Fractran
You are encouraged to solve this task according to the task description, using any language you may know.
FRACTRAN is a Turing-complete esoteric programming language invented by the mathematician John Horton Conway.
A FRACTRAN program is an ordered list of positive fractions , together with an initial positive integer input .
The program is run by updating the integer as follows:
- for the first fraction, , in the list for which is an integer, replace with ;
- repeat this rule until no fraction in the list produces an integer when multiplied by , then halt.
Conway gave a program for primes in FRACTRAN:
- , , , , , , , , , , , , ,
Starting with , this FRACTRAN program will change to , then , generating the following sequence of integers:
- , , , , , , , , , , ,
After 2, this sequence contains the following powers of 2:
- , , , , , , , ,
which are the prime powers of 2.
- Task
Write a program that reads a list of fractions in a natural format from the keyboard or from a string, to parse it into a sequence of fractions (i.e. two integers), and runs the FRACTRAN starting from a provided integer, writing the result at each step. It is also required that the number of step is limited (by a parameter easy to find).
- Extra credit
Use this program to derive the first 20 or so prime numbers.
- See also
For more on how to program FRACTRAN as a universal programming language, see:
- J. H. Conway (1987). Fractran: A Simple Universal Programming Language for Arithmetic. In: Open Problems in Communication and Computation, pages 4–26. Springer.
- J. H. Conway (2010). "FRACTRAN: A simple universal programming language for arithmetic". In Jeffrey C. Lagarias. The Ultimate Challenge: the 3x+1 problem. American Mathematical Society. pp. 249–264. ISBN 978-0-8218-4940-8. Zbl 1216.68068.
- Number Pathology: Fractran by Mark C. Chu-Carroll; October 27, 2006.
Ada
<lang Ada>with Ada.Text_IO;
procedure Fractan is
type Fraction is record Nom: Natural; Denom: Positive; end record;
type Frac_Arr is array(Positive range <>) of Fraction;
function "/" (N: Natural; D: Positive) return Fraction is
Frac: Fraction := (Nom => N, Denom => D);
begin
return Frac;
end "/";
procedure F(List: Frac_Arr; Start: Positive; Max_Steps: Natural) is
N: Positive := Start;
J: Positive;
begin
Ada.Text_IO.Put(" 0:" & Integer'Image(N) & " ");
for I in 1 .. Max_Steps loop
J := List'First; loop if N mod List(J).Denom = 0 then N := (N/List(J).Denom) * List(J).Nom; exit; -- found fraction elsif J >= List'Last then return; -- did try out all fractions else J := J + 1; -- try the next fraction end if; end loop; Ada.Text_IO.Put(Integer'Image(I) & ":" & Integer'Image(N) & " ");
end loop; end F;
begin
-- F((2/3, 7/2, 1/5, 1/7, 1/9, 1/4, 1/8), 2, 100); -- output would be "0: 2 1: 7 2: 1" and then terminate
F((17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23,
77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1),
2, 15);
-- output is "0: 2 1: 15 2: 825 3: 725 ... 14: 132 15: 116"
end Fractan;</lang>
- Output:
0: 2 1: 15 2: 825 3: 725 4: 1925 5: 2275 6: 425 7: 390 8: 330 9: 290 10: 770 11: 910 12: 170 13: 156 14: 132 15: 116
ALGOL 68
<lang algol68># as the numbers required for finding the first 20 primes are quite large, #
- we use Algol 68G's LONG LONG INT with a precision of 100 digits #
PR precision 100 PR
- mode to hold fractions #
MODE FRACTION = STRUCT( INT numerator, INT denominator );
- define / between two INTs to yield a FRACTION #
OP / = ( INT a, b )FRACTION: ( a, b );
- mode to define a FRACTRAN progam #
MODE FRACTRAN = STRUCT( FLEX[0]FRACTION data
, LONG LONG INT n
, BOOL halted
);
- prepares a FRACTRAN program for use - sets the initial value of n and halted to FALSE #
PRIO STARTAT = 1; OP STARTAT = ( REF FRACTRAN f, INT start )REF FRACTRAN: BEGIN
halted OF f := FALSE;
n OF f := start;
f
END;
- sets n OF f to the next number in the sequence or sets halted OF f to TRUE if the sequence has ended #
OP NEXT = ( REF FRACTRAN f )LONG LONG INT:
IF halted OF f
THEN n OF f := 0
ELSE
BOOL found := FALSE;
LONG LONG INT result := 0;
FOR pos FROM LWB data OF f TO UPB data OF f WHILE NOT found DO
LONG LONG INT value = n OF f * numerator OF ( ( data OF f )[ pos ] );
INT denominator = denominator OF ( ( data OF f )[ pos ] );
IF found := ( value MOD denominator = 0 ) THEN result := value OVER denominator FI
OD;
IF NOT found THEN halted OF f := TRUE FI;
n OF f := result
FI ;
- generate and print the sequence of numbers from a FRACTRAN pogram #
PROC print fractran sequence = ( REF FRACTRAN f, INT start, INT limit )VOID: BEGIN
VOID( f STARTAT start );
print( ( "0: ", whole( start, 0 ) ) );
FOR i TO limit
WHILE VOID( NEXT f );
NOT halted OF f
DO
print( ( " " + whole( i, 0 ) + ": " + whole( n OF f, 0 ) ) )
OD;
print( ( newline ) )
END ;
- print the first 16 elements from the primes FRACTRAN program #
FRACTRAN pf := ( ( 17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1 ), 0, FALSE ); print fractran sequence( pf, 2, 15 );
- find some primes using the pf FRACTRAN progam - n is prime for the members in the sequence that are 2^n #
INT primes found := 0; VOID( pf STARTAT 2 ); INT pos := 0; print( ( "seq position prime sequence value", newline ) ); WHILE primes found < 20 AND NOT halted OF pf DO
LONG LONG INT value := NEXT pf;
INT power of 2 := 0;
pos +:= 1;
WHILE value MOD 2 = 0 AND value > 0 DO power of 2 PLUSAB 1; value OVERAB 2 OD;
IF value = 1 THEN
# found a prime #
primes found +:= 1;
print( ( whole( pos, -12 ) + " " + whole( power of 2, -6 ) + " (" + whole( n OF pf, 0 ) + ")", newline ) )
FI
OD</lang>
- Output:
0: 2 1: 15 2: 825 3: 725 4: 1925 5: 2275 6: 425 7: 390 8: 330 9: 290 10: 770 11: 910 12: 170 13: 156 14: 132 15: 116
seq position prime sequence value
19 2 (4)
69 3 (8)
280 5 (32)
707 7 (128)
2363 11 (2048)
3876 13 (8192)
8068 17 (131072)
11319 19 (524288)
19201 23 (8388608)
36866 29 (536870912)
45551 31 (2147483648)
75224 37 (137438953472)
101112 41 (2199023255552)
117831 43 (8796093022208)
152025 47 (140737488355328)
215384 53 (9007199254740992)
293375 59 (576460752303423488)
327020 61 (2305843009213693952)
428553 67 (147573952589676412928)
507519 71 (2361183241434822606848)
AutoHotkey
<lang AutoHotkey>n := 2, steplimit := 15, numerator := [], denominator := [] s := "17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1"
Loop, Parse, s, % A_Space
if (!RegExMatch(A_LoopField, "^(\d+)/(\d+)$", m))
MsgBox, % "Invalid input string (" A_LoopField ")."
