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{{task}}
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 <big><code> = n/2 </code></big>
* If n is '''odd''' then the next n of the sequence <big><code> = (3 * n) + 1 </code></big>
The (unproven) [[wp:Collatz conjecture|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).
;Task:
# Create a routine to generate the hailstone sequence for a number.
# Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with <code>27, 82, 41, 124</code> and ending with <code>8, 4, 2, 1</code>
# Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.<br> (But don't show the actual sequence!)
;See also:
* [http://xkcd.com/710 xkcd] (humourous).
* [https://terrytao.files.wordpress.com/2020/02/collatz.pdf The Notorious Collatz conjecture] Terence Tao, UCLA (Presentation, pdf).
* [https://www.youtube.com/watch?v=094y1Z2wpJg The Simplest Math Problem No One Can Solve] Veritasium (video, sponsored).
<br><br>
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight lang="11l">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[0], m[1]))</syntaxhighlight>
{{out}}
<pre>
Maximum length 351 was found for hailstone(77031) for numbers <100,000
</pre>
=={{header|360 Assembly}}==
<syntaxhighlight lang="360asm">* 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
BR R14 return to caller
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</syntaxhighlight>
{{out}}
<pre>
hailstone( 27)= 112 : 27 82 41 ...... 4 2 1
longest <100000 : hailstone( 77031)= 351
</pre>
=={{header|ABAP}}==
<syntaxhighlight lang="abap">
CLASS lcl_hailstone DEFINITION.
PUBLIC SECTION.
Line 106 ⟶ 258:
cl_demo_output=>write( lcl_hailstone=>get_longest_sequence_upto( 100000 ) ).
cl_demo_output=>display( ).
</syntaxhighlight>
{{out}}
<pre>
Line 123 ⟶ 275:
</pre>
=={{header|ABC}}==
<syntaxhighlight lang="abc">HOW TO RETURN hailstone n:
PUT {} IN seq
WHILE 1=1:
PUT n IN seq[#seq]
SELECT:
n=1: RETURN seq
n mod 2=0: PUT floor(n/2) IN n
n mod 2=1: PUT 3*n+1 IN n
RETURN seq
PUT hailstone 27 IN h27
WRITE "Length of Hailstone sequence for 27:", #h27/
WRITE "First 4 elements:", h27[0], h27[1], h27[2], h27[3]/
WRITE "Last 4 elements:", h27[#h27-4], h27[#h27-3], h27[#h27-2], h27[#h27-1]/
PUT 0, 0 IN longest, length
FOR n IN {1..100000}:
PUT hailstone n IN hn
IF #hn > length:
PUT n, #hn IN longest, length
WRITE longest, "has the longest hailstone sequence < 100,000, of length:", length/</syntaxhighlight>
{{out}}
<pre>Length of Hailstone sequence for 27: 112
First 4 elements: 27 82 41 124
Last 4 elements: 8 4 2 1
77031 has the longest hailstone sequence < 100,000, of length: 351</pre>
=={{header|ACL2}}==
<
(loop for x = len
then (if (evenp x)
Line 139 ⟶ 319:
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))</
{{out}}
Line 153 ⟶ 333:
=={{header|Ada}}==
Similar to [[#C|C method]]:
<
procedure hailstone is
type int_arr is array(Positive range <>) of Integer;
Line 191 ⟶ 371:
end loop;
put_line(Integer'Image(nmax)&" max @ n= "&Integer'Image(stonemax));
end hailstone;</
{{out}}
<pre>
Line 203 ⟶ 383:
hailstones.ads:
<
type Integer_Sequence is array(Positive range <>) of Integer;
function Create_Sequence (N : Positive) return Integer_Sequence;
end Hailstones;</
hailstones.adb:
<
function Create_Sequence (N : Positive) return Integer_Sequence is
begin
Line 222 ⟶ 402:
end if;
end Create_Sequence;
end Hailstones;</
example main.adb:
<
with Hailstones;
Line 266 ⟶ 446:
Ada.Text_IO.Put_Line ("with N =" & Integer'Image (Longest_N));
end;
end Main;</
{{out}}
<pre>Length of 27: 112
Line 275 ⟶ 455:
=={{header|Aime}}==
<
print_hailstone(integer h)
{
Line 282 ⟶ 462:
while (h ^ 1) {
lb_p_integer(l, h);
h =
}
o_form("hailstone sequence for ~ is ~1 ~ ~ ~ .. ~ ~ ~ ~, it is ~ long\n",
l[0], l[1], l[2], l[3], l[-3], l[-2], l[-1], 1,
}
Line 293 ⟶ 473:
{
integer e, i, m;
m = 0;
Line 303 ⟶ 483:
l = 1;
while (h ^ 1) {
if (
l += k;
break;
} else {
l += 1;
h =
}
}
if (m < l) {
Line 322 ⟶ 502:
}
o_form("hailstone sequence length for ~ is ~
}
Line 332 ⟶ 512:
return 0;
}</
{{out}}
<pre>hailstone sequence for 27 is 27 82 41 124 .. 8 4 2 1, it is 112 long
hailstone sequence length for 77031 is 351
=={{header|ALGOL 60}}==
{{works with|A60}}
<syntaxhighlight lang="algol60">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 </syntaxhighlight>
{{out}}
<pre>
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.
</pre>
=={{header|ALGOL 68}}==
Line 342 ⟶ 577:
{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}}
{{works with|ELLA ALGOL 68|Any (with appropriate job cards) - using the ''print'' routine rather than ''printf''}}
<
PROC hailstone = (INT in n, REF[]LINT array)INT:
Line 388 ⟶ 623:
print(("Max",hmax," at j=",jatmax, new line))
#
)</
{{out}}
<pre>
Line 394 ⟶ 629:
[ 77031, 231094, 115547, 346642, ..., 8, 4, 2, 1] len=351
Max 351 at j=77031
</pre>
=={{header|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.
<syntaxhighlight lang="algol">
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?");
READ(N);
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 </syntaxhighlight>
{{out}}
<pre>
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</pre>
=={{header|ALGOL W}}==
<syntaxhighlight lang="algolw">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.</syntaxhighlight>
{{out}}
<pre>
27: length 112, first: [27 82 41 124] last: [8 4 2 1]
Maximum sequence length: 351 for: 77031
</pre>
=={{header|Amazing Hopper}}==
<syntaxhighlight lang="c">
#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,\
" fue encontrada para Hailstone(",valor,\
#(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 '
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|APL}}==
{{works with|Dyalog APL}}
<syntaxhighlight lang="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
}
⍝ tradfn version:
∇seq←hailstone n;next
⍝ Returns the hailstone sequence for a given number
Line 406 ⟶ 870:
n←next[1+2|n] ⍝ Pick the appropriate next step
seq,←n ⍝ Append that to the sequence
:EndWhile
∇
</syntaxhighlight>
{{out}}
<syntaxhighlight lang="apl"> dfnHailstone 5
5 16 8 4 2 1
5↑hailstone 27
27 82 41 124 62
¯5↑hailstone 27
Line 415 ⟶ 884:
112
1↑{⍵[⍒↑(⍴∘hailstone)¨⍵]}⍳100000
77031</
=={{header|AppleScript}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{output}}
<pre>{|length of sequence|:112, |first 4 numbers|:{27, 82, 41, 124}, |last 4 numbers|:{8, 4, 2, 1}}</pre>
<syntaxhighlight lang="applescript">-- 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}</syntaxhighlight>
{{output}}
<pre>{|number(s) giving longest sequence length|:{77031}, |length of sequence|:351}</pre>
=={{header|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 ===
<syntaxhighlight lang="arm assembly"> .org 0x08000000
b ProgramStart
;cartridge header goes here
;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
; 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
adr r1,HailStoneMessage_Init
bl PrintString
bl NewLine
bl ShowHex32
bl NewLine
bl NewLine
bl Hailstone
;function is complete, return the output
adr r1,HailStoneMessage_0
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
adr r1,HailStoneMessage_1
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
adr r1,HailStoneMessage_2
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:
;R1 = scratchpad
;R3 = counter for entries in the sequence.
;R5 = pointer to output ram
stmfd sp!,{r4-r12,lr}
mov r5,#0x02000000
add r5,r5,#4
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
addne r0,r1,r0
addne r0,r0,#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:
.byte "Your input was:",255
.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!</syntaxhighlight>
{{out}}
<pre>
Your input was:
0000001B
First 4 numbers are:
0000001B
00000052
00000029
0000007C
Last 4 numbers are:
00000008
00000004
00000002
00000001
Sequence length is:
00000070
</pre>
[https://ibb.co/WzgWjZ9 Picture of output on VisualBoyAdvance screen]
===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 <code>ResetTextCursors</code> but before the infinite loop:
<syntaxhighlight lang="arm assembly"> 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
add r0,r0,#1
cmp r0,r8
blt loop_getBiggestHailstone
adr r1,HailstoneMessage_0
bl PrintString
bl NewLine
adr r1,HailStoneMessage_1
bl PrintString
bl NewLine
mov r0,r7
bl ShowHex32
bl NewLine
adr r1,HailStoneMessage_2
bl PrintString
bl NewLine
mov r0,r6
bl ShowHex32
bl NewLine</syntaxhighlight>
{{out}}
<pre>
The number that makes the
biggest sequence is:
00012CE7
And that sequence has a length
of:
0000015F
</pre>
[https://ibb.co/hXm7cJ3 Output of second program on VisualBoyAdvance's screen]
=={{header|Arturo}}==
<syntaxhighlight lang="rebol">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]
</syntaxhighlight>
{{out}}
<pre>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
</pre>
=={{header|AutoHotkey}}==
<
; [1] Generate the Hailstone Seq. for a number
Line 457 ⟶ 1,245:
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max</
=={{header|AutoIt}}==
<
$Hail = Hailstone(27)
ConsoleWrite("Sequence-Lenght: "&$Hail&@CRLF)
Line 490 ⟶ 1,279:
Return $Counter
EndFunc ;==>Hailstone
</syntaxhighlight>
{{out}}
<pre>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,
Line 502 ⟶ 1,291:
=={{header|AWK}}==
<
#!/usr/bin/awk -f
function hailstone(v, verbose) {
Line 532 ⟶ 1,321:
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
</syntaxhighlight>
{{out}}
<pre>
Line 542 ⟶ 1,331:
=={{header|BASIC}}==
==={{header|Applesoft BASIC}}===
<
100 N = 27
Line 573 ⟶ 1,362:
470 IF EVEN THEN N=N/2
480 IF NOT EVEN THEN N = (3 * N) + 1
490 NEXT L : STOP</
==={{header|BBC BASIC}}===
<
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
Line 598 ⟶ 1,387:
L% += 1
ENDWHILE
= L% + 1</
{{out}}
<pre>
Line 620 ⟶ 1,409:
Its sequence length is 351
</pre>
==={{header|Commodore BASIC}}===
<syntaxhighlight lang="qbasic">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
</syntaxhighlight>
==={{header|FreeBASIC}}===
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>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</pre>
==={{header|GW-BASIC}}===
<syntaxhighlight lang="gwbasic">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</syntaxhighlight>
==={{header|Liberty BASIC}}===
<
print ""
while hailstone < 1 or hailstone <> int(hailstone)
Line 673 ⟶ 1,618:
print ""
print "The number less than 100,000 with the longest 'Hailstone Sequence' is "; largesthailstone;". It's sequence length is "; max;"."
end</
==={{header|OxygenBasic}}===
<
function Hailstone(sys *n)
Line 715 ⟶ 1,660:
'result 100000, 77031, 351
</syntaxhighlight>
==={{header|PureBasic}}===
<
Procedure.i FillHailstones(n) ; Fills the list & returns the amount of elements in the list
Line 765 ⟶ 1,710:
Print(#CRLF$+#CRLF$+"Press ENTER to exit."): Input()
CloseConsole()
EndIf</
{{out}}
Line 796 ⟶ 1,741:
==={{header|Run BASIC}}===
<
print ""
Line 835 ⟶ 1,780:
end if
wend
END FUNCTION</
==={{header|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.
