Hailstone sequence
You are encouraged to solve this task according to the task description, using any language you may know.
The Hailstone sequence of numbers can be generated from a starting positive integer, n by:
- If n is 1 then the sequence ends.
- If n is even then the next n of the sequence
= n/2 - If n is odd then the next n of the sequence
= (3 * n) + 1
The (unproven) Collatz conjecture is that the hailstone sequence for any starting number always terminates.
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
27, 82, 41, 124and ending with8, 4, 2, 1 - Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.
(But don't show the actual sequence!)
- See also
- xkcd (humourous).
11l
<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]))</lang>
- Output:
Maximum length 351 was found for hailstone(77031) for numbers <100,000
360 Assembly
<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
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</lang>
- Output:
hailstone( 27)= 112 : 27 82 41 ...... 4 2 1 longest <100000 : hailstone( 77031)= 351
ABAP
<lang ABAP> CLASS lcl_hailstone DEFINITION.
PUBLIC SECTION.
TYPES: tty_sequence TYPE STANDARD TABLE OF i
WITH NON-UNIQUE EMPTY KEY,
BEGIN OF ty_seq_len,
start TYPE i,
len TYPE i,
END OF ty_seq_len,
tty_seq_len TYPE HASHED TABLE OF ty_seq_len
WITH UNIQUE KEY start.
CLASS-METHODS:
get_next
IMPORTING
n TYPE i
RETURNING
VALUE(r_next_hailstone_num) TYPE i,
get_sequence
IMPORTING
start TYPE i
RETURNING
VALUE(r_sequence) TYPE tty_sequence,
get_longest_sequence_upto
IMPORTING
limit TYPE i
RETURNING
VALUE(r_longest_sequence) TYPE ty_seq_len.
PRIVATE SECTION.
TYPES: BEGIN OF ty_seq,
start TYPE i,
seq TYPE tty_sequence,
END OF ty_seq.
CLASS-DATA: sequence_buffer TYPE HASHED TABLE OF ty_seq
WITH UNIQUE KEY start.
ENDCLASS.
CLASS lcl_hailstone IMPLEMENTATION.
METHOD get_next.
r_next_hailstone_num = COND #( WHEN n MOD 2 = 0 THEN n / 2
ELSE ( 3 * n ) + 1 ).
ENDMETHOD.
METHOD get_sequence.
INSERT start INTO TABLE r_sequence.
IF start = 1.
RETURN.
ENDIF.
READ TABLE sequence_buffer ASSIGNING FIELD-SYMBOL(<buff>)
WITH TABLE KEY start = start.
IF sy-subrc = 0.
INSERT LINES OF <buff>-seq INTO TABLE r_sequence.
ELSE.
DATA(seq) = get_sequence( get_next( start ) ).
INSERT LINES OF seq INTO TABLE r_sequence.
INSERT VALUE ty_seq( start = start
seq = seq ) INTO TABLE sequence_buffer.
ENDIF.
ENDMETHOD.
METHOD get_longest_sequence_upto.
DATA: max_seq TYPE ty_seq_len,
act_seq TYPE ty_seq_len.
DO limit TIMES.
act_seq-len = lines( get_sequence( sy-index ) ).
IF act_seq-len > max_seq-len.
max_seq-len = act_seq-len.
max_seq-start = sy-index.
ENDIF.
ENDDO.
r_longest_sequence = max_seq. ENDMETHOD.
ENDCLASS.
START-OF-SELECTION.
cl_demo_output=>begin_section( |Hailstone sequence of 27 is: | ).
cl_demo_output=>write( REDUCE string( INIT result = ``
FOR item IN lcl_hailstone=>get_sequence( 27 )
NEXT result = |{ result } { item }| ) ).
cl_demo_output=>write( |With length: { lines( lcl_hailstone=>get_sequence( 27 ) ) }| ).
cl_demo_output=>begin_section( |Longest hailstone sequence upto 100k| ).
cl_demo_output=>write( lcl_hailstone=>get_longest_sequence_upto( 100000 ) ).
cl_demo_output=>display( ).
</lang>
- Output:
Hailstone sequence of 27 is: 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 With length: 112 Longest hailstone sequence upto 100k Structure START LEN 77031 351
ACL2
<lang Lisp>(defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
- Must be tail recursive
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))</lang>
- Output:
> (take 4 (hailstone 27)) (27 82 41 124) > (nthcdr 108 (hailstone 27)) (8 4 2 1) > (len (hailstone 27)) 112 > (max-hailstone-start 100000 0 0) (351 77031)
Ada
Similar to C method: <lang Ada>with Ada.Text_IO; use Ada.Text_IO; procedure hailstone is type int_arr is array(Positive range <>) of Integer; type int_arr_pt is access all int_arr;
function hailstones(num:Integer; pt:int_arr_pt) return Integer is stones : Integer := 1; n : Integer := num; begin if pt /= null then pt(1) := num; end if; while (n/=1) loop stones := stones + 1; if n mod 2 = 0 then n := n/2; else n := (3*n)+1; end if; if pt /= null then pt(stones) := n; end if; end loop; return stones; end hailstones;
nmax,stonemax,stones : Integer := 0; list : int_arr_pt; begin stones := hailstones(27,null); list := new int_arr(1..stones); stones := hailstones(27,list); put(" 27: "&Integer'Image(stones)); new_line; for n in 1..4 loop put(Integer'Image(list(n))); end loop; put(" .... "); for n in stones-3..stones loop put(Integer'Image(list(n))); end loop; new_line; for n in 1..100000 loop stones := hailstones(n,null); if stones>stonemax then nmax := n; stonemax := stones; end if; end loop; put_line(Integer'Image(nmax)&" max @ n= "&Integer'Image(stonemax)); end hailstone;</lang>
- Output:
27: 112 27 82 41 124 .... 8 4 2 1 77031 max @ n= 351
Alternative method
A method without pointers or dynamic memory allocation, but slower for simply counting. This is also used for the "executable library" task Executable library#Ada.
hailstones.ads: <lang Ada>package Hailstones is
type Integer_Sequence is array(Positive range <>) of Integer; function Create_Sequence (N : Positive) return Integer_Sequence;
end Hailstones;</lang> hailstones.adb: <lang Ada>package body Hailstones is
function Create_Sequence (N : Positive) return Integer_Sequence is
begin
if N = 1 then
-- terminate
return (1 => N);
elsif N mod 2 = 0 then
-- even
return (1 => N) & Create_Sequence (N / 2);
else
-- odd
return (1 => N) & Create_Sequence (3 * N + 1);
end if;
end Create_Sequence;
end Hailstones;</lang> example main.adb: <lang Ada>with Ada.Text_IO; with Hailstones;
procedure Main is
package Integer_IO is new Ada.Text_IO.Integer_IO (Integer);
procedure Print_Sequence (X : Hailstones.Integer_Sequence) is
begin
for I in X'Range loop
Integer_IO.Put (Item => X (I), Width => 0);
if I < X'Last then
Ada.Text_IO.Put (", ");
end if;
end loop;
Ada.Text_IO.New_Line;
end Print_Sequence;
Hailstone_27 : constant Hailstones.Integer_Sequence :=
Hailstones.Create_Sequence (N => 27);
begin
Ada.Text_IO.Put_Line ("Length of 27:" & Integer'Image (Hailstone_27'Length));
Ada.Text_IO.Put ("First four: ");
Print_Sequence (Hailstone_27 (Hailstone_27'First .. Hailstone_27'First + 3));
Ada.Text_IO.Put ("Last four: ");
Print_Sequence (Hailstone_27 (Hailstone_27'Last - 3 .. Hailstone_27'Last));
declare
Longest_Length : Natural := 0;
Longest_N : Positive;
Length : Natural;
begin
for I in 1 .. 99_999 loop
Length := Hailstones.Create_Sequence (N => I)'Length;
if Length > Longest_Length then
Longest_Length := Length;
Longest_N := I;
end if;
end loop;
Ada.Text_IO.Put_Line ("Longest length is" & Integer'Image (Longest_Length));
Ada.Text_IO.Put_Line ("with N =" & Integer'Image (Longest_N));
end;
end Main;</lang>
- Output:
Length of 27: 112 First four: 27, 82, 41, 124 Last four: 8, 4, 2, 1 Longest length is 351 with N = 77031
Aime
<lang aime>void print_hailstone(integer h) {
list l;
while (h ^ 1) {
lb_p_integer(l, h);
h = h & 1 ? 3 * h + 1 : h / 2;
}
o_form("hailstone sequence for ~ is ~1 ~ ~ ~ .. ~ ~ ~ ~, it is ~ long\n",
l[0], l[1], l[2], l[3], l[-3], l[-2], l[-1], 1, ~l + 1);
}
void max_hailstone(integer x) {
integer e, i, m; index r;
m = 0;
i = 1;
while (i < x) {
integer h, k, l;
h = i;
l = 1;
while (h ^ 1) {
if (i_j_integer(k, r, h)) {
l += k;
break;
} else {
l += 1;
h = h & 1 ? 3 * h + 1 : h / 2;
}
}
r[i] = l - 1;
if (m < l) {
m = l;
e = i;
}
i += 1; }
o_form("hailstone sequence length for ~ is ~\n", e, m);
}
integer main(void) {
print_hailstone(27); max_hailstone(100000);
return 0;
}</lang>
- Output:
hailstone sequence for 27 is 27 82 41 124 .. 8 4 2 1, it is 112 long hailstone sequence length for 77031 is 351
ALGOL 60
<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 </lang>
- Output:
sequence for 27 : 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 number of elements: 112 number < 100000 with the longest sequence: 77031 , with 351 elements.
ALGOL 68
- note: This specimen retains the original C coding style.
<lang algol68>MODE LINT = # LONG ... # INT;
PROC hailstone = (INT in n, REF[]LINT array)INT: (
INT hs := 1;
INT index := 0;
LINT n := in n;
WHILE n /= 1 DO
hs +:= 1;
IF array ISNT REF[]LINT(NIL) THEN array[index +:= 1] := n FI;
n := IF ODD n THEN 3*n+1 ELSE n OVER 2 FI
OD;
IF array ISNT REF[]LINT(NIL) THEN array[index +:= 1] := n FI;
hs
);
main: (
INT j, hmax := 0;
INT jatmax, n;
INT border = 4;
FOR j TO 100000-1 DO
n := hailstone(j, NIL);
IF hmax < n THEN
hmax := n;
jatmax := j
FI
OD;
[2]INT test := (27, jatmax);
FOR key TO UPB test DO
INT val = test[key];
n := hailstone(val, NIL);
[n]LINT array;
n := hailstone(val, array);
printf(($"[ "n(border)(g(0)", ")" ..."n(border)(", "g(0))"] len="g(0)l$,
array[:border], array[n-border+1:], n))
#;free(array) #
OD;
printf(($"Max "g(0)" at j="g(0)l$, hmax, jatmax))
- ELLA Algol68RS:
print(("Max",hmax," at j=",jatmax, new line))
)</lang>
- Output:
[ 27, 82, 41, 124, ..., 8, 4, 2, 1] len=112 [ 77031, 231094, 115547, 346642, ..., 8, 4, 2, 1] len=351 Max 351 at j=77031
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. <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 </lang>
- Output:
DISPLAY HAILSTONE SEQUENCE FOR WHAT NUMBER?
-> 27
27 82 41 124 62 31 94 47 142 71
214 107 322 161 484 242 121 364 182 91
274 137 412 206 103 310 155 466 233 700
350 175 526 263 790 395 1186 593 1780 890
445 1336 668 334 167 502 251 754 377 1132
566 283 850 425 1276 638 319 958 479 1438
719 2158 1079 3238 1619 4858 2429 7288 3644 1822
911 2734 1367 4102 2051 6154 3077 9232 4616 2308
1154 577 1732 866 433 1300 650 325 976 488
244 122 61 184 92 46 23 70 35 106
53 160 80 40 20 10 5 16 8 4
2 1
SEQUENCE LENGTH = 112
SEARCHING FOR LONGEST SEQUENCE UP TO N = 10000 ...
SEARCH TERMINATED WITH OVERFLOW AT N = 447
MAXIMUM SEQUENCE LENGTH = 144 FOR N = 327
ALGOL W
<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.</lang>
- Output:
27: length 112, first: [27 82 41 124] last: [8 4 2 1] Maximum sequence length: 351 for: 77031
APL
<lang APL>seq←hailstone n;next ⍝ Returns the hailstone sequence for a given number
seq←n ⍝ Init the sequence
- While n≠1
next←(n÷2) (1+3×n) ⍝ Compute both possibilities n←next[1+2|n] ⍝ Pick the appropriate next step seq,←n ⍝ Append that to the sequence
- EndWhile</lang>
- Output:
<lang APL> 5↑hailstone 27 27 82 41 124 62
¯5↑hailstone 27
16 8 4 2 1
⍴hailstone 27
112
1↑{⍵[⍒↑(⍴∘hailstone)¨⍵]}⍳100000
77031</lang>
AppleScript
<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</lang>
- Output:
{|length of sequence|:112, |first 4 numbers|:{27, 82, 41, 124}, |last 4 numbers|:{8, 4, 2, 1}}
<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</lang>
- Output:
{|number(s) giving longest sequence length|:{77031}, |length of sequence|:351}
Arturo
<lang arturo>hailstone: @(n){ ret: #(n) loop n>1 { if [and n 1]=1 { n: 3*n+1 }{ n: n/2 } ret: ret+n } }
print "Hailstone sequence for 27:" print [hailstone 27]
maxHailstoneLength: 0 maxHailstone: 0
loop [range 2 100000] { l: size|hailstone & if l>maxHailstoneLength { maxHailstoneLength: l maxHailstone: & } }
print "max hailstone sequence found (<100000): of length " + maxHailstoneLength + " for " + maxHailstone</lang>
- Output:
Hailstone sequence for 27: #(27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1) max hailstone sequence found (<100000): of length 351 for 77031
AutoHotkey
<lang autohotkey>; Submitted by MasterFocus --- http://tiny.cc/iTunis
- [1] Generate the Hailstone Seq. for a number
List := varNum := 7 ; starting number is 7, not counting elements While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
- [2] Seq. for starting number 27 has 112 elements
Count := 1, List := varNum := 27 ; starting number is 27, counting elements While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
- [3] Find number<100000 with longest seq. and show both values
MaxNum := Max := 0 ; reset the Maximum variables TimesToLoop := 100000 ; limit number here is 100000 Offset := 70000 ; offset - use 0 to process from 0 to 100000 Loop, %TimesToLoop% {
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) ) Break text := "Processing...`n-------------------`n" text .= "Current starting number: " Index "`n" text .= "Current sequence count: " Count text .= "`n-------------------`n" text .= "Maximum starting number: " MaxNum "`n" text .= "Maximum sequence count: " Max " <<" ; text split to avoid long code lines ToolTip, %text% Count := 1 ; going to count the elements, but no "List" required While ( varNum > 1 ) Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) If ( Count > Max ) Max := Count , MaxNum := Index ; set the new maximum values, if necessary
} ToolTip MsgBox % "Number: " MaxNum "`nCount: " Max</lang>
AutoIt
<lang autoit> $Hail = Hailstone(27) ConsoleWrite("Sequence-Lenght: "&$Hail&@CRLF) $Big = -1 $Sequenzlenght = -1 For $I = 1 To 100000 $Hail = Hailstone($i, False) If Number($Hail) > $Sequenzlenght Then $Sequenzlenght = Number($Hail) $Big = $i EndIf Next ConsoleWrite("Longest Sequence : "&$Sequenzlenght&" from number "&$Big&@CRLF) Func Hailstone($int, $sequence = True) $Counter = 0 While True $Counter += 1 If $sequence = True Then ConsoleWrite($int & ",") If $int = 1 Then ExitLoop If Not Mod($int, 2) Then $int = $int / 2 Else $int = 3 * $int + 1 EndIf If Not Mod($Counter, 25) AND $sequence = True Then ConsoleWrite(@CRLF) WEnd If $sequence = True Then ConsoleWrite(@CRLF) Return $Counter EndFunc ;==>Hailstone </lang>
- Output:
27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,412,206,103, 310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132, 566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051, 6154,3077,9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106, 53,160,80,40,20,10,5,16,8,4,2,1, Sequence-Lenght: 112 Longest Sequence : 351 from number 77031
AWK
<lang awk>
- !/usr/bin/awk -f
function hailstone(v, verbose) { n = 1; u = v; while (1) { if (verbose) printf " "u; if (u==1) return(n); n++; if (u%2 > 0 ) u = 3*u+1; else u = u/2; } }
BEGIN { i = 27; printf("hailstone(%i) has %i elements\n",i,hailstone(i,1)); ix=0; m=0; for (i=1; i<100000; i++) { n = hailstone(i,0); if (m<n) { m=n; ix=i; } } printf("longest hailstone sequence is %i and has %i elements\n",ix,m); } </lang>
- Output:
27 82 41 124 ....... 8 4 2 1 hailstone(27) has 112 elements longest hailstone sequence is 77031 and has 351 elements
BASIC
Applesoft BASIC
<lang ApplesoftBasic>10 HOME
100 N = 27 110 GOSUB 400"HAILSTONE 120 DEF FN L(I) = E(I + 4 * (I < 0)) 130IFL=112AND(S(0)=27ANDS(1)=82ANDS(2)=41ANDS(3)=124)AND(FNL(M-3)=8ANDFNL(M-2)=4ANDFNL(M-1)=2ANDFNL(M)=1)THENPRINT"THE HAILSTONE SEQUENCE FOR THE NUMBER 27 HAS 112 ELEMENTS STARTING WITH 27, 82, 41, 124 AND ENDING WITH 8, 4, 2, 1" 140 PRINT 150 V = PEEK(37) + 1
200 N = 1 210 GOSUB 400"HAILSTONE 220 MN = 1 230 ML = L 240 FOR I = 2 TO 99999 250 N = I 260 GOSUB 400"HAILSTONE 270 IFL>MLTHENMN=I:ML=L:VTABV:HTAB1:PRINT "THE NUMBER " MN " HAS A HAILSTONE SEQUENCE LENGTH OF "L" WHICH IS THE LONGEST HAILSTONE SEQUENCE OF NUMBERS LESS THAN ";:Y=PEEK(37)+1:X=PEEK(36)+1 280 IF Y THEN VTAB Y : HTAB X : PRINTI+1; 290 NEXT I
300 END
400 M = 0 410 FOR L = 1 TO 1E38 420 IF L < 5 THEN S(L-1) = N 430 M = (M + 1) * (M < 3) 440 E(M) = N 450 IF N = 1 THEN RETURN 460 EVEN = INT(N/2)=N/2 470 IF EVEN THEN N=N/2 480 IF NOT EVEN THEN N = (3 * N) + 1 490 NEXT L : STOP</lang>
BBC BASIC
<lang bbcbasic> seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1</lang>
- Output:
27 82 41 124 62 31 94 47
142 71 214 107 322 161 484 242
121 364 182 91 274 137 412 206
103 310 155 466 233 700 350 175
526 263 790 395 1186 593 1780 890
445 1336 668 334 167 502 251 754
377 1132 566 283 850 425 1276 638
319 958 479 1438 719 2158 1079 3238
1619 4858 2429 7288 3644 1822 911 2734
1367 4102 2051 6154 3077 9232 4616 2308
1154 577 1732 866 433 1300 650 325
976 488 244 122 61 184 92 46
23 70 35 106 53 160 80 40
20 10 5 16 8 4 2 1
Sequence length = 112
The number with the longest hailstone sequence is 77031
Its sequence length is 351
Commodore BASIC
<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 </lang>
FreeBASIC
<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</lang>
- Output:
sequence for number 27
27 82 41 124 62 31 94 47 142 71
214 107 322 161 484 242 121 364 182 91
274 137 412 206 103 310 155 466 233 700
350 175 526 263 790 395 1186 593 1780 890
445 1336 668 334 167 502 251 754 377 1132
566 283 850 425 1276 638 319 958 479 1438
719 2158 1079 3238 1619 4858 2429 7288 3644 1822
911 2734 1367 4102 2051 6154 3077 9232 4616 2308
1154 577 1732 866 433 1300 650 325 976 488
244 122 61 184 92 46 23 70 35 106
53 160 80 40 20 10 5 16 8 4
2 1
sequence length = 112
-------------------------------------------------------------------------------
The longest sequence is for 77031, it has a sequence length of 351
Liberty BASIC
<lang lb>print "Part 1: Create a routine to generate the hailstone sequence for a number." print "" while hailstone < 1 or hailstone <> int(hailstone)
input "Please enter a positive integer: "; hailstone
wend print "" print "The following is the 'Hailstone Sequence' for your number..." print "" print hailstone while hailstone <> 1
if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1 print hailstone
wend print "" input "Hit 'Enter' to continue to part 2...";dummy$ cls print "Part 2: Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1." print "" print "No. in Seq.","Hailstone Sequence Number for 27" print "" c = 1: hailstone = 27 print c, hailstone while hailstone <> 1
c = c + 1 if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1 print c, hailstone
wend print "" input "Hit 'Enter' to continue to part 3...";dummy$ cls print "Part 3: Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.(But don't show the actual sequence)!" print "" print "Calculating result... Please wait... This could take a little while..." print "" print "Percent Done", "Start Number", "Seq. Length", "Maximum Sequence So Far" print "" for cc = 1 to 99999
hailstone = cc: c = 1
while hailstone <> 1
c = c + 1
if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1
wend
if c > max then max = c: largesthailstone = cc
locate 1, 7
print " "
locate 1, 7
print using("###.###", cc / 99999 * 100);"%", cc, c, max
scan
next cc print "" print "The number less than 100,000 with the longest 'Hailstone Sequence' is "; largesthailstone;". It's sequence length is "; max;"." end</lang>
OxygenBasic
<lang oxygenbasic>
function Hailstone(sys *n) '========================= if n and 1
n=n*3+1
else
n=n>>1
end if end function
function HailstoneSequence(sys n) as sys '======================================= count=1 do
Hailstone n Count++ if n=1 then exit do
end do return count end function
'MAIN '====
maxc=0 maxn=0 e=100000 for n=1 to e
c=HailstoneSequence n if c>maxc maxc=c maxn=n end if
next
print e ", " maxn ", " maxc
'result 100000, 77031, 351 </lang>
PureBasic
<lang PureBasic>NewList Hailstones.i() ; Make a linked list to use as we do not know the numbers of elements needed for an Array
Procedure.i FillHailstones(n) ; Fills the list & returns the amount of elements in the list
Shared Hailstones() ; Get access to the Hailstones-List
ClearList(Hailstones()) ; Remove old data
Repeat
AddElement(Hailstones()) ; Add an element to the list
Hailstones()=n ; Fill current value in the new list element
If n=1
ProcedureReturn ListSize(Hailstones())
ElseIf n%2=0
n/2
Else
n=(3*n)+1
EndIf
ForEver
EndProcedure
If OpenConsole()
Define i, l, maxl, maxi
l=FillHailstones(27)
Print("#27 has "+Str(l)+" elements and the sequence is: "+#CRLF$)
ForEach Hailstones()
If i=6
Print(#CRLF$)
i=0
EndIf
i+1
Print(RSet(Str(Hailstones()),5))
If Hailstones()<>1
Print(", ")
EndIf
Next
i=1
Repeat
l=FillHailstones(i)
If l>maxl
maxl=l
maxi=i
EndIf
i+1
Until i>=100000
Print(#CRLF$+#CRLF$+"The longest sequence below 100000 is #"+Str(maxi)+", and it has "+Str(maxl)+" elements.")