else
numerator[A_Index] := m1, denominator[A_Index] := m2
SetFormat, FloatFast, 0.0 Gui, Add, ListView, R10 W100 -Hdr, | SysGet, VSBW, 2 LV_ModifyCol(1, 95 - VSBW), LV_Add( , 0 ": " n) Gui, Show
Loop, % steplimit {
i := A_Index
Loop, % numerator.MaxIndex()
if (!Mod(nn := n * numerator[A_Index] / denominator[A_Index], 1)) {
LV_Modify(LV_Add( , i ": " (n := nn)), "Vis")
continue, 2
}
break
}</lang>
- Output:
0: 2 1: 15 2: 825 3: 725 4: 1925 5: 2275 6: 425 7: 390 8: 330 9: 290 10: 770 11: 910 12: 170 13: 156 14: 132 15: 116
Batch File
<lang dos>@echo off setlocal enabledelayedexpansion
::Set the inputs set "code=17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1" set "n=2"
::Basic validation of code for %%. in (!code!) do ( echo.%%.|findstr /r /c:"^[0-9][0-9]*/[1-9][0-9]*$">nul||goto error_code ) ::Validate the input set /a "tst=1*!n!" 2>nul if !tst! lss 0 goto error_input if !tst! equ 0 (if not "!n!"=="0" (goto error_input))
::Set the limit outputs set limit=20
::Execute the code echo.Input: echo. !n! echo.Output: for /l %%? in (1,1,!limit!) do ( set shouldwehalt=1 for %%A in (!code!) do ( for /f "tokens=1,2 delims=/" %%B in ("%%A") do ( set /a "tst=!n! %% %%C" if !tst! equ 0 ( if !shouldwehalt! equ 1 ( set shouldwehalt=0 set /a "n=n*%%B/%%C" echo. !n! ) ) ) ) if !shouldwehalt! equ 1 goto halt )
- halt
echo. pause exit /b 0
- error_code
echo.Syntax error in code. echo. pause exit /b 1
- error_input
echo.Invalid input. echo. pause exit /b 1</lang>
- Output:
Input:
2
Output:
15
825
725
1925
2275
425
390
330
290
770
910
170
156
132
116
308
364
68
4
30
Press any key to continue . . .
Bracmat
This program computes the first twenty primes. It has to do almost 430000 iterations to arrive at the twentieth prime, so instead of immediately writing each number to the terminal, it adds it to a list. After the set number of iterations, the list of numbers is written to a text file numbers.lst (21858548 bytes), so you can inspect it. Because it takes some time to do all iterations, its is advisable to write the source code below in a text file 'fractran' and run it in batch mode in the background, instead of starting Bracmat in interactive mode and typing the program at the prompt. The primes, together with the largest number found, are written to a file FRACTRAN.OUT. <lang bracmat>(fractran=
np n fs A Z fi P p N L M
. !arg:(?N,?n,?fs) {Number of iterations, start n, fractions}
& :?P:?L {Initialise accumulators.}
& whl
' ( -1+!N:>0:?N {Stop when counted down to zero.}
& !n !L:?L {Prepend all numbers to result list.}
& (2\L!n:#?p&!P !p:?P|) {If log2(n) is rational, append it to list of primes.}
& !fs:? (/?fi&!n*!fi:~/:?n) ? {This line does the following (See task description):
"for the first fraction, fi, in the list for which
nfi is an integer, replace n by nfi ;"}
)
& :?M
& whl'(!L:%?n ?L&!n !M:?M) {Invert list of numbers. (append to long list is
very expensive. Better to prepend and finally invert.}
& (!M,!P) {Return the two lists}
);
( clk$:?t0 & fractran$(430000, 2, 17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1)
: (?numbers,?primes)
& lst$(numbers,"numbers.lst",NEW) & put$(" FRACTRAN found these primes:"
!primes "\nThe list of numbers is saved in numbers.txt
The biggest number in the list is"
( 0:?max & !numbers:? (>%@!max:?max&~) ? | !max )
str$("\ntime: " flt$(clk$+-1*!t0,4) " sec\n") , "FRACTRAN.OUT",NEW) );</lang> In Linux, run the program as follows (assuming bracmat and the file 'fractran' are in the CWD):
./bracmat 'get$fractran' &
- Output:
in FRACTRAN.OUT
FRACTRAN found these primes: 1 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 The list of numbers is saved in numbers.txt The biggest number in the list is 1842775069354845065175076326808495219647033145169559640049770986129640260031692378106527030467987060546875 time: 1,8668*10E3 sec
C
Using GMP. Powers of two are in brackets.