<syntaxhighlight lang="text"> 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</syntaxhighlight>
{{out}}
<pre>
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.</pre>
=={{header|Batch File}}==
''1. Create a routine to generate the hailstone sequence for a number. ''<br>
''2. Show that the hailstone sequence for the number 27 has 112 elements... ''
<!--The third task is NOT included because task #3 will take several hours in processing...-->
<syntaxhighlight lang="dos">@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
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
set
goto loop</syntaxhighlight>
{{Out}}
<pre>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.</pre>
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.
<syntaxhighlight lang="dos">@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</syntaxhighlight>
{{Out}}
<pre>Number less than 100000 with longest sequence: 77031
With length 351.</pre>
=={{header|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:
<syntaxhighlight lang="beeswax"> >@:N q
>%"d3~@.PNp
d~2~pL~1F{<T_</syntaxhighlight>
'''Returning the sequence for the starting value <code>27</code>'''
<syntaxhighlight lang="beeswax"> >@:N q
>%"d3~@.PNq
d~2~qL~1Ff{<BF3_
{NNgA<</syntaxhighlight>
Output of the sequence, followed by the length of the sequence:
<syntaxhighlight lang="text">
27
82
41
124
62
31
94
47
...
2158
1079
3238
1619
4858
2429
7288
3644
1822
...
16
8
4
2
1
112</syntaxhighlight>
'''Number below 100,000 with the longest hailstone sequence, and the length of that sequence:'''
<syntaxhighlight lang="beeswax"> >@: 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</syntaxhighlight>
Output:
<syntaxhighlight lang="text">77031 351</syntaxhighlight>
=={{header|Befunge}}==
<
> :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<
>"@"*+.@
</syntaxhighlight>
{{out}}
<pre>
Line 885 ⟶ 2,045:
77031
351</pre>
=={{header|BQN}}==
Works in: [[CBQN]]
<syntaxhighlight lang="bqn">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</syntaxhighlight>
<syntaxhighlight lang="bqn">⟨ 5 16 8 4 2 1 ⟩
⟨ 77031 351 ⟩</syntaxhighlight>
=={{header|Bracmat}}==
<
( hailstone
= L len
Line 929 ⟶ 2,107:
)
)
);</
=={{header|Brainf***}}==
{{incomplete}}
<syntaxhighlight lang="brainf***">>>>>>>,>,>,<<
[
.[-<+>]
]
>
[
.[-<+>]
]
>
[
.[-<+>]
]
<<<<
>------------------------------------------------[<<+>>-]>
[
<<<
[<+>-]<
[>++++++++++<-]>
>>>
------------------------------------------------
[<<<+>>>-]>
[
------------------------------------------------
[<<
]
<
<<<[>+<<<+>>-]>[-<+>]>>>>>>>>>++++[>+++++++++++<-]++++[>>++++++++<<-]<<<<<<<<<<
[
>>>>>>>>>>+>.>.<<<<<<<<<<<<
>>+>+<<<
[-[->]<]+
>>>[>]
<[-<]<[-]<
[>+>+<<-]>[<+>-]+
>[
<<<[->>>>+>+>+<<<<<<]>>>>>>
[-<<<<<<+>>>>>>]<[-<<<<<+>>>>>]<[-<<<<+>>>>]
<<<<+>>
-
>[-]]
<<[-]>[
<<[-<+>[-<->>>>>+>]<<<<<]>>>>[-<<<<+>>>>]<<
-]
<<[->+>+<<]>[-<+>]>
[>>+>+<<<-]>>>[<<<+>>>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]
++++++++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[->>++++++++[<++++++>-]]<[.[-]<]<
-[+>]<
]
[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. ]
</syntaxhighlight>
=={{header|Brat}}==
<
sequence = [num]
while { num != 1 }
Line 982 ⟶ 2,228:
}
p "Longest was starting from #{longest[:number]} and was of length #{longest[:length]}"</
{{out}}
<pre>Sequence for 27 is correct
Line 992 ⟶ 2,238:
Longest so far: 77031 @ 351 elements
Longest was starting from 77031 and was of length 351</pre>
=={{header|Bruijn}}==
<syntaxhighlight lang="bruijn">
:import std/Combinator .
:import std/List .
:import std/Math M
:import std/Number/Binary .
# hailstone sequence using binary shifts
hailstone y [[(0 =? (+1b)) {}0 go]]
go 0 : (=²?0 (1 /²0) (1 (↑¹0 + 0)))
# --- tests ---
seq-27 hailstone (+27b)
:test (∀seq-27) ((+112))
:test (take (+4) seq-27) ((+27b) : ((+82b) : ((+41b) : {}(+124b))))
:test (take (+4) <~>seq-27) ((+1b) : ((+2b) : ((+4b) : {}(+8b))))
below-100000 [0 : ∀(hailstone 0)] <$> seq
seq take (+99999) (iterate ++‣ (+1b))
main [head (max-by (M.compare ⋔ tail) below-100000)]
</syntaxhighlight>
=={{header|Burlesque}}==
<
blsq ) 27{^^^^2.%{3.*1.+}\/{2./}\/ie}{1!=}w!bx{\/+]}{\/isn!}w!L[
112
</syntaxhighlight>
=={{header|C}}==
<
#include <stdlib.h>
Line 1,041 ⟶ 2,313:
return 0;
}</
{{out}}
<pre>[ 27, 82, 41, 124, ...., 8, 4, 2, 1] len= 112
Line 1,048 ⟶ 2,320:
===With caching===
Much faster if you want to go over a million or so.
<
#define N 10000000
Line 1,078 ⟶ 2,350:
printf("max below %d: %d, length %d\n", N, mi, max);
return 0;
}</
=={{header|C sharp|C#}}==
<
using System.Collections.Generic;
using System.Linq;
Line 1,124 ⟶ 2,396:
}
}
}</
<pre>
112 Elements
Line 1,133 ⟶ 2,405:
===With caching===
As with the [[#C|C example]], much faster if you want to go over a million or so.
<
using System.Collections.Generic;
Line 1,174 ⟶ 2,446:
}
}
}</
<pre>
max below 100000: 77031 (351 steps)
Line 1,180 ⟶ 2,452:
=={{header|C++}}==
<
#include <vector>
#include <utility>
Line 1,229 ⟶ 2,501:
return 0;
}</
{{out}}
Line 1,237 ⟶ 2,509:
the longest hailstone sequence under 100,000 is 77031 with 351 elements.
</pre>
=== {{libheader|Qt}} ===
{{uses from|library|Qt}}
Templated solution works for all of Qt's sequential container classes (QLinkedList, QList, QVector).
<syntaxhighlight lang="cpp">
#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();
}
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Ceylon}}==
<syntaxhighlight lang="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``");
}</syntaxhighlight>
=={{header|CLIPS}}==
<
(slot bound) ; upper bound for the range of values to check
(slot next (default 2)) ; next value that needs to be checked
Line 1,314 ⟶ 2,665:
(printout t "sequence is " ?start " with a length of " ?len "." crlf)
(printout t crlf)
)</
{{out}}
Line 1,326 ⟶ 2,677:
=={{header|Clojure}}==
<
{:pre [(pos? n)]}
(lazy-seq
Line 1,342 ⟶ 2,693:
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))</
=={{header|CLU}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|COBOL}}==
Testing with GnuCOBOL
<syntaxhighlight lang="cobol"> 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
add 1 to stones
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
add 1 to stones
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.</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|CoffeeScript}}==
Recursive version:
<
if n is 1
[n]
Line 1,371 ⟶ 2,887:
maxnums = [i]
console.log "Max length: #{maxlength}; numbers generating sequences of this length: #{maxnums}"</
<pre>hailstone(27) = 27,82,41,124 ... 8,4,2,1 (length: 112)
Max length: 351; numbers generating sequences of this length: 77031</pre>
=={{header|Common Lisp}}==
<
(cond ((= n 1) '(1))
((evenp n) (cons n (hailstone (/ n 2))))
Line 1,386 ⟶ 2,902:
(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:
<pre>ROSETTA> (length (hailstone 27))
Line 1,397 ⟶ 2,913:
Longest hailstone sequence under 100000 for 77031, having length 351.
NIL</pre>
=={{header|Cowgol}}==
<syntaxhighlight lang="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[113];
var h27len := hailstone(27, &h27[0]);
# 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();</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Crystal}}==
<syntaxhighlight lang="ruby">
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]]
</syntaxhighlight>
=={{header|D}}==
===Basic Version===
<
auto hailstone(uint n) pure nothrow {
Line 1,422 ⟶ 3,029:
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</
{{out}}
<pre>hailstone(27)= [27, 82, 41, 124] ... [8, 4, 2, 1]
Line 1,430 ⟶ 3,037:
===Lazy Version===
Same output.
<
auto hailstone(uint m) pure nothrow @nogc {
Line 1,449 ⟶ 3,056:
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</
===Faster Lazy Version===
Same output.
<
uint n;
bool empty() const pure nothrow @nogc { return n == 0; }
Line 1,475 ⟶ 3,082:
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</
===Lazy Version With Caching===
Faster, same output.
<
struct Hailstone(size_t cacheSize = 500_000) {
Line 1,511 ⟶ 3,118:
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</
===Generator Range Version===
<
auto hailstone(size_t n) {
Line 1,537 ⟶ 3,144:
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</
=={{header|Dart}}==
<syntaxhighlight lang="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)");
}
</syntaxhighlight>
{{out}}
<pre>h(1) = [1]
h(2) = [2, 1]
h(3) = [3, 10, 5, 16, 8, 4, 2, 1]
h(4) = [4, 2, 1]
h(5) = [5, 16, 8, 4, 2, 1]
h(6) = [6, 3, 10, 5, 16, 8, 4, 2, 1]
h(7) = [7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
h(8) = [8, 4, 2, 1]
h(9) = [9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
h(10) = [10, 5, 16, 8, 4, 2, 1]
first 4 elements of h(27): [27, 82, 41, 124]
last 4 elements of h(27): [8, 4, 2, 1]
length of sequence h(27): 112
up to 100000 the sequence h(77031) has the largest length (351)</pre>
=={{header|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.
