %d %d ... %d %d %d %d\nnumber of elements: %d\n', x(1:4), x(end-3:end), numel(x)) Output: hailstone(27): 27 82 41 124 ... 8 4 2 1 number of elements: 112 ### Longest Hailstone Sequence Under N Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length: #### Basic Version (use the above routine) N = 1e5; maxLen = 0; for k = 1:N kLen = numel(hailstone(k)); if kLen > maxLen maxLen = kLen; n = k; end end Output: n = 77031 maxLen = 351 #### Faster Version function [n, maxLen] = longestHailstone(N) maxLen = 0; for k = 1:N a = k; kLen = 1; while a > 1 if a ~= floor(a / 2) * 2 a = a * 3 + 1; else a = a / 2; end kLen = kLen + 1; end if kLen > maxLen maxLen = kLen; n = k; end end Output: >> [n, maxLen] = longestHailstone(1e5) n = 77031 maxLen = 351 #### Much Faster Version With Caching function [n, maxLen] = longestHailstone(N) lenList(N) = 0; lenList(1) = 1; maxLen = 0; for k = 2:N a = k; kLen = 0; while a >= k if a == floor(a / 2) * 2 a = a / 2; else a = a * 3 + 1; end kLen = kLen + 1; end kLen = kLen + lenList(a); lenList(k) = kLen; if kLen > maxLen maxLen = kLen; n = k; end end Output: >> [n, maxLen] = longestHailstone(1e5) n = 77031 maxLen = 351 ## 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] */

## 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.
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 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.

## MiniScript

### Non-cached version

Calculates sequence without using previous calculated sequences.

getSequence = function(n)
results = [n]
while n > 1
if n % 2 then
n = 3 * n + 1
else
n = n / 2
end if
results.push n
end while
return results
end function

h = getSequence(27)
print "The hailstone sequence for 27 has 112 elements starting with"
print h[:4]
print "and ending with"
print h[-4:]

maxSeqLen = 0
maxSeqVal = 0
for i in range(1,100000)
h = getSequence(i)
if h.len > maxSeqLen then
maxSeqLen = h.len
maxSeqVal = i
end if
end for
print
print "The number < 100,000 which has the longest hailstone sequence is " + maxSeqVal + "."
print "This sequence has " + maxSeqLen + " elements."
Output:
The hailstone sequence for 27 has 112 elements starting with
[27, 82, 41, 124]
and ending with
[8, 4, 2, 1]

The number < 100,000 which has the longest hailstone sequence is 77031.
This sequence has 351 elements.

### Cached version

Calculations are stored for used in later calculations.

cache = {}
calc = function(n)
if cache.hasIndex(n) then return
items = [n]
origNum = n
while n > 1 and not cache.hasIndex(n)
if n % 2 then
n = 3 * n + 1
else
n = n /2
end if
items.push n
end while
cache[origNum] = {"len": items.len,"items":items}
end function

getLen = function(n)
if not cache.hasIndex(n) then calc n
if n == 1 then return 1
return cache[n].len + getLen(cache[n].items[-1]) - 1
end function

getSequence = function(n)
if not cache.hasIndex(n) then calc n
if n == 1 then return 
return cache[n].items[:-1] + getSequence(cache[n].items[-1])
end function

h = getSequence(27)
print "The hailstone sequence for 27 has " + h.len + " elements starting with"
print h[:4]
print "and ending with"
print h[-4:]

longSeq = 0
longSeqVal =0
for i in range(2, 100000)
seq = getLen(i)
if longSeq < seq then
longSeq = seq
longSeqVal = i
end if
end for
print "The number < 100,000 which has the longest hailstone sequence is " + longSeqVal + "."
print "This sequence has " + longSeq + " elements."
Output:
Output is the same as the non-cached version above.
The hailstone sequence for 27 has 112 elements starting with
[27, 82, 41, 124]
and ending with
[8, 4, 2, 1]

The number < 100,000 which has the longest hailstone sequence is 77031.
This sequence has 351 elements.

## ML

### MLite

fun hail (x = 1) = 
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)

fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| 	(x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| 	(x :: xs) = hailstorm (x :: xs, 1, 0, 0)

;

val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print  len (h27);
print " elements starting with ";
print  sub (h27, 0, 4);
print " and ending with ";
print  sub (h27, len(h27)-4, len h27);
println ".";

val biggest = hailstorm  iota (100000 - 1);

print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print  ref (biggest, 0);
print " and is of length ";
println  ref (biggest, 1);
Output:
hailstone sequence for the number 27 has 112 elements starting with [27, 82, 41, 124] and ending with [8, 4, 2, 1].
The number less than 100,000 which has the longest hailstone sequence is at element 77031 and is of length 351

## 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. ## 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

## Nanoquery

def hailstone(n)
seq = list()

while (n > 1)
append seq n
if (n % 2)=0
n = int(n / 2)
else
n = int((3 * n) + 1)
end
end
append seq n
return seq
end

h = hailstone(27)
println "hailstone(27)"
println "total elements: " + len(hailstone(27))
print   h + ", " + h + ", " + h + ", " + h + ", ..., "
println h[-4] + ", " + h[-3] + ", " + h[-2] + ", " + h[-1]

max = 0
maxLoc = 0
for i in range(1,99999)
result = len(hailstone(i))
if (result > max)
max = result
maxLoc = i
end
end
print   "\nThe number less than 100,000 with the longest sequence is "
println maxLoc + " with a length of " + max
Output:
hailstone(27)
total elements: 112
27, 82, 41, 124, ..., 8, 4, 2, 1

The number less than 100,000 with the longest sequence is 77031 with a length of 351

## 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


## Nim

proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2

when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."

var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
Output:
Hailstone sequence for number 27 has 112 elements.
This sequence begins with 27, 82, 41, 124 and ends with 8, 4, 2, 1.

For numbers < 100_000, maximum length 351 was found for Hailstone(77031).

## 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.

## 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"] *)

## Oforth

: hailstone   // n -- [n]
| l |
ListBuffer new ->l
while(dup 1 <>) [ dup l add dup isEven ifTrue: [ 2 / ] else: [ 3 * 1+ ] ]
l add l dup freeze ;

hailstone(27) dup size println dup left(4) println right(4) println
100000 seq map(#[ dup hailstone size swap Pair new ]) reduce(#maxKey) println
Output:
112
[27, 82, 41, 124]
[8, 4, 2, 1]
[351, 77031]