For extra credit, pipe the output through | less -S.
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <gmp.h>
typedef struct frac_s *frac; struct frac_s { int n, d; frac next; };
frac parse(char *s) { int offset = 0; struct frac_s h = {0}, *p = &h;
while (2 == sscanf(s, "%d/%d%n", &h.n, &h.d, &offset)) { s += offset; p = p->next = malloc(sizeof *p); *p = h; p->next = 0; }
return h.next; }
int run(int v, char *s) { frac n, p = parse(s); mpz_t val; mpz_init_set_ui(val, v);
loop: n = p; if (mpz_popcount(val) == 1) gmp_printf("\n[2^%d = %Zd]", mpz_scan1(val, 0), val); else gmp_printf(" %Zd", val);
for (n = p; n; n = n->next) { // assuming the fractions are not reducible if (!mpz_divisible_ui_p(val, n->d)) continue;
mpz_divexact_ui(val, val, n->d); mpz_mul_ui(val, val, n->n); goto loop; }
gmp_printf("\nhalt: %Zd has no divisors\n", val);
mpz_clear(val); while (p) { n = p->next; free(p); p = n; }
return 0; }
int main(void) { run(2, "17/91 78/85 19/51 23/38 29/33 77/29 95/23 " "77/19 1/17 11/13 13/11 15/14 15/2 55/1");
return 0; }</lang>
C++
<lang cpp>
- include <iostream>
- include <sstream>
- include <iterator>
- include <vector>
- include <cmath>
using namespace std;
class fractran { public:
void run( std::string p, int s, int l )
{
start = s; limit = l;
istringstream iss( p ); vector<string> tmp;
copy( istream_iterator<string>( iss ), istream_iterator<string>(), back_inserter<vector<string> >( tmp ) );
string item; vector< pair<float, float> > v;
pair<float, float> a; for( vector<string>::iterator i = tmp.begin(); i != tmp.end(); i++ ) { string::size_type pos = ( *i ).find( '/', 0 ); if( pos != std::string::npos ) { a = make_pair( atof( ( ( *i ).substr( 0, pos ) ).c_str() ), atof( ( ( *i ).substr( pos + 1 ) ).c_str() ) ); v.push_back( a ); } }
exec( &v );
}
private:
void exec( vector< pair<float, float> >* v )
{
int cnt = 0; while( cnt < limit ) { cout << cnt << " : " << start << "\n"; cnt++; vector< pair<float, float> >::iterator it = v->begin(); bool found = false; float r; while( it != v->end() ) { r = start * ( ( *it ).first / ( *it ).second ); if( r == floor( r ) ) { found = true; break; } ++it; }
if( found ) start = ( int )r; else break; }
} int start, limit;
}; int main( int argc, char* argv[] ) {
fractran f; f.run( "17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2, 15 ); cin.get(); return 0;
} </lang>
- Output:
0 : 2 1 : 15 2 : 825 3 : 725 4 : 1925 5 : 2275 6 : 425 7 : 390 8 : 330 9 : 290 10 : 770 11 : 910 12 : 170 13 : 156 14 : 132
Common Lisp
<lang lisp>(defun fractran (n frac-list)
(lambda ()
(prog1
n
(when n
(let ((f (find-if (lambda (frac)
(integerp (* n frac)))
frac-list)))
(when f (setf n (* f n))))))))
- test
(defvar *primes-ft* '(17/91 78/85 19/51 23/38 29/33 77/29 95/23
77/19 1/17 11/13 13/11 15/14 15/2 55/1))
(loop with fractran-instance = (fractran 2 *primes-ft*)
repeat 20
for next = (funcall fractran-instance)
until (null next)
do (print next))</lang>
- Output:
2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116 308 364 68 4
D
Simple Version
<lang d>import std.stdio, std.algorithm, std.conv, std.array;
void fractran(in string prog, int val, in uint limit) {
const fracts = prog.split.map!(p => p.split("/").to!(int[])).array;
foreach (immutable n; 0 .. limit) {
writeln(n, ": ", val);
const found = fracts.find!(p => val % p[1] == 0);
if (found.empty)
break;
val = found.front[0] * val / found.front[1];
}
}
void main() {
fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23
77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2, 15);
}</lang>
- Output:
0: 2 1: 15 2: 825 3: 725 4: 1925 5: 2275 6: 425 7: 390 8: 330 9: 290 10: 770 11: 910 12: 170 13: 156 14: 132
Lazy Version
<lang d>import std.stdio, std.algorithm, std.conv, std.array, std.range;
struct Fractran {
int front; bool empty = false; const int[][] fracts;
this(in string prog, in int val) {
this.front = val;
fracts = prog.split.map!(p => p.split("/").to!(int[])).array;
}
void popFront() {
const found = fracts.find!(p => front % p[1] == 0);
if (found.empty)
empty = true;
else
front = found.front[0] * front / found.front[1];
}
}
void main() {
Fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23
77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2)
.take(15).writeln;
}</lang>
- Output:
[2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132]
Elixir
<lang elixir>defmodule Fractran do
use Bitwise
defp binary_to_ratio(b) do
[_, num, den] = Regex.run(~r/(\d+)\/(\d+)/, b)
{String.to_integer(num), String.to_integer(den)}
end
def load(program) do
String.split(program) |> Enum.map(&binary_to_ratio(&1))
end
defp step(_, []), do: :halt
defp step(n, [f|fs]) do
{p, q} = mulrat(f, {n, 1})
case q do
1 -> p
_ -> step(n, fs)
end
end
def exec(k, n, program) do
exec(k-1, n, fn (_) -> true end, program, [n]) |> Enum.reverse
end
def exec(k, n, pred, program) do
exec(k-1, n, pred, program, [n]) |> Enum.reverse
end
defp exec(0, _, _, _, steps), do: steps
defp exec(k, n, pred, program, steps) do
case step(n, program) do
:halt -> steps
m -> if pred.(m), do: exec(k-1, m, pred, program, [m|steps]),
else: exec(k, m, pred, program, steps)
end
end
def is_pow2(n), do: band(n, n-1) == 0
def lowbit(n), do: lowbit(n, 0)
defp lowbit(n, k) do
case band(n, 1) do
0 -> lowbit(bsr(n, 1), k + 1)
1 -> k
end
end
# rational multiplication
defp mulrat({a, b}, {c, d}) do
{p, q} = {a*c, b*d}
g = gcd(p, q)
{div(p, g), div(q, g)}
end
defp gcd(a, 0), do: a
defp gcd(a, b), do: gcd(b, rem(a, b))
end
primegen = Fractran.load("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1") IO.puts "The first few states of the Fractran prime automaton are:\n#{inspect Fractran.exec(20, 2, primegen)}\n" prime = Fractran.exec(26, 2, &Fractran.is_pow2/1, primegen)
|> Enum.map(&Fractran.lowbit/1)
|> tl
IO.puts "The first few primes are:\n#{inspect prime}"</lang>
- Output:
The first few states of the Fractran prime automaton are: [2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132, 116, 308, 364, 68, 4] The first few primes are: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
Erlang
The exec() function can be passed a predicate which filters steps that satisfy a condition, which for the prime automata is a check to see if the number is a power of 2. <lang erlang>#! /usr/bin/escript
-mode(native). -import(lists, [map/2, reverse/1]).