<syntaxhighlight lang="dc">27
[[--: ]nzpq]sq
[d 2/ p]se
[d 3*1+ p]so
[d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx</syntaxhighlight>
{{out}}
<pre>
82
41
124
62
(omitted)
8
4
2
1
--: 112
</pre>
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.
<syntaxhighlight lang="dc">0dsLsT1st
[dsLltsT]sM
[[zdlL<M q]sq
[d 2/]se
[d 3*1+ ]so
[d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx]ss
[lt1+dstlsxc lt100000>l]dslx
lTn[:]nlLp
</syntaxhighlight>
{{out}} (Takes quite some time on a decent machine)
<pre>77031:351</pre>
=={{header|DCL}}==
<
$ i = 1
$ loop:
Line 1,561 ⟶ 3,268:
$ 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</
{{out}}
<pre>$ @hailstone 27
sequence has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1</pre>
<
$ i = 1
$ max_so_far = 0
Line 1,579 ⟶ 3,286:
$ 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 )
Line 1,604 ⟶ 3,312:
$ sequence_length == I
$ exit
$ endsubroutine</
{{out}}
<pre>$ @longest_hailstone 100000
77031 is the number less than 100000 which has the longest hailstone sequence which is 351 in length</pre>
=={{header|Delphi}}==
=== Using List<Integer> ===
<syntaxhighlight lang="delphi">program ShowHailstoneSequence;
{$APPTYPE CONSOLE}
uses SysUtils, Generics.Collections;
procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>);
var
n: Integer;
begin
aHailstoneList.Clear;
aHailstoneList.Add(aStartingNumber);
n := aStartingNumber;
while n <> 1 do
begin
if Odd(n) then
n := (3 * n) + 1
else
n := n div 2;
aHailstoneList.Add(n);
end;
end;
var
i: Integer;
lList: TList<Integer>;
lMaxSequence: Integer;
lMaxLength: Integer;
begin
lList := TList<Integer>.Create;
try
GetHailstoneSequence(27, lList);
Writeln(Format('27: %d elements', [lList.Count]));
Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
[lList[0], lList[1], lList[2], lList[3],
lList[lList.Count - 4], lList[lList.Count - 3], lList[lList.Count - 2], lList[lList.Count - 1]]));
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
for i := 1 to 100000 do
begin
GetHailstoneSequence(i, lList);
if lList.Count > lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lList.Count;
end;
end;
Writeln(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength]));
finally
lList.Free;
end;
Readln;
end.</syntaxhighlight>
{{out}}
<pre>27: 112 elements
[27 82 41 124 ... 8 4 2 1]
Longest sequence under 100,000: 77031 with 351 elements</pre>
=== Using Boost.Algorithm and TParallel.For ===
{{libheader| System.SysUtils}}
{{libheader| System.Types}}
{{libheader| System.Threading}}
{{libheader| System.SyncObjs}}
{{libheader| Boost.Algorithm}}
{{libheader| Boost.Int}}[https://github.com/MaiconSoft/DelphiBoostLib]
{{libheader| System.Diagnostics}}
<syntaxhighlight lang="delphi">
program ShowHailstoneSequence;
{$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.
</syntaxhighlight>
{{out}}
<pre>
27: 112 elements
[27, 82, 41, 124 ... 8, 4, 2, 1]
Longest sequence under 100,000: 77031 with 351 elements in 520 ms
</pre>
=={{header|Déjà Vu}}==
<
swap [ over ]
while < 1 dup:
Line 1,645 ⟶ 3,491:
!print( "number: " to-str max ", length: " to-str maxlen )
else:
@hailstone</
{{out}}
<pre>true
Line 1,658 ⟶ 3,504:
number: 77031, length: 351</pre>
=={{header|
<syntaxhighlight lang=easylang>
proc hailstone n . list[] .
list[] = [ ]
while n <> 1
list[] &= n
if n mod 2 = 0
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
</syntaxhighlight>
=={{header|EchoLisp}}==
<syntaxhighlight lang="scheme">
(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))
</syntaxhighlight>
{{out}}
<syntaxhighlight lang="scheme">
(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
</syntaxhighlight>
=={{header|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.
<syntaxhighlight lang="edsac">
[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]
[11] 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]
[12] 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]
[21] T D [store]
[22] 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]
[38] T F [clear acc]
A 8 D [load term]
R D [term/2]
[Common code, acc = new term]
[41] 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]
[46] E F [return with acc = 0]
[47] 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]
[51] T 1024 F [for checking valid address]
[52] H 682 DT 682 D [(2^34 - 1)/3]
[54] P DP F [1]
[56] 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"]
[0] & 848 F PF [maximum n = 100000]
[2] P 13 D PF [n = 27 as demo of sequence]
[4] P D PF [1]
[Double variables]
[6] P F P F [n, start of Collatz sequence]
[8] P F P F [n with maximum count]
[Single constants]
[10] P 400 F [where to store sequence]
[11] P 2 F [to change addresses by 2]
[12] @ F [carriage return]
[13] & F [line feed]
[14] K 4096 F [null char]
[15] A D [used for maiking 'A addr D' order]
[16] P 8 F [ used for adding 8 to address]
[Single variables]
[17] P F [maximum number of terms]
[18] P F [temporary store]
[19] P F [marks end of printing]
[Subroutine to print 4 numbers starting at address in 6F.
Prints new line (CR, LF) at end.]
[20] 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]
[27] U 31 @ [plant 'A addr D' order in code]
S 19 @ [test for end]
E 38 @ [out if so]
T F [clear acc]
[31] A D [load number]
T D [to 0D for printing]
[33] 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]
[38] O 12 @ [here when done, print CR LF]
O 13 @
[40] E F [return]
[Enter with acc = 0]
[PART 1]
[41] 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]
[45] 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]
[50] 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]
[63] A 63 @ [print demo n]
G V
[65] 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]
[81] A 81 @
G V
[83] 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]
[115] A 115 @ [print first 4 terms]
G 20 @
A 18 @ [retrieve address of last 4]
T 6 F [pass as parameter]
[119] 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]
[124] 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]
[130] 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]
[140] 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.]
[144] 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]
[157] A 157 @ [call print subroutine]
G V
[159] 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]
[175] A 175 @ [call print subroutine]
G V
[177] 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]
[192] A 192 @ [call print subroutine]
G V
[194] 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.]
[198] 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]
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Egel}}==
<syntaxhighlight lang="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)) ]
[ N -> (N, length (hailstone N)) ]
def hailmax =
[ (N, NMAX), (M, MMAX) -> if (NMAX < MMAX) then (M, MMAX) else (N, NMAX) ]
[ 1 -> (1, 1)
let M0 = hailpair N 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)
</syntaxhighlight>
=={{header|Eiffel}}==
<syntaxhighlight lang="eiffel">
class
APPLICATION
Line 1,873 ⟶ 3,975:
hailstone_sequence (n: INTEGER): LINKED_LIST [INTEGER]
-- Members of the Hailstone Sequence
require
n_is_positive: n > 0
Line 1,898 ⟶ 4,000:
end
</syntaxhighlight>
{{out}}
<pre>
Line 1,905 ⟶ 4,007:
The last 4 elements are: 8 4 2 1
The longest sequence for numbers below 100000 is 351 for the number 77031.
</pre>
=={{header|Elena}}==
ELENA 6.x :
<syntaxhighlight lang="elena">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)
}</syntaxhighlight>
{{out}}
<pre>
max bellow 100000: 77031 (351 steps)
</pre>
=={{header|Elixir}}==
<
require Integer
Line 1,930 ⟶ 4,089:
end
Hailstone.run</
{{out}}
Line 1,936 ⟶ 4,095:
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.
</pre>
=={{header|EMal}}==
<syntaxhighlight lang="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 + ".")
</syntaxhighlight>
{{out}}
<pre>
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.
</pre>
=={{header|Erlang}}==
<
-import(io).
-export([main/0]).
Line 1,962 ⟶ 4,158:
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]).</
{{out}}
<pre>Eshell V5.8.4 (abort with ^G)
Line 1,976 ⟶ 4,172:
ok</pre>
'''Erlang 2'''
This version has one collatz function for just calculating totals (just for fun) and the second generating lists.
<syntaxhighlight lang="erlang">
-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) -> [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 ) ] ).
</syntaxhighlight>
'''Output'''
<pre>
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
</pre>
=={{header|ERRE}}==
In Italy it's known also as "Ulam conjecture".
<syntaxhighlight lang="erre">
PROGRAM ULAM
Line 2,010 ⟶ 4,253:
PRINT("Max. number is";NMAX;" with";MAX_COUNT;"elements")
END PROGRAM
</syntaxhighlight>
{{out}}
<pre>
Line 2,042 ⟶ 4,285:
=={{header|Euler Math Toolbox}}==
<syntaxhighlight lang="euler math toolbox">
>function hailstone (n) ...
$ v=[n];
Line 2,087 ⟶ 4,330:
351
77031
</syntaxhighlight>
=={{header|Euphoria}}==
<
sequence s
s = {n}
Line 2,135 ⟶ 4,378:
printf(1,"The longest hailstone sequence under 100,000 is %d with %d elements.\n",
{imax,max})</
{{out}}
<pre>hailstone(27) =
Line 2,156 ⟶ 4,399:
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
=={{header|Ezhil}}==
Ezhil is a Tamil programming language, see [http://en.wikipedia.org/wiki/Ezhil_%28programming_language%29 | Wikipedia] entry.