Print(#CRLF$+#CRLF$+"Press ENTER to exit."): Input()
CloseConsole()
EndIf</lang>
- Output:
#27 has 112 elements and the sequence is:
27, 82, 41, 124, 62, 31,
94, 47, 142, 71, 214, 107,
322, 161, 484, 242, 121, 364,
182, 91, 274, 137, 412, 206,
103, 310, 155, 466, 233, 700,
350, 175, 526, 263, 790, 395,
1186, 593, 1780, 890, 445, 1336,
668, 334, 167, 502, 251, 754,
377, 1132, 566, 283, 850, 425,
1276, 638, 319, 958, 479, 1438,
719, 2158, 1079, 3238, 1619, 4858,
2429, 7288, 3644, 1822, 911, 2734,
1367, 4102, 2051, 6154, 3077, 9232,
4616, 2308, 1154, 577, 1732, 866,
433, 1300, 650, 325, 976, 488,
244, 122, 61, 184, 92, 46,
23, 70, 35, 106, 53, 160,
80, 40, 20, 10, 5, 16,
8, 4, 2, 1
The longest sequence found up to 100000 is #77031 which has 351 elements.
Press ENTER to exit.
Run BASIC
<lang runbasic>print "Part 1: Create a routine to generate the hailstone sequence for a number." print ""
while hailstone < 1 or hailstone <> int(hailstone)
input "Please enter a positive integer: "; hailstone
wend count = doHailstone(hailstone,"Y")
print: print "Part 2: Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1." count = doHailstone(27,"Y")
print: print "Part 3: Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.(But don't show the actual sequence)!" print "Calculating result... Please wait... This could take a little while..." print "Stone Percent Count" for i = 1 to 99999
count = doHailstone(i,"N")
if count > maxCount then theBigStone = i maxCount = count
print using("#####",i);" ";using("###.#", i / 99999 * 100);"% ";using("####",count)
end if
next i end
'--------------------------------------------- ' pass number and print (Y/N) FUNCTION doHailstone(hailstone,prnt$) if prnt$ = "Y" then
print print "The following is the 'Hailstone Sequence' for number:";hailstone
end if while hailstone <> 1
if (hailstone and 1) then hailstone = (hailstone * 3) + 1 else hailstone = hailstone / 2 doHailstone = doHailstone + 1 if prnt$ = "Y" then print hailstone;chr$(9); if (doHailstone mod 10) = 0 then print end if
wend END FUNCTION</lang>
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.
<lang> 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</lang>
- Output:
Enter a positive integer 27 1 27 2 82 3 41 .... 110 4 111 2 112 1 Testing for longest chain for n<100000... Uh oh, getting an overflow for 32766x2 + 12139 at step number 72 trying to triple 32766x15980 + 7565 The longest sequence starts at 32766x2 + 11499 And goes for 351 steps.
Batch File
1. Create a routine to generate the hailstone sequence for a number.
2. Show that the hailstone sequence for the number 27 has 112 elements...
<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 seq=%1 set sav=%1 set cnt=0
- loop
set /a cnt+=1 if !num! equ 1 goto :eof set /a isodd=%num%%%2 if !isodd! equ 0 goto divideby2
set /a num=(3*%num%)+1 set seq=!seq! %num% goto loop
- divideby2
set /a num/=2 set seq=!seq! %num% goto loop</lang>
- Output:
Task #1: (Start:111) 111 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Sequence has 70 elements. Task #2: (Start:27) 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Sequence has 112 elements.
The script above could only be used in smaller inputs. Thus, for the third task, a slightly different script will be used. However, this script is still slow. I tried this on a fast computer and it took about 40-45 minutes to complete. <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</lang>
- Output:
Number less than 100000 with longest sequence: 77031 With length 351.
beeswax
This approach reuses the main hailstone sequence function for all three tasks.
The pure hailstone sequence function, returning the sequence for any number entered in the console:
<lang beeswax> >@:N q >%"d3~@.PNp d~2~pL~1F{<T_</lang>
Returning the sequence for the starting value 27
<lang beeswax> >@:N q >%"d3~@.PNq d~2~qL~1Ff{<BF3_ {NNgA<</lang>
Output of the sequence, followed by the length of the sequence:
<lang> 27 82 41 124 62 31 94 47
...
2158 1079 3238 1619 4858 2429 7288 3644 1822
...
16 8 4 2 1
112</lang>
Number below 100,000 with the longest hailstone sequence, and the length of that sequence:
<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</lang>
Output:
<lang>77031 351</lang>
Befunge
<lang befunge>93*:. v > :2%v > v+1*3_2/ >" ",:.v v< <v v-1:< < +1\_$1+v^ \ v .,+94<>^>::v >" "03pv :* p v67:" "< 0: 1 >p78p25 *^*p0
v!-1: <<*^<
9$_:0\ ^-^< v v01g00:< 1 4 >g"@"*+`v^ <+ v01/"@":_ $ ^, >p"@"%00p\$:^. vg01g00 ,+49< >"@"*+.@ </lang>
- Output:
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 112 77031 351
Bracmat
<lang bracmat>(
( hailstone
= L len
. !arg:?L
& whl
' ( !arg:~1
& (!arg*1/2:~/|3*!arg+1):?arg
& !arg !L:?L
)
& (!L:? [?len&!len.!L)
)
& ( reverse
= L e
. :?L
& whl'(!arg:%?e ?arg&!e !L:?L)
& !L
)
& hailstone$27:(?len.?list) & reverse$!list:?first4 [4 ? [-5 ?last4 & put$"Hailstone sequence starting with " & put$!first4 & put$(str$(" has " !len " elements and ends with ")) & put$(!last4 \n) & 1:?N & 0:?max:?Nmax & whl
' ( !N+1:<100000:?N
& hailstone$!N
: ( >!max:?max&!N:?Nmax
| ?
. ?
)
)
& out
$ ( str
$ ( "The number <100000 with the longest hailstone sequence is "
!Nmax
" with "
!max
" elements."
)
)
);</lang>
Brainf***
<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. ] </lang>
Brat
<lang brat>hailstone = { num |
sequence = [num]
while { num != 1 }
{ true? num % 2 == 0
{ num = num / 2 }
{ num = num * 3 + 1 }
sequence << num
}
sequence
}
- Check sequence for 27
seq = hailstone 27 true? (seq[0,3] == [27 82 41 124] && seq[-1, -4] == [8 4 2 1])
{ p "Sequence for 27 is correct" }
{ p "Sequence for 27 is not correct!" }
- Find longest sequence for numbers < 100,000
longest = [number: 0 length: 0]
1.to 99999 { index |
seq = hailstone index
true? seq.length > longest[:length]
{ longest[:length] = seq.length
longest[:number] = index
p "Longest so far: #{index} @ #{longest[:length]} elements"
}
index = index + 1 }
p "Longest was starting from #{longest[:number]} and was of length #{longest[:length]}"</lang>
- Output:
Sequence for 27 is correct Longest so far: 1 @ 1 elements Longest so far: 2 @ 2 elements Longest so far: 3 @ 8 elements ... Longest so far: 52527 @ 340 elements Longest so far: 77031 @ 351 elements Longest was starting from 77031 and was of length 351
Burlesque
<lang burlesque> blsq ) 27{^^^^2.%{3.*1.+}\/{2./}\/ie}{1!=}w!bx{\/+]}{\/isn!}w!L[ 112 </lang>
C
<lang C>#include <stdio.h>
- include <stdlib.h>
int hailstone(int n, int *arry) {
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main() {
int j, hmax = 0; int jatmax, n; int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}</lang>
- Output:
[ 27, 82, 41, 124, ...., 8, 4, 2, 1] len= 112 Max 351 at j= 77031
With caching
Much faster if you want to go over a million or so. <lang c>#include <stdio.h>
- define N 10000000
- define CS N /* cache size */
typedef unsigned long ulong; ulong cache[CS] = {0};
ulong hailstone(ulong n) { int x; if (n == 1) return 1; if (n < CS && cache[n]) return cache[n];
x = 1 + hailstone((n & 1) ? 3 * n + 1 : n / 2); if (n < CS) cache[n] = x; return x; }
int main() { int i, l, max = 0, mi; for (i = 1; i < N; i++) { if ((l = hailstone(i)) > max) { max = l; mi = i; } } printf("max below %d: %d, length %d\n", N, mi, max); return 0; }</lang>
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Text;
namespace Hailstone {
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}</lang>
112 Elements Starting with : 27,82,41,124 and ending with : 8,4,2,1 Number < 100000 with longest Hailstone seq.: 77031 with length of 351
With caching
As with the C example, much faster if you want to go over a million or so. <lang csharp>using System; using System.Collections.Generic;
namespace ConsoleApplication1 {
class Program
{
public static void Main()
{
int longestChain = 0, longestNumber = 0;
var recursiveLengths = new Dictionary<int, int>();
const int maxNumber = 100000;
for (var i = 1; i <= maxNumber; i++)
{
var chainLength = Hailstone(i, recursiveLengths);
if (longestChain >= chainLength)
continue;
longestChain = chainLength;
longestNumber = i;
}
Console.WriteLine("max below {0}: {1} ({2} steps)", maxNumber, longestNumber, longestChain);
}
private static int Hailstone(int num, Dictionary<int, int> lengths)
{
if (num == 1)
return 1;
while (true)
{
if (lengths.ContainsKey(num))
return lengths[num];
lengths[num] = 1 + ((num%2 == 0) ? Hailstone(num/2, lengths) : Hailstone((3*num) + 1, lengths));
}
}
}
}</lang>
max below 100000: 77031 (351 steps)
C++
<lang cpp>#include <iostream>
- include <vector>
- include <utility>
std::vector<int> hailstone(int i) {
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n) {
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
// Use the routine to show that the hailstone sequence for the number 27
std::vector<int> h27; h27 = hailstone(27);
// has 112 elements
int l = h27.size(); std::cout << "length of hailstone(27) is " << l;
// starting with 27, 82, 41, 124 and
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
// ending with 8, 4, 2, 1
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}</lang>
- Output:
length of hailstone(27) is 112 first four elements of hailstone(27) are 27 82 41 124 last four elements of hailstone(27) are 8 4 2 1 the longest hailstone sequence under 100,000 is 77031 with 351 elements.
Templated solution works for all of Qt's sequential container classes (QLinkedList, QList, QVector). <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();
} </lang>
- Output:
hailstone(27): length: 112 elements first 4 elements: QVector(27, 82, 41, 124) last 4 elements: QVector(8, 4, 2, 1) longest sequence with starting element under 100000: length: 351 elements starting element: 77031
Ceylon
<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``"); }</lang>
CLIPS
<lang clips>(deftemplate longest
(slot bound) ; upper bound for the range of values to check (slot next (default 2)) ; next value that needs to be checked (slot start (default 1)) ; starting value of longest sequence (slot len (default 1)) ; length of longest sequence
)
(deffacts startup
(query 27) (longest (bound 100000))
)
(deffunction hailstone-next
(?n) (if (evenp ?n) then (div ?n 2) else (+ (* 3 ?n) 1) )
)
(defrule extend-sequence
?hail <- (hailstone $?sequence ?tail&:(> ?tail 1)) => (retract ?hail) (assert (hailstone ?sequence ?tail (hailstone-next ?tail)))
)
(defrule start-query
(query ?num) => (assert (hailstone ?num))
)
(defrule result-query
(query ?num) (hailstone ?num $?sequence 1) => (bind ?sequence (create$ ?num ?sequence 1)) (printout t "Hailstone sequence starting with " ?num ":" crlf) (bind ?len (length ?sequence)) (printout t " Length: " ?len crlf) (printout t " First four: " (implode$ (subseq$ ?sequence 1 4)) crlf) (printout t " Last four: " (implode$ (subseq$ ?sequence (- ?len 3) ?len)) crlf) (printout t crlf)
)
(defrule longest-create-next-hailstone
(longest (bound ?bound) (next ?next)) (test (<= ?next ?bound)) (not (hailstone ?next $?)) => (assert (hailstone ?next))
)
(defrule longest-check-next-hailstone
?longest <- (longest (bound ?bound) (next ?next) (start ?start) (len ?len)) (test (<= ?next ?bound)) ?hailstone <- (hailstone ?next $?sequence 1) => (retract ?hailstone) (bind ?thislen (+ 2 (length ?sequence))) (if (> ?thislen ?len) then (modify ?longest (start ?next) (len ?thislen) (next (+ ?next 1))) else (modify ?longest (next (+ ?next 1))) )
)
(defrule longest-finished
(longest (bound ?bound) (next ?next) (start ?start) (len ?len)) (test (> ?next ?bound)) => (printout t "The number less than " ?bound " that has the largest hailstone" crlf) (printout t "sequence is " ?start " with a length of " ?len "." crlf) (printout t crlf)
)</lang>
- Output:
The number less than 100000 that has the largest hailstone sequence is 77031 with a length of 351. Hailstone sequence starting with 27: Length: 112 First four: 27 82 41 124 Last four: 8 4 2 1
Clojure
<lang clojure>(defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert) (->> hseq (take 4) (= [27 82 41 124]) assert) (->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))</lang>
COBOL
Testing with GnuCOBOL <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.</lang>
- Output:
prompt$ cobc -x hailstones.cob prompt$ ./hailstones +0000000027: 27, 82, 41, 124, ..., 8, 4, 2, 1: +0000000112 elements Longest at: +0000077031 with +0000000351 elements prompt$ ./hailstones 42 +0000000042: 42, 21, 64, 32, ..., 8, 4, 2, 1: +0000000009 elements Longest at: +0000077031 with +0000000351 elements
CoffeeScript
Recursive version: <lang coffeescript>hailstone = (n) ->
if n is 1 [n] else if n % 2 is 0 [n].concat hailstone n/2 else [n].concat hailstone (3*n) + 1
h27 = hailstone 27 console.log "hailstone(27) = #{h27[0..3]} ... #{h27[-4..]} (length: #{h27.length})"
maxlength = 0 maxnums = []
for i in [1..100000]
seq = hailstone i if seq.length is maxlength maxnums.push i else if seq.length > maxlength maxlength = seq.length maxnums = [i]
console.log "Max length: #{maxlength}; numbers generating sequences of this length: #{maxnums}"</lang>
hailstone(27) = 27,82,41,124 ... 8,4,2,1 (length: 112) Max length: 351; numbers generating sequences of this length: 77031
Common Lisp
<lang lisp>(defun hailstone (n)
(cond ((= n 1) '(1))
((evenp n) (cons n (hailstone (/ n 2)))) (t (cons n (hailstone (+ (* 3 n) 1))))))
(defun longest (n)
(let ((k 0) (l 0)) (loop for i from 1 below n do
(let ((len (length (hailstone i)))) (when (> len l) (setq l len k i))) finally (format t "Longest hailstone sequence under ~A for ~A, having length ~A." n k l))))</lang> Sample session:
ROSETTA> (length (hailstone 27)) 112 ROSETTA> (subseq (hailstone 27) 0 4) (27 82 41 124) ROSETTA> (last (hailstone 27) 4) (8 4 2 1) ROSETTA> (longest-hailstone 100000) Longest hailstone sequence under 100000 for 77031, having length 351. NIL
Crystal
<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]]
</lang>
D
Basic Version
<lang d>import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</lang>
- Output:
hailstone(27)= [27, 82, 41, 124] ... [8, 4, 2, 1] Length hailstone(27)= 112 Longest sequence in [1,100000]= 77031 with len 351
Lazy Version
Same output. <lang d>import std.stdio, std.algorithm, std.typecons, std.range;
auto hailstone(uint m) pure nothrow @nogc {
return m
.recurrence!q{ a[n - 1] & 1 ? a[n - 1] * 3 + 1 : a[n - 1]/2}
.until!q{ a == 1 }(OpenRight.no);
}
void main() {
enum M = 27;
immutable h = M.hailstone.array;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.walkLength, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</lang>
Faster Lazy Version
Same output. <lang d>struct Hailstone {
uint n;
bool empty() const pure nothrow @nogc { return n == 0; }
uint front() const pure nothrow @nogc { return n; }
void popFront() pure nothrow @nogc {
n = n == 1 ? 0 : (n & 1 ? (n * 3 + 1) : n / 2);
}
}
void main() {
import std.stdio, std.algorithm, std.range, std.typecons;
enum M = 27;
immutable h = M.Hailstone.array;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.Hailstone.walkLength, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</lang>
Lazy Version With Caching
Faster, same output. <lang d>import std.stdio, std.algorithm, std.range, std.typecons;
struct Hailstone(size_t cacheSize = 500_000) {
size_t n; __gshared static size_t[cacheSize] cache;
bool empty() const pure nothrow @nogc { return n == 0; }
size_t front() const pure nothrow @nogc { return n; }
void popFront() nothrow {
if (n >= cacheSize) {
n = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
} else if (cache[n]) {
n = cache[n];
} else {
immutable n2 = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
n = cache[n] = n2;
}
}
}
void main() {
enum M = 27;
const h = M.Hailstone!().array;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.Hailstone!().walkLength, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</lang>
Generator Range Version
<lang d>import std.stdio, std.algorithm, std.range, std.typecons, std.concurrency;
auto hailstone(size_t n) {
return new Generator!size_t({
yield(n);
while (n > 1) {
n = (n & 1) ? (3 * n + 1) : (n / 2);
yield(n);
}
});
}
void main() {
enum M = 27;
const h = M.hailstone.array;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.walkLength, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}</lang>
Dart
<lang dart>List<int> hailstone(int n) {
if(n<=0) {
throw new IllegalArgumentException("start value must be >=1)");
}
Queue<int> seq=new Queue<int>();
seq.add(n);
while(n!=1) {
n=n%2==0?(n/2).toInt():3*n+1;
seq.add(n);
}
return new List<int>.from(seq);
}
// apparently List is missing toString() String iterableToString(Iterable seq) {
String str="[";
Iterator i=seq.iterator();
while(i.hasNext()) {
str+=i.next();
if(i.hasNext()) {
str+=",";
}
}
return str+"]";
}
main() {
for(int i=1;i<=10;i++) {
print("h($i)="+iterableToString(hailstone(i)));
}
List<int> h27=hailstone(27);
List<int> first4=h27.getRange(0,4);
print("first 4 elements of h(27): "+iterableToString(first4));
Expect.listEquals([27,82,41,124],first4);
List<int> last4=h27.getRange(h27.length-4,4);
print("last 4 elements of h(27): "+iterableToString(last4));
Expect.listEquals([8,4,2,1],last4);
print("length of sequence h(27): "+h27.length);
Expect.equals(112,h27.length);
int seq,max=0;
for(int i=1;i<=100000;i++) {
List<int> h=hailstone(i);
if(h.length>max) {
max=h.length;
seq=i;
}
}
print("up to 100000 the sequence h($seq) has the largest length ($max)");
}</lang>
- Output:
h(1)=[1] h(2)=[2,1] h(3)=[3,10,5,16,8,4,2,1] h(4)=[4,2,1] h(5)=[5,16,8,4,2,1] h(6)=[6,3,10,5,16,8,4,2,1] h(7)=[7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1] h(8)=[8,4,2,1] h(9)=[9,28,14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1] h(10)=[10,5,16,8,4,2,1] first 4 elements of h(27): [27,82,41,124] last 4 elements of h(27): [8,4,2,1] length of sequence h(27): 112 up to 100000 the sequence h(77031) has the largest length (351)