binary_to_ratio(B) ->
{match, [_, Num, Den]} = re:run(B, "([0-9]+)/([0-9]+)"),
{binary_to_integer(binary:part(B, Num)),
binary_to_integer(binary:part(B, Den))}.
load(Program) ->
map(fun binary_to_ratio/1, re:split(Program, "[ ]+")).
step(_, []) -> halt; step(N, [F|Fs]) ->
{P, Q} = mulrat(F, {N, 1}),
case Q of
1 -> P;
_ -> step(N, Fs)
end.
exec(K, N, Program) -> reverse(exec(K - 1, N, fun (_) -> true end, Program, [N])). exec(K, N, Pred, Program) -> reverse(exec(K - 1, N, Pred, Program, [N])).
exec(0, _, _, _, Steps) -> Steps; exec(K, N, Pred, Program, Steps) ->
case step(N, Program) of
halt -> Steps;
M -> case Pred(M) of
true -> exec(K - 1, M, Pred, Program, [M|Steps]);
false -> exec(K, M, Pred, Program, Steps)
end
end.
is_pow2(N) -> N band (N - 1) =:= 0.
lowbit(N) -> lowbit(N, 0). lowbit(N, K) ->
case N band 1 of
0 -> lowbit(N bsr 1, K + 1);
1 -> K
end.
main(_) ->
PrimeGen = load("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1"),
io:format("The first few states of the Fractran prime automaton are: ~p~n~n", [exec(20, 2, PrimeGen)]),
io:format("The first few primes are: ~p~n", [tl(map(fun lowbit/1, exec(26, 2, fun is_pow2/1, PrimeGen)))]).
% rational multiplication
mulrat({A, B}, {C, D}) ->
{P, Q} = {A*C, B*D},
G = gcd(P, Q),
{P div G, Q div G}.
gcd(A, 0) -> A; gcd(A, B) -> gcd(B, A rem B). </lang>
- Output:
└─ $ ▶ ./fractran.erl
The first few states of the Fractran prime automaton are: [2,15,825,725,1925,
2275,425,390,330,
290,770,910,170,
156,132,116,308,
364,68,4]
The first few primes are: [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,
67,71,73,79,83,89,97]
Fortran
The Plan
As ever, how long is a piece of string: what might be the upper limit for the list of fractions? Similarly, obtaining the source code to be interpreted is a nuisance since as ever, one doesn't know how long it might be and performing a scan to find out will require more messing about than the activity of interest. One could compile in the specified example, but this is not very flexible even though a re-compilation will be swift. So, a pre-emptive requirement: first specify the number of fractions to be read in. Then reading the fractions has its annoyances also (aside from copying from the example text carrying along all the .html formatting) since given 2/3 a slash is not a suitable delimiter for free-format input, indeed it signifies end-of-input. Again, one can devise an input procedure involving a suitable scan, with more code drowning the stuff specific to the task. So, replace the slashes with a space, an acceptable free-format delimiter. As a comma is also acceptable, use those between the fractions, so, 17 91, 78 85, 19 51, 23 38, 29 33, 77 29, 95 23, 77 19, 1 17, 11 13, 13 11, 15 14, 15 2, 55 1 Then, all that remains is to specify the starting number, and as well, a step limit. Naturally, the output can be in a more gracious form. It is presented as the input is read so that should something go awry there would be some indication of what was going on.
The Code
The source style is F77 except for the use of the I0 format code, though not all F77 compilers will offer INTEGER*8. By not using the MODULE scheme, array parameters can't be declared via P(:) which implies a secret additional parameter giving the size of the array and which can be accessed via the likes of UBOUND(P, DIM = 1) Instead, the old-style specification involves no additional parameters and can be given as P(*) meaning "no statement" as to the upper bound, or P(M) which may be interpreted as the upper bound being the value of M in the compilers that allow this. The actual upper bound of the parameter is unknown and unchecked, so the older style of P(12345) or similar might be used. Rather to my surprise, this compiler (Compaq F90/95) complained if parameter M was declared after the arrays P(M),Q(M) as it is my habit to declare parameters in the order of their appearance. <lang Fortran>C:\Nicky\RosettaCode\FRACTRAN\FRACTRAN.for(6) : Warning: This name has not been given an explicit type. [M]
INTEGER P(M),Q(M)!The terms of the fractions.</lang>
So much for multi-pass compilers!