<
நிரல்பாகம் hailstone ( எண் )
பதிப்பி "=> ",எண் #hailstone seq
Line 2,180 ⟶ 4,424:
பதிப்பி "**********************************************"
முடி
</syntaxhighlight>
=={{header|F_Sharp|F#}}==
<syntaxhighlight lang="fsharp">let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI</syntaxhighlight>
{{out}}
<pre>Maximum length 351 was found for hailstone(77031)</pre>
=={{header|Factor}}==
<
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
Line 2,213 ⟶ 4,475:
PRIVATE>
MAIN: main</
{{out}}
<pre>$ ./factor -run=rosetta.hailstone
Line 2,224 ⟶ 4,486:
=={{header|FALSE}}==
<
[[$." "$1>][n;!]#%]s:
[1\[$1>][\1+\n;!]#%]c:
Line 2,231 ⟶ 4,493:
0m:0f:
1[$100000\>][$c;!$m;>[m:$f:0]?%1+]#%
f;." has hailstone sequence length "m;.</
=={{header|Fermat}}==
<syntaxhighlight lang="fermat">Array g[2]
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[1] is the number, g[2] 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[1]:=champ;
g[2]:=record;
.</syntaxhighlight>
=={{header|Forth}}==
<
dup 1 and if 3 * 1+ else 2/ then ;
: .hail ( n -- )
Line 2,252 ⟶ 4,542:
swap . ." has hailstone sequence length " . ;
100000 longest-hail</
=={{header|Fortran}}==
{{works with|Fortran|95 and later}}
<
implicit none
Line 2,304 ⟶ 4,594:
end subroutine
end program</
{{out}}
<pre>
Line 2,317 ⟶ 4,607:
{{Works with|Frege|3.21.586-g026e8d7}}
<
import Data.List (maximumBy)
Line 2,336 ⟶ 4,626:
let t4 = show $ drop (length h27 - 4) h27
putStrLn ("hailstone 27: " ++ h4 ++ " ... " ++ t4)
putStrLn $ show $ maximumBy (comparing fst) $ map (withResult (length . hailstone)) [1..100000]</
{{out}}
Line 2,347 ⟶ 4,637:
</pre>
=={{header|
<syntaxhighlight lang="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[1]
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"]
</syntaxhighlight>
=={{header|FunL}}==
<
hailstone( 1 ) = [1]
hailstone( n ) = n # hailstone( if 2|n then n/2 else n*3 + 1 )
Line 2,376 ⟶ 4,682:
val (n, len) = maxBy( snd, [(i, hailstone( i ).length()) | i <- 1:100000] )
println( n, len )</
{{out}}
Line 2,383 ⟶ 4,689:
77031, 351
</pre>
=={{header|Futhark}}==
<syntaxhighlight lang="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[0] = 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)))))
</syntaxhighlight>
=={{header|FutureBasic}}==
<syntaxhighlight lang="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
</syntaxhighlight>
{{output}}
<pre>
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.
</pre>
=={{header|Fōrmulæ}}==
{{FormulaeEntry|page=https://formulae.org/?script=examples/Collatz_conjecture}}
'''Part 1.''' Create a routine to generate the hailstone sequence for a number
[[File:Fōrmulæ - Collatz conjecture 01.png]]
'''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}
[[File:Fōrmulæ - Collatz conjecture 02.png]]
[[File:Fōrmulæ - Collatz conjecture 03.png]]
[[File:Fōrmulæ - Collatz conjecture 04.png]]
[[File:Fōrmulæ - Collatz conjecture 05.png]]
'''Part 3.''' Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length
[[File:Fōrmulæ - Collatz conjecture 06.png]]
[[File:Fōrmulæ - Collatz conjecture 07.png]]
The sequence for the number 77,031 is the longest one, with 351 terms
[[File:Fōrmulæ - Collatz conjecture 08.png]]
[[File:Fōrmulæ - Collatz conjecture 09.png]]
'''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
[[File:Fōrmulæ - Collatz conjecture 10.png]]
[[File:Fōrmulæ - Collatz conjecture 11.png]]
The sequence for the number 77,671 has the highest term: 1,570,824,736
[[File:Fōrmulæ - Collatz conjecture 12.png]]
[[File:Fōrmulæ - Collatz conjecture 13.png]]
=={{header|GAP}}==
<
local v;
v := [ n ];
Line 2,444 ⟶ 4,914:
# [ 77031, 351 ]
CollatzMax(1, 1000000);
# [ 837799, 525 ]</
=={{header|Go}}==
<
import "fmt"
Line 2,481 ⟶ 4,951:
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}</
{{out}}
<pre>
Line 2,495 ⟶ 4,965:
Output is the same as version above.
<
import "fmt"
Line 2,562 ⟶ 5,032:
}
fmt.Printf("hs(%d): %d elements\n", nMaxLen, computedLen[nMaxLen])
}</
=={{header|Groovy}}==
<
def sequence = []
while (start != 1) {
Line 2,572 ⟶ 5,042:
}
sequence << start
}</
Test Code
<
assert sequence.size() == 112
assert sequence[0..3] == [27, 82, 41, 124]
Line 2,580 ⟶ 5,050:
def results = (1..100000).collect { [n:it, size:hailstone(it).size()] }.max { it.size }
println results</
{{out}}
<pre>[n:77031, size:351]</pre>
=={{header|Haskell}}==
<
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
--------------------------- 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)
]</syntaxhighlight>
{{Out}}
<pre>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</pre>
The following is an older version, which some of the language examples on this page are translated from:
<
import Data.
hailstone :: Int -> [Int]
hailstone 1
hailstone n
| even
| otherwise = n : hailstone (n * 3 + 1)
withResult :: (
withResult f x = (f x, x)
h27 :: [Int]
h27 = hailstone 27
main :: IO ()
main =
mapM_
putStrLn
, "hailstone 27: " ++
intercalate " ... " (show <$> [take 4 h27, drop (length h27 - 4) h27])
, show $
maximumBy (comparing fst) $
withResult (length . hailstone) <$> [1 .. 100000]
]</syntaxhighlight>
{{out}}
<pre>112
hailstone 27: [27,82,41,124] ... [8,4,2,1]
(351,77031)</pre>
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.<br>
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:
<syntaxhighlight lang="haskell">import Data.List (unfoldr)
-------------------- HAILSTONE SEQUENCE ------------------
hailStones :: Int -> [Int]
hailStones = (<> [1]) . 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
]</syntaxhighlight>
{{Out}}
<pre>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</pre>
=={{header|HicEst}}==
<
H27 = hailstone(27)
Line 2,668 ⟶ 5,229:
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;
<br /> number=77031; longest_sequence=351;
Line 2,674 ⟶ 5,235:
=={{header|Icon}} and {{header|Unicon}}==
A simple solution that <i>generates</i> (in the Icon sense) the sequence is:
<
while n > 1 do {
suspend n
Line 2,680 ⟶ 5,241:
}
suspend 1
end</
and a test program for this solution is:
<
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.
<
static cache
initial {
Line 2,697 ⟶ 5,258:
/cache[n] := [n] ||| hailstone(if n%2 = 0 then n/2 else 3*n+1)
return cache[n]
end</
A test program is:
<
n := integer(!args) | 27
task2(n)
Line 2,722 ⟶ 5,283:
}
write(maxN," has a sequence of ",maxHS," values")
end</
A sample run is:
<pre>
Line 2,739 ⟶ 5,300:
->
</pre>
=={{header|Inform 7}}==
This solution uses a cache to speed up the length calculation for larger numbers.
{{works with|Glulx virtual machine}}
<
To decide which list of numbers is the hailstone sequence for (N - number):
Line 2,853 ⟶ 5,356:
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.</
{{out}}
<pre>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.</pre>
=={{header|Io}}==
Here is a simple, brute-force approach:
<syntaxhighlight lang="io">
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.")
</syntaxhighlight>
{{out}}
<pre>
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.
</pre>
=={{header|Ioke}}==
{{needs-review|Ioke|Calculates the Hailstone sequence but might not complete everything from task description.}}
<syntaxhighlight lang="ioke">collatz = method(n,
n println
unless(n <= 1,
if(n even?, collatz(n / 2), collatz(n * 3 + 1)))
)</syntaxhighlight>
=={{header|J}}==
'''Solution:'''
<
'''Usage:'''
<
112
4 _4 {."0 1 hailseq 27 NB. first & last 4 numbers in sequence
Line 2,869 ⟶ 5,431:
8 4 2 1
(>:@(i. >./) , >./) #@hailseq }.i. 1e5 NB. number < 100000 with max seq length & its seq length
77031 351</
See also the [[j:Essays/Collatz Conjecture|Collatz Conjecture essay on the J wiki]].
=={{header|Java}}==
{{works with|Java|1.5+}}
<
import java.util.HashMap;
import java.util.List;
Line 2,969 ⟶ 5,531:
return;
}
}</
{{out}}
<pre>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]
Line 2,978 ⟶ 5,540:
=={{header|JavaScript}}==
===ES5===
====Imperative====
<syntaxhighlight lang="javascript">function hailstone (n) {
var seq = [n];
while (n > 1) {
Line 3,000 ⟶ 5,564:
}
}
print("longest sequence: " + max + " numbers for starting point " + n);</
{{out}}
<pre>sequence 27 is (27, 82, 41, 124 ... 8, 4, 2, 1). length: 112
longest sequence: 351 numbers for starting point 77031</pre>
====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).
<syntaxhighlight lang="javascript">(function () {
// Hailstone Sequence
// n -> [n]
function hailstone(n) {
return n === 1 ? [1] : (
[n].concat(
hailstone(n % 2 ? n * 3 + 1 : n / 2)
)
)
}
var lstCollatz27 = hailstone(27);
return {
length: lstCollatz27.length,
sequence: lstCollatz27
};
})();</syntaxhighlight>
{{out}}
<syntaxhighlight lang="javascript">{"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]}</syntaxhighlight>
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.
<syntaxhighlight lang="javascript">(function () {
function memoizedHailstone() {
var dctMemo = {};
return function hailstone(n) {
var value = dctMemo[n];
if (typeof value === "undefined") {
dctMemo[n] = value = (n === 1) ?
[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);
})();</syntaxhighlight>
{{out}}
<syntaxhighlight lang="javascript">// Number, length of sequence
{"n":77031, "l":351}</syntaxhighlight>
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.
<syntaxhighlight lang="javascript">(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)[0]) {
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);
})();</syntaxhighlight>
{{out}}
<syntaxhighlight lang="javascript">[
{"n":77031, "l":351}, // 100,000
{"n":837799, "l":525}, // 1,000,000
{"n":8400511, "l":686} // 10,000,000
]</syntaxhighlight>
<syntaxhighlight lang="javascript">longestBelow(100000000)
-> {"n":63728127, "l":950}</syntaxhighlight>
===ES6===
<syntaxhighlight lang="javascript">(() => {
// hailstones :: Int -> [Int]
const hailstones = x => {
const collatz = memoized(n =>
even(n) ? div(n, 2) : (3 * n) + 1);
return reverse(until(
xs => xs[0] === 1,
xs => cons(collatz(xs[0]), xs), [x]
));
};
// collatzLength :: Int -> Int
const collatzLength = n =>
until(
xi => xi[0] === 1,
([x, i]) => [(x % 2 ? 3 * x + 1 : x / 2), i + 1], //
[n, 1]
)[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[0] : 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[0])
) : 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 ${maxLen}.`
]);
})();</syntaxhighlight>
{{Out}}
(Run in the Atom editor, through the Script package)
<pre>Collatz sequence for 27:
27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,
274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,
1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,850,425,1276,
638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,
911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,1732,866,433,
1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,40,20,
10,5,16,8,4,2,1
The number 77031 has the longest hailstone sequence
for any starting number under 100000.
The length of that sequence is 351.