Dc
Firstly, this code takes the value from the stack, computes and prints the corresponding Hailstone sequence, and the length of the sequence. The q procedure is for counting the length of the sequence. The e and o procedure is for even and odd number respectively. The x procedure is for overall control. <lang Dc>27 [[--: ]nzpq]sq [d 2/ p]se [d 3*1+ p]so [d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx</lang>
- Output:
82 41 124 62 (omitted) 8 4 2 1 --: 112
Then we could wrap the procedure x with a new procedure s, and call it with l which is loops the value of t from 1 to 100000, and cleaning up the stack after each time we finish up with a number. Register L for the length of the longest sequence and T for the corresponding number. Also, procedure q is slightly modified for storing L and T if needed, and all printouts in procedure e and o are muted. <lang Dc>0dsLsT1st [dsLltsT]sM [[zdlL<M q]sq [d 2/]se [d 3*1+ ]so [d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx]ss [lt1+dstlsxc lt100000>l]dslx lTn[:]nlLp </lang>
- Output:
(Takes quite some time on a decent machine)
77031:351
DCL
<lang DCL>$ n = f$integer( p1 ) $ i = 1 $ loop: $ if p2 .nes. "QUIET" then $ s'i = n $ if n .eq. 1 then $ goto done $ i = i + 1 $ if .not. n $ then $ n = n / 2 $ else $ if n .gt. 715827882 then $ exit ! avoid overflowing $ n = 3 * n + 1 $ endif $ goto loop $ done: $ if p2 .nes. "QUIET" $ then $ penultimate_i = i - 1 $ antepenultimate_i = i - 2 $ preantepenultimate_i = i - 3 $ write sys$output "sequence has ", i, " elements starting with ", s1, ", ", s2, ", ", s3, ", ", s4, " and ending with ", s'preantepenultimate_i, ", ", s'antepenultimate_i, ", ", s'penultimate_i, ", ", s'i $ endif $ sequence_length == i</lang>
- Output:
$ @hailstone 27 sequence has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1
<lang DCL>$ limit = f$integer( p1 ) $ i = 1 $ max_so_far = 0 $ loop: $ call hailstone 'i quiet $ if sequence_length .gt. max_so_far $ then $ max_so_far = sequence_length $ current_record_holder = i $ endif $ i = i + 1 $ if i .lt. limit then $ goto loop $ write sys$output current_record_holder, " is the number less than ", limit, " which has the longest hailstone sequence which is ", max_so_far, " in length" $ exit $ $ hailstone: subroutine $ n = f$integer( p1 ) $ i = 1 $ loop: $ if p2 .nes. "QUIET" then $ s'i = n $ if n .eq. 1 then $ goto done $ i = i + 1 $ if .not. n $ then $ n = n / 2 $ else $ if n .gt. 715827882 then $ exit ! avoid overflowing $ n = 3 * n + 1 $ endif $ goto loop $ done: $ if p2 .nes. "QUIET" $ then $ penultimate_i = i - 1 $ antepenultimate_i = i - 2 $ preantepenultimate_i = i - 3 $ write sys$output "sequence has ", i, " elements starting with ", s1, ", ", s2, ", ", s3, ", ", s4, " and ending with ", s'preantepenultimate_i, ", ", s'antepenultimate_i, ", ", s'penultimate_i, ", ", s'i $ endif $ sequence_length == I $ exit $ endsubroutine</lang>
- Output:
$ @longest_hailstone 100000 77031 is the number less than 100000 which has the longest hailstone sequence which is 351 in length
Delphi
Using List<Integer>
<lang Delphi>program ShowHailstoneSequence;
{$APPTYPE CONSOLE}
uses SysUtils, Generics.Collections;
procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>); var
n: Integer;
begin
aHailstoneList.Clear; aHailstoneList.Add(aStartingNumber); n := aStartingNumber;
while n <> 1 do
begin
if Odd(n) then
n := (3 * n) + 1
else
n := n div 2;
aHailstoneList.Add(n);
end;
end;
var
i: Integer; lList: TList<Integer>; lMaxSequence: Integer; lMaxLength: Integer;
begin
lList := TList<Integer>.Create;
try
GetHailstoneSequence(27, lList);
Writeln(Format('27: %d elements', [lList.Count]));
Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
[lList[0], lList[1], lList[2], lList[3],
lList[lList.Count - 4], lList[lList.Count - 3], lList[lList.Count - 2], lList[lList.Count - 1]]));
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
for i := 1 to 100000 do
begin
GetHailstoneSequence(i, lList);
if lList.Count > lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lList.Count;
end;
end;
Writeln(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength]));
finally
lList.Free;
end;
Readln;
end.</lang>
- Output:
27: 112 elements [27 82 41 124 ... 8 4 2 1] Longest sequence under 100,000: 77031 with 351 elements
Using Boost.Algorithm and TParallel.For
<lang Delphi> program ShowHailstoneSequence;
{$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.
</lang>
- Output:
27: 112 elements [27, 82, 41, 124 ... 8, 4, 2, 1] Longest sequence under 100,000: 77031 with 351 elements in 520 ms
Déjà Vu
<lang dejavu>local hailstone: swap [ over ] while < 1 dup: if % over 2: #odd ++ * 3 else: #even / swap 2 swap push-through rot dup drop
if = (name) :(main): local :h27 hailstone 27 !. = 112 len h27 !. = 27 h27! 0 !. = 82 h27! 1 !. = 41 h27! 2 !. = 124 h27! 3 !. = 8 h27! 108 !. = 4 h27! 109 !. = 2 h27! 110 !. = 1 h27! 111
local :max 0 local :maxlen 0 for i range 1 99999: dup len hailstone i if < maxlen: set :maxlen set :max i else: drop !print( "number: " to-str max ", length: " to-str maxlen ) else: @hailstone</lang>
- Output:
true true true true true true true true true number: 77031, length: 351
EchoLisp
<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))
</lang>
- Output:
<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
</lang>
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. <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]
</lang>
- Output:
START 27
LENGTH 112
FIRST AND LAST FOUR
27 82 41 124
8 4 2 1
MAX N 100000
MAX LEN 351
AT N = 77031
Egel
<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)) ]
def hailpair =
[ N -> (N, length (hailstone N)) ]
def hailmax =
[ (N, NMAX), (M, MMAX) -> if (NMAX < MMAX) then (M, MMAX) else (N, NMAX) ]
def largest =
[ 1 -> (1, 1)
| N ->
let M0 = hailpair N in
let M1 = largest (N - 1) in
hailmax M0 M1 ]
)
using System using List using Hailstone
def task0 = let H27 = hailstone 27 in length H27
def task1 =
let H27 = hailstone 27 in
let L = length H27 in
(take 4 H27, drop (L - 4) H27)
def task2 = largest 100000
def main = (task0, task1, task2) </lang>
Eiffel
<lang Eiffel> class APPLICATION
create make
feature
make local test: LINKED_LIST [INTEGER] count, number, te: INTEGER do create test.make test := hailstone_sequence (27) io.put_string ("There are " + test.count.out + " elements in the sequence for the number 27.") io.put_string ("%NThe first 4 elements are: ") across 1 |..| 4 as t loop io.put_string (test [t.item].out + "%T") end io.put_string ("%NThe last 4 elements are: ") across (test.count - 3) |..| test.count as t loop io.put_string (test [t.item].out + "%T") end across 1 |..| 99999 as c loop test := hailstone_sequence (c.item) te := test.count if te > count then count := te number := c.item end end io.put_string ("%NThe longest sequence for numbers below 100000 is " + count.out + " for the number " + number.out + ".") end
hailstone_sequence (n: INTEGER): LINKED_LIST [INTEGER] -- Members of the Hailstone Sequence starting from 'n'. require n_is_positive: n > 0 local seq: INTEGER do create Result.make from seq := n until seq = 1 loop Result.extend (seq) if seq \\ 2 = 0 then seq := seq // 2 else seq := ((3 * seq) + 1) end end Result.extend (seq) ensure sequence_terminated: Result.last = 1 end
end </lang>
- Output:
There are 112 elements in the sequence for the number 27. The first 4 elements are: 27 82 41 124 The last 4 elements are: 8 4 2 1 The longest sequence for numbers below 100000 is 351 for the number 77031.
Elena
ELENA 4.x : <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)
}</lang>
- Output:
max bellow 100000: 77031 (351 steps)
Elixir
<lang elixir>defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has #{len27} elements: #{repr}"
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with #{start} and has #{len} elements."
end
end
Hailstone.run</lang>
- Output:
Hailstone(27) has 112 elements: [27, 82, 41, 124, ..., 8, 4, 2, 1] Longest sequence starting under 100000 begins with 77031 and has 351 elements.
Erlang
<lang erlang>-module(hailstone). -import(io). -export([main/0]).
hailstone(1) -> [1]; hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)]; hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).</lang>
- Output:
Eshell V5.8.4 (abort with ^G)
1> c(hailstone).
{ok,hailstone}
2> hailstone:main().
hailstone(4): [4,2,1]
hailstone(27) length: 112
hailstone(27) first 4: [27,82,41,124]
hailstone(27) last 4: [8,4,2,1]
finding maximum hailstone(N) length for 1 <= N <= 100000... done.
hailstone(77031) length: 351
ok
Erlang 2
This version has one collatz function for just calculating totals (just for fun) and the second generating lists.
<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 ) ] ).
</lang> Output
64> collatz:main(). collatz(4) non-list total: 3 coll(4) with lists [4,2,1] coll(27) length: 112 coll(27) first 4: [27,82,41,124] collatz(27) last 4: [8,4,2,1] maximum N <= 100000...Max: 77031 Total: 351 ok
ERRE
In Italy it's known also as "Ulam conjecture". <lang ERRE> PROGRAM ULAM
!$DOUBLE
PROCEDURE HAILSTONE(X,PRT%->COUNT)
COUNT=1
IF PRT% THEN PRINT(X,) END IF
REPEAT
IF X/2<>INT(X/2) THEN
X=X*3+1
ELSE
X=X/2
END IF
IF PRT% THEN PRINT(X,) END IF
COUNT=COUNT+1
UNTIL X=1
IF PRT% THEN PRINT END IF
END PROCEDURE
BEGIN
HAILSTONE(27,TRUE->COUNT)
PRINT("Sequence length for 27:";COUNT)
MAX_COUNT=2
NMAX=2
FOR I=3 TO 100000 DO
HAILSTONE(I,FALSE->COUNT)
IF COUNT>MAX_COUNT THEN NMAX=I MAX_COUNT=COUNT END IF
END FOR
PRINT("Max. number is";NMAX;" with";MAX_COUNT;"elements")
END PROGRAM </lang>
- Output:
27 82 41 124 62
31 94 47 142 71
214 107 322 161 484
242 121 364 182 91
274 137 412 206 103
310 155 466 233 700
350 175 526 263 790
395 1186 593 1780 890
445 1336 668 334 167
502 251 754 377 1132
566 283 850 425 1276
638 319 958 479 1438
719 2158 1079 3238 1619
4858 2429 7288 3644 1822
911 2734 1367 4102 2051
6154 3077 9232 4616 2308
1154 577 1732 866 433
1300 650 325 976 488
244 122 61 184 92
46 23 70 35 106
53 160 80 40 20
10 5 16 8 4
2 1
Sequence length for 27: 112
Max. number is 77031 with 351 elements
Euler Math Toolbox
<lang Euler Math Toolbox> >function hailstone (n) ... $ v=[n]; $ repeat $ if mod(n,2) then n=3*n+1; $ else n=n/2; $ endif; $ v=v|n; $ until n==1; $ end; $ return v; $ endfunction >hailstone(27), length(%)
[ 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 ] 112
>function hailstonelength (n) ... $ v=zeros(1,n); $ v[1]=4; v[2]=2; $ loop 3 to n; $ count=1; $ n=#; $ repeat $ if mod(n,2) then n=3*n+1; $ else n=n/2; $ endif; $ if n<=cols(v) and v[n] then $ v[#]=v[n]+count; $ break; $ endif; $ count=count+1; $ end; $ end; $ return v; $ endfunction >h=hailstonelength(100000); >ex=extrema(h); ex[3], ex[4]
351 77031
</lang>
Euphoria
<lang euphoria>function hailstone(atom n)
sequence s
s = {n}
while n != 1 do
if remainder(n,2)=0 then
n /= 2
else
n = 3*n + 1
end if
s &= n
end while
return s
end function
function hailstone_count(atom n)
integer count
count = 1
while n != 1 do
if remainder(n,2)=0 then
n /= 2
else
n = 3*n + 1
end if
count += 1
end while
return count
end function
sequence s s = hailstone(27) puts(1,"hailstone(27) =\n") ? s printf(1,"len = %d\n\n",length(s))
integer max,imax,count max = 0 for i = 2 to 1e5-1 do
count = hailstone_count(i)
if count > max then
max = count
imax = i
end if
end for
printf(1,"The longest hailstone sequence under 100,000 is %d with %d elements.\n",
{imax,max})</lang>
- Output:
hailstone(27) =
{27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,
91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,
1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,
850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,
7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,
577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,
106,53,160,80,40,20,10,5,16,8,4,2,1}
len = 112
The longest hailstone sequence under 100,000 is 77031 with 351 elements.
Excel
In cell A1, place the starting number. In cell A2 enter this formula =IF(MOD(A1,2)=0,A1/2,A1*3+1) Drag and copy the formula down until 4, 2, 1
Ezhil
Ezhil is a Tamil programming language, see | Wikipedia entry.
<lang src="Python"> நிரல்பாகம் hailstone ( எண் )
பதிப்பி "=> ",எண் #hailstone seq
@( எண் == 1 ) ஆனால் பின்கொடு எண் முடி
@( (எண்%2) == 1 ) ஆனால் hailstone( 3*எண் + 1)
இல்லை
hailstone( எண்/2 )
முடி
முடி
எண்கள் = [5,17,19,23,37]
@(எண்கள் இல் இவ்வெண்) ஒவ்வொன்றாக
பதிப்பி "****** calculating hailstone seq for ",இவ்வெண்," *********" hailstone( இவ்வெண் ) பதிப்பி "**********************************************"
முடி </lang>
F#
<lang fsharp>let rec hailstone n = seq {
match n with | 1 -> yield 1 | n when n % 2 = 0 -> yield n; yield! hailstone (n / 2) | n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq assert (Array.length hailstone27 = 112) assert (hailstone27.[..3] = [|27;82;41;124|]) assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i} printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI</lang>
- Output:
Maximum length 351 was found for hailstone(77031)
Factor
<lang factor>! rosetta/hailstone/hailstone.factor USING: arrays io kernel math math.ranges prettyprint sequences vectors ; IN: rosetta.hailstone
- hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
- main ( -- )
27 hailstone dup dup "The hailstone sequence from 27:" print " has length " write length . " starts with " write 4 head [ unparse ] map ", " join print " ends with " write 4 tail* [ unparse ] map ", " join print
! Maps n => { length n }, and reduces to longest Hailstone sequence.
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main</lang>
- Output:
$ ./factor -run=rosetta.hailstone Loading resource:work/rosetta/hailstone/hailstone.factor The hailstone sequence from 27: has length 112 starts with 27, 82, 41, 124 ends with 8, 4, 2, 1 The hailstone sequence from 77031 has length 351.
FALSE
<lang false>[$1&$[%3*1+0~]?~[2/]?]n: [[$." "$1>][n;!]#%]s: [1\[$1>][\1+\n;!]#%]c: 27s;! 27c;!." " 0m:0f: 1[$100000\>][$c;!$m;>[m:$f:0]?%1+]#% f;." has hailstone sequence length "m;.</lang>
Forth
<lang forth>: hail-next ( n -- n )
dup 1 and if 3 * 1+ else 2/ then ;
- .hail ( n -- )
begin dup . dup 1 > while hail-next repeat drop ;
- hail-len ( n -- n )
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr 27 .hail cr
- longest-hail ( max -- )
0 0 rot 1+ 1 do ( n length )
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail</lang>
Fortran
<lang fortran>program Hailstone
implicit none
integer :: i, maxn integer :: maxseqlen = 0, seqlen integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number integer, intent(out) :: length integer, optional, intent(inout) :: seqArray(:) integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program</lang>
- Output:
Hailstone sequence for 27 has 112 elements Sequence = 27, 82, 41, 124, ...., 8, 4, 2, 1 Longest sequence under 100000 is for 77031 with 351 elements
Frege
<lang frege>module Hailstone where
import Data.List (maximumBy)
hailstone :: Int -> [Int] hailstone 1 = [1] hailstone n | even n = n : hailstone (n `div` 2)
| otherwise = n : hailstone (n * 3 + 1)
withResult :: (t -> t1) -> t -> (t1, t) withResult f x = (f x, x)
main :: IO () main = do
let h27 = hailstone 27
putStrLn $ show $ length h27
let h4 = show $ take 4 h27
let t4 = show $ drop (length h27 - 4) h27
putStrLn ("hailstone 27: " ++ h4 ++ " ... " ++ t4)
putStrLn $ show $ maximumBy (comparing fst) $ map (withResult (length . hailstone)) [1..100000]</lang>
- Output:
112 hailstone 27: [27, 82, 41, 124] ... [8, 4, 2, 1] (351, 77031) runtime 0.969 wallclock seconds.
Frink
<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"] </lang>
FunL
<lang funl>def
hailstone( 1 ) = [1] hailstone( n ) = n # hailstone( if 2|n then n/2 else n*3 + 1 )
if _name_ == '-main-'
h27 = hailstone( 27 ) assert( h27.length() == 112 and h27.startsWith([27, 82, 41, 124]) and h27.endsWith([8, 4, 2, 1]) )
val (n, len) = maxBy( snd, [(i, hailstone( i ).length()) | i <- 1:100000] )
println( n, len )</lang>
- Output:
77031, 351
Futhark
<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)))))
</lang>
Fōrmulæ
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
GAP
<lang gap>CollatzSequence := function(n)
local v;
v := [ n ];
while n > 1 do
if IsEvenInt(n) then
n := QuoInt(n, 2);
else
n := 3*n + 1;
fi;
Add(v, n);
od;
return v;
end;
CollatzLength := function(n)
local m;
m := 1;
while n > 1 do
if IsEvenInt(n) then
n := QuoInt(n, 2);
else
n := 3*n + 1;
fi;
m := m + 1;
od;
return m;
end;
CollatzMax := function(a, b)
local n, len, nmax, lmax;
lmax := 0;
for n in [a .. b] do
len := CollatzLength(n);
if len > lmax then
nmax := n;
lmax := len;
fi;
od;
return [ nmax, lmax ];
end;
CollatzSequence(27);
- [ 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206,
- 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502,
- 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429,
- 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300,
- 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 ]
CollatzLength(27);
- 112
CollatzMax(1, 100);
- [ 97, 119 ]
CollatzMax(1, 1000);
- [ 871, 179 ]
CollatzMax(1, 10000);
- [ 6171, 262 ]
CollatzMax(1, 100000);
- [ 77031, 351 ]
CollatzMax(1, 1000000);
- [ 837799, 525 ]</lang>
Go
<lang go>package main
import "fmt"
// 1st arg is the number to generate the sequence for. // 2nd arg is a slice to recycle, to reduce garbage. func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}</lang>
- Output:
hs(27): 112 elements: [27 82 41 124 ... 8 4 2 1] hs(77031): 351 elements
Alternate solution (inspired both by recent news of a new proof submitted for publication and by recent chat on #rosettacode about generators.)