## 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


## 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_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.

## Oz

declare
fun {HailstoneSeq N}
N > 0 = true %% assert
if N == 1 then         
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)  ## PARI/GP ### Version #1. 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 ### Version #2. Works with: PARI/GP version 2.7.4 and above Different kind of PARI scripts for Collatz sequences you can find in OEIS, e.g.: A070165 \\ Get vector with Collatz sequence for the specified starting number. \\ Limit vector to the lim length, or less, if 1 (one) term is reached (when lim=0). \\ 3/26/2016 aev Collatz(n,lim=0)={ my(c=n,e=0,L=List(n)); if(lim==0, e=1; lim=n*10^6); for(i=1,lim, if(c%2==0, c=c/2, c=3*c+1); listput(L,c); if(e&&c==1, break)); return(Vec(L)); } Collatzmax(ns,nf)={ my(V,vn,mxn=1,mx,im=1); print("Search range: ",ns,"..",nf); for(i=ns,nf, V=Collatz(i); vn=#V; if(vn>mxn, mxn=vn; im=i); kill(V)); print("Hailstone/Collatz(",im,") has the longest length = ",mxn); } { \\ Required tests: print("Required tests:"); my(Vr,vrn); Vr=Collatz(27); vrn=#Vr; print("Hailstone/Collatz(27): ",Vr[1..4]," ... ",Vr[vrn-3..vrn],"; length = ",vrn); Collatzmax(1,100000); } Output: Required tests: Hailstone/Collatz(27): [27, 82, 41, 124] ... [8, 4, 2, 1]; length = 112 Search range: 1..100000 Hailstone/Collatz(77031) has the longest length = 351 (15:32) gp > ## *** last result computed in 15,735 ms.  ## Pascal See Delphi or try this transformed Delphi version without generics.Use of a static array. program ShowHailstoneSequence; {$IFDEF FPC}
{$MODE delphi} //or objfpc {$Else}
{$Apptype Console} // for delphi {$ENDIF}
uses
SysUtils;// format
const
maxN = 10*1000*1000;// for output 1000*1000*1000

type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr    : tiaArr
end;
tpiaArr = ^tiaArr;

function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
//ensure n to be odd
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;

IF n > 1 then
repeat
//now n == odd -> so two steps in one can be made
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
//now n == even -> so only one step can be made
repeat
n := n shr 1;      inc(result);
until odd(n);
until n = 1;
end;

procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n  := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos  := maxPos;
end;

var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);//319804831
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d  elements',[iaArr,Limit+1]));
write(iaArr,',',iaArr,',',iaArr,',',iaArr,'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;

lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit      : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d    |  %4d   |      %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
end.
Output:
sequence of 27 has 112  elements
27,82,41,124..8,4,2,1

Limit      : number with max length | average length
10 :         9    |    20   |        5.490
100 :        97    |   119   |       27.504
1000 :       871    |   179   |       50.683
10000 :      6171    |   262   |       71.119
100000 :     77031    |   351   |       89.137
1000000 :    837799    |   525   |      108.613
10000000 :   8400511    |   686   |      127.916
100000000 :  63728127    |   950   |      147.337
1000000000 : 670617279    |   987   |      166.780

real  6m22.968s // 32-bit compiled
real  3m56.573s // 64-bit compiled

## Perl

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

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


## Phix

Copy of Euphoria

with javascript_semantics
function hailstone(atom n)
sequence 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 = 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 = hailstone(27)
printf(1,"hailstone(27) = %v\n",{shorten(s,"numbers",4)})

integer hmax = 1, imax = 1,count
for i=2 to 1e5-1 do
count = hailstone_count(i)
if count>hmax then
hmax = count
imax = i
end if
end for

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

Output:
hailstone(27) = {27,82,41,124,"...",8,4,2,1," (112 numbers)"}
The longest hailstone sequence under 100,000 is 77031 with 351 elements.


## 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  ## Picat import util. go => println("H27:"), H27 = hailstoneseq(27), H27Len = H27.len, println(len=H27.len), println(take(H27,4)++['...']++drop(H27,H27Len-4)), nl, println("Longest sequence < 100_000:"), longest_seq(99_999), nl. % The Hailstone value of a number hailstone(N) = N // 2, N mod 2 == 0 => true. hailstone(N) = 3*N+1, N mod 2 == 1 => true. % Sequence for a number hailstoneseq(N) = Seq => Seq := [N], while (N > 1) N := hailstone(N), Seq := Seq ++ [N] end. % % Use a map to cache the lengths. % Here we don't care about the actual sequence. % longest_seq(Limit) => Lens = new_map(), % caching the lengths MaxLen = 0, MaxN = 1, foreach(N in 1..Limit-1) M = N, CLen = 1, while (M > 1) if Lens.has_key(M) then CLen := CLen + Lens.get(M) - 1, M := 1 else M := hailstone(M), % call the CLen := CLen + 1 end end, Lens.put(N, CLen), if CLen > MaxLen then MaxLen := CLen, MaxN := N end end, println([maxLen=MaxLen, maxN=MaxN]), nl. Output: H27: len = 112 [27,82,41,124,...,8,4,2,1] Longest sequence < 100_000: [maxLen = 351,maxN = 77031] ### Mode-directed tabling If we just want to get the length of the longest sequence - and are not forced to use the same Hailstone function as for the H27 task - then this version using model-directed tabling is faster than longest_seq/1: 0.055s vs 0.127s. (Original idea by Neng-Fa Zhou.) go2 => time(max_chain(MaxN,MaxLen)), printf("MaxN=%w,MaxLen=%w%n",MaxN,MaxLen). table (-,max) max_chain(N,Len) => between(2,99_999,N), gen(N,Len). table (+,-) gen(1,Len) => Len=1. gen(N,Len), N mod 2 == 0 => gen(N div 2,Len1), Len=Len1+1. gen(N,Len) => gen(3*N+1,Len1), Len=Len1+1. ## 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 ## 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  ## 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  ### PL/I-80 hailstone_demo: proc options (main); %replace true by '1'b, false by '0'b; dcl (slen, longest) fixed bin(15), (n, n_longest,limit) fixed decimal(12), answer char(1); put skip list ('Display hailstone sequence for what number? '); get list (n); slen = hailstone(n, true); put skip list ('Sequence length = ', slen); put skip(2) list ('Search for longest sequence (y/n)? '); get list (answer); if ((answer ^= 'y') & (answer ^= 'Y')) then stop; put list ('Search to what limit? '); get list (limit); longest = 1; n = 2; do while (n < limit); slen = hailstone(n, false); if slen > longest then do; longest = slen; n_longest = n; end; n = n + 1; end; put skip edit ('Longest sequence =',longest,' for n =',n_longest) (a,f(4),a,f(6)); /* compute, and optionally display, hailstone sequence for n */ hailstone: procedure(n, show) returns (fixed binary); dcl (len, col) fixed binary, (n, k) fixed decimal(12), show bit(1); /* make local copy since n is passed by reference */ k = n; col = 1; len = 1; do while ((k ^= 1) & (k > 0)); if (show) then /* print 8 columns across */ do; put edit (k) (f(8)); col = col + 1; if col > 8 then do; put skip; col = 1; end; end; if (mod(k,2) = 0) then k = k / 2; else k = k * 3 + 1; len = len + 1; end; if (show) then put edit (k) (f(8)); return (len); end hailstone; end hailstone_demo; Output: Display hailstone sequence for what number? 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 1136 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 Search for longest sequence (y/n)? y Search to what limit? 100000 Longest sequence = 351 for n = 77031  ## Plain TeX The following code works with any TeX engine. \newif\ifprint \newcount\itercount \newcount\currentnum \def\hailstone#1{\itercount=0 \currentnum=#1 \hailstoneaux} \def\hailstoneaux{% \advance\itercount1 \ifprint\number\currentnum\space\space\fi \ifnum\currentnum>1 \ifodd\currentnum \multiply\currentnum3 \advance\currentnum1 \else \divide\currentnum2 \fi \expandafter\hailstoneaux \fi } \parindent=0pt \printtrue\hailstone{27} Length = \number\itercount \bigbreak \newcount\ii \ii=1 \printfalse \def\lenmax{0} \def\seed{0} \loop \ifnum\ii<100000 \hailstone\ii \ifnum\itercount>\lenmax\relax \edef\lenmax{\number\itercount}% \edef\seed{\number\ii}% \fi \advance\ii1 \repeat Seed max = \seed, length = \lenmax \bye pdf or dvi 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 Length = 112 Seed max = 77031, length = 351  ## Pointless output = println(format(fmt, [seqLength, initSeq, tailSeq] ++ toList(longestPair) )) fmt = """getSeq(27) (length): {} getSeq(27) (first 4): {} getSeq(27) (last 4): {} max length {} for n = {}""" ----------------------------------------------------------- seq = getSeq(27) seqLength = length(seq) initSeq = take(4, seq) tailSeq = drop(seqLength - 4, seq) ----------------------------------------------------------- longestPair = range(1, 99999) |> map(n => (length(getSeq(n)), n)) |> argmax(at(0)) ----------------------------------------------------------- -- generate full sequence getSeq(n) = iterate(step, n) |> takeUntil(eq(1)) ----------------------------------------------------------- -- get the next number in a sequence step(n) = if n % 2 == 0 then round(n / 2) else n * 3 + 1 Output: getSeq(27) (length): 112 getSeq(27) (first 4): [27, 82, 41, 124] getSeq(27) (last 4): [8, 4, 2, 1] max length 351 for n = 77031 ## PowerShell Works with: PowerShell version 3.0+ 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)}
}
}