[Finished in 1.139s]
</pre>
=={{header|jq}}==
{{works with|jq|1.4}}
<
def hailstone:
recurse( if . > 1 then
Line 3,022 ⟶ 5,882:
([0,0];
($i | count(hailstone)) as $l
| if $l > .[1] then [$i, $l] else . end);</
'''Examples''':
<
| "[27|hailstone]|length is \($h|length)",
"The first four numbers: \($h[0:4])",
Line 3,030 ⟶ 5,890:
"",
max_hailstone(100000) as $m
| "Maximum length for n|hailstone for n in 1..100000 is \($m[1]) (n == \($m[0]))"</
{{out}}
<
[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)</
=={{header|Julia}}==
{{works with|Julia|0.6 and 1.0+}}
===Dynamic solution===
<syntaxhighlight lang="julia">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</syntaxhighlight>
{{out}}
<pre>
hailstonelength(27) = 112
findmax((hailstonelength(i) for i = 1:100000)) = (351, 77031)
</pre>
===Solution with iterator===
====Julia 1.0====
{{works with|Julia|1.0+}}
<syntaxhighlight lang="julia">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)))</syntaxhighlight>
{{out}}
<pre>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...}</pre>
====Julia 0.6====
{{works with|Julia|0.6}}
<syntaxhighlight lang="julia">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
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)))</syntaxhighlight>
{{out}}
<pre>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...)</pre>
=={{header|K}}==
<
seqn: hail 27
Line 3,071 ⟶ 6,032:
{m,x@s?m:|/s:{#hail x}'x}{x@&x!2}!:1e5
351 77031</
=={{header|Kotlin}}==
<syntaxhighlight lang
fun hailstone(start: Int) = generateSequence(start) { n ->
when {
n == 1 -> null
n % 2 == 0 -> n / 2
else -> n * 3 + 1
}
}
fun main() {
val hail27 = hailstone(27).toList()
println("The hailstone sequence for 27 has ${hail27.size} elements:\n$hail27")
val (n, length) = (1..100000).asSequence()
.map { it to hailstone(it).count() }
.maxBy { it.second }
println("The number between 1 and 100000 with the longest hailstone sequence is $n, of length $length")
}
</syntaxhighlight>
Alternative, doing it manually:
<syntaxhighlight lang="kotlin">import java.util.ArrayDeque
fun hailstone(n
val hails = when {
n == 1 -> ArrayDeque<Int>()
Line 3,082 ⟶ 6,066:
else -> hailstone(3 * n + 1)
}
hails
return hails
}
fun main(args
val hail27 = hailstone(27)
fun showSeq(s
println("Hailstone sequence for 27 is " + showSeq(hail27.take(3)) + " ... "
+ showSeq(hail27.drop(hail27.size - 3)) + " with length ${hail27.size}.")
var longestHail = hailstone(1)
for (x in 1
longestHail =
println("${longestHail.first} is the number less than 100000 with " +
"
}</syntaxhighlight>
{{out}}
Line 3,109 ⟶ 6,088:
=={{header|Lasso}}==
<syntaxhighlight lang="lasso">[
define_tag("hailstone", -required="n", -type="integer", -copy);
local("sequence") = array(#n);
Line 3,140 ⟶ 6,119:
"<br/>";
"Number with the longest sequence under 100,000: " #longest_index + ", with " + #longest_sequence + " elements.";
]</
=={{header|Limbo}}==
<syntaxhighlight lang="text">implement Hailstone;
include "sys.m"; sys: Sys;
Line 3,221 ⟶ 6,175:
return i :: hailstone((big 3 * i) + big 1);
}
</syntaxhighlight>
{{out}}
Line 3,227 ⟶ 6,181:
hailstone(77031) has length 351
</pre>
=={{header|Lingo}}==
<syntaxhighlight lang="lingo">on hailstone (n, sequenceList)
len = 1
repeat while n<>1
if listP(sequenceList) then sequenceList.add(n)
if n mod 2 = 0 then
n = n / 2
else
n = 3 * n + 1
end if
len = len + 1
end repeat
if listP(sequenceList) then sequenceList.add(n)
return len
end</syntaxhighlight>
Usage:
<syntaxhighlight lang="lingo">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</syntaxhighlight>
=={{header|Logo}}==
<syntaxhighlight lang="logo">to hail.next :n
output ifelse equal? 0 modulo :n 2 [:n/2] [3*:n + 1]
end
to hail.seq :n
if :n = 1 [output [1]]
output fput :n hail.seq hail.next :n
end
show hail.seq 27
show count hail.seq 27
to max.hail :n
localmake "max.n 0
localmake "max.length 0
repeat :n [if greater? count hail.seq repcount :max.length [
make "max.n repcount
make "max.length count hail.seq repcount
] ]
(print :max.n [has hailstone sequence length] :max.length)
end
max.hail 100000</syntaxhighlight>
=={{header|Logtalk}}==
<
:- public(generate_sequence/2).
Line 3,321 ⟶ 6,334:
).
:- end_object.</
Testing:
<
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
Line 3,333 ⟶ 6,346:
| ?- hailstone::longest_sequence(1, 100000, N, Length).
N = 77031, Length = 351
true</
=={{header|LOLCODE}}==
There is presently no way to query a <tt>BUKKIT</tt> for the existence of a given key, thus making memoization infeasible. This solution takes advantage of prior knowledge to run in reasonable time.
<
HOW IZ I hailin YR stone
Line 3,385 ⟶ 6,398:
VISIBLE "len(hail(" max ")) = " len
KTHXBYE</
{{out}}
<pre>hail(27) = 27 82 41 124 ... 8 4 2 1, length = 112
Line 3,391 ⟶ 6,404:
=={{header|Lua}}==
<
local n_iter = 1
Line 3,420 ⟶ 6,433:
end
print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )</
=={{header|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
<syntaxhighlight lang="m2000 interpreter">
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
</syntaxhighlight>
{{out}}
<pre>
27 82 41 124
8 4 2 1
counter:112
Number 77031 has then longest hailstone sequence of length 351
</pre >
=={{header|Maple}}==
Define the procedure:
<syntaxhighlight lang="maple">
hailstone := proc( N )
local n := N, HS := Array([n]);
Line 3,437 ⟶ 6,512:
HS;
end proc;
</syntaxhighlight>
Run the command and show the appropriate portion of the result;
<syntaxhighlight lang="maple">
> r := hailstone(27):
[ 1..112 1-D Array ]
Line 3,447 ⟶ 6,522:
> r(1..4) ... r(-4..);
[27, 82, 41, 124] .. [8, 4, 2, 1]
</syntaxhighlight>
Compute the first 100000 sequences:
<syntaxhighlight lang="maple">
longest := 0; n := 0;
for i from 1 to 100000 do
Line 3,459 ⟶ 6,534:
od:
printf("The longest Hailstone sequence in the first 100k is n=%d, with %d terms\n",n,longest);
</syntaxhighlight>
{{out}}
<pre>
Line 3,465 ⟶ 6,540:
</pre>
=={{header|Mathematica}} / {{header|Wolfram Language}}==
Here are
=== Nested function call formulation ===
<syntaxhighlight lang="mathematica">HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]</syntaxhighlight>
This is probably the most readable and shortest implementation.
=== Fixed-Point formulation ===
<
=== Recursive formulation
<
HailstoneR[
HailstoneR[
=== Procedural implementation ===
<
=== Validation ===
I use this version to do the validation:
<
NestWhileList[
c27 = Hailstone@27;
Print["Hailstone sequence for n = 27: [", c27[[;; 4]], "...", c27[[-4 ;;]], "]"]
Line 3,494 ⟶ 6,576:
{i, 100000}
]
Print["Longest Hailstone sequence at n = ", longest, "\nwith length = ", comp]; </
{{out}}
<pre>
Line 3,504 ⟶ 6,586:
I think the fixed-point and the recursive piece-wise function formulations are more idiomatic for Mathematica
==== Sequence 27 ====
<syntaxhighlight lang="mathematica">With[{seq = HailstoneFP[27]}, { Length[seq], Take[seq, 4], Take[seq, -4]}]</syntaxhighlight>
{{out}}
<pre>{112, {27, 82, 41, 124}, {8, 4, 2, 1}}</pre>
Alternatively,
<syntaxhighlight lang="mathematica">Short[HailstoneFP[27],0.45]</syntaxhighlight>
{{out}}
<pre>{27, 82, 41, 124, <<104>>, 8, 4, 2, 1}</pre>
==== Longest sequence length ====
<syntaxhighlight lang="mathematica">MaximalBy[Table[{i, Length[HailstoneFP[i]]}, {i, 100000}], Last]</syntaxhighlight>
{{out}}
<pre>{{77031, 351}}</pre>
=={{header|MATLAB}} / {{header|Octave}}==
===Hailstone Sequence For N===
<syntaxhighlight lang="matlab">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</syntaxhighlight>
Show sequence of hailstone(27) and number of elements:
<syntaxhighlight lang="matlab">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))</syntaxhighlight>
{{out}}
<pre>hailstone(27): 27 82 41 124 ... 8 4 2 1
number of elements: 112</pre>
===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)====
<syntaxhighlight lang="matlab">N = 1e5;
maxLen = 0;
for k = 1:N
kLen = numel(hailstone(k));
if kLen > maxLen
maxLen = kLen;
n = k;
end
end</syntaxhighlight>
{{out}}
<pre>n = 77031
maxLen = 351</pre>
====Faster Version====
<syntaxhighlight lang="matlab">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</syntaxhighlight>
{{out}}
<syntaxhighlight lang="matlab">>> [n, maxLen] = longestHailstone(1e5)
n = 77031
maxLen = 351</syntaxhighlight>
====Much Faster Version With Caching====
<syntaxhighlight lang="matlab">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</syntaxhighlight>
{{out}}
<syntaxhighlight lang="matlab">>> [n, maxLen] = longestHailstone(1e5)
n = 77031
maxLen = 351</syntaxhighlight>
=={{header|Maxima}}==
<
(n: if evenp(n) then n/2 else 3*n + 1, L: endcons(n, L)), L)$
Line 3,566 ⟶ 6,702:
length(%); /* 112 */
collatz_length(27); /* 112 */
collatz_max(100000); /* [77031, 351] */</
=={{header|Mercury}}==
The actual calculation (including module ceremony)
providing both a function and a predicate implementation:
<
:- interface.
Line 3,589 ⟶ 6,725:
; hailstone(3 * N + 1, S) ).
:- end_module hailstone.</
The mainline test driver (making use of [http://en.wikipedia.org/wiki/Unification_(computer_science) unification] for more succinct tests):
<
:- interface.
Line 3,627 ⟶ 6,763:
; io.write_string("At least one test failed.\n", !IO) ).
:- end_module test_hailstone.</
{{out}} of running this program is:
Line 3,634 ⟶ 6,770:
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 <code>longest/3</code> …
<
… you see that the second and third parameters are '''output''' parameters.
Line 3,644 ⟶ 6,780:
Thus we know that the correct sequences and values were generated
without bothering to print them out.
=={{header|MiniScript}}==
===Non-cached version===
Calculates sequence without using previous calculated sequences.
<syntaxhighlight lang="miniscript">
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."