This solution interprets the wording of the task differently, and takes the word "generate" to mean use a generator. This has the advantage of not storing the whole sequence in memory at once. Elements are generated one at a time, counted and discarded. A time optimization added for task 3 is to store the sequence lengths computed so far.
Output is the same as version above. <lang go>package main
import "fmt"
// Task 1 implemented with a generator. Calling newHg will "create // a routine to generate the hailstone sequence for a number." func newHg(n int) func() int {
return func() (n0 int) {
n0 = n
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
return
}
}
func main() {
// make generator for sequence starting at 27
hg := newHg(27)
// save first four elements for printing later
s1, s2, s3, s4 := hg(), hg(), hg(), hg()
// load next four elements in variables to use as shift register.
e4, e3, e2, e1 := hg(), hg(), hg(), hg()
// 4+4= 8 that we've generated so far
ec := 8
// until we get to 1, generate another value, shift, and increment.
// note that intermediate elements--those shifted off--are not saved.
for e1 > 1 {
e4, e3, e2, e1 = e3, e2, e1, hg()
ec++
}
// Complete task 2:
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
ec, s1, s2, s3, s4, e4, e3, e2, e1)
// Task 3: strategy is to not store sequences, but just the length // of each sequence. as soon as the sequence we're currently working on // dips into the range that we've already computed, we short-circuit // to the end by adding the that known length to whatever length // we've accumulated so far.
var nMaxLen int // variable holds n with max length encounted so far
// slice holds sequence length for each n as it is computed
var computedLen [1e5]int
computedLen[1] = 1
for n := 2; n < 1e5; n++ {
var ele, lSum int
for hg := newHg(n); ; lSum++ {
ele = hg()
// as soon as we get an element in the range we have already
// computed, we're done...
if ele < n {
break
}
}
// just add the sequence length already computed from this point.
lSum += computedLen[ele]
// save the sequence length for this n
computedLen[n] = lSum
// and note if it's the maximum so far
if lSum > computedLen[nMaxLen] {
nMaxLen = n
}
}
fmt.Printf("hs(%d): %d elements\n", nMaxLen, computedLen[nMaxLen])
}</lang>
Groovy
<lang groovy>def hailstone = { long start ->
def sequence = []
while (start != 1) {
sequence << start
start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
}
sequence << start
}</lang> Test Code <lang groovy>def sequence = hailstone(27) assert sequence.size() == 112 assert sequence[0..3] == [27, 82, 41, 124] assert sequence[-4..-1] == [8, 4, 2, 1]
def results = (1..100000).collect { [n:it, size:hailstone(it).size()] }.max { it.size } println results</lang>
- Output:
[n:77031, size:351]
Haskell
<lang haskell>import Data.List (maximumBy) import Data.Ord (comparing)
collatz :: Int -> Int collatz n
| even n = n `div` 2 | otherwise = 1 + 3 * n
hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz
longestChain :: Int longestChain =
fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000]
TEST -------------------------
main :: IO () main =
mapM_ putStrLn [ "Collatz sequence for 27: " , (show . hailstone) 27 , "The number " <> show longestChain , "has the longest hailstone sequence for any number less then 100000. " , "The sequence has length: " <> (show . length . hailstone $ longestChain) ]</lang>
- Output:
Collatz sequence for 27: [27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2] The number 77031 has the longest hailstone sequence for any number less then 100000. The sequence has length: 350
The following is an older version, which some of the language examples on this page are translated from: <lang haskell>import Data.Ord (comparing) import Data.List (maximumBy, intercalate)
hailstone :: Int -> [Int] hailstone 1 = [1] hailstone n
| even n = n : hailstone (n `div` 2) | otherwise = n : hailstone (n * 3 + 1)
withResult :: (Int -> Int) -> Int -> (Int, Int) withResult f x = (f x, x)
h27 :: [Int] h27 = hailstone 27
main :: IO () main =
mapM_
putStrLn
[ (show . length) h27
, "hailstone 27: " ++
intercalate " ... " (show <$> [take 4 h27, drop (length h27 - 4) h27])
, show $
maximumBy (comparing fst) $
withResult (length . hailstone) <$> [1 .. 100000]
]</lang>
- Output:
112 hailstone 27: [27,82,41,124] ... [8,4,2,1] (351,77031)
Or, going back to basics, we can observe that the hailstone sequence is an 'anamorphism' – it builds up a list structure from a single integer value, which makes unfoldr the obvious first thing to reach for the first main task.
In turn, deriving the longest sequence for starting values below 100000 essentially involves a 'catamorphism' – it takes a list of hailstone sequences (or at least a list of their seed values and their lengths), and strips that structure down to a single (N, length) pair. This makes foldr the obvious recursion scheme to start with for the second main task.
One approach to using unfoldr and then foldr might be:
<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 ]</lang>
- Output:
Sequence 27 length: 112 Sequence 27 start: [27,82,41,124] Sequence 27 end: [8,4,2,1] N with longest sequence where N <= 100000 77031 length of this sequence: 351
HicEst
<lang HicEst>DIMENSION stones(1000)
H27 = hailstone(27) ALIAS(stones,1, first4,4) ALIAS(stones,H27-3, last4,4) WRITE(ClipBoard, Name) H27, first4, "...", last4
longest_sequence = 0 DO try = 1, 1E5
elements = hailstone(try)
IF(elements >= longest_sequence) THEN
number = try
longest_sequence = elements
WRITE(StatusBar, Name) number, longest_sequence
ENDIF
ENDDO WRITE(ClipBoard, Name) number, longest_sequence END
FUNCTION hailstone( n )
USE : stones
stones(1) = n
DO i = 1, LEN(stones)
IF(stones(i) == 1) THEN
hailstone = i
RETURN
ELSEIF( MOD(stones(i),2) ) THEN
stones(i+1) = 3*stones(i) + 1
ELSE
stones(i+1) = stones(i) / 2
ENDIF
ENDDO
END</lang>
H27=112; first4(1)=27; first4(2)=82; first4(3)=41; first4(4)=124; ...; last4(1)=8; last4(2)=4; last4(3)=2; last4(4)=1;
number=77031; longest_sequence=351;
Icon and Unicon
A simple solution that generates (in the Icon sense) the sequence is: <lang icon>procedure hailstone(n)
while n > 1 do {
suspend n
n := if n%2 = 0 then n/2 else 3*n+1
}
suspend 1
end</lang> and a test program for this solution is: <lang icon>procedure main(args)
n := integer(!args) | 27
every writes(" ",hailstone(n))
end</lang> but this solution is computationally expensive when run repeatedly (task 3).
The following solution uses caching to improve performance on task 3 at the expense of space. <lang icon>procedure hailstone(n)
static cache
initial {
cache := table()
cache[1] := [1]
}
/cache[n] := [n] ||| hailstone(if n%2 = 0 then n/2 else 3*n+1)
return cache[n]
end</lang>
A test program is: <lang icon>procedure main(args)
n := integer(!args) | 27 task2(n) write() task3()
end
procedure task2(n)
count := 0
every writes(" ",right(!(sequence := hailstone(n)),5)) do
if (count +:= 1) % 15 = 0 then write()
write()
write(*sequence," value",(*sequence=1,"")|"s"," in the sequence.")
end
procedure task3()
maxHS := 0
every n := 1 to 100000 do {
count := *hailstone(n)
if maxHS <:= count then maxN := n
}
write(maxN," has a sequence of ",maxHS," values")
end</lang> A sample run is:
->hs
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484
242 121 364 182 91 274 137 412 206 103 310 155 466 233 700
350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167
502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438
719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051
6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488
244 122 61 184 92 46 23 70 35 106 53 160 80 40 20
10 5 16 8 4 2 1
112 values in the sequence.
77031 has a sequence of 351 values
->
Inform 7
This solution uses a cache to speed up the length calculation for larger numbers.
<lang inform7>Home is a room.
To decide which list of numbers is the hailstone sequence for (N - number): let result be a list of numbers; add N to result; while N is not 1: if N is even, let N be N / 2; otherwise let N be (3 * N) + 1; add N to result; decide on result.
Hailstone length cache relates various numbers to one number.
To decide which number is the hailstone sequence length for (N - number): let ON be N; let length so far be 0; while N is not 1: if N relates to a number by the hailstone length cache relation: let result be length so far plus the number to which N relates by the hailstone length cache relation; now the hailstone length cache relation relates ON to result; decide on result; if N is even, let N be N / 2; otherwise let N be (3 * N) + 1; increment length so far; let result be length so far plus 1; now the hailstone length cache relation relates ON to result; decide on result.
To say first and last (N - number) entry/entries in (L - list of values of kind K): let length be the number of entries in L; if length <= N * 2: say L; else: repeat with M running from 1 to N: if M > 1, say ", "; say entry M in L; say " ... "; repeat with M running from length - N + 1 to length: say entry M in L; if M < length, say ", ".
When play begins: let H27 be the hailstone sequence for 27; say "Hailstone sequence for 27 has [number of entries in H27] element[s]: [first and last 4 entries in H27]."; let best length be 0; let best number be 0; repeat with N running from 1 to 99999: let L be the hailstone sequence length for N; if L > best length: let best length be L; let best number be N; say "The number under 100,000 with the longest hailstone sequence is [best number] with [best length] element[s]."; end the story.</lang>
- Output:
Hailstone sequence for 27 has 112 elements: 27, 82, 41, 124 ... 8, 4, 2, 1. The number under 100,000 with the longest hailstone sequence is 77031 with 351 elements.
Io
Here is a simple, brute-force approach: <lang io> makeItHail := method(n,
stones := list(n)
while (n != 1,
if(n isEven,
n = n / 2,
n = 3 * n + 1
)
stones append(n)
)
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.") </lang>
- Output:
For the sequence beginning at 27, the number of elements generated is 112. The first four elements generated are 27 82 41 124 . The last four elements generated are 8 4 2 1 . For numbers less than or equal to 100,000, 77031 has the longest sequence of 351 elements.
Ioke
<lang ioke>collatz = method(n,
n println unless(n <= 1, if(n even?, collatz(n / 2), collatz(n * 3 + 1)))
)</lang>
J
Solution: <lang j>hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0</lang> Usage: <lang j> # hailseq 27 NB. sequence length 112
4 _4 {."0 1 hailseq 27 NB. first & last 4 numbers in sequence
27 82 41 124
8 4 2 1 (>:@(i. >./) , >./) #@hailseq }.i. 1e5 NB. number < 100000 with max seq length & its seq length
77031 351</lang> See also the Collatz Conjecture essay on the J wiki.
Java
<lang java5>import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
// Simple way
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
// More memory efficient way
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
// Efficient for analyzing all sequences
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}</lang>
- Output:
Sequence for 27 has 112 elements: [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1] Method 1, number 77031 has the longest sequence, with a length of 351 Method 2, number 77031 has the longest sequence, with a length of 351 Method 3, number 77031 has the longest sequence, with a length of 351
JavaScript
ES5
Imperative
<lang javascript>function hailstone (n) {
var seq = [n];
while (n > 1) {
n = n % 2 ? 3 * n + 1 : n / 2;
seq.push(n);
}
return seq;
}
// task 2: verify the sequence for n = 27 var h = hailstone(27), hLen = h.length; print("sequence 27 is (" + h.slice(0, 4).join(", ") + " ... "
+ h.slice(hLen - 4, hLen).join(", ") + "). length: " + hLen);
// task 3: find the longest sequence for n < 100000 for (var n, max = 0, i = 100000; --i;) {
var seq = hailstone(i), sLen = seq.length;
if (sLen > max) {
n = i;
max = sLen;
}
} print("longest sequence: " + max + " numbers for starting point " + n);</lang>
- Output:
sequence 27 is (27, 82, 41, 124 ... 8, 4, 2, 1). length: 112 longest sequence: 351 numbers for starting point 77031
Functional
This simple problem turns out to be a good test of the constraints on composing (ES5) JavaScript code in a functional style.
The first sub-problem falls easily within reach of a basic recursive definition (translating one of the Haskell solutions).
<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
};
})();</lang>
- Output:
<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]}</lang>
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.
<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);
})();</lang>
- Output:
<lang JavaScript>// Number, length of sequence {"n":77031, "l":351}</lang>
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.
<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);
})();</lang>
- Output:
<lang JavaScript>[
{"n":77031, "l":351}, // 100,000
{"n":837799, "l":525}, // 1,000,000
{"n":8400511, "l":686} // 10,000,000
]</lang>
<lang JavaScript>longestBelow(100000000) -> {"n":63728127, "l":950}</lang>
ES6
<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}.`
]);
})();</lang>
- Output:
(Run in the Atom editor, through the Script package)
Collatz sequence for 27: 27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91, 274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593, 1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,850,425,1276, 638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822, 911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,1732,866,433, 1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,40,20, 10,5,16,8,4,2,1 The number 77031 has the longest hailstone sequence for any starting number under 100000. The length of that sequence is 351. [Finished in 1.139s]
jq
<lang jq># Generate the hailstone sequence as a stream to save space (and time) when counting def hailstone:
recurse( if . > 1 then
if . % 2 == 0 then ./2|floor else 3*. + 1 end
else empty
end );
def count(g): reduce g as $i (0; .+1);
- return [i, length] for the first maximal-length hailstone sequence where i is in [1 .. n]
def max_hailstone(n):
# state: [i, length]
reduce range(1; n+1) as $i
([0,0];
($i | count(hailstone)) as $l
| if $l > .[1] then [$i, $l] else . end);</lang>
Examples: <lang jq>[27|hailstone] as $h | "[27|hailstone]|length is \($h|length)",
"The first four numbers: \($h[0:4])", "The last four numbers: \($h|.[length-4:length])", "", max_hailstone(100000) as $m | "Maximum length for n|hailstone for n in 1..100000 is \($m[1]) (n == \($m[0]))"</lang>
- Output:
<lang sh>$ jq -M -r -n -f hailstone.jq [27|hailstone]|length is 112 The first four numbers: [27,82,41,124] The last four numbers: [8,4,2,1]
Maximum length for n|hailstone for n in 1..100000 is 351 (n == 77031)</lang>
Julia
Dynamic solution
<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</lang>
- Output:
hailstonelength(27) = 112 findmax((hailstonelength(i) for i = 1:100000)) = (351, 77031)
Solution with iterator
Julia 1.0
<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)))</lang>
- Output:
Collection of the Hailstone sequence from 27: HailstoneSeq{27, 82, 411, 124, 62...}
First 5 elements: 27, 82, 41, 124, 62
Last 5 elements: 16, 8, 4, 2, 1
The number with the longest sequence under 100,000 is: HailstoneSeq{777031, 231094, 115547, 346642, 173321...}
Julia 0.6
<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 s1, (s1, s2) end
function Base.length(hs::HailstoneSeq) r = 0 for _ in hs r += 1 end return r end
function Base.show(io::IO, hs::HailstoneSeq) f5 = collect(Iterators.take(hs, 5)) print(io, "HailstoneSeq(", join(f5, ", "), "...)") end
hs = HailstoneSeq(27) println("Collection of the Hailstone sequence from 27: $hs") cl = collect(hs) println("First 5 elements: ", join(cl[1:5], ", ")) println("Last 5 elements: ", join(cl[end-4:end], ", "))
Base.isless(h::HailstoneSeq, s::HailstoneSeq) = length(h) < length(s) println("The number with the longest sequence under 100,000 is: ", maximum(HailstoneSeq.(1:100_000)))</lang>
- Output:
Collection of the Hailstone sequence from 27: HailstoneSeq(27, 82, 41, 124, 62...) First 5 elements: 27, 82, 41, 124, 62 Last 5 elements: 16, 8, 4, 2, 1 The number with the longest sequence under 100,000 is: HailstoneSeq(77031, 231094, 115547, 346642, 173321...)
K
<lang k> hail: (1<){:[x!2;1+3*x;_ x%2]}\
seqn: hail 27
#seqn
112
4#seqn
27 82 41 124
-4#seqn
8 4 2 1
{m,x@s?m:|/s:{#hail x}'x}{x@&x!2}!:1e5
351 77031</lang>
Kotlin
<lang kotlin>import java.util.ArrayDeque
fun hailstone(n: Int): ArrayDeque<Int> {
val hails = when {
n == 1 -> ArrayDeque<Int>()
n % 2 == 0 -> hailstone(n / 2)
else -> hailstone(3 * n + 1)
}
hails.addFirst(n)
return hails
}
fun main(args: Array<String>) {
val hail27 = hailstone(27)
fun showSeq(s: List<Int>) = s.map { it.toString() }.reduce { a, b -> a + ", " + b }
println("Hailstone sequence for 27 is " + showSeq(hail27.take(3)) + " ... "
+ showSeq(hail27.drop(hail27.size - 3)) + " with length ${hail27.size}.")
var longestHail = hailstone(1)
for (x in 1..99999)
longestHail = arrayOf(hailstone(x), longestHail).maxBy { it.size } ?: longestHail
println("${longestHail.first} is the number less than 100000 with " +
"the longest sequence, having length ${longestHail.size}.")
}</lang>
- Output:
Hailstone sequence for 27 is 27, 82, 41 ... 4, 2, 1 with length 112. 77031 is the number less than 100000 with the longest sequence, having length 351.
Lasso
<lang Lasso>[ define_tag("hailstone", -required="n", -type="integer", -copy); local("sequence") = array(#n); while(#n != 1); ((#n % 2) == 0) ? #n = (#n / 2) | #n = (#n * 3 + 1); #sequence->insert(#n); /while; return(#sequence); /define_tag;
local("result"); #result = hailstone(27); while(#result->size > 8); #result->remove(5); /while; #result->insert("...",5);
"Hailstone sequence for n = 27 -> { " + #result->join(", ") + " }";
local("longest_sequence") = 0; local("longest_index") = 0; loop(-from=1, -to=100000); local("length") = hailstone(loop_count)->size; if(#length > #longest_sequence); #longest_index = loop_count; #longest_sequence = #length; /if; /loop;
"
";
"Number with the longest sequence under 100,000: " #longest_index + ", with " + #longest_sequence + " elements.";
]</lang>
Limbo
<lang>implement Hailstone;
include "sys.m"; sys: Sys; include "draw.m";
Hailstone: module { init: fn(ctxt: ref Draw->Context, args: list of string); };
init(nil: ref Draw->Context, nil: list of string) { sys = load Sys Sys->PATH;
seq := hailstone(big 27); l := len seq;
sys->print("hailstone(27): "); for(i := 0; i < 4; i++) { sys->print("%bd, ", hd seq); seq = tl seq; } sys->print("⋯");
while(len seq > 4) seq = tl seq;
while(seq != nil) { sys->print(", %bd", hd seq); seq = tl seq; } sys->print(" (length %d)\n", l);
max := 1; maxn := big 1; for(n := big 2; n < big 100000; n++) { cur := len hailstone(n); if(cur > max) { max = cur; maxn = n; } } sys->print("hailstone(%bd) has length %d\n", maxn, max); }
hailstone(i: big): list of big { if(i == big 1) return big 1 :: nil; if(i % big 2 == big 0) return i :: hailstone(i / big 2); return i :: hailstone((big 3 * i) + big 1); } </lang>
- Output:
hailstone(27): 27, 82, 41, 124, ⋯, 8, 4, 2, 1 (length 112) hailstone(77031) has length 351
Lingo
<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</lang> Usage: <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</lang>
Logo
<lang logo>to hail.next :n
output ifelse equal? 0 modulo :n 2 [:n/2] [3*:n + 1]
end
to hail.seq :n
if :n = 1 [output [1]] output fput :n hail.seq hail.next :n
end
show hail.seq 27 show count hail.seq 27
to max.hail :n
localmake "max.n 0 localmake "max.length 0 repeat :n [if greater? count hail.seq repcount :max.length [ make "max.n repcount make "max.length count hail.seq repcount ] ] (print :max.n [has hailstone sequence length] :max.length)
end
max.hail 100000</lang>
Logtalk
<lang logtalk>:- object(hailstone).
:- public(generate_sequence/2). :- mode(generate_sequence(+natural, -list(natural)), zero_or_one). :- info(generate_sequence/2, [ comment is 'Generates the Hailstone sequence that starts with its first argument. Fails if the argument is not a natural number.', argnames is ['Start', 'Sequence'] ]).