Similarly, without the MODULE protocol, in all calling routines function FRACTRAN would be deemed floating-point so a type declaration is needed in each. <lang Fortran> INTEGER FUNCTION FRACTRAN(N,P,Q,M) !Notion devised by J. H. Conway. Careful: the rule is N*P/Q being integer. N*6/3 is integer always because this is N*2/1, but 3 may not divide N. Could check GCD(P,Q), dividing out the common denominator so MOD(N,Q) works.
INTEGER*8 N !The work variable. Modified!
INTEGER M !The number of fractions supplied.
INTEGER P(M),Q(M)!The terms of the fractions.
INTEGER I !A stepper.
DO I = 1,M !Search the supplied fractions, P(i)/Q(i).
IF (MOD(N,Q(I)).EQ.0) THEN !Does the denominator divide N?
N = N/Q(I)*P(I) !Yes, compute N*P/Q but trying to dodge overflow.
FRACTRAN = I !Report the hit.
RETURN !Done!
END IF !Otherwise,
END DO !Try the next fraction in the order supplied.
FRACTRAN = 0 !No hit.
END FUNCTION FRACTRAN !That's it! Even so, "Turing complete"...
PROGRAM POKE
INTEGER FRACTRAN !Not the default type of function.
INTEGER P(66),Q(66) !Holds the fractions as P(i)/Q(i).
INTEGER*8 N !The working number.
INTEGER I,IT,L,M !Assistants.
WRITE (6,1) !Announce.
1 FORMAT ("Interpreter for J.H. Conway's FRACTRAN language.")
Chew into an example programme.
OPEN (10,FILE = "Fractran.txt",STATUS="OLD",ACTION="READ") !Rather than compiled-in stuff.
READ (10,*) L !I need to know this without having to scan the input.
WRITE (6,2) L !Reveal in case of trouble.
2 FORMAT (I0," fractions, as follow:") !Should the input evoke problems.
READ (10,*) (P(I),Q(I),I = 1,L) !Ask for the specified number of P,Q pairs.
WRITE (6,3) (P(I),Q(I),I = 1,L) !Show what turned up.
3 FORMAT (24(I0,"/",I0:", ")) !As P(i)/Q(i) pairs. The colon means that there will be no trailing comma.
READ (10,*) N,M !The start value, and the step limit.
CLOSE (10) !Finished with input.
WRITE (6,4) N,M !Hopefully, all went well.
4 FORMAT ("Start with N = ",I0,", step limit ",I0)
Commence.
WRITE (6,10) 0,N !Splat a heading.
10 FORMAT (/," Step #F: N",/,I6,4X,": ",I0) !Matched FORMAT 11.
DO I = 1,M !Here we go!
IT = FRACTRAN(N,P,Q,L) !Do it!
WRITE (6,11) I,IT,N !Show it!
11 FORMAT (I6,I4,": ",I0) !N last, as it may be big.
IF (IT.LE.0) EXIT !No hit, so quit.
END DO !The next step.
END !Whee!
</lang>
The Results
Output:
Interpreter for J.H. Conway's FRACTRAN language.
14 fractions, as follow:
17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1
Start with N = 2, step limit 28
Step #F: N
0 : 2
1 13: 15
2 14: 825
3 5: 725
4 6: 1925
5 11: 2275
6 1: 425
7 2: 390
8 10: 330
9 5: 290
10 6: 770
11 11: 910
12 1: 170
13 2: 156
14 10: 132
15 5: 116
16 6: 308
17 11: 364
18 1: 68
19 9: 4
20 13: 30
21 13: 225
22 14: 12375
23 5: 10875
24 6: 28875
25 5: 25375
26 6: 67375
27 11: 79625
28 1: 14875
Later Fortrans might offer the library function POPCNT(n) which returns the number of on-bits in an integer, most convenient for detecting a straight power of two in a binary computer. Adjusting the interpretation loop to be <lang Fortran> DO I = 1,M !Here we go!
IT = FRACTRAN(N,P,Q,L) !Do it!
IF (POPCNT(N).EQ.1) WRITE (6,11) I,IT,N !Show it!
11 FORMAT (I6,I4,": ",I0) !N last, as it may be big.
IF (IT.LE.0) EXIT !No hit, so quit.
IF (N.LE.0) THEN !Otherwise, worry about overflow.
WRITE (6,*) "Integer overflow!" !Justified. The test is not certain.
WRITE (6,11) I,IT,N !Alas, the step failed.
EXIT !Give in.
END IF !So much for overflow.
END DO !The next step.
</lang> Leads to the following output:
Interpreter for J.H. Conway's FRACTRAN language.
14 fractions, as follow:
17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1
Start with N = 2, step limit 246810
Step #F: N
0 : 2
19 9: 4
69 9: 8
280 9: 32
707 9: 128
2363 9: 2048
Integer overflow!
2863 6: -3816591145242754741
And no, fraction #9 is not always the one with the power of two only, for instance step 130 with N = 24x7. So, even a 64-bit integer is not enough, and with a 32-bit integer,
Integer overflow! 296 6: -1406264171
In the absence of a proper test such as the IF OVERFLOW ... of First Fortran (1957), the check for overflow relies on the improper appearance of negative numbers during two's-complement binary arithmetic that should only produce positive numbers. This will catch about half the events: N*M might or might not leave the sign bit on when the result overflows.
Revised Plan
One could introduce a scheme for multi-precision (or "bignum") arithmetic, but there is no need to maintain a unified number N because there are no additions or subtractions, only multiplications and division without remainders. In other words, why not mess about with collections of prime factors? That is, represent N, P, and Q as the list of prime numbers with their powers for each case. To calculate N*P, just add the list of prime number powers for P to that for N, and similarly, for N/Q subtract. At no stage (except perhaps to decorate output or for debugging) need the list be multiplied out to give the unified number and so there is no need for multi-precision multiplies and divides. To determine if N is divisible by Q (that is, if N*fraction = N*P/Q is integral), check only that the primes listed by Q have powers no greater than those of the corresponding primes for N.