</syntaxhighlight>
{{out}}
<pre>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.</pre>
===Cached version===
Calculations are stored for used in later calculations.
<syntaxhighlight lang="miniscript">
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 [1]
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."
</syntaxhighlight>
{{out}}Output is the same as the non-cached version above.
<pre>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.</pre>
=={{header|ML}}==
==={{header|MLite}}===
<
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
Line 3,665 ⟶ 6,906:
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
Line 3,681 ⟶ 6,922:
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);</
{{out}}
<pre>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</pre>
=={{header|Modula-2}}==
<
IMPORT InOut;
Line 3,759 ⟶ 7,000:
FindMax (100000);
InOut.WriteString ("Done."); InOut.WriteLn
END hailst.</
Producing:
<pre>jan@Beryllium:~/modula/rosetta$ hailst
Line 3,787 ⟶ 7,028:
=={{header|MUMPS}}==
<
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</
=={{header|Nanoquery}}==
<syntaxhighlight lang="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[0] + ", " + h[1] + ", " + h[2] + ", " + h[3] + ", ..., "
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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|NetRexx}}==
<
options replace format comments java crossref savelog symbols binary
Line 3,838 ⟶ 7,119:
hs = hs' 'hn
return hs.strip</
{{out}}
<pre>
Line 3,848 ⟶ 7,129:
=={{header|Nim}}==
<syntaxhighlight lang="nim">proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
Line 3,859 ⟶ 7,139:
result.add n
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})."</syntaxhighlight>
{{out}}
<pre>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).</pre>
=={{header|Oberon-2}}==
<
IMPORT Out;
Line 3,951 ⟶ 7,241:
Out.String ("Done.");
Out.Ln
END hailst.</
Producing
<pre>
Line 3,979 ⟶ 7,269:
=={{header|OCaml}}==
<
open Num;;
Line 4,041 ⟶ 7,331:
"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"] *)</
=={{header|Oforth}}==
<
| l |
ListBuffer new ->l
while(
hailstone(27) dup size println dup left(4) println right(4) println
100000 seq map(#[ dup hailstone size swap Pair new ]) reduce(#maxKey) println</
{{out}}
Line 4,062 ⟶ 7,351:
[351, 77031]
</pre>
=={{header|Onyx (wasm)}}==
<syntaxhighlight lang="C">
use core { * }
hailstone :: (n: u32) -> [..]u32 {
seq: [..]u32;
array.push(&seq, n);
while n > 1 {
n = n/2 if n%2 == 0 else (n*3)+1;
array.push(&seq, n);
}
return seq;
}
Longest :: struct { num, len : u32; }
main :: () {
// -------
// task 1:
// -------
// "Create a routine to generate the hailstone
// sequence for a number."
i := 27;
seq := hailstone(i);
printf("Task 1:\n{}: {}\n\n",
i,
seq
);
// -------
// task 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"
slice_size := 4;
len := seq.length;
slice_first := seq[0..slice_size];
slice_last := seq[seq.length-slice_size..seq.length];
printf("Task 2:\nlength: {}, first: {}, last: {}\n\n",
len,
slice_first,
slice_last
);
// -------
// task 3:
// -------
// "Show the number less than 100,000
// which has the longest hailstone sequence
// together with that sequences length."
l : Longest;
for i in 1..100000 {
seq := hailstone(i);
if l.len < seq.length { l = .{num = i, len = seq.length}; }
}
printf("Task 3:\nLongest Num: {}, Sequence Length: {}\n", l.num, l.len);
}
</syntaxhighlight>
=={{header|ooRexx}}==
<syntaxhighlight lang="oorexx">
sequence = hailstone(27)
say "Hailstone sequence for 27 has" sequence~items "elements and is ["sequence~toString('l', ", ")"]"
Line 4,092 ⟶ 7,446:
end
return sequence
</syntaxhighlight>
{{out}}
<pre>
Line 4,101 ⟶ 7,455:
=={{header|Order}}==
To display the length, and first and last elements, of the hailstone sequence for 27, we could do this:
<
#define ORDER_PP_DEF_8hailstone ORDER_PP_FN( \
Line 4,117 ⟶ 7,471:
8(starts with:) 8seq_take(4, 8H) 8comma 8space
8(ends with:) 8seq_drop(8minus(8S, 4), 8H))
) )</
{{out}}
<syntaxhighlight lang="text">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:
<
8fn(8M, 8P, \
8if(8is_0(8M), \
Line 4,134 ⟶ 7,488:
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):
<
8let((8P,
8seq_head(
Line 4,145 ⟶ 7,499:
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 <code>8nat</code>. <code>8to_lit</code> converts a digit sequence back to a readable number.
=={{header|Oz}}==
<
fun {HailstoneSeq N}
N > 0 = true %% assert
Line 4,173 ⟶ 7,527:
MaxI#MaxLen = {List.foldL Pairs MaxBy2nd 0#0}
{System.showInfo
"Maximum length "#MaxLen#" was found for hailstone("#MaxI#")"}</
{{out}}
<pre>
Line 4,180 ⟶ 7,534:
=={{header|PARI/GP}}==
===Version #1.===
<syntaxhighlight lang="parigp">show(n)={
my(t=1);
while(n>1,
Line 4,210 ⟶ 7,566:
show(27)
r=0;for(n=1,1e5,t=len(n);if(t>r,r=t;ra=n));print(ra"\t"r)</
{{out}}
<pre>27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,4
Line 4,220 ⟶ 7,576:
and
<pre>77031 351</pre>
===Version #2.===
{{Works with|PARI/GP|2.7.4 and above}}
Different kind of PARI scripts for Collatz sequences you can find in OEIS, e.g.:
[http://oeis.org/A070165 A070165]
<syntaxhighlight lang="parigp">
\\ 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);
}
</syntaxhighlight>
{{Output}}
<pre>
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.
</pre>
=={{header|Pascal}}==
See [[Hailstone_sequence#Delphi | Delphi]]
or try this transformed Delphi version without generics.Use of a static array.
<syntaxhighlight lang="pascal">program ShowHailstoneSequence;
{$IFDEF FPC}
{$MODE delphi} //or objfpc
Line 4,231 ⟶ 7,629:
{$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 :
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:
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
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:
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);//319804831
with lList
begin
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i :=
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
inc(lAverageLength, lgth);
begin
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.</syntaxhighlight>
{{out}}
<pre>sequence of 27 has 112 elements
27,82,41,124..8,4,2,1
Limit : number with max length | average length
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
real 3m56.573s // 64-bit compiled</pre>
=={{header|Perl}}==
=== Straightforward ===
<
use warnings;
Line 4,373 ⟶ 7,788:
return @sequence;
}</
{{out}}
Line 4,384 ⟶ 7,799:
=== Compact ===
A more compact version:
<
use strict;
Line 4,398 ⟶ 7,813:
@h = ();
for (1 .. 99_999) { @h = ($_, $h[2]) if ($h[2] = hailstone($_)) > $h[1] }
printf "%d: (%d)\n", @h;</
{{out}}
Line 4,411 ⟶ 7,822:
</pre>
=={{header|
Copy of [[Hailstone_sequence#Euphoria|Euphoria]]
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">hailstone</span><span style="color: #0000FF;">(</span><span style="color: #004080;">atom</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">s</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">n</span><span style="color: #0000FF;">}</span>
<span style="color: #008080;">while</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">1</span> <span style="color: #008080;">do</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">remainder</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">)=</span><span style="color: #000000;">0</span> <span style="color: #008080;">then</span>
<span style="color: #000000;">n</span> <span style="color: #0000FF;">/=</span> <span style="color: #000000;">2</span>
<span style="color: #008080;">else</span>
<span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">3</span><span style="color: #0000FF;">*</span><span style="color: #000000;">n</span><span style="color: #0000FF;">+</span><span style="color: #000000;">1</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #000000;">s</span> <span style="color: #0000FF;">&=</span> <span style="color: #000000;">n</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">while</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">s</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">hailstone_count</span><span style="color: #0000FF;">(</span><span style="color: #004080;">atom</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">count</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">while</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">1</span> <span style="color: #008080;">do</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">remainder</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">)=</span><span style="color: #000000;">0</span> <span style="color: #008080;">then</span>
<span style="color: #000000;">n</span> <span style="color: #0000FF;">/=</span> <span style="color: #000000;">2</span>
<span style="color: #008080;">else</span>
<span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">3</span><span style="color: #0000FF;">*</span><span style="color: #000000;">n</span><span style="color: #0000FF;">+</span><span style="color: #000000;">1</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #000000;">count</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">while</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">count</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">s</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">hailstone</span><span style="color: #0000FF;">(</span><span style="color: #000000;">27</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"hailstone(27) = %v\n"</span><span style="color: #0000FF;">,{</span><span style="color: #7060A8;">shorten</span><span style="color: #0000FF;">(</span><span style="color: #000000;">s</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"numbers"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">4</span><span style="color: #0000FF;">)})</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">hmax</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">imax</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">count</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">2</span> <span style="color: #008080;">to</span> <span style="color: #000000;">1e5</span><span style="color: #0000FF;">-</span><span style="color: #000000;">1</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">count</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">hailstone_count</span><span style="color: #0000FF;">(</span><span style="color: #000000;">i</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">count</span><span style="color: #0000FF;">></span><span style="color: #000000;">hmax</span> <span style="color: #008080;">then</span>
<span style="color: #000000;">hmax</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">count</span>
<span style="color: #000000;">imax</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">i</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"The longest hailstone sequence under 100,000 is %d with %d elements.\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">imax</span><span style="color: #0000FF;">,</span><span style="color: #000000;">hmax</span><span style="color: #0000FF;">})</span>
<!--</syntaxhighlight>-->
{{out}}
<pre>
hailstone(27) = {27,82,41,124,"...",8,4,2,1," (112 numbers)"}
The longest hailstone sequence under 100,000 is 77031 with 351 elements.
</pre>
=={{header|PHP}}==
<
$sequence = $seq;
$sequence[] = $n;
Line 4,447 ⟶ 7,898:
foreach($maxResult as $key => $val){
echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val;
}</
<pre>
112 Elements.
Line 4,453 ⟶ 7,904:
Number < 100000 with longest Hailstone seq.: 77031 with length of 351
</pre>
=={{header|Picat}}==
<syntaxhighlight lang="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.</syntaxhighlight>
{{out}}
<pre>H27:
len = 112
[27,82,41,124,...,8,4,2,1]
Longest sequence < 100_000:
[maxLen = 351,maxN = 77031]</pre>
===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.)
<syntaxhighlight lang="picat">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.