:- public(write_sequence/1). :- mode(write_sequence(+natural), zero_or_one). :- info(write_sequence/1, [ comment is 'Writes to the standard output the Hailstone sequence that starts with its argument. Fails if the argument is not a natural number.', argnames is ['Start'] ]).
:- public(sequence_length/2). :- mode(sequence_length(+natural, -natural), zero_or_one). :- info(sequence_length/2, [ comment is 'Calculates the length of the Hailstone sequence that starts with its first argument. Fails if the argument is not a natural number.', argnames is ['Start', 'Length'] ]).
:- public(longest_sequence/4). :- mode(longest_sequence(+natural, +natural, -natural, -natural), zero_or_one). :- info(longest_sequence/4, [ comment is 'Calculates the longest Hailstone sequence in the interval [Start, End]. Fails if the interval is not valid.', argnames is ['Start', 'End', 'N', 'Length'] ]).
generate_sequence(Start, Sequence) :- integer(Start), Start >= 1, sequence(Start, Sequence).
sequence(1, [1]) :- !. sequence(N, [N| Sequence]) :- ( N mod 2 =:= 0 -> M is N // 2 ; M is (3 * N) + 1 ), sequence(M, Sequence).
write_sequence(Start) :- integer(Start), Start >= 1, sequence(Start).
sequence(1) :- !, write(1), nl. sequence(N) :- write(N), write(' '), ( N mod 2 =:= 0 -> M is N // 2 ; M is (3 * N) + 1 ), sequence(M).
sequence_length(Start, Length) :- integer(Start), Start >= 1, sequence_length(Start, 1, Length).
sequence_length(1, Length, Length) :- !. sequence_length(N, Length0, Length) :- Length1 is Length0 + 1, ( N mod 2 =:= 0 -> M is N // 2 ; M is (3 * N) + 1 ), sequence_length(M, Length1, Length).
longest_sequence(Start, End, N, Length) :- integer(Start), integer(End), Start >= 1, Start =< End, longest_sequence(Start, End, 1, N, 1, Length).
longest_sequence(Current, End, N, N, Length, Length) :- Current > End, !. longest_sequence(Current, End, N0, N, Length0, Length) :- sequence_length(Current, 1, CurrentLength), Next is Current + 1, ( CurrentLength > Length0 -> longest_sequence(Next, End, Current, N, CurrentLength, Length) ; longest_sequence(Next, End, N0, N, Length0, Length) ).
- - end_object.</lang>
Testing: <lang logtalk>| ?- hailstone::write_sequence(27). 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 true
| ?- hailstone::sequence_length(27, Length). Length = 112 true
| ?- hailstone::longest_sequence(1, 100000, N, Length). N = 77031, Length = 351 true</lang>
LOLCODE
There is presently no way to query a BUKKIT for the existence of a given key, thus making memoization infeasible. This solution takes advantage of prior knowledge to run in reasonable time. <lang LOLCODE>HAI 1.3
HOW IZ I hailin YR stone
I HAS A sequence ITZ A BUKKIT sequence HAS A length ITZ 1 sequence HAS A SRS 0 ITZ stone
IM IN YR stoner
BOTH SAEM stone AN 1, O RLY?
YA RLY, FOUND YR sequence
OIC
MOD OF stone AN 2, O RLY?
YA RLY, stone R SUM OF PRODUKT OF stone AN 3 AN 1
NO WAI, stone R QUOSHUNT OF stone AN 2
OIC
sequence HAS A SRS sequence'Z length ITZ stone
sequence'Z length R SUM OF sequence'Z length AN 1
IM OUTTA YR stoner
IF U SAY SO
I HAS A hail27 ITZ I IZ hailin YR 27 MKAY VISIBLE "hail(27) = "!
IM IN YR first4 UPPIN YR i TIL BOTH SAEM i AN 4
VISIBLE hail27'Z SRS i " "!
IM OUTTA YR first4 VISIBLE "..."!
IM IN YR last4 UPPIN YR i TIL BOTH SAEM i AN 4
VISIBLE " " hail27'Z SRS SUM OF 108 AN i!
IM OUTTA YR last4 VISIBLE ", length = " hail27'Z length
I HAS A max, I HAS A len ITZ 0
BTW, DIS IZ RLY NOT FAST SO WE ONLY CHEK N IN [75000, 80000) IM IN YR maxer UPPIN YR n TIL BOTH SAEM n AN 5000
I HAS A n ITZ SUM OF n AN 75000
I HAS A seq ITZ I IZ hailin YR n MKAY
BOTH SAEM len AN SMALLR OF len AN seq'Z length, O RLY?
YA RLY, max R n, len R seq'Z length
OIC
IM OUTTA YR maxer
VISIBLE "len(hail(" max ")) = " len
KTHXBYE</lang>
- Output:
hail(27) = 27 82 41 124 ... 8 4 2 1, length = 112 len(hail(77031)) = 351
Lua
<lang lua>function hailstone( n, print_numbers )
local n_iter = 1
while n ~= 1 do
if print_numbers then print( n ) end
if n % 2 == 0 then
n = n / 2
else
n = 3 * n + 1
end
n_iter = n_iter + 1
end
if print_numbers then print( n ) end
return n_iter;
end
hailstone( 27, true )
max_i, max_iter = 0, 0 for i = 1, 100000 do
num = hailstone( i, false )
if num >= max_iter then
max_i = i
max_iter = num
end
end
print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )</lang>
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 <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 </lang>
- Output:
27 82 41 124
8 4 2 1
counter:112
Number 77031 has then longest hailstone sequence of length 351
Maple
Define the procedure: <lang Maple> hailstone := proc( N )
local n := N, HS := Array([n]);
while n > 1 do
if type(n,even) then
n := n/2;
else
n := 3*n+1;
end if;
HS(numelems(HS)+1) := n;
end do;
HS;
end proc; </lang> Run the command and show the appropriate portion of the result; <lang Maple> > r := hailstone(27):
[ 1..112 1-D Array ]
r := [ Data Type: anything ]
[ Storage: rectangular ]
[ Order: Fortran_order ]
> r(1..4) ... r(-4..);
[27, 82, 41, 124] .. [8, 4, 2, 1]
</lang> Compute the first 100000 sequences: <lang Maple> longest := 0; n := 0; for i from 1 to 100000 do
len := numelems(hailstone(i));
if len > longest then
longest := len;
n := i;
end if;
od: printf("The longest Hailstone sequence in the first 100k is n=%d, with %d terms\n",n,longest); </lang>
- Output:
The longest Hailstone sequence in the first 100k is n=77031, with 351 terms
Mathematica / Wolfram Language
Here are four ways to generate the sequence.
Nested function call formulation
<lang Mathematica>HailstoneF[n_] := NestWhileList[If[OddQ@#, 3 # + 1, #/2] &, n, # > 1 &]</lang>
This is probably the most readable and shortest implementation.
Fixed-Point formulation
<lang Mathematica>HailstoneFP[n_] := Most@FixedPointList[Switch[#, 1, 1, _?OddQ , 3# + 1, _, #/2] &, n]</lang>
Recursive formulation
<lang Mathematica>HailstoneR[1] = {1} HailstoneR[n_?OddQ] := Prepend[HailstoneR[3 n + 1], n] HailstoneR[n_] := Prepend[HailstoneR[n/2], n] </lang>
Procedural implementation
<lang Mathematica>HailstoneP[n_] := Module[{x = {n}, s = n},
While[s > 1, x = {x, s = If[OddQ@s, 3 s + 1, s/2]}]; Flatten@x] </lang>
Validation
I use this version to do the validation: <lang Mathematica>Hailstone[n_] :=
NestWhileList[Which[Mod[#, 2] == 0, #/2, True, ( 3*# + 1) ] &, n, # != 1 &];
c27 = Hailstone@27;
Print["Hailstone sequence for n = 27: [", c27;; 4, "...", c27-4 ;;, "]"]
Print["Length Hailstone[27] = ", Length@c27]
longest = -1; comp = 0; Do[temp = Length@Hailstone@i;
If[comp < temp, comp = temp; longest = i],
{i, 100000}
]
Print["Longest Hailstone sequence at n = ", longest, "\nwith length = ", comp]; </lang>
- Output:
Hailstone sequence for n = 27: [{27,82,41,124}...{8,4,2,1}]
Length Hailstone[27] = 112
Longest Hailstone sequence at n = 77031
with length = 351
I think the fixed-point and the recursive piece-wise function formulations are more idiomatic for Mathematica
Sequence 27
<lang Mathematica>With[{seq = HailstoneFP[27]}, { Length[seq], Take[seq, 4], Take[seq, -4]}]</lang>
- Output:
{112, {27, 82, 41, 124}, {8, 4, 2, 1}}
Alternatively, <lang Mathematica>Short[HailstoneFP[27],0.45]</lang>
- Output:
{27, 82, 41, 124, <<104>>, 8, 4, 2, 1}
Longest sequence length
<lang Mathematica>MaximalBy[Table[{i, Length[HailstoneFP[i]]}, {i, 100000}], Last]</lang>
- Output:
{{77031, 351}}
MATLAB / Octave
Hailstone Sequence For N
<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</lang>
Show sequence of hailstone(27) and number of elements: <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))</lang>
- Output:
hailstone(27): 27 82 41 124 ... 8 4 2 1 number of elements: 112
Longest Hailstone Sequence Under N
Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length:
Basic Version (use the above routine)
<lang Matlab>N = 1e5; maxLen = 0; for k = 1:N
kLen = numel(hailstone(k)); if kLen > maxLen maxLen = kLen; n = k; end
end</lang>
- Output:
n = 77031 maxLen = 351
Faster Version
<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</lang>
- Output:
<lang matlab>>> [n, maxLen] = longestHailstone(1e5) n = 77031 maxLen = 351</lang>
Much Faster Version With Caching
<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</lang>
- Output:
<lang matlab>>> [n, maxLen] = longestHailstone(1e5) n = 77031 maxLen = 351</lang>
Maxima
<lang maxima>collatz(n) := block([L], L: [n], while n > 1 do (n: if evenp(n) then n/2 else 3*n + 1, L: endcons(n, L)), L)$
collatz_length(n) := block([m], m: 1, while n > 1 do (n: if evenp(n) then n/2 else 3*n + 1, m: m + 1), m)$
collatz_max(n) := block([j, m, p], m: 0, for i from 1 thru n do
(p: collatz_length(i), if p > m then (m: p, j: i)),
[j, m])$
collatz(27); /* [27, 82, 41, ..., 4, 2, 1] */ length(%); /* 112 */ collatz_length(27); /* 112 */ collatz_max(100000); /* [77031, 351] */</lang>
Mercury
The actual calculation (including module ceremony) providing both a function and a predicate implementation: <lang mercury>:- module hailstone.
- - interface.
- - import_module int, list.
- - func hailstone(int) = list(int).
- - pred hailstone(int::in, list(int)::out) is det.
- - implementation.
hailstone(N) = S :- hailstone(N, S).
hailstone(N, [N|S]) :-
( N = 1 -> S = [] ; N mod 2 = 0 -> hailstone(N/2, S) ; hailstone(3 * N + 1, S) ).
- - end_module hailstone.</lang>
The mainline test driver (making use of unification for more succinct tests): <lang mercury>:- module test_hailstone.
- - interface.
- - import_module io.
- - pred main(io.state::di, io.state::uo) is det.
- - implementation.
- - import_module int, list.
- - import_module hailstone.
- - pred longest(int::in, int::out, int::out) is det.
- - pred longest(int::in, int::in, int::in, int::out, int::out) is det.
longest(M, N, L) :- longest(M, 0, 0, N, L).
longest(N, CN, CL, MN, ML) :-
( N > 1 ->
L = list.length(hailstone(N)),
( L > CL -> longest(N - 1, N, L, MN, ML)
; longest(N - 1, CN, CL, MN, ML) )
; MN = CN, ML = CL ).
main(!IO) :-
S = hailstone(27),
( list.length(S) = 112,
list.append([27, 82, 41, 124], _, S),
list.remove_suffix(S, [8, 4, 2, 1], _),
longest(100000, 77031, 351) ->
io.write_string("All tests succeeded.\n", !IO)
; io.write_string("At least one test failed.\n", !IO) ).
- - end_module test_hailstone.</lang>
- Output:
of running this program is
All tests succeeded.
For those unused to logic programming languages it seems that nothing has been proved in terms of confirming anything, but if you look at the predicate declaration for longest/3 …
<lang mercury>:- pred longest(int::in, int::out, int::out) is det.</lang>
… you see that the second and third parameters are output parameters.
This by calling longest(100000, 77031, 351) you prove,
through unification, that the longest sequence is with the
number 77031 and that it is 351 cycles long.
Similarly, using list.append([27, 82, 41, 124], _, S) automatically proves that the generated sequence begins with the provided sequence, etc.
Thus we know that the correct sequences and values were generated
without bothering to print them out.
ML
MLite
<lang ocaml>fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm ([], i, largest, largest_at) = (largest_at, largest) | (x :: xs, i, largest, largest_at) = let val k = len (hail x) in if k > largest then hailstorm (xs, i + 1, k, i) else hailstorm (xs, i + 1, largest, largest_at) end | (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27; print "hailstone sequence for the number 27 has "; print ` len (h27); print " elements starting with "; print ` sub (h27, 0, 4); print " and ending with "; print ` sub (h27, len(h27)-4, len h27); println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest "; print "hailstone sequence is at element "; print ` ref (biggest, 0); print " and is of length "; println ` ref (biggest, 1);</lang>
- Output:
hailstone sequence for the number 27 has 112 elements starting with [27, 82, 41, 124] and ending with [8, 4, 2, 1]. The number less than 100,000 which has the longest hailstone sequence is at element 77031 and is of length 351
Modula-2
<lang modula2>MODULE hailst;
IMPORT InOut;
CONST maxCard = MAX (CARDINAL) DIV 3; TYPE action = (List, Count, Max); VAR a : CARDINAL;
PROCEDURE HailStone (start : CARDINAL; type : action) : CARDINAL;
VAR n, max, count : CARDINAL;
BEGIN
count := 1;
n := start;
max := n;
LOOP
IF type = List THEN
InOut.WriteCard (n, 12);
IF count MOD 6 = 0 THEN InOut.WriteLn END
END;
IF n = 1 THEN EXIT END;
IF ODD (n) THEN
IF n < maxCard THEN
n := 3 * n + 1;
IF n > max THEN max := n END
ELSE
InOut.WriteString ("Exceeding max value for type CARDINAL at count = ");
InOut.WriteCard (count, 10);
InOut.WriteString (" for intermediate value ");
InOut.WriteCard (n, 10);
InOut.WriteString (". Aborting.");
HALT
END
ELSE
n := n DIV 2
END;
INC (count)
END;
IF type = Max THEN RETURN max ELSE RETURN count END
END HailStone;
PROCEDURE FindMax (num : CARDINAL);
VAR val, maxCount, maxVal, cnt : CARDINAL;
BEGIN
maxCount := 0;
maxVal := 0;
FOR val := 2 TO num DO
cnt := HailStone (val, Count);
IF cnt > maxCount THEN
maxVal := val;
maxCount := cnt
END
END;
InOut.WriteString ("Longest sequence below "); InOut.WriteCard (num, 1);
InOut.WriteString (" is "); InOut.WriteCard (HailStone (maxVal, Count), 1);
InOut.WriteString (" for n = "); InOut.WriteCard (maxVal, 1);
InOut.WriteString (" with an intermediate maximum of ");
InOut.WriteCard (HailStone (maxVal, Max), 1);
InOut.WriteLn
END FindMax;
BEGIN
a := HailStone (27, List);
InOut.WriteLn;
InOut.WriteString ("Iterations total = "); InOut.WriteCard (HailStone (27, Count), 12);
InOut.WriteString (" max value = "); InOut.WriteCard (HailStone (27, Max) , 12);
InOut.WriteLn;
FindMax (100000);
InOut.WriteString ("Done."); InOut.WriteLn
END hailst.</lang> Producing:
jan@Beryllium:~/modula/rosetta$ hailst
27 82 41 124 62 31
94 47 142 71 214 107
322 161 484 242 121 364
182 91 274 137 412 206
103 310 155 466 233 700
350 175 526 263 790 395
1186 593 1780 890 445 1336
668 334 167 502 251 754
377 1132 566 283 850 425
1276 638 319 958 479 1438
719 2158 1079 3238 1619 4858
2429 7288 3644 1822 911 2734
1367 4102 2051 6154 3077 9232
4616 2308 1154 577 1732 866
433 1300 650 325 976 488
244 122 61 184 92 46
23 70 35 106 53 160
80 40 20 10 5 16
8 4 2 1
Iterations total = 112 max value = 9232
Longest sequence below 100000 is 351 for n = 77031 with an intermediate maximum of 21933016
Done.When trying the same for all values below 1 million:
Exceeding max value for type CARDINAL at n = 159487 , count = 60 and intermediate value 1699000271. Aborting.
MUMPS
<lang MUMPS>hailstone(n) ; If n=1 Quit n If n#2 Quit n_" "_$$hailstone(3*n+1) Quit n_" "_$$hailstone(n\2) Set x=$$hailstone(27) Write !,$Length(x," ")," terms in ",x,! 112 terms in 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1</lang>
Nanoquery
<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</lang>
- Output:
hailstone(27) total elements: 112 27, 82, 41, 124, ..., 8, 4, 2, 1 The number less than 100,000 with the longest sequence is 77031 with a length of 351
NetRexx
<lang NetRexx>/* NetRexx */
options replace format comments java crossref savelog symbols binary
do
start = 27 hs = hailstone(start) hsCount = hs.words say 'The number' start 'has a hailstone sequence comprising' hsCount 'elements' say ' its first four elements are:' hs.subword(1, 4) say ' and last four elements are:' hs.subword(hsCount - 3)
hsMax = 0
hsCountMax = 0
llimit = 100000
loop x_ = 1 to llimit - 1
hs = hailstone(x_)
hsCount = hs.words
if hsCount > hsCountMax then do
hsMax = x_
hsCountMax = hsCount
end
end x_
say 'The number' hsMax 'has the longest hailstone sequence in the range 1 to' llimit - 1 'with a sequence length of' hsCountMax
catch ex = Exception
ex.printStackTrace
end
return
method hailstone(hn = long) public static returns Rexx signals IllegalArgumentException
hs = Rexx()
if hn <= 0 then signal IllegalArgumentException('Invalid start point. Must be a positive integer greater than 0')
loop label n_ while hn > 1
hs = hs' 'hn
if hn // 2 \= 0 then hn = hn * 3 + 1
else hn = hn % 2
end n_
hs = hs' 'hn
return hs.strip</lang>
- Output:
The number 27 has a hailstone sequence comprising 112 elements its first four elements are: 27 82 41 124 and last four elements are: 8 4 2 1 The number 77031 has the longest hailstone sequence in the range 1 to 99999 with a sequence length of 351
Nim
<lang nim>proc hailstone(n): auto =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
let h = hailstone 27 assert h.len == 112 and h[0..3] == @[27,82,41,124] and h[h.high-3..h.high] == @[8,4,2,1] var m, mi = 0 for i in 1 .. <100_000:
let n = hailstone(i).len if n > m: m = n mi = i
echo "Maximum length ", m, " was found for hailstone(", mi, ") for numbers <100,000"</lang>
- Output:
Maximum length 351 was found for hailstone(77031) for numbers <100,000
Oberon-2
<lang oberon2>MODULE hailst;
IMPORT Out;
CONST maxCard = MAX (INTEGER) DIV 3;
List = 1;
Count = 2;
Max = 3;
VAR a : INTEGER;
PROCEDURE HailStone (start, type : INTEGER) : INTEGER;
VAR n, max, count : INTEGER;
BEGIN
count := 1;
n := start;
max := n;
LOOP
IF type = List THEN
Out.Int (n, 12);
IF count MOD 6 = 0 THEN Out.Ln END
END;
IF n = 1 THEN EXIT END;
IF ODD (n) THEN
IF n < maxCard THEN
n := 3 * n + 1;
IF n > max THEN max := n END
ELSE
Out.String ("Exceeding max value for type INTEGER at: ");
Out.String (" n = "); Out.Int (start, 12);
Out.String (" , count = "); Out.Int (count, 12);
Out.String (" and intermediate value ");
Out.Int (n, 1);
Out.String (". Aborting.");
Out.Ln;
HALT (2)
END
ELSE
n := n DIV 2
END;
INC (count)
END;
IF type = Max THEN RETURN max ELSE RETURN count END
END HailStone;
PROCEDURE FindMax (num : INTEGER);
VAR val, maxCount, maxVal, cnt : INTEGER;
BEGIN
maxCount := 0;
maxVal := 0;
FOR val := 2 TO num DO
cnt := HailStone (val, Count);
IF cnt > maxCount THEN
maxVal := val;
maxCount := cnt
END
END;
Out.String ("Longest sequence below "); Out.Int (num, 1);
Out.String (" is "); Out.Int (HailStone (maxVal, Count), 1);
Out.String (" for n = "); Out.Int (maxVal, 1);
Out.String (" with an intermediate maximum of ");
Out.Int (HailStone (maxVal, Max), 1);
Out.Ln
END FindMax;
BEGIN
a := HailStone (27, List);
Out.Ln;
Out.String ("Iterations total = "); Out.Int (HailStone (27, Count), 12);
Out.String (" max value = "); Out.Int (HailStone (27, Max) , 12);
Out.Ln;
FindMax (1000000);
Out.String ("Done.");
Out.Ln
END hailst.</lang> Producing
27 82 41 124 62 31
94 47 142 71 214 107
322 161 484 242 121 364
182 91 274 137 412 206
103 310 155 466 233 700
350 175 526 263 790 395
1186 593 1780 890 445 1336
668 334 167 502 251 754
377 1132 566 283 850 425
1276 638 319 958 479 1438
719 2158 1079 3238 1619 4858
2429 7288 3644 1822 911 2734
1367 4102 2051 6154 3077 9232
4616 2308 1154 577 1732 866
433 1300 650 325 976 488
244 122 61 184 92 46
23 70 35 106 53 160
80 40 20 10 5 16
8 4 2 1
Iterations total = 112 max value = 9232
Exceeding max value for type INTEGER at: n = 113383 , count = 120 and intermediate value 827370449. Aborting.