To facilitate access without searching the list of primes for N, instead its list is represented by an array of powers, NPPOW, with most entries zero. Thus, NPPOW(i) has the power for PRIME(i) in the factorisation of N, and for N = 2, NPPOW(1) = 1 with all other elements zero. But for FP (the factorisation of P) and FQ (for Q) there is a proper list provided via type FACTORED, whereby FP.PLIST(0) is the count of prime factors and FP.PLIST(1:FP.PLIST(0)) fingers the prime numbers that are factors, with FP.PPOW(i) having their corresponding powers. Thus, FP.PLIST(2) has the index of the second prime factor of P (should it have so many), which is PRIME(FP.PLIST(2)), and its power is FP.PPOW(2). Accordingly, to determine if N (as NPPOW) is divisible by one of the fractions FQ, the appropriate elements of NPPOW (that give its powers) must be compared to the corresponding powers in FQ.PPOW, and if ALL of the fingered powers in NPPOW are greater than or equal to those in FQ.PPOW, then a hit! Note that the prime number factorisation of one has no elements in its list, and happily, the ALL operation applied for no tests yields true as desired, because one divides any number. Thus, in the FRACTRAN op-code system, a fraction P/1 is always a match, and no fractions beyond it will ever be considered.
As a part of the preparation, for each fraction the greatest common divisor is divided out to remove the possibility that converting the test N times fraction being integral via fraction = P/Q to Q divides N will behave differently. For example, N*6/3 will always be integral, but N may not be divisible by three. Reducing 6/3 to 2/1 however will work as will reducing 55/25 to 11/5. The example contains no such occasions, but the possibility nags.
For output, the value of N will not be shown multiplied out but via a schedule showing the powers of the first few prime numbers that form its factorisation. Rather than staring in puzzlement at opaque monster strings of digits, one can view each separate prime factor's power counting up and down as the calculation proceeds. A simple scan of all the factorisations soon determines the highest prime employed, and this never changes. An extension of this checks for which primes are omitted, and in this example, none are. However, a further extension modifies the output of the schedule of powers so as to blank out those that are zero. This could be achieved by replacing the likes of WRITE (...) NPPOW by WRITE (...) I6FMT(NPPOW) where function I6FMT writes out its integer with I6 format if the value is positive, and supplies six spaces if not, were it not that few Fortran systems allow such re-entrant usage of the system for formatted output. So, prepare the output with a straightforward WRITE to a CHARACTER variable, blank out the portions where unwanted values appear, and write the result.
Revised Code
Because this scheme requires a supply of prime numbers, it is convenient to employ the routines prepared for the extensible prime generator via module PRIMEBAG. So, this means escalating to the F90 style, and given that, some compound data structures can be used (for better mnemonics) in place of collections of arrays. <lang Fortran> MODULE CONWAYSIDEA !Notion devised by J. H. Conway.
USE PRIMEBAG !This is a common need.
INTEGER LASTP,ENUFF !Some size allowances.
PARAMETER (LASTP = 66, ENUFF = 66) !Should suffice for the example in mind.
INTEGER NPPOW(1:LASTP) !Represent N as a collection of powers of prime numbers.
TYPE FACTORED !But represent P and Q of freaction = P/Q
INTEGER PNUM(0:LASTP) !As a list of prime number indices with PNUM(0) the count.
INTEGER PPOW(LASTP) !And the powers. for the fingered primes.
END TYPE FACTORED !Rather than as a simple number multiplied out.
TYPE(FACTORED) FP(ENUFF),FQ(ENUFF) !Thus represent a factored fraction, P(i)/Q(i).
INTEGER PLIVE(ENUFF),NL !Helps subroutine SHOWN display NPPOW.
CONTAINS !Now for the details.
SUBROUTINE SHOWFACTORS(N) !First, to show an internal data structure.
TYPE(FACTORED) N !It is supplied as a list of prime factors.
INTEGER I !A stepper.
DO I = 1,N.PNUM(0) !Step along the list.
IF (I.GT.1) WRITE (MSG,"('x',$)") !Append a glyph for "multiply".
WRITE (MSG,"(I0,$)") PRIME(N.PNUM(I)) !The prime fingered in the list.
IF (N.PPOW(I).GT.1) WRITE (MSG,"('^',I0,$)") N.PPOW(I) !With an interesting power?
END DO !On to the next element in the list.
WRITE (MSG,1) N.PNUM(0) !End the line
1 FORMAT (": Factor count ",I0) !With a count of prime factors.
END SUBROUTINE SHOWFACTORS !Hopefully, this will not be needed often.
TYPE(FACTORED) FUNCTION FACTOR(IT) !Into a list of primes and their powers.
INTEGER IT,N !The number and a copy to damage.
INTEGER P,POW !A stepper and a power.
INTEGER F,NF !A factor and a counter.
IF (IT.LE.0) STOP "Factor only positive numbers!" !Or else...
N = IT !A copy I can damage.
NF = 0 !No factors found.
P = 0 !Because no primes have been tried.
PP:DO WHILE (IT.GT.1) !Step through the possibilities.
P = P + 1 !Another prime impends.
F = PRIME(P) !Grab a possible factor.
POW = 0 !It has no power yet.
FP:DO WHILE(MOD(N,F).EQ.0) !Well?
POW = POW + 1 !Count a factor..
N = N/F !Reduce the number.
END DO FP !The P'th prime's power's produced.
IF (POW.GT.0) THEN !So, was it a factor?
IF (NF.GE.LASTP) THEN !Yes. Have I room in the list?
WRITE (MSG,1) IT,LASTP !Alas.
1 FORMAT ("Factoring ",I0," but with provision for only ",
1 I0," prime factors!")
FACTOR.PNUM(0) = NF !Place the count so far,
CALL SHOWFACTORS(FACTOR)!So this can be invoked.