</syntaxhighlight>
=={{header|PicoLisp}}==
<
(make
(until (= 1 (link N))
Line 4,467 ⟶ 8,005:
(let N (maxi '((N) (length (hailstone N))) (range 1 100000))
(println N (length (hailstone N))) )</
{{out}}
<pre>27 112 (27 82 41 124) - (8 4 2 1)
Line 4,473 ⟶ 8,011:
=={{header|Pike}}==
<
int next(int n)
Line 4,511 ⟶ 8,049:
}
write("longest sequence starting at %d has %d elements\n", longest->start, longest->length);
}</
{{out}}
Line 4,520 ⟶ 8,058:
=={{header|PL/I}}==
<
declare (longest, n) fixed (15);
declare flag bit (1);
Line 4,557 ⟶ 8,095:
end hailstones;
end test;</
{{out}}
<pre style="height:30ex;overflow:scroll">
Line 4,676 ⟶ 8,214:
77031 has the longest sequence of 351
</pre>
==={{header|PL/I-80}}===
<syntaxhighlight lang="pl/i">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;
</syntaxhighlight>
{{out}}
<pre>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
</pre>
=={{header|plainTeX}}==
The following code works with any TeX engine.
<syntaxhighlight lang="tex">\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</syntaxhighlight>
pdf or dvi output:
<pre>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
</pre>
=={{header|Pointless}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|PowerShell}}==
{{works with|PowerShell|3.0+}}
<syntaxhighlight lang="powershell">
function Get-HailStone {
Line 4,693 ⟶ 8,416:
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}</syntaxhighlight>
{{out}}
Line 4,728 ⟶ 8,446:
=={{header|Prolog}}==
1. Create a routine to generate the hailstone sequence for a number.
<
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.
<
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.
<
longesthailstone(1, Cn, Cl, Mn, Ml):- Mn = Cn,
Ml = Cl.
Line 4,750 ⟶ 8,468:
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.
<
to get the following result
<pre>
Line 4,763 ⟶ 8,481:
Works with SWI-Prolog and module '''chr''' written by '''Tom Schrijvers''' and '''Jan Wielemaker''' <br>
<
:- chr_option(debug, off).
:- chr_option(optimize, full).
Line 4,779 ⟶ 8,497:
% Hailstone loop
hailstone(1) ==> true.
hailstone(N) ==> N \= 1 | collatz(N, H), hailstone(H).</
Code for task one :
<
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)).</
{{out}}
<pre> ?- task1.
Line 4,793 ⟶ 8,511:
true.</pre>
Code for task two :
<
seq(2, 100000, 1-[1], Len-V),
format('For ~w sequence has ~w len ! ~n', [V, Len]).
Line 4,814 ⟶ 8,532:
findall(hailstone(X), find_chr_constraint(hailstone(X)), L),
length(L, Len),
clean.</
{{out}}
<pre> ?- longest_sequence.
Line 4,822 ⟶ 8,540:
=={{header|Pure}}==
<
type odd x::int = x mod 2;
type even x::int = ~odd x;
Line 4,850 ⟶ 8,568:
(foldr (\ (a,b) (c,d) -> if (b > d) then (a,b) else (c,d))
(0,0)
(map (\ x -> (x, # hailstone x)) (1..100000)));</
{{out}}
<pre>
Line 4,858 ⟶ 8,576:
=={{header|Python}}==
===Procedural===
<syntaxhighlight lang="python">def hailstone(n):
seq = [n]
while n > 1:
n = 3 * n + 1 if n & 1 else n // 2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert (len(h)
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")</syntaxhighlight>
{{out}}
<pre>Maximum length 351 was found for hailstone(77031) for numbers <100,000</pre>
===Using a generator===
<syntaxhighlight lang="python">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")</syntaxhighlight>
{{out}}
<pre>Maximum length 351 was found for hailstone(77031) for numbers <100,000</pre>
===Composition of pure functions===
{{Works with|Python|3.7}}
<syntaxhighlight lang="python">'''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))) + [1]
# 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[1]
# 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()</syntaxhighlight>
{{Out}}
<pre>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.</pre>
=={{header|Quackery}}==
<syntaxhighlight lang="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</syntaxhighlight>
'''Output:'''
<syntaxhighlight lang="quackery">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.
</syntaxhighlight>
=={{header|R}}==
===Iterative solution===
<syntaxhighlight lang="rsplus">### PART 1:
makeHailstone <- function(n){
hseq <- n
Line 4,908 ⟶ 8,831:
cat("Between ", lower.bound, " and ", upper.bound, ", the input of ",
max.index, " gives the longest hailstone sequence, which has length ",
max.length, ". \n", sep="")</
{{out}}
Line 4,926 ⟶ 8,849:
Between 1 and 1e+05, the input of 77031 gives the longest hailstone sequence,
which has length 351.</pre>
===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:
<syntaxhighlight lang="rsplus">###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)).</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Racket}}==
<syntaxhighlight lang="racket">
#lang racket
Line 4,949 ⟶ 8,907:
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
</syntaxhighlight>
{{out}}
Line 4,956 ⟶ 8,914:
for x<=100000, 77031 has the longest sequence with 351 items
</pre>
=={{header|Raku}}==
(formerly Perl 6)
<syntaxhighlight lang="raku" line>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({$m.value}) for numbers < 100_000";</syntaxhighlight>
{{out}}
<pre>
Length of hailstone(27) = 112
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
Max length 351 was found for hailstone(77031) for numbers < 100_000
</pre>
=={{header|REBOL}}==
<syntaxhighlight lang="rebol">
hail: func [
"Returns the hailstone sequence for n"
n [integer!]
/local seq
] [
seq: copy reduce [n]
while [n <> 1] [
append seq n: either n % 2 == 0 [n / 2] [3 * n + 1]
]
seq
]
hs27: hail 27
print [
"the hail sequence of 27 has length" length? hs27
"and has the form " copy/part hs27 3 "..."
back back back tail hs27
]
maxN: maxLen: 0
repeat n 99999 [
if (len: length? hail n) > maxLen [
maxN: n
maxLen: len
]
]
print [
"the number less than 100000 with the longest hail sequence is"
maxN "with length" maxLen
]</syntaxhighlight>
{{out}}
<pre>the hail sequence of 27 has length 112 and has the form 27 82 41 ... 4 2 1
the number less than 100000 with the longest hail sequence is 77031 with length 351</pre>
=={{header|Refal}}==
<syntaxhighlight lang="refal">$ENTRY Go {
= <ShowHailstone 27>
<ShowLongest 100000>;
}
Hailstone {
1 = 1;
s.N, <Mod s.N 2>: {
0 = s.N <Hailstone <Div s.N 2>>;
1 = s.N <Hailstone <+ 1 <* 3 s.N>>>;
};
};
ShowHailstone {
s.N, <Hailstone s.N>: e.Seq,
<Lenw e.Seq>: s.Len s.1 s.2 s.3 s.4 e.X s.D4 s.D3 s.D2 s.D1
= <Prout 'The hailstone sequence for the number '
<Symb s.N> ' has ' <Symb s.Len> ' elements,\n'
'starting with ' s.1 s.2 s.3 s.4
'and ending with ' s.D4 s.D3 s.D2 <Symb s.D1>'.'>;
}
FindLongest {
s.Max = <FindLongest s.Max 1 1 1>;
s.Max s.Max s.Long s.Len = s.Long s.Len;
s.Max s.Cur s.Long s.Len,
<Hailstone s.Cur>: e.CurSeq,
<Lenw e.CurSeq>: s.CurLen e.X,
<+ s.Cur 1>: s.Next,
<Compare s.CurLen s.Len>: {
'+' = <FindLongest s.Max s.Next s.Cur s.CurLen>;
s.X = <FindLongest s.Max s.Next s.Long s.Len>;
};
};
ShowLongest {
s.Max, <FindLongest s.Max>: s.Long s.Len
= <Prout 'The number < ' <Symb s.Max> ' which has the longest'
' hailstone sequence is ' <Symb s.Long> '.\n'
'The length of its Hailstone sequence is '
<Symb s.Len> '.'>;
};</syntaxhighlight>
{{out}}
<pre>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 < 100000 which has the longest hailstone sequence is 77031.
The length of its Hailstone sequence is 351.</pre>
=={{header|REXX}}==
===non-optimized===
<syntaxhighlight lang
numeric digits 20 /*be able to handle
parse arg x y . /*get optional arguments from
if x=='' | x==
if y=='' | y==
$= hailstone(x) /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 1▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say x ' has a hailstone sequence of ' words($)
say
say
if y==0 then exit /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 2▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say
w= 0; do j=1 for y; call hailstone
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 */
{{out|output|text= when using the default inputs:}}
<pre>
27 has a hailstone sequence of 112
(between
</pre>
===optimized===
This
It makes use of:
::::* previously calculated Collatz sequences (memoization)
::::* a faster method of determining if an integer is even
<syntaxhighlight lang
!.=0; !.0=1; !.2=1; !.4=1; !.6=1; !.8=1 /*assign even
numeric digits 20; @.= 0 /*handle big
parse arg x y z .; !.h= y /*get optional arguments from
if x=='' | x==
if y=='' | y==
if z=='' | z==
hm= max(y, 500000) /*use memoization (maximum num for @.)*/
$= hailstone(x) /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 1▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say
say
say ' and ends with: ∙∙∙' subword($, max(z+1,
if y==0 then exit /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 2▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say
w= 0; do j=1 for y; $= hailstone(j)
#hs= words($) /*find the length of the
if #hs<=w then iterate /*Not big
bigJ= j;
end /*j*/
say '(between
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
if @._\==0 then leave /*Was number already found? Return S.*/
{{out|output|text= when using the default inputs:}}
<pre>
27 has a hailstone sequence of 112
(between
</pre>
{{out|output|text= when using the inputs: <tt> , 1000000 </tt>}}
<pre>
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
</pre>
=={{header|Ring}}==
<syntaxhighlight lang="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
</syntaxhighlight>
=={{header|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|4.2.7}}
{| class="wikitable"
! 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
{{out}}
<pre>
2: 127
1: 77031
</pre>
=={{header|Ruby}}==
This program uses new methods (Integer#even? and Enumerable#max_by) from Ruby 1.8.7.
{{works with|Ruby|1.8.7}}
<
seq = [n]
until n == 1
Line 5,067 ⟶ 9,221:
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}"</
{{out}}
<pre>
Line 5,081 ⟶ 9,235:
This avoids recomputing the end of the sequence.
{{works with|Ruby|1.8.7}}
<
ListNode = Struct.new(:value, :size, :succ) do
def each
Line 5,118 ⟶ 9,272:
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.