OCaml
<lang ocaml>#load "nums.cma";; open Num;;
(* generate Hailstone sequence *) let hailstone n =
let one = Int 1
and two = Int 2
and three = Int 3 in
let rec g s x =
if x =/ one
then x::s
else g (x::s) (if mod_num x two =/ one
then three */ x +/ one
else x // two)
in
g [] (Int n)
(* compute only sequence length *) let haillen n =
let one = Int 1
and two = Int 2
and three = Int 3 in
let rec g s x =
if x =/ one
then s+1
else g (s+1) (if mod_num x two =/ one
then three */ x +/ one
else x // two)
in
g 0 (Int n)
(* max length for starting values in 1..n *) let hailmax =
let rec g idx len = function
| 0 -> (idx, len)
| i ->
let a = haillen i in
if a > len
then g i a (i-1)
else g idx len (i-1)
in
g 0 0
hailmax 100000 ;; (* - : int * int = (77031, 351) *)
List.rev_map string_of_num (hailstone 27) ;;
(* - : string list = ["27"; "82"; "41"; "124"; "62"; "31"; "94"; "47"; "142"; "71"; "214"; "107";
"322"; "161"; "484"; "242"; "121"; "364"; "182"; "91"; "274"; "137"; "412"; "206"; "103"; "310"; "155"; "466"; "233"; "700"; "350"; "175"; "526"; "263"; "790"; "395"; "1186"; "593"; "1780"; "890"; "445"; "1336"; "668"; "334"; "167"; "502"; "251"; "754"; "377"; "1132"; "566"; "283"; "850"; "425"; "1276"; "638"; "319"; "958"; "479"; "1438"; "719"; "2158"; "1079"; "3238"; "1619"; "4858"; "2429"; "7288"; "3644"; "1822"; "911"; "2734"; "1367"; "4102"; "2051"; "6154"; "3077"; "9232"; "4616"; "2308"; "1154"; "577"; "1732"; "866"; "433"; "1300"; "650"; "325"; "976"; "488"; "244"; "122"; "61"; "184"; "92"; "46"; "23"; "70"; "35"; "106"; "53"; "160"; "80"; "40"; "20"; "10"; "5"; "16"; "8"; "4"; "2"; "1"] *)</lang>
Oforth
<lang Oforth>: hailstone // n -- [n] | l |
ListBuffer new ->l while(dup 1 <>) [ dup l add dup isEven ifTrue: [ 2 / ] else: [ 3 * 1+ ] ] l add l dup freeze ;
hailstone(27) dup size println dup left(4) println right(4) println 100000 seq map(#[ dup hailstone size swap Pair new ]) reduce(#maxKey) println</lang>
- Output:
112 [27, 82, 41, 124] [8, 4, 2, 1] [351, 77031]
ooRexx
<lang ooRexx> sequence = hailstone(27) say "Hailstone sequence for 27 has" sequence~items "elements and is ["sequence~toString('l', ", ")"]"
highestNumber = 1 highestCount = 1
loop i = 2 to 100000
sequence = hailstone(i)
count = sequence~items
if count > highestCount then do
highestNumber = i
highestCount = count
end
end say "Number" highestNumber "has the longest sequence with" highestCount "elements"
-- short routine to generate a hailstone sequence
- routine hailstone
use arg n
sequence = .array~of(n)
loop while n \= 1
if n // 2 == 0 then n = n / 2
else n = 3 * n + 1
sequence~append(n)
end
return sequence
</lang>
- Output:
Hailstone sequence for 27 has 112 elements and is [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 77, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 102, 051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 0, 40, 20, 10, 5, 16, 8, 4, 2, 1] Number 77031 has the longest sequence with 351 elements
Order
To display the length, and first and last elements, of the hailstone sequence for 27, we could do this: <lang c>#include <order/interpreter.h>
- define ORDER_PP_DEF_8hailstone ORDER_PP_FN( \
8fn(8N, \
8cond((8equal(8N, 1), 8seq(1)) \
(8is_0(8remainder(8N, 2)), \
8seq_push_front(8N, 8hailstone(8quotient(8N, 2)))) \
(8else, \
8seq_push_front(8N, 8hailstone(8inc(8times(8N, 3))))))) )
ORDER_PP(
8lets((8H, 8seq_map(8to_lit, 8hailstone(27)))
(8S, 8seq_size(8H)),
8print(8(h(27) - length:) 8to_lit(8S) 8comma 8space
8(starts with:) 8seq_take(4, 8H) 8comma 8space
8(ends with:) 8seq_drop(8minus(8S, 4), 8H))
) )</lang>
- Output:
<lang>h(27) - length:112, starts with:(27)(82)(41)(124), ends with:(8)(4)(2)(1)</lang>
Unfortunately, the C preprocessor not really being designed with large amounts of garbage collection in mind, trying to compute the hailstone sequences up to 100000 is almost guaranteed to run out of memory (and take a very, very long time). If we wanted to try, we could add this to the program, which in most languages would use relatively little memory: <lang c>#define ORDER_PP_DEF_8h_longest ORDER_PP_FN( \ 8fn(8M, 8P, \
8if(8is_0(8M), \
8P, \
8let((8L, 8seq_size(8hailstone(8M))), \
8h_longest(8dec(8M), \
8if(8greater(8L, 8tuple_at_1(8P)), \
8pair(8M, 8L), 8P))))) )
ORDER_PP(
8let((8P, 8h_longest(8nat(1,0,0,0,0,0), 8pair(0, 0))),
8pair(8to_lit(8tuple_at_0(8P)), 8to_lit(8tuple_at_1(8P))))
)</lang>
...or even this "more elegant" version, which will run out of memory very quickly indeed (but in practice seems to work better for smaller ranges): <lang c>ORDER_PP(
8let((8P,
8seq_head(
8seq_sort(8fn(8P, 8Q, 8greater(8tuple_at_1(8P),
8tuple_at_1(8Q))),
8seq_map(8fn(8N,
8pair(8N, 8seq_size(8hailstone(8N)))),
8seq_iota(1, 8nat(1,0,0,0,0,0)))))),
8pair(8to_lit(8tuple_at_0(8P)), 8to_lit(8tuple_at_1(8P)))) )</lang>
Notice that large numbers (>100) must be entered as digit sequences with 8nat. 8to_lit converts a digit sequence back to a readable number.
Oz
<lang oz>declare
fun {HailstoneSeq N}
N > 0 = true %% assert
if N == 1 then [1]
elseif {IsEven N} then N|{HailstoneSeq N div 2}
else N|{HailstoneSeq 3*N+1}
end
end
HSeq27 = {HailstoneSeq 27}
{Length HSeq27} = 112
{List.take HSeq27 4} = [27 82 41 124]
{List.drop HSeq27 108} = [8 4 2 1]
fun {MaxBy2nd A=A1#A2 B=B1#B2}
if B2 > A2 then B else A end
end
Pairs = {Map {List.number 1 99999 1}
fun {$ I} I#{Length {HailstoneSeq I}} end}
MaxI#MaxLen = {List.foldL Pairs MaxBy2nd 0#0}
{System.showInfo
"Maximum length "#MaxLen#" was found for hailstone("#MaxI#")"}</lang>
- Output:
Maximum length 351 was found for hailstone(77031)
PARI/GP
Version #1.
<lang parigp>show(n)={
my(t=1);
while(n>1,
print1(n",");
n=if(n%2,
3*n+1
,
n/2
);
t++
);
print(1);
t
};
len(n)={
my(t=1);
while(n>1,
if(n%2,
t+=2;
n+=(n>>1)+1
,
t++;
n>>=1
)
);
t
};
show(27) r=0;for(n=1,1e5,t=len(n);if(t>r,r=t;ra=n));print(ra"\t"r)</lang>
- Output:
27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,4 12,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,133 6,668,334,167,502,251,754,377,1132,566,283,850,425,1276,638,319,958,479,1438,719 ,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077, 9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,2 3,70,35,106,53,160,80,40,20,10,5,16,8,4,2,1
and
77031 351
Version #2.
Different kind of PARI scripts for Collatz sequences you can find in OEIS, e.g.: A070165
<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); } </lang>
- Output:
Required tests: Hailstone/Collatz(27): [27, 82, 41, 124] ... [8, 4, 2, 1]; length = 112 Search range: 1..100000 Hailstone/Collatz(77031) has the longest length = 351 (15:32) gp > ## *** last result computed in 15,735 ms.
Pascal
See Delphi or try this transformed Delphi version without generics.Use of a static array. <lang pascal>program ShowHailstoneSequence; {$IFDEF FPC}
{$MODE delphi} //or objfpc
{$Else}
{$Apptype Console} // for delphi
{$ENDIF} uses
SysUtils;// format
const
maxN = 10*1000*1000;// for output 1000*1000*1000
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt; begin
result := 0; //ensure n to be odd while not(ODD(n)) do Begin inc(result); n := n shr 1; end;
IF n > 1 then
repeat
//now n == odd -> so two steps in one can be made
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
//now n == even -> so only one step can be made
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr); var
maxPos: NativeInt; n: UInt64; pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt; lList: tIntArr; lAverageLength:Uint64; lMaxSequence: NativeInt; lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);//319804831
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[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 := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
end.</lang>
- Output:
sequence of 27 has 112 elements
27,82,41,124..8,4,2,1
Limit : number with max length | average length
10 : 9 | 20 | 5.490
100 : 97 | 119 | 27.504
1000 : 871 | 179 | 50.683
10000 : 6171 | 262 | 71.119
100000 : 77031 | 351 | 89.137
1000000 : 837799 | 525 | 108.613
10000000 : 8400511 | 686 | 127.916
100000000 : 63728127 | 950 | 147.337
1000000000 : 670617279 | 987 | 166.780
real 6m22.968s // 32-bit compiled
real 3m56.573s // 64-bit compiled
Perl
Straightforward
<lang Perl>#!/usr/bin/perl
use warnings; use strict;
my @h = hailstone(27); print "Length of hailstone(27) = " . scalar @h . "\n"; print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0); for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n; }
return @sequence;
}</lang>
- Output:
Length of hailstone(27) = 112 [27, 82, 41, 124, ..., 8, 4, 2, 1] Max length 351 was found for hailstone(77031) for numbers < 100_000
Compact
A more compact version: <lang Perl>#!/usr/bin/perl use strict;
sub hailstone {
@_ = local $_ = shift; push @_, $_ = $_ % 2 ? 3 * $_ + 1 : $_ / 2 while $_ > 1; @_;
}
my @h = hailstone($_ = 27); print "$_: @h[0 .. 3] ... @h[-4 .. -1] (".@h.")\n";
@h = (); for (1 .. 99_999) { @h = ($_, $h[2]) if ($h[2] = hailstone($_)) > $h[1] } printf "%d: (%d)\n", @h;</lang>
- Output:
27: 27 82 41 124 ... 8 4 2 1 (112) 77031: (351)
Phix
Copy of Euphoria <lang Phix>function hailstone(atom n) sequence s = {n}
while n!=1 do
if remainder(n,2)=0 then
n /= 2
else
n = 3*n+1
end if
s &= n
end while
return s
end function
function hailstone_count(atom n) integer count = 1
while n!=1 do
if remainder(n,2)=0 then
n /= 2
else
n = 3*n+1
end if
count += 1
end while
return count
end function
sequence s = hailstone(27) integer ls = length(s) s[5..-5] = {".."} puts(1,"hailstone(27) = ") ? s printf(1,"length = %d\n\n",ls)
integer hmax = 1, imax = 1,count for i=2 to 1e5-1 do
count = hailstone_count(i)
if count>hmax then
hmax = count
imax = i
end if
end for
printf(1,"The longest hailstone sequence under 100,000 is %d with %d elements.\n",{imax,hmax})</lang>
- Output:
hailstone(27) = {27,82,41,124,"..",8,4,2,1}
length = 112
The longest hailstone sequence under 100,000 is 77031 with 351 elements.
PHP
<lang php>function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } }
$result = hailstone(27);
echo count($result) . ' Elements.
';
echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '
';
$maxResult = array(0);
for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }</lang>
112 Elements. Starting with : 27,82,41,124 and ending with : 8,4,2,1 Number < 100000 with longest Hailstone seq.: 77031 with length of 351
PicoLisp
<lang PicoLisp>(de hailstone (N)
(make
(until (= 1 (link N))
(setq N
(if (bit? 1 N)
(inc (* N 3))
(/ N 2) ) ) ) ) )
(let L (hailstone 27)
(println 27 (length L) (head 4 L) '- (tail 4 L)) )
(let N (maxi '((N) (length (hailstone N))) (range 1 100000))
(println N (length (hailstone N))) )</lang>
- Output:
27 112 (27 82 41 124) - (8 4 2 1) 77031 351
Pike
<lang Pike>#!/usr/bin/env pike
int next(int n) {
if (n==1)
return 0;
if (n%2)
return 3*n+1;
else
return n/2;
}
array(int) hailstone(int n) {
array seq = ({ n });
while (n=next(n))
seq += ({ n });
return seq;
}
void main() {
array(int) two = hailstone(27);
if (equal(two[0..3], ({ 27, 82, 41, 124 })) && equal(two[<3..], ({ 8,4,2,1 })))
write("sizeof(({ %{%d, %}, ... %{%d, %} }) == %d\n", two[0..3], two[<3..], sizeof(two));
mapping longest = ([ "length":0, "start":0 ]);
foreach(allocate(100000); int start; )
{
int length = sizeof(hailstone(start));
if (length > longest->length)
{
longest->length = length;
longest->start = start;
}
}
write("longest sequence starting at %d has %d elements\n", longest->start, longest->length);
}</lang>
- Output:
sizeof(({ 27, 82, 41, 124, , ... 8, 4, 2, 1, }) == 112
longest sequence starting at 77031 has 351 elements
PL/I
<lang pli>test: proc options (main);
declare (longest, n) fixed (15); declare flag bit (1); declare (i, value) fixed (15);
/* Task 1: */
flag = '1'b;
put skip list ('The sequence for 27 is');
i = hailstones(27);
/* Task 2: */
flag = '0'b;
longest = 0;
do i = 1 to 99999;
if longest < hailstones(i) then
do; longest = hailstones(i); value = i; end;
end;
put skip edit (value, ' has the longest sequence of ', longest) (a);
hailstones: procedure (n) returns ( fixed (15));
declare n fixed (15) nonassignable; declare (m, p) fixed (15);
m = n;
p = 1;
if flag then put skip list (m);
do p = 1 by 1 while (m > 1);
if iand(m, 1) = 0 then
m = m/2;
else
m = 3*m + 1;
if flag then put skip list (m);
end;
if flag then put skip list ('The hailstone sequence has length' || p);
return (p);
end hailstones;
end test;</lang>
- Output:
The sequence for 27 is
27
82
41
124
62
31
94
47
142
71
214
107
322
161
484
242
121
364
182
91
274
137
412
206
103
310
155
466
233
700
350
175
526
263
790
395
1186
593
1780
890
445
1336
668
334
167
502
251
754
377
1132
566
283
850
425
1276
638
319
958
479
1438
719
2158
1079
3238
1619
4858
2429
7288
3644
1822
911
2734
1367
4102
2051
6154
3077
9232
4616
2308
1154
577
1732
866
433
1300
650
325
976
488
244
122
61
184
92
46
23
70
35
106
53
160
80
40
20
10
5
16
8
4
2
1
The hailstone sequence has length 112
77031 has the longest sequence of 351
Plain TeX
The following code works with any TeX engine. <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</lang>
pdf or dvi output:
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Length = 112 Seed max = 77031, length = 351
Pointless
<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</lang>
- Output:
getSeq(27) (length): 112 getSeq(27) (first 4): [27, 82, 41, 124] getSeq(27) (last 4): [8, 4, 2, 1] max length 351 for n = 77031
PowerShell
<lang Powershell>
function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}</lang>
- Output:
PS C:\> Get-HailStone 27 27 82 41 ... 8 4 2 1 PS C:\> (Get-HailStone 27).count 112 PS C:\> Get-HailStoneBelowLimit 100000 | Sort Count -Descending | Select -first 1 Number Count ------ ----- 77031 351
Prolog
1. Create a routine to generate the hailstone sequence for a number. <lang prolog>hailstone(1,[1]) :- !. hailstone(N,[N|S]) :- 0 is N mod 2, N1 is N / 2, hailstone(N1,S). hailstone(N,[N|S]) :- 1 is N mod 2, N1 is (3 * N) + 1, hailstone(N1, S).</lang>
2. Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1.
The following query performs the test. <lang prolog>hailstone(27,X), length(X,112), append([27, 82, 41, 124], _, X), append(_, [8, 4, 2, 1], X).</lang>
3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequences length. <lang prolog>longestHailstoneSequence(M, Seq, Len) :- longesthailstone(M, 1, 1, Seq, Len). longesthailstone(1, Cn, Cl, Mn, Ml):- Mn = Cn, Ml = Cl. longesthailstone(N, _, Cl, Mn, Ml) :- hailstone(N, X),
length(X, L),
Cl < L,
N1 is N-1,
longesthailstone(N1, N, L, Mn, Ml).
longesthailstone(N, Cn, Cl, Mn, Ml) :- N1 is N-1,
longesthailstone(N1, Cn, Cl, Mn, Ml).</lang>
run this query. <lang prolog>longestHailstoneSequence(100000, Seq, Len).</lang> to get the following result
Seq = 77031, Len = 351
Constraint Handling Rules
CHR is a programming language created by Professor Thom Frühwirth.
Works with SWI-Prolog and module chr written by Tom Schrijvers and Jan Wielemaker
<lang Prolog>:- use_module(library(chr)).
- - chr_option(debug, off).
- - chr_option(optimize, full).
- - chr_constraint collatz/2, hailstone/1, clean/0.
% to remove all constraints hailstone/1 after computation clean @ clean \ hailstone(_) <=> true. clean @ clean <=> true.
% compute Collatz number init @ collatz(1,X) <=> X = 1 | true. collatz @ collatz(N, C) <=> (N mod 2 =:= 0 -> C is N / 2; C is 3 * N + 1).
% Hailstone loop hailstone(1) ==> true. hailstone(N) ==> N \= 1 | collatz(N, H), hailstone(H).</lang>
Code for task one : <lang Prolog>task1 :- hailstone(27), findall(X, find_chr_constraint(hailstone(X)), L), clean, % check the requirements ( (length(L, 112), append([27, 82, 41, 124 | _], [8,4,2,1], L)) -> writeln(ok); writeln(ko)).</lang>
- Output:
?- task1. ok true.
Code for task two : <lang Prolog>longest_sequence :- seq(2, 100000, 1-[1], Len-V), format('For ~w sequence has ~w len ! ~n', [V, Len]).