function Get-HailStoneBelowLimit {
param($UpperLimit) for ($i = 1; $i -lt$UpperLimit; $i++) { [pscustomobject]@{ 'Number' =$i
'Count' = (Get-HailStone $i).count } } } 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 ## Prolog 1. Create a routine to generate the hailstone sequence for a number. hailstone(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  ### 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-, 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  sequence has 351 len ! true.  ## 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 = ; 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  ## Python ### Procedural 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]) max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000)) print(f"Maximum length {max_length} was found for hailstone({n}) " f"for numbers <100,000") Output: Maximum length 351 was found for hailstone(77031) for numbers <100,000 ### Using a generator from itertools import islice def hailstone(n): yield n while n > 1: n = 3 * n + 1 if n & 1 else n // 2 yield n if __name__ == '__main__': h = hailstone(27) assert list(islice(h, 4)) == [27, 82, 41, 124] for _ in range(112 - 4 * 2): next(h) assert list(islice(h, 4)) == [8, 4, 2, 1] max_length, n = max((sum(1 for _ in hailstone(i)), i) for i in range(1, 100_000)) print(f"Maximum length {max_length} was found for hailstone({n}) " f"for numbers <100,000") Output: Maximum length 351 was found for hailstone(77031) for numbers <100,000 ### Composition of pure functions Works with: Python version 3.7 '''Hailstone sequences''' from itertools import (islice, takewhile) # hailstone :: Int -> [Int] def hailstone(x): '''Hailstone sequence starting with x.''' def p(n): return 1 != n return list(takewhile(p, iterate(collatz)(x))) +  # collatz :: Int -> Int def collatz(n): '''Next integer in the hailstone sequence.''' return 3 * n + 1 if 1 & n else n // 2 # ------------------------- TEST ------------------------- # main :: IO () def main(): '''Tests.''' n = 27 xs = hailstone(n) print(unlines([ f'The hailstone sequence for {n} has {len(xs)} elements,', f'starting with {take(4)(xs)},', f'and ending with {drop(len(xs) - 4)(xs)}.\n' ])) (a, b) = (1, 99999) (i, x) = max( enumerate( map(compose(len)(hailstone), enumFromTo(a)(b)) ), key=snd ) print(unlines([ f'The number in the range {a}..{b} ' f'which produces the longest sequence is {1 + i},', f'generating a hailstone sequence of {x} integers.' ])) # ----------------------- GENERIC ------------------------ # compose (<<<) :: (b -> c) -> (a -> b) -> a -> c def compose(g): '''Function composition.''' return lambda f: lambda x: g(f(x)) # drop :: Int -> [a] -> [a] # drop :: Int -> String -> String def drop(n): '''The sublist of xs beginning at (zero-based) index n. ''' def go(xs): if isinstance(xs, (list, tuple, str)): return xs[n:] else: take(n)(xs) return xs return go # enumFromTo :: (Int, Int) -> [Int] def enumFromTo(m): '''Integer enumeration from m to n.''' return lambda n: range(m, 1 + n) # iterate :: (a -> a) -> a -> Gen [a] def iterate(f): '''An infinite list of repeated applications of f to x. ''' def go(x): v = x while True: yield v v = f(v) return go # snd :: (a, b) -> b def snd(tpl): '''Second component of a tuple.''' return tpl # take :: Int -> [a] -> [a] # take :: Int -> String -> String def take(n): '''The prefix of xs of length n, or xs itself if n > length xs. ''' def go(xs): return ( xs[0:n] if isinstance(xs, (list, tuple)) else list(islice(xs, n)) ) return go # unlines :: [String] -> String def unlines(xs): '''A single newline-delimited string derived from a list of strings.''' return '\n'.join(xs) if __name__ == '__main__': main() Output: The hailstone sequence for 27 has 112 elements, starting with [27, 82, 41, 124], and ending with [8, 4, 2, 1]. The number in the range 1..99999 which produces the longest sequence is 77031, generating a hailstone sequence of 351 integers. ## Quackery [ 1 & ] is odd ( n --> b ) [ [] [ over join swap dup 1 > while dup odd iff [ 3 * 1 + ] else [ 2 / ] swap again ] drop ] is hailstone ( n --> [ ) [ stack ] is longest ( --> s ) [ stack ] is length ( --> s ) 27 hailstone say "The hailstone sequence for 27 has " dup size echo say " elements." cr say "It starts with" dup 4 split drop witheach [ sp echo ] say " and ends with" -4 split nip witheach [ sp echo ] say "." cr cr 0 longest put 0 length put 99999 times [ i^ 1+ hailstone size dup length share > if [ dup length replace i^ 1+ longest replace ] drop ] longest take echo say " has the longest sequence of any number less than 100000." cr say "It is " length take echo say " elements long." cr Output: The hailstone sequence for 27 has 112 elements. It starts with 27 82 41 124 and ends with 8 4 2 1. 77031 has the longest sequence of any number less than 100000. It is 351 elements long. ## R ### Iterative solution ### 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
   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