STOP "Not enough storage!" !Quite.
END IF !But normally,
NF = NF + 1 !Admit another factor.
FACTOR.PNUM(NF) = P !Identify the prime.
FACTOR.PPOW(NF) = POW !Place its power.
END IF !So much for that factor.
IF (N.LE.1) EXIT PP !Perhaps nothing remains?
END DO PP !Try another prime.
FACTOR.PNUM(0) = NF !Place the count.
END FUNCTION FACTOR !Thus, a list of primes and their powers.
INTEGER FUNCTION GCD(I,J) !Greatest common divisor.
INTEGER I,J !Of these two integers.
INTEGER N,M,R !Workers.
N = MAX(I,J) !Since I don't want to damage I or J,
M = MIN(I,J) !These copies might as well be the right way around.
1 R = MOD(N,M) !Divide N by M to get the remainder R.
IF (R.GT.0) THEN !Remainder zero?
N = M !No. Descend a level.
M = R !M-multiplicity has been removed from N.
IF (R .GT. 1) GO TO 1 !No point dividing by one.
END IF !If R = 0, M divides N.
GCD = M !There we are.
END FUNCTION GCD !Euclid lives on!
INTEGER FUNCTION FRACTRAN(L) !Applies Conway's idea to a list of fractions.
Could abandon all parameters since global variables have the details...
INTEGER L !The last fraction to consider.
INTEGER I,NF !Assistants.
DO I = 1,L !Step through the fractions in the order they were given.
NF = FQ(I).PNUM(0) !How many factors are listed in FQ(I)?
IF (ALL(NPPOW(FQ(I).PNUM(1:NF)) !Can N (as NPPOW) be divided by Q (as FQ)?
1 .GE. FQ(I).PPOW(1:NF))) THEN !By comparing the supplies of prime factors.
FRACTRAN = I !Yes!
NPPOW(FQ(I).PNUM(1:NF)) = NPPOW(FQ(I).PNUM(1:NF)) !Remove prime powers from N
1 - FQ(I).PPOW(1:NF) !Corresponding to Q.
NF = FP(I).PNUM(0) !Add powers to N
NPPOW(FP(I).PNUM(1:NF)) = NPPOW(FP(I).PNUM(1:NF)) !Corresponding to P.
1 + FP(I).PPOW(1:NF) !Thus, N = N/Q*P.
RETURN !That's all it takes! No multiplies nor divides!
END IF !So much for that fraction.
END DO !This relies on ALL(zero tests) yielding true, as when Q = 1.
FRACTRAN = 0 !No hit.
END FUNCTION FRACTRAN !No massive multi-precision arithmetic!
SUBROUTINE SHOWN(S,F) !Service routine to show the state after a step is calculated.
Could imaging a function I6FMT(23) that returns " 23" and " " for non-positive numbers. Can't do it, as if this were invoked via a WRITE statement, re-entrant use of WRITE usually fails.
INTEGER S,F !Step number, Fraction number.
INTEGER I !A stepper.
CHARACTER*(9+4+1 + NL*6) ALINE !A scratchpad matching FORMAT 103.
WRITE (ALINE,103) S,F,NPPOW(PLIVE(1:NL)) !Show it!
103 FORMAT (I9,I4,":",<NL>I6) !As a sequence of powers of primes.
IF (F.LE.0) ALINE(10:13) = "" !Scrub when no fraction is fingered.
DO I = 1,NL !Step along the live primes.
IF (NPPOW(PLIVE(I)).GT.0) CYCLE !Ignoring the empowered ones.
ALINE(15 + (I - 1)*6:14 + I*6) = "" !Blank out zero powers.
END DO !On to the next.
WRITE (MSG,"(A)") ALINE !Reveal at last.
END SUBROUTINE SHOWN !A struggle.
END MODULE CONWAYSIDEA !Simple...
PROGRAM POKE
USE CONWAYSIDEA !But, where does he get his ideas from?
INTEGER P(ENUFF),Q(ENUFF) !Holds the fractions as P(i)/Q(i).
INTEGER N !The working number.
INTEGER LF !Last fraction given.
INTEGER LP !Last prime needed.
INTEGER MS !Maximum number of steps.
INTEGER I,IT !Assistants.
LOGICAL*1 PUSED(ENUFF) !Track the usage of prime numbers,
MSG = 6 !Standard output.
WRITE (6,1) !Announce.
1 FORMAT ("Interpreter for J. H. Conway's FRACTRAN language.")
Chew into an example programme.
10 OPEN (10,FILE = "Fractran.txt",STATUS="OLD",ACTION="READ") !Rather than compiled-in stuff.
READ (10,*) LF !I need to know this without having to scan the input.
WRITE (MSG,11) LF !Reveal in case of trouble.
11 FORMAT (I0," fractions, as follow:") !Should the input evoke problems.
READ (10,*) (P(I),Q(I),I = 1,LF) !Ask for the specified number of P,Q pairs.
WRITE (MSG,12) (P(I),Q(I),I = 1,LF) !Show what turned up.
12 FORMAT (24(I0,"/",I0:", ")) !As P(i)/Q(i) pairs. The colon means that there will be no trailing comma.
READ (10,*) N,MS !The start value, and the step limit.
CLOSE (10) !Finished with input.
WRITE (MSG,13) N,MS !Hopefully, all went well.
13 FORMAT ("Start with N = ",I0,", step limit ",I0)
IF (.NOT.GRASPPRIMEBAG(66)) STOP "Gan't grab my file of primes!" !Attempt in hope.
Convert the starting number to a more convenient form, an array of powers of successive prime numbers.
20 FP(1) = FACTOR(N) !Borrow one of the factor list variables.
NPPOW = 0 !Clear all prime factor counts.
DO I = 1,FP(1).PNUM(0) !Now find what they are.
NPPOW(FP(1).PNUM(I)) = FP(1).PPOW(I) !Convert from a variable-length list
END DO !To a fixed-length random-access array.