% walk through 2 to 100000 and compute the length of the sequences
% memorize the longest
seq(N, Max, Len-V, Len-V) :- N is Max + 1, !.
seq(N, Max, CLen - CV, FLen - FV) :-
len_seq(N, Len - N),
( Len > CLen -> Len1 = Len, V1 = [N]
; Len = CLen -> Len1 = Len, V1 = [N | CV]
; Len1 = CLen, V1 = CV),
N1 is N+1,
seq(N1, Max, Len1 - V1, FLen - FV).
% compute the len of the Hailstone sequence for a number len_seq(N, Len - N) :- hailstone(N), findall(hailstone(X), find_chr_constraint(hailstone(X)), L), length(L, Len), clean.</lang>
- Output:
?- longest_sequence. For [77031] sequence has 351 len ! true.
Pure
<lang pure>// 1. Create a routine to generate the hailstone sequence for a number. type odd x::int = x mod 2; type even x::int = ~odd x; odd x = typep odd x; even x = typep even x;
hailstone 1 = [1]; hailstone n::even = n:hailstone (n div 2); hailstone n::odd = n:hailstone (3*n + 1);
// 2. Use the routine to show that the hailstone sequence for the number 27 // has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1 n = 27; hs = hailstone n; l = # hs; using system;
printf
("the hailstone sequence for the number %d has %d elements " +
"starting with %s and ending with %s\n")
(n, l, __str__ (hs!!(0..3)), __str__ ( hs!!((l-4)..l)));
// 3. Show the number less than 100,000 which has the longest hailstone // sequence together with that sequences length. printf ("the number under 100,000 with the longest sequence is %d " +
"with a sequence length of %d\n")
(foldr (\ (a,b) (c,d) -> if (b > d) then (a,b) else (c,d))
(0,0)
(map (\ x -> (x, # hailstone x)) (1..100000)));</lang>
- Output:
the hailstone sequence for the number 27 has 112 elements starting with [27,82,41,124] and ending with [8,4,2,1] the number under 100,000 with the longest sequence is 77031 with a sequence length of 351
Python
Procedural
<lang python>def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))</lang>
- Output:
Maximum length 351 was found for hailstone(77031) for numbers <100,000
Composition of pure functions
<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):
return xs[n:]
else:
take(n)(xs)
return xs
return lambda xs: go(xs)
- enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
Integer enumeration from m to n. return lambda n: list(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 lambda x: go(x)
- 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.
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
- 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()</lang>
- Output:
The hailstone sequence for 27 has 112 elements, starting with [27, 82, 41, 124], and ending with [8, 4, 2, 1]. The number in the range 1..99999 which produces the longest sequence is 77031, generating a hailstone sequence of 351 integers.
Quackery
<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</lang>
Output: <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. </lang>
R
Iterative solution
<lang r>### PART 1: makeHailstone <- function(n){
hseq <- n
while (hseq[length(hseq)] > 1){
current.value <- hseq[length(hseq)]
if (current.value %% 2 == 0){
next.value <- current.value / 2
} else {
next.value <- (3 * current.value) + 1
}
hseq <- append(hseq, next.value)
}
return(list(hseq=hseq, seq.length=length(hseq)))
}
- PART 2:
twenty.seven <- makeHailstone(27) twenty.seven$hseq twenty.seven$seq.length
- PART 3:
max.length <- 0; lower.bound <- 1; upper.bound <- 100000
for (index in lower.bound:upper.bound){
current.hseq <- makeHailstone(index)
if (current.hseq$seq.length > max.length){
max.length <- current.hseq$seq.length
max.index <- index
}
}
cat("Between ", lower.bound, " and ", upper.bound, ", the input of ",
max.index, " gives the longest hailstone sequence, which has length ", max.length, ". \n", sep="")</lang>
- Output:
> twenty.seven$hseq [1] 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 [16] 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 [31] 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 [46] 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 [61] 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 [76] 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 [91] 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 [106] 10 5 16 8 4 2 1 > twenty.seven$seq.length [1] 112 Between 1 and 1e+05, the input of 77031 gives the longest hailstone sequence, which has length 351.
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. Observe how short the following code is: <lang r>###Task 1: collatz<-function(n) {
output<-c(n)
lastEntry<-output[1]
while(lastEntry!=1)
{
if(lastEntry%%2==0){output<-c(output,lastEntry%/%2)}
else{output<-c(output,3*lastEntry+1)}
lastEntry<-output[length(output)]
}
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])
- Task 3:
- Notice how a several line long loop can be avoided with R's sapply or Vectorize:
seqLenghts<-sapply(1:99999,function(x) length(collatz(x))) longest<-which.max(seqLenghts) print(paste0("The longest sequence before the 100000th is found at n=",longest,". It has length ",seqLenghts[longest],"."))
- Equivalently, line 1 could have been: seqLenghts<-sapply(Vectorize(collatz)(1:99999),length)</lang>
- Output:
> twentySeven<-collatz(27)
> cat("The first four elements are:", twentySeven[1:4],"and the last four are:", twentySeven[length(twentySeven)-3:0])
The first four elements are: 27 82 41 124 and the last four are: 8 4 2 1
> seqLenghts<-sapply(1:99999,function(x) length(collatz(x)))
> longest<-which.max(seqLenghts)
> print(paste0("The longest sequence before the 100000th is found at n=",longest,". It has length ",seqLenghts[longest],"."))
[1] "The longest sequence before the 100000th is found at n=77031. It has length 351."
Racket
<lang Racket>
- lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27)) (printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000) (define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))]) (define h (hailstone i)) (if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
</lang>
- Output:
h(27) = (27 82 41 124 ... 8 4 2 1), 112 items for x<=100000, 77031 has the longest sequence with 351 items
Raku
(formerly Perl 6)
<lang perl6>sub hailstone($n) { $n, { $_ %% 2 ?? $_ div 2 !! $_ * 3 + 1 } ... 1 }
my @h = hailstone(27); say "Length of hailstone(27) = {+@h}"; say ~@h;
my $m = max ( (1..99_999).race.map: { +hailstone($_) => $_ } ); say "Max length {$m.key} was found for hailstone({$m.value}) for numbers < 100_000";</lang>
- Output:
Length of hailstone(27) = 112 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Max length 351 was found for hailstone(77031) for numbers < 100_000
REBOL
<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 ]</lang>
- Output:
the hail sequence of 27 has length 112 and has the form 27 82 41 ... 4 2 1 the number less than 100000 with the longest hail sequence is 77031 with length 351
REXX
non-optimized
<lang rexx>/*REXX program tests a number and also a range for hailstone (Collatz) sequences. */ numeric digits 20 /*be able to handle gihugeic numbers. */ parse arg x y . /*get optional arguments from the C.L. */ if x== | x=="," then x= 27 /*No 1st argument? Then use default.*/ if y== | y=="," then y= 100000 - 1 /* " 2nd " " " " */ $= hailstone(x) /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 1▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/ say x ' has a hailstone sequence of ' words($) say ' and starts with: ' subword($, 1, 4) " ∙∙∙" say ' and ends with: ∙∙∙' subword($, max(5, words($)-3)) if y==0 then exit /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 2▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/ say w= 0; do j=1 for y; call hailstone j /*traipse through the range of numbers.*/
if #hs<=w then iterate /*Not big 'nuff? Then keep traipsing.*/
bigJ= j; w= #hs /*remember what # has biggest hailstone*/
end /*j*/
say '(between 1 ──►' y") " bigJ ' has the longest hailstone sequence: ' w exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ hailstone: procedure expose #hs; parse arg n 1 s /*N and S: are set to the 1st argument.*/
do #hs=1 while n\==1 /*keep loop while N isn't unity. */
if n//2 then n= n * 3 + 1 /*N is odd ? Then calculate 3*n + 1 */
else n= n % 2 /*" " even? Then calculate fast ÷ */
s= s n /* [↑] % is REXX integer division. */
end /*#hs*/ /* [↑] append N to the sequence list*/
return s /*return the S string to the invoker.*/</lang>
- output when using the default inputs:
27 has a hailstone sequence of 112
and starts with: 27 82 41 124 ∙∙∙
and ends with: ∙∙∙ 8 4 2 1
(between 1 ──► 99999) 77031 has the longest hailstone sequence: 351
optimized
This version is about 7 times faster than the previous (unoptimized) version.
It makes use of:
- previously calculated Collatz sequences (memoization)
- a faster method of determining if an integer is even
<lang rexx>/*REXX program tests a number and also a range for hailstone (Collatz) sequences. */ !.=0; !.0=1; !.2=1; !.4=1; !.6=1; !.8=1 /*assign even numerals to be "true". */ numeric digits 20; @.= 0 /*handle big numbers; initialize array.*/ parse arg x y z .; !.h= y /*get optional arguments from the C.L. */ if x== | x=="," then x= 27 /*No 1st argument? Then use default.*/ if y== | y=="," then y= 100000 - 1 /* " 2nd " " " " */ if z== | z=="," then z= 12 /*head/tail number? " " " */ hm= max(y, 500000) /*use memoization (maximum num for @.)*/ $= hailstone(x) /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 1▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/ say x ' has a hailstone sequence of ' words($) say ' and starts with: ' subword($, 1, z) " ∙∙∙" say ' and ends with: ∙∙∙' subword($, max(z+1, words($)-z+1)) if y==0 then exit /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 2▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/ say w= 0; do j=1 for y; $= hailstone(j) /*traipse through the range of numbers.*/
#hs= words($) /*find the length of the hailstone seq.*/
if #hs<=w then iterate /*Not big enough? Then keep traipsing.*/
bigJ= j; w= #hs /*remember what # has biggest hailstone*/
end /*j*/
say '(between 1 ──►' y") " bigJ ' has the longest hailstone sequence: ' w exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ hailstone: procedure expose @. !. hm; parse arg n 1 s 1 o,@.1 /*N,S,O: are the 1st arg*/
do while @.n==0 /*loop while the residual is unknown. */
parse var n -1 L /*extract the last decimal digit of N.*/
if !.L then n= n % 2 /*N is even? Then calculate fast ÷ */
else n= n * 3 + 1 /*" " odd ? " " 3*n + 1 */
s= s n /* [↑] %: is the REXX integer division*/
end /*while*/ /* [↑] append N to the sequence list*/
s= s @.n /*append the number to a sequence list.*/
@.o= subword(s, 2); parse var s _ r /*use memoization for this hailstone #.*/
do while r\==; parse var r _ r /*obtain the next hailstone sequence. */
if @._\==0 then leave /*Was number already found? Return S.*/
if _>hm then iterate /*Is number out of range? Ignore it.*/
@._= r /*assign subsequence number to array. */
end /*while*/; return s</lang>
- output when using the default inputs:
27 has a hailstone sequence of 112
and starts with: 27 82 41 124 62 31 94 47 142 71 214 107 ∙∙∙
and ends with: ∙∙∙ 53 160 80 40 20 10 5 16 8 4 2 1
(between 1 ──► 99999) 77031 has the longest hailstone sequence: 351
- output when using the inputs: , 1000000
27 has a hailstone sequence of 112
and starts with: 27 82 41 124 62 31 94 47 142 71 214 107 ∙∙∙
and ends with: ∙∙∙ 53 160 80 40 20 10 5 16 8 4 2 1
(between 1 ──► 1000000) 837799 has the longest hailstone sequence: 525
Ring
<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
</lang>
Ruby
This program uses new methods (Integer#even? and Enumerable#max_by) from Ruby 1.8.7.
<lang ruby>def hailstone n
seq = [n] until n == 1 n = (n.even?) ? (n / 2) : (3 * n + 1) seq << n end seq
end
puts "for n = 27, show sequence length and first and last 4 elements" hs27 = hailstone 27 p [hs27.length, hs27[0..3], hs27[-4..-1]]
- find the longest sequence among n less than 100,000
n = (1 ... 100_000).max_by{|n| hailstone(n).length} puts "#{n} has a hailstone sequence length of #{hailstone(n).length}" puts "the largest number in that sequence is #{hailstone(n).max}"</lang>
- Output:
for n = 27, show sequence length and first and last 4 elements [112, [27, 82, 41, 124], [8, 4, 2, 1]] 77031 has a hailstone sequence length of 351 the largest number in that sequence is 21933016
This version builds some linked lists with shared structure. Hailstone::ListNode is an adaptation of ListNode from Singly-linked list/Element definition#Ruby. When two sequences contain the same value, those two lists share a tail. This avoids recomputing the end of the sequence.
<lang ruby>module Hailstone
ListNode = Struct.new(:value, :size, :succ) do
def each
node = self
while node
yield node.value
node = node.succ
end
end
end
@@sequence = {1 => ListNode[1,1]}
module_function
def sequence(n)
unless @@sequence[n]
m, ary = n, []
until succ = @@sequence[m]
ary << m
m = m.even? ? (m / 2) : (3 * m + 1)
end
ary.reverse_each do |m|
@@sequence[m] = succ = ListNode[m, succ.size + 1, succ]
end
end
@@sequence[n]
end
end
puts "for n = 27, show sequence length and first and last 4 elements" hs27 = Hailstone.sequence(27).entries p [hs27.size, hs27[0..3], hs27[-4..-1]]
- find the longest sequence among n less than 100,000
n = (1 ... 100_000).max_by{|n| Hailstone.sequence(n).size} puts "#{n} has a hailstone sequence length of #{Hailstone.sequence(n).size}" puts "the largest number in that sequence is #{Hailstone.sequence(n).max}"</lang> output is the same as the above.
Rust
<lang rust>fn hailstone(start : u32) -> Vec<u32> {
let mut res = Vec::new(); let mut next = start;
res.push(start);
while next != 1 {
next = if next % 2 == 0 { next/2 } else { 3*next+1 };
res.push(next);
}
res
}
fn main() {
let test_num = 27; let test_hailseq = hailstone(test_num);
println!("For {} number of elements is {} ", test_num, test_hailseq.len());
let fst_slice = test_hailseq[0..4].iter()
.fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " });
let last_slice = test_hailseq[test_hailseq.len()-4..].iter()
.fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " });
println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice);
let max_range = 100000;
let mut max_len = 0;
let mut max_seed = 0;
for i_seed in 1..max_range {
let i_len = hailstone(i_seed).len();
if i_len > max_len {
max_len = i_len;
max_seed = i_seed;
}
}
println!("Longest sequence is {} element long for seed {}", max_len, max_seed);
}</lang>
- Output:
For 27 number of elements is 112 hailstone starting with 27, 82, 41, 124, ending with 8, 4, 2, 1, Longest sequence is 351 element long for seed 77031
S-lang
<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 (har0:3)
() = printf("%d, ", $1);
() = printf("\nand ending with:\n\t"); foreach $1 (harlength(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");</lang>
- Output:
Hailstone(27) has 112 elements starting with:
27, 82, 41, 124,
and ending with:
8, 4, 2, 1
longest sequence started w/77031 and had 351 elements
SAS
<lang SAS>
- Create a routine to generate the hailstone sequence for one number;
%macro gen_seq(n);
data hailstone;
array hs_seq(100000);
n=&n;
do until (n=1);
seq_size + 1;
hs_seq(seq_size) = n;
if mod(n,2)=0 then n=n/2;
else n=(3*n)+1;
end;
seq_size + 1;
hs_seq(seq_size)=n;
call symputx('seq_length',seq_size);
run;
proc sql;
title "First and last elements of Hailstone Sequence for number &n";
select seq_size as sequence_length, hs_seq1, hs_seq2, hs_seq3, hs_seq4 %do i=&seq_length-3 %to &seq_length; , hs_seq&i %end; from hailstone; quit; %mend;
- Use the routine to output the first and last four numbers in the sequence for 27;
%gen_seq(27);
- Show the number less than 100,000 which has the longest hailstone sequence, and what that length is ;
%macro longest_hailstone(start_num, end_num); data hailstone_analysis; do start=&start_num to &end_num; n=start; length_of_sequence=1; do while (n>1); length_of_sequence+1; if mod(n,2)=0 then n=n/2; else n=(3*n) + 1; end; output; end; run;
proc sort data=hailstone_analysis; by descending length_of_sequence; run;
proc print data=hailstone_analysis (obs=1) noobs; title "Number from &start_num to &end_num with longest Hailstone sequence"; var start length_of_sequence; run; %mend; %longest_hailstone(1,99999); </lang>
- Output:
First and last elements of Hailstone Sequence for number 27
sequence_
length hs_seq1 hs_seq2 hs_seq3 hs_seq4 hs_seq109 hs_seq110 hs_seq111 hs_seq112
-------------------------------------------------------------------------------------------------
112 27 82 41 124 8 4 2 1
Number from 1 to 99999 with longest Hailstone sequence
length_of_
start sequence
77031 351
S-BASIC
<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 </lang>
- Output:
Display hailstone sequence for what number? 27
27 82 41 124 62 31 94 47 142 71
214 107 322 161 484 242 121 364 182 91
274 137 412 206 103 310 155 466 233 700
350 175 526 263 790 395 1186 593 1780 890
445 1336 668 334 167 502 251 754 377 1132
566 283 850 425 1276 638 319 958 479 1438
719 2158 1079 3238 1619 4858 2429 7288 3644 1822
911 2734 1367 4102 2051 6154 3077 9232 4616 2308
1154 577 1732 866 433 1300 650 325 976 488
244 122 61 184 92 46 23 70 35 106
53 160 80 40 20 10 5 16 8 4
2 1
Sequence length = 112
Searching for longest sequence up to N = 1000 ...
Search terminated with overflow at 447
Maximum sequence length = 144 for N = 327
Scala
<lang Scala>object HailstoneSequence extends App {
def hailstone(n: Int): Stream[Int] = n #:: (if (n == 1) Stream.empty else hailstone(if (n % 2 == 0) n / 2 else n * 3 + 1))
val nr = args.headOption.map(_.toInt).getOrElse(27)
val collatz = hailstone(nr)
println(s"Use the routine to show that the hailstone sequence for the number: $nr.")
println(collatz.toList)
println(s"It has ${collatz.length} elements.")
println
println(
"Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length.")
val (n, len) = (1 until 100000).map(n => (n, hailstone(n).length)).maxBy(_._2)
println(s"Longest hailstone sequence length= $len occurring with number $n.")
}</lang>
- Output:
Use the routine to show that the hailstone sequence for the number: 27. List(27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1) It has 112 elements. Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length. Longest hailstone sequence length= 351 occurring with number 77031.
Scheme
<lang scheme>(define (collatz n) (if (= n 1) '(1) (cons n (collatz (if (even? n) (/ n 2) (+ 1 (* 3 n)))))))
(define (collatz-length n) (let aux ((n n) (r 1)) (if (= n 1) r (aux (if (even? n) (/ n 2) (+ 1 (* 3 n))) (+ r 1)))))
(define (collatz-max a b) (let aux ((i a) (j 0) (k 0)) (if (> i b) (list j k) (let ((h (collatz-length i))) (if (> h k) (aux (+ i 1) i h) (aux (+ i 1) j k))))))
(collatz 27)
- (27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182
- 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395
- 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283
- 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429
- 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154
- 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35
- 106 53 160 80 40 20 10 5 16 8 4 2 1)
(collatz-length 27)
- 112
(collatz-max 1 100000)
- (77031 351)</lang>
Scilab
<lang>function x=hailstone(n)
// iterative definition
// usage: global verbose; verbose=%T; hailstone(27)
global verbose
x=0; loop=%T
while(loop)
x=x+1
if verbose then
printf('%i ',n)
end
if n==1 then
loop=%F
elseif modulo(n,2)==1 then
n=3*n+1
else
n=n/2
end
end
endfunction
global verbose; verbose=1; N=hailstone(27); printf('\n\n%i\n',N);
global verbose; verbose=0; N=100000; M=zeros(N,1); for k=1:N
M(k)=hailstone(k);
end; [maxLength,n]=max(M)</lang>
- Output:
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 112 n = 77031. maxLength = 351.