> twenty.seven$seq.length  112 Between 1 and 1e+05, the input of 77031 gives the longest hailstone sequence, which has length 351. ### Vectorization solution The previous solution is entirely satisfactory and may be more efficient than the following solution. However, problems like these are a great chance to show off the strength of R's vectorization. Also, this lets us show off how the <- syntax can do multiple variable assignments in one line. Observe how short the following code is: ###Task 1: collatz <- function(n) { lastIndex <- 1 output <- lastEntry <- n while(lastEntry != 1) { #Each branch updates lastEntry, lastIndex, and appends a new element to the end of output. #Note that the return value of lastIndex <- lastIndex + 1 is lastIndex + 1. #You may be surprised that output can be appended to despite starting as just a single number. #If so, recall that R's numerics are vectors, meaning that output<-n created a vector of length 1. #It's ugly, but efficient. if(lastEntry %% 2) lastEntry <- output[lastIndex <- lastIndex + 1] <- 3 * lastEntry + 1 else lastEntry <- output[lastIndex <- lastIndex + 1] <- lastEntry %/% 2 } output } ###Task 2: #Notice how easy it is to access the required elements: twentySeven <- collatz(27) cat("The first four elements are:", twentySeven[1:4], "and the last four are:", twentySeven[length(twentySeven) - 3:0], "\n") ###Task 3: #Notice how a several line long loop can be avoided with R's sapply or Vectorize: seqLenghts <- sapply(seq_len(99999), function(x) length(collatz(x))) longest <- which.max(seqLenghts) cat("The longest sequence before the 100000th is found at n =", longest, "and it has length", seqLenghts[longest], "\n") #Equivalently, line 1 could have been: seqLenghts <- sapply(Vectorize(collatz)(1:99999), length). #Another good option would be seqLenghts <- lengths(Vectorize(collatz)(1:99999)). Output: The first four elements are: 27 82 41 124 and the last four are: 8 4 2 1 The longest sequence before the 100000th is found at n = 77031 and it has length 351 ## 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  ## Raku (formerly 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 $m = max ( (1..99_999).race.map: { +hailstone($_) => $_ } ); say "Max length {$m.key} was found for hailstone({$m.value}) for numbers < 100_000"; Output: Length of hailstone(27) = 112 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Max length 351 was found for hailstone(77031) for numbers < 100_000  ## REBOL hail: func [ "Returns the hailstone sequence for n" n [integer!] /local seq ] [ seq: copy reduce [n] while [n <> 1] [ append seq n: either n % 2 == 0 [n / 2] [3 * n + 1] ] seq ] hs27: hail 27 print [ "the hail sequence of 27 has length" length? hs27 "and has the form " copy/part hs27 3 "..." back back back tail hs27 ] maxN: maxLen: 0 repeat n 99999 [ if (len: length? hail n) > maxLen [ maxN: n maxLen: len ] ] print [ "the number less than 100000 with the longest hail sequence is" maxN "with length" maxLen ] Output: the hail sequence of 27 has length 112 and has the form 27 82 41 ... 4 2 1 the number less than 100000 with the longest hail sequence is 77031 with length 351 ## REXX ### non-optimized /*REXX program tests a number and also a range for hailstone (Collatz) sequences. */ numeric digits 20 /*be able to handle gihugeic numbers. */ parse arg x y . /*get optional arguments from the C.L. */ if x=='' | x=="," then x= 27 /*No 1st argument? Then use default.*/ if y=='' | y=="," then y= 100000 - 1 /* " 2nd " " " " */$= hailstone(x)     /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 1▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say  x   ' has a hailstone sequence of '      words($) say ' and starts with: ' subword($, 1, 4)    " ∙∙∙"
say      '    and  ends  with:  ∙∙∙'          subword($, max(5, words($)-3))
if y==0  then exit  /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 2▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say
w= 0;         do j=1  for y;  call hailstone j   /*traipse through the range of numbers.*/
if #hs<=w  then iterate            /*Not big 'nuff?   Then keep traipsing.*/
bigJ= j;   w= #hs                  /*remember what # has biggest hailstone*/
end   /*j*/
say '(between 1 ──►'   y") "       bigJ      ' has the longest hailstone sequence: '   w
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
hailstone: procedure expose #hs; parse arg n 1 s /*N and S: are set to the 1st argument.*/
do #hs=1   while  n\==1     /*keep loop while   N   isn't  unity.  */
if n//2  then n= n * 3  + 1 /*N is odd ?   Then calculate  3*n + 1 */
else n= n % 2      /*"  " even?   Then calculate  fast ÷  */
s= s n                      /* [↑]  %   is REXX integer division.  */
end   /*#hs*/               /* [↑]  append  N  to the sequence list*/
return s                              /*return the  S  string to the invoker.*/
output   when using the default inputs:
27  has a hailstone sequence of  112
and starts with:  27 82 41 124  ∙∙∙
and  ends  with:  ∙∙∙ 8 4 2 1

(between 1 ──► 99999)  77031  has the longest hailstone sequence:  351


### optimized

This version is about   7   times faster than the previous (unoptimized) version.

It makes use of:

•   previously calculated Collatz sequences (memoization)
•   a faster method of determining if an integer is even
/*REXX program tests a  number  and also a  range for  hailstone  (Collatz)  sequences. */
!.=0;     !.0=1;  !.2=1;  !.4=1;  !.6=1;  !.8=1  /*assign even numerals to be  "true".  */
numeric digits 20;  @.= 0                        /*handle big numbers; initialize array.*/
parse arg x y z .;  !.h= y                       /*get optional arguments from the C.L. */
if x=='' | x==","   then x=     27               /*No  1st  argument?  Then use default.*/
if y=='' | y==","   then y= 100000 - 1           /* "  2nd      "        "   "     "    */
if z=='' | z==","   then z=     12               /*head/tail number?     "   "     "    */
hm= max(y, 500000)                               /*use memoization (maximum num for  @.)*/
$= hailstone(x) /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 1▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/ say x ' has a hailstone sequence of ' words($)
say      '    and starts with: '               subword($, 1, z) " ∙∙∙" say ' and ends with: ∙∙∙' subword($, max(z+1, words($)-z+1)) if y==0 then exit /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 2▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/ say w= 0; do j=1 for y;$= hailstone(j)     /*traipse through the range of numbers.*/
#hs= words($) /*find the length of the hailstone seq.*/ if #hs<=w then iterate /*Not big enough? Then keep traipsing.*/ bigJ= j; w= #hs /*remember what # has biggest hailstone*/ end /*j*/ say '(between 1 ──►' y") " bigJ ' has the longest hailstone sequence: ' w exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ hailstone: procedure expose @. !. hm; parse arg n 1 s 1 o,@.1 /*N,S,O: are the 1st arg*/ do while @.n==0 /*loop while the residual is unknown. */ parse var n '' -1 L /*extract the last decimal digit of N.*/ if !.L then n= n % 2 /*N is even? Then calculate fast ÷ */ else n= n * 3 + 1 /*" " odd ? " " 3*n + 1 */ s= s n /* [↑] %: is the REXX integer division*/ end /*while*/ /* [↑] append N to the sequence list*/ s= s @.n /*append the number to a sequence list.*/ @.o= subword(s, 2); parse var s _ r /*use memoization for this hailstone #.*/ do while r\==''; parse var r _ r /*obtain the next hailstone sequence. */ if @._\==0 then leave /*Was number already found? Return S.*/ if _>hm then iterate /*Is number out of range? Ignore it.*/ @._= r /*assign subsequence number to array. */ end /*while*/; return s output when using the default inputs: 27 has a hailstone sequence of 112 and starts with: 27 82 41 124 62 31 94 47 142 71 214 107 ∙∙∙ and ends with: ∙∙∙ 53 160 80 40 20 10 5 16 8 4 2 1 (between 1 ──► 99999) 77031 has the longest hailstone sequence: 351  output when using the inputs: , 1000000 27 has a hailstone sequence of 112 and starts with: 27 82 41 124 62 31 94 47 142 71 214 107 ∙∙∙ and ends with: ∙∙∙ 53 160 80 40 20 10 5 16 8 4 2 1 (between 1 ──► 1000000) 837799 has the longest hailstone sequence: 525  ## Ring size = 27 aList = [] hailstone(size) func hailstone n add(aList,n) while n != 1 if n % 2 = 0 n = n / 2 else n = 3 * n + 1 ok add(aList, n) end see "first 4 elements : " for i = 1 to 4 see "" + aList[i] + " " next see nl see "last 4 elements : " for i = len(aList) - 3 to len(aList) see "" + aList[i] + " " next ## RPL HP-28 emulator's timedog preventing from longlasting code execution, the third item of the task is achieved by calculating the 1-100,000 sequence in increments of 5,000 numbers, with a derivative of the generator that only provides the length of the sequence to increase execution speed. Works with: Halcyon Calc version 4.2.7 Code Comments ≪ DUP 1 →LIST SWAP WHILE DUP 1 ≠ REPEAT IF DUP 2 MOD THEN 3 * 1 + ELSE 2 / END SWAP OVER + SWAP END DROP ≫ 'SYRAQ' STO ≪ 1 SWAP WHILE DUP 1 ≠ REPEAT IF DUP 2 MOD THEN 3 * 1 + ELSE 2 / END SWAP 1 + SWAP END DROP ≫ 'SYRAF' STO ≪ SyMax SYRAF SWAP DUP 5000 + DUP 4 ROLLD FOR n IF DUP n SYRAF < THEN DROP n SYRAF n 'SyMax' STO END NEXT DROP ≫ 'SYR5K' STO   ( n -- { sequence } ) initialize sequence with n Calculate next element Store it into sequence Forget n ( n -- sequence_length ) Initialize counter Calculate next element Increment counter Forget n ( n -- n+10000 ) Initialize loop If sequence length greater than previous ones... ... memorize n  The following instructions deliver what is required: 27 SYRAQ SIZE 1 SYR5K … SYR5K [20 times] DROP SyMax RCL  Output: 2: 127 1: 77031  ## 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 puts "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 = (1 ... 100_000).max_by{|n| hailstone(n).length} puts "#{n} has a hailstone sequence length of #{hailstone(n).length}" puts "the largest number in that sequence is #{hailstone(n).max}" Output: for n = 27, show sequence length and first and last 4 elements [112, [27, 82, 41, 124], [8, 4, 2, 1]] 77031 has a hailstone sequence length of 351 the largest number in that sequence is 21933016  ### With shared structure This version builds some linked lists with shared structure. Hailstone::ListNode is an adaptation of ListNode from Singly-linked list/Element definition#Ruby. When two sequences contain the same value, those two lists share a tail. This avoids recomputing the end of the sequence. Works with: Ruby version 1.8.7 module Hailstone ListNode = Struct.new(:value, :size, :succ) do def each node = self while node yield node.value node = node.succ end end end @@sequence = {1 => ListNode[1,1]} module_function def sequence(n) unless @@sequence[n] m, ary = 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[m, succ.size + 1, succ] end end @@sequence[n] end end puts "for n = 27, show sequence length and first and last 4 elements" hs27 = Hailstone.sequence(27).entries p [hs27.size, hs27[0..3], hs27[-4..-1]] # find the longest sequence among n less than 100,000 n = (1 ... 100_000).max_by{|n| Hailstone.sequence(n).size} puts "#{n} has a hailstone sequence length of #{Hailstone.sequence(n).size}" puts "the largest number in that sequence is #{Hailstone.sequence(n).max}" output is the same as the above. ## Rust fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); } Output: For 27 number of elements is 112 hailstone starting with 27, 82, 41, 124, ending with 8, 4, 2, 1, Longest sequence is 351 element long for seed 77031 ## S-lang % lst=1, return list of elements; lst=0 just return length define hailstone(n, lst) { variable l; if (lst) l = {n}; else l = 1; while (n > 1) { if (n mod 2) n = 3 * n + 1; else n /= 2; if (lst) list_append(l, n); else l++; % if (prn) () = printf("%d, ", n); } % if (prn) () = printf("\n"); return l; } variable har = list_to_array(hailstone(27, 1)), more = 0; () = printf("Hailstone(27) has %d elements starting with:\n\t", length(har)); foreach$1 (har[[0:3]])
() = printf("%d, ", $1); () = printf("\nand ending with:\n\t"); foreach$1 (har[[length(har)-4:]]) {
if (more) () = printf(", ");
more = printf("%d", $1); } () = printf("\ncalculating...\r"); variable longest, longlen = 0, h; _for$1 (2, 99999, 1) {
$2 = hailstone($1, 0);
if ($2 > longlen) { longest =$1;
longlen = $2; () = printf("longest sequence started w/%d and had %d elements \r", longest, longlen); } } () = printf("\n"); Output: Hailstone(27) has 112 elements starting with: 27, 82, 41, 124, and ending with: 8, 4, 2, 1 longest sequence started w/77031 and had 351 elements ## SAS * Create a routine to generate the hailstone sequence for one number; %macro gen_seq(n); data hailstone; array hs_seq(100000); n=&n; do until (n=1); seq_size + 1; hs_seq(seq_size) = n; if mod(n,2)=0 then n=n/2; else n=(3*n)+1; end; seq_size + 1; hs_seq(seq_size)=n; call symputx('seq_length',seq_size); run; proc sql; title "First and last elements of Hailstone Sequence for number &n"; select seq_size as sequence_length, hs_seq1, hs_seq2, hs_seq3, hs_seq4 %do i=&seq_length-3 %to &seq_length; , hs_seq&i %end; from hailstone; quit; %mend; * Use the routine to output the first and last four numbers in the sequence for 27; %gen_seq(27); * Show the number less than 100,000 which has the longest hailstone sequence, and what that length is ; %macro longest_hailstone(start_num, end_num); data hailstone_analysis; do start=&start_num to &end_num; n=start; length_of_sequence=1; do while (n>1); length_of_sequence+1; if mod(n,2)=0 then n=n/2; else n=(3*n) + 1; end; output; end; run; proc sort data=hailstone_analysis; by descending length_of_sequence; run; proc print data=hailstone_analysis (obs=1) noobs; title "Number from &start_num to &end_num with longest Hailstone sequence"; var start length_of_sequence; run; %mend; %longest_hailstone(1,99999); Output:  First and last elements of Hailstone Sequence for number 27 sequence_ length hs_seq1 hs_seq2 hs_seq3 hs_seq4 hs_seq109 hs_seq110 hs_seq111 hs_seq112 ------------------------------------------------------------------------------------------------- 112 27 82 41 124 8 4 2 1 Number from 1 to 99999 with longest Hailstone sequence length_of_ start sequence 77031 351  ## S-BASIC comment Compute and display "hailstone" (i.e., Collatz) sequence for a given number and find the longest sequence in the range permitted by S-BASIC's 16-bit integer data type. end$lines