=={{header|Rust}}==
<syntaxhighlight lang="rust">fn hailstone(start : u32) -> Vec<u32> {
let mut res = Vec::new();
let mut next = start;
res.push(start);
while
}
}
fn main() {
let
let
println!("
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
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!("
}</
{{out}}
<pre>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</pre>
=={{header|S-lang}}==
<syntaxhighlight lang="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");</syntaxhighlight>
{{out}}
<pre>
27, 82, 41, 124,
and ending with:
8, 4, 2, 1
longest sequence started w/77031 and had 351 elements</pre>
=={{header|SAS}}==
<syntaxhighlight lang="sas">
* Create a routine to generate the hailstone sequence for one number;
%macro gen_seq(n);
Line 5,209 ⟶ 9,430:
%mend;
%longest_hailstone(1,99999);
</syntaxhighlight>
{{out}}
<pre>
First and last elements of Hailstone Sequence for number 27
Line 5,223 ⟶ 9,444:
start sequence
77031 351
</pre>
=={{header|S-BASIC}}==
<syntaxhighlight lang="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
</syntaxhighlight>
{{out}}
<pre>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
</pre>
Line 5,228 ⟶ 9,533:
{{libheader|Scala}}
{{works with|Scala|2.10.2}}
<
def hailstone(n: Int): Stream[Int] =
n #:: (if (n == 1) Stream.empty else hailstone(if (n % 2 == 0) n / 2 else n * 3 + 1))
Line 5,242 ⟶ 9,547:
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.")
}</
{{Out}}
<pre>Use the routine to show that the hailstone sequence for the number: 27.
Line 5,252 ⟶ 9,557:
=={{header|Scheme}}==
<
(if (= n 1) '(1)
(cons n (collatz (if (even? n) (/ n 2) (+ 1 (* 3 n)))))))
Line 5,279 ⟶ 9,584:
(collatz-max 1 100000)
; (77031 351)</
=={{header|Scilab}}==
{{trans|MATLAB}}
<syntaxhighlight lang="text">function x=hailstone(n)
// iterative definition
// usage: global verbose; verbose=%T; hailstone(27)
Line 5,315 ⟶ 9,620:
M(k)=hailstone(k);
end;
[maxLength,n]=max(M)</
{{out}}
<pre>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
Line 5,323 ⟶ 9,628:
=={{header|Seed7}}==
<
const func array integer: hailstone (in var integer: n) is func
Line 5,380 ⟶ 9,685:
writeln(" length=" <& length(h27));
writeln("Maximum length " <& maxLength <& " at number=" <& numberOfMaxLength);
end func;</
{{out}}
<pre>
hailstone(27):
Line 5,387 ⟶ 9,693:
Maximum length 351 at number=77031
</pre>
=={{header|SETL}}==
<syntaxhighlight lang="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;</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Sidef}}==
<
var
while (n > 1) {
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
# 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[1] && (
h = [i, l]
)
}
printf("%d: (%d)\n", h...)
=={{header|Smalltalk}}==
{{works with|GNU Smalltalk}}
<
Sequences class >> hailstone: n [
|seq|
Line 5,435 ⟶ 9,775:
ifFalse: [ ^ Sequences hailstoneCount: ( (3*n) + 1) num: (m + 1) ].
]
].</
<
r := Sequences hailstone: 27. "hailstone 'from' 27"
(r size) displayNl. "its length"
Line 5,456 ⟶ 9,796:
('Sequence generator %1, sequence length %2' % { (longest at: 1) . (longest at: 2) })
displayNl.</
=={{header|SNUSP}}==
Line 5,469 ⟶ 9,809:
\+<-/!< ----------.++++++++++/ # +/! +/! +/! +/! +/
</pre>
=={{header|Swift}}==
<syntaxhighlight lang="swift">
func hailstone(var n:Int) -> [Int] {
Line 5,504 ⟶ 9,845:
}
println("Longest sequence for numbers under 100,000 is with \(longest.n). Which has \(longest.len) items.")</
{{out}}
<pre>
Line 5,513 ⟶ 9,854:
=={{header|Tcl}}==
The core looping structure is an example of an [[Loops/N plus one half|n-plus-one-half loop]], except the loop is officially infinite here.
<
while 1 {
lappend seq $n
Line 5,531 ⟶ 9,872:
if {$l>$maxlen} {set maxlen $l;set max $i}
}
puts "max is $max, with length $maxlen"</
{{out}}
<pre>
h27 len=112
Line 5,542 ⟶ 9,884:
=={{header|TI-83 BASIC}}==
===Task 1===
<
N→M: 0→X: 1→L
While L=1
Line 5,556 ⟶ 9,898:
End
End
{N,X}</
{{out}}
<pre> 10
Line 5,568 ⟶ 9,910:
===Task 2===
As the calculator is quite slow, so the output is for N=200
<
0→A:0→B
for(I,1,N)
Line 5,588 ⟶ 9,930:
Disp {I,X}
End
{A,B}</
{{out}}
<pre>{171,125}</pre>
=={{header|Transd}}==
{{trans|Python}}
<syntaxhighlight lang="scheme">#lang transd
MainModule: {
hailstone: (λ n Int()
(with seq Vector<Int>([n])
(while (> n 1)
(= n (if (mod n 2) (+ (* 3 n) 1)
else (/ n 2)))
(append seq n)
)
(ret seq)
)
),
_start: (λ
(with h (hailstone 27) l 0 n 0 t 0
(lout "Length of (27): " (size h))
(lout "First 4 of (27): " Range(in: h 0 4))
(lout "Last 4 of (27): " Range(in: h -4 -0))
(for i in Range(100000) do
(= t (size (hailstone (to-Int i)))) (if (> t l) (= l t) (= n i))
)
(lout "For n < 100.000 the max. sequence length is " l " for " n)
)
)
}</syntaxhighlight>{{out}}
<pre>
Length of (27): 112
First 4 of (27): [27, 82, 41, 124]
Last 4 of (27): [8, 4, 2, 1]
For n < 100.000 the max. sequence length is 351 for 77031
</pre>
=={{header|TXR}}==
<
(cons n
(gen (not (eq n 1))
Line 5,618 ⟶ 9,995:
(set max len)
(set maxi i))))
(format t "longest sequence is ~a for n = ~a\n" max maxi)))</
<pre>$ txr -l hailstone.txr
longest sequence is 351 for n = 77031</pre>
=={{header|uBasic/4tH}}==
{{Trans|FreeBASIC}}
<syntaxhighlight lang="text">' ------=< 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@)</syntaxhighlight>
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.
{{out}}<pre>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
</pre>
=={{header|UNIX Shell}}==
The best way is to use a shell with built-in arrays and arithmetic, such as Bash.
{{works with|Bash}}
<
# seq is the array genereated by hailstone
# index is used for seq
Line 5,662 ⟶ 10,132:
done
echo "${max} has a hailstone sequence length of ${maxlen}"</
{{out}}
<pre>27: 112
27 82 41 124 ... 8 4 2 1
Line 5,670 ⟶ 10,140:
===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}}
<
# Clobbers $n.
hailstone() {
Line 5,704 ⟶ 10,176:
i=`expr $i + 1`
done
echo "Hailstone sequence from $max has $maxlen elements."</
==={{header|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.
<
# Clobbers $n.
alias hailstone eval \''@ n = \!:1:q \\
Line 5,759 ⟶ 10,232:
@ i += 1
end
echo "Hailstone sequence from $max has $maxlen elements."</
{{out}}
<pre>$ 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.</pre>
=={{header|Ursa}}==
===Implementation===
<b>hailstone.u</b>
<syntaxhighlight lang="ursa">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</syntaxhighlight>
===Usage===
{{out}}
<pre>> 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
> _</pre>
=={{header|Ursala}}==
<
#import nat
Line 5,778 ⟶ 10,292:
<
^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 <code>hail</code> function computes the sequence as follows.
* Given a number as an argument, <code>@iNC</code> makes a list containing only that number before passing it to the rest of the function. The <code>i</code> in the expression stands for the identity function, <code>N</code> for the constant null function, and <code>C</code> for the cons operator.
Line 5,796 ⟶ 10,310:
* The maximizing operator (<code>$^</code>) with respect to the natural less or equal relation (<code>nleq</code>) applied to the right sides (<code>&r</code>) of its pair of arguments extracts the number with the maximum length sequence.
{{out}}
<pre><27,82,41,124>...<8,4,2,1> has length 112
77031 has sequence length 351</pre>
=={{header|VBA}}==
{{trans|Phix}}<syntaxhighlight lang="vb">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</syntaxhighlight>{{out}}<pre>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.</pre>
=={{header|VBScript}}==
<syntaxhighlight lang="vb">
'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
</syntaxhighlight>
{{Out}}
<pre>
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
</pre>
=={{header|Visual Basic}}==
{{trans|PL/I}}
{{works with|Visual Basic|VB6 Standard}}
<
Dim flag As Boolean ' true to print values
Sub main()
Line 5,859 ⟶ 10,486:
End If
hailstones = p
End Function 'hailstones</
{{Out}}
<pre>The sequence for 27 is: 27 82 41 124 ... 8 4 2 1
Line 5,867 ⟶ 10,494:
=={{header|Visual Basic .NET}}==
{{works with|Visual Basic .NET|2005+}}
<
Sub Main()
' Checking sequence of 27.
Line 5,908 ⟶ 10,535:
End Function
End Module</
{{out}}
Line 5,914 ⟶ 10,541:
27, 82, 41, 124, ... , 8, 4, 2, 1
Max elements in sequence for number below 100k: 77031 with 351 elements.
</pre>
=={{header|V (Vlang)}}==
{{trans|go}}
<syntaxhighlight lang="v (vlang)">// 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[0]} ${seq[1]} ${seq[2]} ${seq[3]} ... ${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")
}</syntaxhighlight>
{{out}}
<pre>hs(27): 112 elements: [27 82 41 124 ... 8 4 2 1]
hs(77031): 351 elements</pre>
=={{header|Wren}}==
<syntaxhighlight lang="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
h.add(n)
}
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)")</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|XPL0}}==
<
int Seq(1000); \more than enough for longest sequence
Line 5,946 ⟶ 10,653:
];
IntOut(0, SN); Text(0, "'s Hailstone length = "); IntOut(0, MaxLen);
]</
{{out}}
<pre>
27's Hailstone length = 112
Sequence = 27 82 41 124 ... 8 4 2 1
77031's Hailstone length = 351
</pre>
=={{header|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.
<syntaxhighlight lang="z80">;;;;;;;;;;;;;;;;;;; HEADER ;;;;;;;;;;;;;;;;;;;
read "\SrcCPC\winape_macros.asm"
read "\SrcCPC\MemoryMap.asm"
read "\SrcALL\winapeBuildCompat.asm"
;;;;;;;;;;;;;;;;;;; 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 ;;;;;;;;;;;;;;;;;;;
read "\SrcCPC\winape_stringop.asm"
read "\SrcCPC\winape_showhex.asm"
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
add hl,de ;hl = de*3
ld de,1
add hl,de
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
;you can stop reading here.
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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|zkl}}==
<
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.
{{out}}
Line 5,970 ⟶ 10,849:
</pre>
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.
<
collatz, // generate Collatz sequence(n)
fcn(c,n){ // if new longest sequence, save length/C, return longest
if(c.len()>n[0]) n.clear(c.len(),c[0]); n}.fp1(L(0,0)))</
{{out}}
<pre>
L(351,77031) // length, hailstone
</pre>
=={{header|ZX Spectrum Basic}}==
{{trans|BBC_BASIC}}
<syntaxhighlight lang="zxbasic">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</syntaxhighlight>
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