Seed7
<lang seed7>$ include "seed7_05.s7i";
const func array integer: hailstone (in var integer: n) is func
result
var array integer: hSequence is 0 times 0;
begin
while n <> 1 do
hSequence &:= n;
if odd(n) then
n := 3 * n + 1;
else
n := n div 2;
end if;
end while;
hSequence &:= n;
end func;
const func integer: hailstoneSequenceLength (in var integer: n) is func
result
var integer: sequenceLength is 1;
begin
while n <> 1 do
incr(sequenceLength);
if odd(n) then
n := 3 * n + 1;
else
n := n div 2;
end if;
end while;
end func;
const proc: main is func
local
var integer: number is 0;
var integer: length is 0;
var integer: maxLength is 0;
var integer: numberOfMaxLength is 0;
var array integer: h27 is 0 times 0;
begin
for number range 1 to 99999 do
length := hailstoneSequenceLength(number);
if length > maxLength then
maxLength := length;
numberOfMaxLength := number;
end if;
end for;
h27 := hailstone(27);
writeln("hailstone(27):");
for number range 1 to 4 do
write(h27[number] <& ", ");
end for;
write("....");
for number range length(h27) -3 to length(h27) do
write(", " <& h27[number]);
end for;
writeln(" length=" <& length(h27));
writeln("Maximum length " <& maxLength <& " at number=" <& numberOfMaxLength);
end func;</lang>
- Output:
hailstone(27): 27, 82, 41, 124, ...., 8, 4, 2, 1 length=112 Maximum length 351 at number=77031
Sidef
<lang ruby>func hailstone (n) {
var sequence = [n]
while (n > 1) {
sequence << (
n.is_even ? n.div!(2)
: n.mul!(3).add!(1)
)
}
return(sequence)
}
- The hailstone sequence for the number 27
var arr = hailstone(var nr = 27) say "#{nr}: #{arr.first(4)} ... #{arr.last(4)} (#{arr.len})"
- The longest hailstone sequence for a number less than 100,000
var h = [0, 0] for i (1 .. 99_999) {
(var l = hailstone(i).len) > h[1] && (
h = [i, l]
)
} printf("%d: (%d)\n", h...)</lang>
Smalltalk
<lang smalltalk>Object subclass: Sequences [
Sequences class >> hailstone: n [
|seq|
seq := OrderedCollection new.
seq add: n.
(n = 1) ifTrue: [ ^seq ].
(n even) ifTrue: [ seq addAll: (Sequences hailstone: (n / 2)) ]
ifFalse: [ seq addAll: (Sequences hailstone: ( (3*n) + 1 ) ) ].
^seq.
]
Sequences class >> hailstoneCount: n [
^ (Sequences hailstoneCount: n num: 1)
]
"this 'version' avoids storing the sequence, it just counts
its length - no memoization anyway"
Sequences class >> hailstoneCount: n num: m [
(n = 1) ifTrue: [ ^m ].
(n even) ifTrue: [ ^ Sequences hailstoneCount: (n / 2) num: (m + 1) ]
ifFalse: [ ^ Sequences hailstoneCount: ( (3*n) + 1) num: (m + 1) ].
]
].</lang>
<lang smalltalk>|r| r := Sequences hailstone: 27. "hailstone 'from' 27" (r size) displayNl. "its length"
"test 'head' ..." ( (r first: 4) = #( 27 82 41 124 ) asOrderedCollection ) displayNl.
"... and 'tail'" ( ( (r last: 4 ) ) = #( 8 4 2 1 ) asOrderedCollection) displayNl.
|longest| longest := OrderedCollection from: #( 1 1 ). 2 to: 100000 do: [ :c |
|l|
l := Sequences hailstoneCount: c.
(l > (longest at: 2) ) ifTrue: [ longest replaceFrom: 1 to: 2 with: { c . l } ].
].
('Sequence generator %1, sequence length %2' % { (longest at: 1) . (longest at: 2) })
displayNl.</lang>
SNUSP
/@+@@@+++# 27
| halve odd /===count<<\ /recurse\ #/?\ zero
$>@/===!/===-?\==>?!/-<+++\ \!/=!\@\>?!\@/<@\.!\-/
/+<-\!>\?-<+>/++++<\?>+++/*6+4 | | \=/ \=itoa=@@@+@+++++#
\=>?/<=!=\ | | ! /+ !/+ !/+ !/+ \ mod10
|//!==/========\ | /<+> -\!?-\!?-\!?-\!?-\!
/=>?\<=/\<+>!\->+>+<<?/>>=print@/\ln \?!\-?!\-?!\-?!\-?!\-?/\ div10
\+<-/!< ----------.++++++++++/ # +/! +/! +/! +/! +/
Swift
<lang Swift> func hailstone(var n:Int) -> [Int] {
var arr = [n]
while n != 1 {
if n % 2 == 0 {
n /= 2
} else {
n = (3 * n) + 1
}
arr.append(n) } return arr
}
let n = hailstone(27)
println("hailstone(27): \(n[0...3]) ... \(n[n.count-4...n.count-1]) for a count of \(n.count).")
var longest = (n: 1, len: 1)
for i in 1...100_000 {
let new = hailstone(i)
if new.count > longest.len {
longest = (i, new.count)
}
}
println("Longest sequence for numbers under 100,000 is with \(longest.n). Which has \(longest.len) items.")</lang>
- Output:
hailstone(27): [27, 82, 41, 124] ... [8, 4, 2, 1] for a count of 112 Longest sequence for numbers under 100,000 is with 77031. Which has 351 items.
Tcl
The core looping structure is an example of an n-plus-one-half loop, except the loop is officially infinite here. <lang tcl>proc hailstone n {
while 1 {
lappend seq $n if {$n == 1} {return $seq} set n [expr {$n & 1 ? $n*3+1 : $n/2}]
}
}
set h27 [hailstone 27] puts "h27 len=[llength $h27]" puts "head4 = [lrange $h27 0 3]" puts "tail4 = [lrange $h27 end-3 end]"
set maxlen [set max 0] for {set i 1} {$i<100000} {incr i} {
set l [llength [hailstone $i]]
if {$l>$maxlen} {set maxlen $l;set max $i}
} puts "max is $max, with length $maxlen"</lang>
- Output:
h27 len=112 head4 = 27 82 41 124 tail4 = 8 4 2 1 max is 77031, with length 351
TI-83 BASIC
Task 1
<lang ti83b>prompt N N→M: 0→X: 1→L While L=1 X+1→X Disp M If M=1 Then: 0→L Else If remainder(M,2)=1 Then: 3*M+1→M Else: M/2→M End End End {N,X}</lang>
- Output:
10
5
16
8
4
2
1
{27,112}
Task 2
As the calculator is quite slow, so the output is for N=200 <lang ti83b>prompt N 0→A:0→B for(I,1,N) I→M: 0→X: 1→L While L=1 X+1→X If M=1 Then: 0→L Else If remainder(M,2)=1 Then: 3*M+1→M Else: M/2→M End End End If X>B: Then I→A:X→B End Disp {I,X} End {A,B}</lang>
- Output:
{171,125}
TXR
<lang txr>@(do (defun hailstone (n)
(cons n
(gen (not (eq n 1))
(set n (if (evenp n)
(trunc n 2)
(+ (* 3 n) 1)))))))
@(next :list @(mapcar* (fun tostring) (hailstone 27))) 27 82 41 124 @(skip) 8 4 2 1 @(eof) @(do (let ((max 0) maxi)
(each* ((i (range 1 99999))
(h (mapcar* (fun hailstone) i))
(len (mapcar* (fun length) h)))
(if (> len max)
(progn
(set max len)
(set maxi i))))
(format t "longest sequence is ~a for n = ~a\n" max maxi)))</lang>
$ txr -l hailstone.txr longest sequence is 351 for n = 77031
uBasic/4tH
<lang>' ------=< 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
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@)</lang> uBasic is an interpreted language. Doing a sequence up to 100,000 would take over an hour, so we did up to 10,000 here.
- Output:
sequence for number 2727 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 1sequence length = 112
--------------------------------------------------------------------------------
The longest sequence is for 6171, it has a sequence length of 262
UNIX Shell
The best way is to use a shell with built-in arrays and arithmetic, such as Bash.
<lang bash>#!/bin/bash
- seq is the array genereated by hailstone
- index is used for seq
declare -a seq declare -i index
- Create a routine to generate the hailstone sequence for a number
hailstone () {
unset seq index seq[$((index++))]=$((n=$1)) while [ $n -ne 1 ]; do [ $((n % 2)) -eq 1 ] && ((n=n*3+1)) || ((n=n/2)) seq[$((index++))]=$n done
}
- Use the routine to show that the hailstone sequence for the number 27
- has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1
i=27 hailstone $i echo "$i: ${#seq[@]}" echo "${seq[@]:0:4} ... ${seq[@]:(-4):4}"
- Show the number less than 100,000 which has the longest hailstone
- sequence together with that sequences length.
- (But don't show the actual sequence)!
max=0 maxlen=0 for ((i=1;i<100000;i++)); do
hailstone $i
if [ $((len=${#seq[@]})) -gt $maxlen ]; then
max=$i
maxlen=$len
fi
done
echo "${max} has a hailstone sequence length of ${maxlen}"</lang>
- Output:
27: 112 27 82 41 124 ... 8 4 2 1 77031 has a hailstone sequence of 351
Bourne Shell
This script follows tradition for the Bourne Shell; its hailstone() function writes the sequence to standard output, so the shell can capture or pipe this output. This script is very slow because it forks many processes. Each `command substitution` forks a subshell, and each expr(1) command forks a process.
- Therefore, this script only examines sequences from 1 to 1000, not 100000. A fast computer might run this script in 45 to 120 seconds, using most time to run system calls in kernel mode. If the script went to 100000, it would need several hours.
<lang bash># Outputs a hailstone sequence from $1, with one element per line.
- Clobbers $n.
hailstone() { n=`expr "$1" + 0` eval "test $? -lt 2 || return $?" # $n must be integer.
echo $n while test $n -ne 1; do if expr $n % 2 >/dev/null; then n=`expr 3 \* $n + 1` else n=`expr $n / 2` fi echo $n done }
set -- `hailstone 27` echo "Hailstone sequence from 27 has $# elements:" first="$1, $2, $3, $4" shift `expr $# - 4` echo " $first, ..., $1, $2, $3, $4"
i=1 max=0 maxlen=0 while test $i -lt 1000; do len=`hailstone $i | wc -l | tr -d ' '` test $len -gt $maxlen && max=$i maxlen=$len i=`expr $i + 1` done echo "Hailstone sequence from $max has $maxlen elements."</lang>
C Shell
This script is several times faster than the previous Bourne Shell script, because it uses C Shell expressions, not the expr(1) command. This script is slow, but it can reach 100000, and a fast computer might run it in less than 15 minutes.
<lang csh># Outputs a hailstone sequence from !:1, with one element per line.
- Clobbers $n.
alias hailstone eval \@ n = \!:1:q \\ echo $n \\ while ( $n != 1 ) \\ if ( $n % 2 ) then \\ @ n = 3 * $n + 1 \\ else \\ @ n /= 2 \\ endif \\ echo $n \\ end \\ '\'
set sequence=(`hailstone 27`) echo "Hailstone sequence from 27 has $#sequence elements:" @ i = $#sequence - 3 echo " $sequence[1-4] ... $sequence[$i-]"
- hailstone-length $i
- acts like
- @ len = `hailstone $i | wc -l | tr -d ' '`
- but without forking any subshells.
alias hailstone-length eval \@ n = \!:1:q \\ @ len = 1 \\ while ( $n != 1 ) \\ if ( $n % 2 ) then \\ @ n = 3 * $n + 1 \\ else \\ @ n /= 2 \\ endif \\ @ len += 1 \\ end \\ '\'
@ i = 1 @ max = 0 @ maxlen = 0 while ($i < 100000) # XXX - I must run hailstone-length in a subshell, because my # C Shell has a bug when it runs hailstone-length inside this # while ($i < 1000) loop: it forgets about this loop, and # reports an error <<end: Not in while/foreach.>> @ len = `hailstone-length $i; echo $len` if ($len > $maxlen) then @ max = $i @ maxlen = $len endif @ i += 1 end echo "Hailstone sequence from $max has $maxlen elements."</lang>
- Output:
$ csh -f hailstone.csh Hailstone sequence from 27 has 112 elements: 27 82 41 124 ... 8 4 2 1 Hailstone sequence from 77031 has 351 elements.
Ursa
Implementation
hailstone.u <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</lang>
Usage
- Output:
> import "hailstone.u"
> out (hailstone 27) endl console
class java.lang.Integer<27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1>
> out (size (hailstone 27)) endl console
112
> decl int i max maxLoc
> for (set i 1) (< i 100000) (inc i)
.. decl int result
.. set result (size (hailstone i))
..
.. if (> result max)
.. set max result
.. set maxLoc i
.. end if
..end for
> out "hailstone(" maxLoc ")= " max endl console
hailstone(77031)= 351
> _
Ursala
<lang Ursala>#import std
- import nat
hail = @iNC ~&h~=1->x ^C\~& @h ~&h?\~&t successor+ sum@iNiCX
- show+
main =
<
^T(@ixX take/$4; %nLP~~lrxPX; ^|TL/~& :/'...',' has length '--@h+ %nP+ length) hail 27, ^|TL(~&,:/' has sequence length ') %nP~~ nleq$^&r ^(~&,length+ hail)* nrange/1 100000></lang>
The hail function computes the sequence as follows.
- Given a number as an argument,
@iNCmakes a list containing only that number before passing it to the rest of the function. Theiin the expression stands for the identity function,Nfor the constant null function, andCfor the cons operator. - The iteration combinator (
->) is used with a predicate of~&h~=lwhich tests the condition that the head (~&h) of its argument is not equal (~=) to 1. Iteration of the rest of the function continues while this predicate holds. - The
xsuffix says to return the reversal of the list after the iteration finishes. - The function being iterated builds a list using the cons operator (
^C) with the identity function (~&) of the argument for the tail, and the result of the rest of the line for the head. - The
@hoperator says that the function following will be applied to the head of the list. - The conditional operator (
?) has the head function (~&h) as its predicate, which tests whether the head of its argument is non-null. - In this case, the argument is a natural number, but naturals are represented as lists of booleans, so taking the head of a number is the same as testing the least significant bit.
- If the condition is not met, the number has a 0 least significant bit, and therefore is even. In this case, the conditional predicate calls for taking its tail (
~&t), effectively dividing it by 2 using a bit shift. - If the condition is met, the number is odd, so the rest of the function computes the successor of the number multiplied by three.
- Rather than multiplying the hard way, the function
sum@iNiCXcomputes the sum of the pair (X) of numbers given by the identity function (i) of the argument, and the doubling of the argument (NiC), also obtained by a bit shift, with a zero bit (N) consed (C) with the identity (i).
Most of the main expression pertains to less interesting printing and formatting, but the part that searches for the longest sequence in the range is nleq$^&r ^(~&,length+ hail)* nrange/1 100000.
- The expression
nrange/1 100000evaluates to the list of the first 100000 positive integers. - The map operator (
*) causes a list to be made of the results of its operand applied to each number. - The operand to the map operator, applied to an individual number in the list, constructs a pair (
^) with the identity function (~&) of the number on the left, and the length of thehailsequence on the right. - The maximizing operator (
$^) with respect to the natural less or equal relation (nleq) applied to the right sides (&r) of its pair of arguments extracts the number with the maximum length sequence.
- Output:
<27,82,41,124>...<8,4,2,1> has length 112 77031 has sequence length 351
VBA
<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</lang>
- Output:
hailstone(27) = 27, 82, 41, 124, ... 16, 8, 4, 2, 1length = 112
The longest hailstone sequence under 100,000 is 77031 with 351 elements.
VBScript
<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 </lang>
- Output:
Sequence for 27: 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 Number of Elements: 112 Number less than 100k with the longest sequence: Number: 77031 Length: 351
Visual Basic
<lang vb>Option Explicit Dim flag As Boolean ' true to print values Sub main()
Dim longest As Long, n As Long
Dim i As Long, value As Long
' Task 1:
flag = True
i = 27
Debug.Print "The hailstone sequence has length of "; i; " is "; hailstones(i)
' Task 2:
flag = False
longest = 0
For i = 1 To 99999
If longest < hailstones(i) Then
longest = hailstones(i)
value = i
End If
Next i
Debug.Print value; " has the longest sequence of "; longest
End Sub 'main Function hailstones(n As Long) As Long
Dim m As Long, p As Long
Dim m1 As Long, m2 As Long, m3 As Long, m4 As Long
If flag Then Debug.Print "The sequence for"; n; "is: ";
p = 1
m = n
If flag Then Debug.Print m;
While m > 1
p = p + 1
If (m Mod 2) = 0 Then
m = m / 2
Else
m = 3 * m + 1
End If
If p <= 4 Then If flag Then Debug.Print m;
m4 = m3
m3 = m2
m2 = m1
m1 = m
Wend
If flag Then
If p <= 4 Then
Debug.Print
ElseIf p = 5 Then
Debug.Print m1
ElseIf p = 6 Then
Debug.Print m2; m1
ElseIf p = 7 Then
Debug.Print m3; m2; m1
ElseIf p = 8 Then
Debug.Print m4; m3; m2; m1
Else
Debug.Print "..."; m4; m3; m2; m1
End If
End If
hailstones = p
End Function 'hailstones</lang>
- Output:
The sequence for 27 is: 27 82 41 124 ... 8 4 2 1 The hailstone sequence has length of 27 is 112 77031 has the longest sequence of 351
Visual Basic .NET
<lang vbnet>Module HailstoneSequence
Sub Main()
' Checking sequence of 27.
Dim l As List(Of Long) = HailstoneSequence(27)
Console.WriteLine("27 has {0} elements in sequence:", l.Count())
For i As Integer = 0 To 3 : Console.Write("{0}, ", l(i)) : Next
Console.Write("... ")
For i As Integer = l.Count - 4 To l.Count - 1 : Console.Write(", {0}", l(i)) : Next
Console.WriteLine()
' Finding longest sequence for numbers below 100000.
Dim max As Integer = 0
Dim maxCount As Integer = 0
For i = 1 To 99999
l = HailstoneSequence(i)
If l.Count > maxCount Then
max = i
maxCount = l.Count
End If
Next
Console.WriteLine("Max elements in sequence for number below 100k: {0} with {1} elements.", max, maxCount)
Console.ReadLine()
End Sub
Private Function HailstoneSequence(ByVal n As Long) As List(Of Long)
Dim valList As New List(Of Long)()
valList.Add(n)
Do Until n = 1
n = IIf(n Mod 2 = 0, n / 2, (3 * n) + 1)
valList.Add(n)
Loop
Return valList End Function
End Module</lang>
- Output:
27 has 112 elements in sequence: 27, 82, 41, 124, ... , 8, 4, 2, 1 Max elements in sequence for number below 100k: 77031 with 351 elements.
Wren
<lang ecmascript>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)")</lang>
- Output:
For the Hailstone sequence starting with n = 27: Number of elements = 112 First four elements = [27, 82, 41, 124] Final four elements = [8, 4, 2, 1] The Hailstone sequence for n < 100,000 with the longest length is: Longest = 77031 Length = 351
XPL0
<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations int Seq(1000); \more than enough for longest sequence
func Hailstone(N); \Return length of Hailstone sequence starting at N int N; \ also fills Seq array with sequence int I; [I:= 0; loop [Seq(I):= N; I:= I+1;
if N=1 then return I;
N:= if N&1 then N*3+1 else N/2;
];
];
int N, SN, Len, MaxLen; [Len:= Hailstone(27); Text(0, "27's Hailstone length = "); IntOut(0, Len); CrLf(0);
Text(0, "Sequence = "); for N:= 0 to 3 do [IntOut(0, Seq(N)); ChOut(0, ^ )]; Text(0, "... "); for N:= Len-4 to Len-1 do [IntOut(0, Seq(N)); ChOut(0, ^ )]; CrLf(0);
MaxLen:= 0; for N:= 1 to 100_000-1 do
[Len:= Hailstone(N); if Len > MaxLen then [MaxLen:= Len; SN:= N]; \save N with max length ];
IntOut(0, SN); Text(0, "'s Hailstone length = "); IntOut(0, MaxLen); ]</lang>
- Output:
27's Hailstone length = 112 Sequence = 27 82 41 124 ... 8 4 2 1 77031's Hailstone length = 351
zkl
<lang zkl>fcn collatz(n,z=L()){ z.append(n); if(n==1) return(z);
if(n.isEven) return(self.fcn(n/2,z)); return(self.fcn(n*3+1,z)) }</lang>
This uses tail recursion and thus is stack efficient.
- Output:
var n=collatz(27) n.len() 112 n[0,4] L(27,82,41,124) n[-4,*] L(8,4,2,1)
Rather than write a function that calculates the length, just roll through all 100,000 sequences and save the largest (length,sequence start) pair. Creating all those Collatz lists isn't quick. This works by using a [mutable] list to hold state as the pump does the basic looping. <lang zkl>[2..0d100_000].pump(Void, // loop n from 2 to 100,000
collatz, // generate Collatz sequence(n)
fcn(c,n){ // if new longest sequence, save length/C, return longest
if(c.len()>n[0]) n.clear(c.len(),c[0]); n}.fp1(L(0,0)))</lang>
- Output:
L(351,77031) // length, hailstone
ZX Spectrum Basic
<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</lang>
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