$constant false = 0$constant true = FFFFH

rem - compute p mod q
function mod(p, q = integer) = integer
end = p - q * (p/q)

comment
Compute, and optionally display, hailstone sequence for n.
Return length of sequence or zero on overflow
end
function hailstone(n, display = integer) = integer
var length = integer
length = 1
while (n <> 1) and (n > 0) do
begin
if display then print using "#####   ", n;
if mod(n,2) = 0 then
n = n / 2
else
n = (n * 3) + 1
length = length + 1
end
if display then print using "#####   ", n
rem - return 0 on overflow
if n < 0 then length = 0
end = length

var n, limit, slen, longest, n_longest = integer

input "Display hailstone sequence for what number"; n
slen = hailstone(n, true)
print "Sequence length = "; slen

rem - find longest sequence before overflow
n = 2
longest = 1
slen = 1
limit = 1000;
print "Searching for longest sequence up to N =", limit," ..."
while (n < limit) and (slen <> 0) do
begin
slen = hailstone(n, false)
if slen > longest then
begin
longest = slen
n_longest = n
end
n = n + 1
end
if slen = 0 then print "Search terminated with overflow at";n-1
print "Maximum sequence length =";longest;" for N =";n_longest

end
Output:
Display hailstone sequence for what number? 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
Sequence length = 112
Searching for longest sequence up to N = 1000 ...
Search terminated with overflow at 447
Maximum sequence length = 144 for N = 327


## 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 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.

## 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)

## Scilab

Translation of: MATLAB
function x=hailstone(n)
// iterative definition
// usage: global verbose; verbose=%T; hailstone(27)
global verbose
x=0; loop=%T
while(loop)
x=x+1
if verbose then
printf('%i ',n)
end
if n==1 then
loop=%F
elseif modulo(n,2)==1 then
n=3*n+1
else
n=n/2
end
end
endfunction

global verbose;
verbose=1;
N=hailstone(27);
printf('\n\n%i\n',N);

global verbose;
verbose=0;
N=100000;
M=zeros(N,1);
for k=1:N
M(k)=hailstone(k);
end;
[maxLength,n]=max(M)
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
112
n  =      77031.
maxLength  =      351.  

$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  ## SETL program hailstone_sequence; hail27 := hailstone(27); print("The hailstone sequence for the number 27 has", #hail27, "elements,"); print("starting with", hail27(..4), "and ending with", hail27(#hail27-3..)); sizes := [#hailstone(n) : n in [1..99999]]; maxsize := max/sizes; maxelem := [n : n in [1..#sizes] | sizes(n) = maxsize](1); print("The number < 100,000 with the longest hailstone sequence is",maxelem); print("The length of its sequence is",sizes(maxelem)); proc hailstone(n); seq := []; loop doing seq with:= n; while n/=1 do if even n then n div:= 2; else n := 3*n + 1; end if; end loop; return seq; end proc; end program; 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 < 100,000 with the longest hailstone sequence is 77031 The length of its sequence is 351 ## Sidef func hailstone (n) { var sequence = [n] while (n > 1) { sequence << ( n.is_even ? n.div!(2) : n.mul!(3).add!(1) ) } return(sequence) } # The hailstone sequence for the number 27 var arr = hailstone(var nr = 27) say "#{nr}: #{arr.first(4)} ... #{arr.last(4)} (#{arr.len})" # The longest hailstone sequence for a number less than 100,000 var h = [0, 0] for i (1 .. 99_999) { (var l = hailstone(i).len) > h && ( h = [i, l] ) } printf("%d: (%d)\n", h...) ## 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. ## SNUSP  /@+@@@+++# 27 | halve odd /===count<<\ /recurse\ #/?\ zero$>@/===!/===-?\==>?!/-<+++\    \!/=!\@\>?!\@/<@\.!\-/
/+<-\!>\?-<+>/++++<\?>+++/*6+4  |    |   \=/  \=itoa=@@@+@+++++#
\=>?/<=!=\   |                  |    !     /+ !/+ !/+ !/+   \    mod10
|//!==/========\         |    /<+> -\!?-\!?-\!?-\!?-\!
/=>?\<=/\<+>!\->+>+<<?/>>=print@/\ln \?!\-?!\-?!\-?!\-?!\-?/\    div10
\+<-/!<     ----------.++++++++++/      #  +/! +/! +/! +/! +/


## Swift

func hailstone(var n:Int) -> [Int] {

var arr = [n]

while n != 1 {

if n % 2 == 0 {
n /= 2
} else {
n = (3 * n) + 1
}

arr.append(n)
}

return arr
}

let n = hailstone(27)

println("hailstone(27): \(n[0...3]) ... \(n[n.count-4...n.count-1]) for a count of \(n.count).")

var longest = (n: 1, len: 1)

for i in 1...100_000 {

let new = hailstone(i)

if new.count > longest.len {
longest = (i, new.count)
}
}

println("Longest sequence for numbers under 100,000 is with \(longest.n). Which has \(longest.len) items.")
Output:
hailstone(27): [27, 82, 41, 124] ... [8, 4, 2, 1] for a count of 112
Longest sequence for numbers under 100,000 is with 77031. Which has 351 items.


## 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  ## TI-83 BASIC ### Task 1 prompt N N→M: 0→X: 1→L While L=1 X+1→X Disp M If M=1 Then: 0→L Else If remainder(M,2)=1 Then: 3*M+1→M Else: M/2→M End End End {N,X} Output:  10 5 16 8 4 2 1 {27,112} ### Task 2 As the calculator is quite slow, so the output is for N=200 prompt N 0→A:0→B for(I,1,N) I→M: 0→X: 1→L While L=1 X+1→X If M=1 Then: 0→L Else If remainder(M,2)=1 Then: 3*M+1→M Else: M/2→M End End End If X>B: Then I→A:X→B End Disp {I,X} End {A,B} Output: {171,125} ## Transd Translation of: Python #lang transd MainModule: { hailstone: (λ n Int() (with seq Vector<Int>([n]) (while (> n 1) (= n (if (mod n 2) (+ (* 3 n) 1) else (/ n 2))) (append seq n) ) (ret seq) ) ), _start: (λ (with h (hailstone 27) l 0 n 0 t 0 (lout "Length of (27): " (size h)) (lout "First 4 of (27): " Range(in: h 0 4)) (lout "Last 4 of (27): " Range(in: h -4 -0)) (for i in Range(100000) do (= t (size (hailstone (to-Int i)))) (if (> t l) (= l t) (= n i)) ) (lout "For n < 100.000 the max. sequence length is " l " for " n) ) ) } Output: Length of (27): 112 First 4 of (27): [27, 82, 41, 124] Last 4 of (27): [8, 4, 2, 1] For n < 100.000 the max. sequence length is 351 for 77031  ## 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

## uBasic/4tH

Translation of: FreeBASIC
' ------=< MAIN >=------

m = 0
Proc _hailstone_print(27)
Print

For x = 1 To 10000
n = Func(_hailstone(x))
If n > m Then
t = x
m = n
EndIf
Next

Print  "The longest sequence is for "; t; ", it has a sequence length of "; m

End

_hailstone_print Param (1)
' print the number and sequence

Local (1)
b@ = 1

Print "sequence for number "; a@
Print Using "________"; a@;   'starting number

Do While a@ # 1
If (a@ % 2 ) = 1 Then
a@ = a@ * 3 + 1   ' n * 3 + 1
Else
a@ = a@ / 2       ' n / 2
EndIf

b@ = b@ + 1
Print Using "________"; a@;

If (b@ % 10) = 0 Then Print
Loop

Print : Print
Print "sequence length = "; b@
Print

For b@ = 0 To 79
Print "-";
Next

Print
Return

_hailstone Param (1)
' normal version
' only counts the sequence

Local (1)
b@ = 1

Do While a@ # 1
If (a@ % 2) = 1 Then
a@ = a@ * 3 + 1  ' n * 3 + 1
Else
a@ = a@ / 2      ' divide number by 2
EndIf

b@ = b@ + 1
Loop

Return (b@)

uBasic is an interpreted language. Doing a sequence up to 100,000 would take over an hour, so we did up to 10,000 here.

Output:
sequence for number 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

sequence length = 112
--------------------------------------------------------------------------------
The longest sequence is for 6171, it has a sequence length of 262



## 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

### 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."

### 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."
Output:
$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. ## Ursa ### Implementation hailstone.u import "math" def hailstone (int n) decl int<> seq while (> n 1) append n seq if (= (mod n 2) 0) set n (floor (/ n 2)) else set n (int (+ (* 3 n) 1)) end if end while append n seq return seq end hailstone ### Usage Output: > import "hailstone.u" > out (hailstone 27) endl console class java.lang.Integer<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> > out (size (hailstone 27)) endl console 112 > decl int i max maxLoc > for (set i 1) (< i 100000) (inc i) .. decl int result .. set result (size (hailstone i)) .. .. if (> result max) .. set max result .. set maxLoc i .. end if ..end for > out "hailstone(" maxLoc ")= " max endl console hailstone(77031)= 351 > _ ## 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 ## VBA Translation of: Phix Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub Output: hailstone(27) = 27, 82, 41, 124, ... 16, 8, 4, 2, 1 length = 112 The longest hailstone sequence under 100,000 is 77031 with 351 elements. ## VBScript 'function arguments: "num" is the number to sequence and "return" is the value to return - "s" for the sequence or '"e" for the number elements. Function hailstone_sequence(num,return) n = num sequence = num elements = 1 Do Until n = 1 If n Mod 2 = 0 Then n = n / 2 Else n = (3 * n) + 1 End If sequence = sequence & " " & n elements = elements + 1 Loop Select Case return Case "s" hailstone_sequence = sequence Case "e" hailstone_sequence = elements End Select End Function 'test driving. 'show sequence for 27 WScript.StdOut.WriteLine "Sequence for 27: " & hailstone_sequence(27,"s") WScript.StdOut.WriteLine "Number of Elements: " & hailstone_sequence(27,"e") WScript.StdOut.WriteBlankLines(1) 'show the number less than 100k with the longest sequence count = 1 n_elements = 0 n_longest = "" Do While count < 100000 current_n_elements = hailstone_sequence(count,"e") If current_n_elements > n_elements Then n_elements = current_n_elements n_longest = "Number: " & count & " Length: " & n_elements End If count = count + 1 Loop WScript.StdOut.WriteLine "Number less than 100k with the longest sequence: " WScript.StdOut.WriteLine n_longest Output: Sequence for 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 Number of Elements: 112 Number less than 100k with the longest sequence: Number: 77031 Length: 351  ## Visual Basic Translation of: PL/I Works with: Visual Basic version VB6 Standard Option Explicit Dim flag As Boolean ' true to print values Sub main() Dim longest As Long, n As Long Dim i As Long, value As Long ' Task 1: flag = True i = 27 Debug.Print "The hailstone sequence has length of "; i; " is "; hailstones(i) ' Task 2: flag = False longest = 0 For i = 1 To 99999 If longest < hailstones(i) Then longest = hailstones(i) value = i End If Next i Debug.Print value; " has the longest sequence of "; longest End Sub 'main Function hailstones(n As Long) As Long Dim m As Long, p As Long Dim m1 As Long, m2 As Long, m3 As Long, m4 As Long If flag Then Debug.Print "The sequence for"; n; "is: "; p = 1 m = n If flag Then Debug.Print m; While m > 1 p = p + 1 If (m Mod 2) = 0 Then m = m / 2 Else m = 3 * m + 1 End If If p <= 4 Then If flag Then Debug.Print m; m4 = m3 m3 = m2 m2 = m1 m1 = m Wend If flag Then If p <= 4 Then Debug.Print ElseIf p = 5 Then Debug.Print m1 ElseIf p = 6 Then Debug.Print m2; m1 ElseIf p = 7 Then Debug.Print m3; m2; m1 ElseIf p = 8 Then Debug.Print m4; m3; m2; m1 Else Debug.Print "..."; m4; m3; m2; m1 End If End If hailstones = p End Function 'hailstones Output: The sequence for 27 is: 27 82 41 124 ... 8 4 2 1 The hailstone sequence has length of 27 is 112 77031 has the longest sequence of 351  ## 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.  ## V (Vlang) Translation of: go // 1st arg is the number to generate the sequence for. // 2nd arg is a slice to recycle, to reduce garbage. fn hs(nn int, recycle []int) []int { mut n := nn mut s := recycle[..0] s << n for n > 1 { if n&1 == 0 { n /= 2 } else { n = 3*n + 1 } s << n } return s } fn main() { mut seq := hs(27, []) println("hs(27):$seq.len elements: [${seq}${seq} ${seq}${seq} ... ${seq[seq.len-4]}${seq[seq.len-3]} ${seq[seq.len-2]}${seq[seq.len-1]}]")

mut max_n, mut max_len := 0,0
for n in 1..100000 {
seq = hs(n, seq)
if seq.len > max_len {
max_n = n
max_len = seq.len
}
}
println("hs($max_n):$max_len elements")
}
Output:
hs(27): 112 elements: [27 82 41 124 ... 8 4 2 1]
hs(77031): 351 elements

## Wren

var hailstone = Fn.new { |n|
if (n < 1) Fiber.abort("Parameter must be a positive integer.")
var h = [n]
while (n != 1) {
n = (n%2 == 0) ? (n/2).floor : 3*n + 1
}
return h
}

var h = hailstone.call(27)
System.print("For the Hailstone sequence starting with n = 27:")
System.print("   Number of elements  = %(h.count)")
System.print("   First four elements = %(h[0..3])")
System.print("   Final four elements = %(h[-4..-1])")

System.print("\nThe Hailstone sequence for n < 100,000 with the longest length is:")
var longest = 0
var longlen = 0
for (n in 1..99999) {
var h = hailstone.call(n)
var c = h.count
if (c > longlen) {
longest = n
longlen = c
}
}
System.print("   Longest = %(longest)")
System.print("   Length  = %(longlen)")
Output:
For the Hailstone sequence starting with n = 27:
Number of elements  = 112
First four elements = [27, 82, 41, 124]
Final four elements = [8, 4, 2, 1]

The Hailstone sequence for n < 100,000 with the longest length is:
Longest = 77031
Length  = 351


## 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


## Z80 Assembly

This task will be split into two parts, in order to fit all the output of each on one Amstrad CPC screen.

### Show The Sequence with n=27

Output is in hexadecimal but is otherwise correct.

;;;;;;;;;;;;;;;;;;; HEADER   ;;;;;;;;;;;;;;;;;;;
;;;;;;;;;;;;;;;;;;; PROGRAM  ;;;;;;;;;;;;;;;;;;;

org &8000
ld de,27
call doHailstone
;returns length of sequence, and writes each entry in the sequence
;	to RAM

;print the sequence length (in hex)
ld a,h
call ShowHex
ld a,l
ld (memdump_smc),a
;just to prove I didn't need to know the sequence length at
;	compile time, I'll store the calculated length as the operand
;	of "doMemDump" which normally takes a constant embedded after
;	it as the number of bytes to display.

;      If that doesn't make sense, don't worry.
;      This has nothing to do with calculating the hailstone sequence, just showing the results.
call ShowHex

call NewLine		;prints CRLF
call NewLine

call doMemDump
memdump_smc:
byte 0			;operand of "doMemDump" (gets overwritten with the sequence length)
word HailstoneBuffer	;operand of "doMemDump"

ret

;;;;;;;;;;;;;;;;;;; LIBRARY  ;;;;;;;;;;;;;;;;;;;

doHailstone:
;you need the proper input for the function "hailstone"
;returns addr. of last element in IX.
call hailstone
ld de,HailstoneBuffer
or a	;clear carry
push ix
pop hl		;returns element count in HL.
sbc hl,de	;subtract the two to get the length of the array.

SRL H
RR L		;divide array size by 2, since each entry is 2 bytes.
INC L
ret nz		;if no carry, don't increment H.
INC H
ret

hailstone:
;input - de = n
ld ix,HailstoneBuffer
ld a,d
or e
ret z	;zero is not allowed.
loop_hailstone:
ld (IX+0),e
ld (IX+1),d
ld a,e
cp 1
jr nz,continue_hailstone
ld a,d
or a
ret z	;if de = 1, stop.
continue_hailstone:
bit 0,e
jr z,DE_IS_EVEN
;de is odd
push de
pop hl		;ld hl,de

SLA E
RL D

ld de,1

push hl
pop de		;ld de,hl

inc ix
inc ix
jr loop_hailstone

DE_IS_EVEN:
SRL D		;A/2
RR E
inc ix
inc ix
jr loop_hailstone

doMemDump:
;show the hailstone sequence to the screen. This is just needed to display the data, if you don't care about that
pop hl		;get PC
ld b,(hl)	;get byte count
inc hl
ld e,(hl)	;get low byte of start addr.
inc hl
ld d,(hl)	;get high byte of start addr.
inc hl
push hl		;now when we return we'll skip the data block.
ex de,hl

call NewLine

;we'll dump 8 words per line.

ld c,8
loop_doMemDump:
inc hl
ld a,(hl)
call ShowHex
dec hl
ld a,(hl)
call ShowHex
ld a,' '
call PrintChar
inc hl
inc hl
dec c
ld a,c
and %00001111
jr nz,continueMemdump
ld c,8
continueMemdump:
djnz loop_doMemDump
ret

HailstoneBuffer:
ds 512,0
Output:
call &8000
0070

001B 0052 0029 007C 003E 001F 005E 002F
008E 0047 00D6 006B 0142 00A1 01E4 00F2
0079 016C 00B6 005B 0112 0089 019C 00CE
0067 0136 009B 01D2 00E9 02BC 015E 00AF
01BD 0538 029C 014E 00A7 01F6 00FB 02F2
0179 046C 0236 011B 0352 01A9 04FC 027E
013F 03BE 01DF 059E 02CF 086E 0437 0CA6
0653 12FA 097D 1C78 0E3C 071E 038F 0AAE
0557 1006 0803 180A 0C05 2410 1208 0904
0482 0241 06C4 0362 01B1 0514 028A 0145
03D0 01E8 00F4 007A 003D 00B8 005C 002E
0017 0046 0023 006A 0035 00A0 0050 0028
0014 000A 0005 0010 0008 0004 0002 0001


## 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) n.clear(c.len(),c); n}.fp1(L(0,0)))
Output:
L(351,77031)  // length, hailstone


## ZX Spectrum Basic

Translation of: BBC_BASIC
10 LET n=27: LET s=1
20 GO SUB 1000
30 PRINT '"Sequence length = ";seqlen
40 LET maxlen=0: LET s=0
50 FOR m=2 TO 100000
60 LET n=m
70 GO SUB 1000
80 IF seqlen>maxlen THEN LET maxlen=seqlen: LET maxnum=m
90 NEXT m
100 PRINT "The number with the longest hailstone sequence is ";maxnum
110 PRINT "Its sequence length is ";maxlen
120 STOP
1000 REM Hailstone
1010 LET l=0
1020 IF s THEN PRINT n;"  ";
1030 IF n=1 THEN LET seqlen=l+1: RETURN
1040 IF FN m(n,2)=0 THEN LET n=INT (n/2): GO TO 1060
1050 LET n=3*n+1
1060 LET l=l+1
1070 GO TO 1020
2000 DEF FN m(a,b)=a-INT (a/b)*b`