PUSED = NPPOW.GT.0 !Note which primes have been used.
LP = FP(1).PNUM(FP(1).PNUM(0)) !Recall the last prime required. More later.
Convert the supplied P(i)/Q(i) fractions to lists of prime number factors and powers in FP(i) and FQ(i).
DO I = 1,LF !Step through the fractions.
IT = GCD(P(I),Q(I)) !Suspicion.
IF (IT.GT.1) THEN !Justified?
WRITE (MSG,21) I,P(I),Q(I),IT !Alas. Complain. The rule is N*(P/Q) being integer.
21 FORMAT ("Fraction ",I3,", ",I0,"/",I0,!N*6/3 is integer always because this is N*2/1, but 3 may not divide N.
1 " has common factor ",I0,"!") !By removing IT,
P(I) = P(I)/IT !The test need merely check if N is divisible by Q.
Q(I) = Q(I)/IT !And, as N is factorised in NPPOW
END IF !And Q in FQ, subtractions of powers only is needed.
FP(I) = FACTOR(P(I)) !Righto, form the factor list for P.
PUSED(FP(I).PNUM(1:FP(I).PNUM(0))) = .TRUE. !Mark which primes it fingers.
LP = MAX(LP,FP(I).PNUM(FP(I).PNUM(0))) !One has no prime factors: PNUM(0) = 0.
FQ(I) = FACTOR(Q(I)) !And likewise for Q.
PUSED(FQ(I).PNUM(1:FQ(I).PNUM(0))) = .TRUE. !Some primes may be omitted.
LP = MAX(LP,FQ(I).PNUM(FQ(I).PNUM(0))) !If no prime factors, PNUM(0) fingers element zero, which is zero.
END DO !All this messing about saves on multiplication and division.
Check which primes are in use, preparing an index of live primes..
NL = 0 !No live primes.
DO I = 1,LP !Check up to the last prime.
IF (PUSED(I)) THEN !This one used?
NL = NL + 1 !Yes. Another.
PLIVE(NL) = I !Fingered.
END IF !So much for that prime.
END DO !On to the next.
WRITE (MSG,22) NL,LP,PRIME(LP) !Remark on usage.
22 FORMAT ("Require ",I0," primes only, up to Prime(",I0,") = ",I0) !Presume always more than one prime.
IF (LP.GT.LASTP) STOP "But, that's too many for array NPPOW!"
Cast forth a heading.
100 WRITE (MSG,101) (PRIME(PLIVE(I)), I = 1,NL) !Splat a heading.
101 FORMAT (/,14X,"N as powers of prime factors",/, !The prime heading,
1 5X,"Step F#:",<LP>I6) !With primes beneath.
CALL SHOWN(0,0) !Initial state of N as NPPOW. Step zero, no fraction.
Commence!
DO I = 1,MS !Here we go!
IT = FRACTRAN(LF) !Do it!
CALL SHOWN(I,IT) !Show it!
IF (IT.LE.0) EXIT !Quit it?
END DO !The next step.
Complete!
END !Whee!
</lang>
Revised Results
Edited in are >>> markers for the prime powers of two.
Interpreter for J. H. Conway's FRACTRAN language.
14 fractions, as follow:
17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1
Start with N = 2, step limit 100
Require 10 primes only, up to Prime(10) = 29
N as powers of prime factors
Step F#: 2 3 5 7 11 13 17 19 23 29
0 : 1
1 13: 1 1
2 14: 1 2 1
3 5: 2 1
4 6: 2 1 1
5 11: 2 1 1
6 1: 2 1
7 2: 1 1 1 1
8 10: 1 1 1 1
9 5: 1 1 1
10 6: 1 1 1 1
11 11: 1 1 1 1
12 1: 1 1 1
13 2: 2 1 1
14 10: 2 1 1
15 5: 2 1
16 6: 2 1 1
17 11: 2 1 1
18 1: 2 1
>>> 19 9: 2
20 13: 1 1 1
21 13: 2 2
22 14: 2 3 1
23 5: 1 3 1
24 6: 1 3 1 1
25 5: 3 1 1
26 6: 3 2 1
27 11: 3 2 1
28 1: 3 1 1
29 2: 1 1 2 1 1
30 1: 1 1 2 1
31 2: 2 2 1 1
32 10: 2 2 1 1
33 5: 2 1 1 1
34 6: 2 1 1 1 1
35 5: 2 1 1 1
36 6: 2 1 2 1
37 11: 2 1 2 1
38 1: 2 1 1 1
39 2: 3 1 1 1
40 1: 3 1 1
41 3: 3 1
42 4: 2 1
43 7: 2 1 1
44 4: 1 1 1
45 7: 1 2 1
46 4: 2 1
47 7: 3 1
48 8: 3 1 1
49 11: 3 1 1
50 1: 3 1
51 2: 1 1 2 1
52 10: 1 1 2 1
53 5: 1 2 1
54 6: 1 2 1 1
55 11: 1 2 1 1
56 1: 1 2 1
57 2: 2 1 1 1
58 10: 2 1 1 1
59 5: 2 1 1
60 6: 2 1 1 1
61 11: 2 1 1 1
62 1: 2 1 1
63 2: 3 1 1
64 10: 3 1 1
65 5: 3 1
66 6: 3 1 1
67 11: 3 1 1
68 1: 3 1
>>> 69 9: 3
70 13: 2 1 1
71 13: 1 2 2
72 13: 3 3
73 14: 3 4 1
74 5: 2 4 1
75 6: 2 4 1 1
76 5: 1 4 1 1
77 6: 1 4 2 1
78 5: 4 2 1
79 6: 4 3 1
80 11: 4 3 1
81 1: 4 2 1
82 2: 1 1 3 2 1
83 1: 1 1 3 1 1
84 2: 2 2 2 1 1
85 1: 2 2 2 1
86 2: 3 3 1 1
87 10: 3 3 1 1
88 5: 3 2 1 1
89 6: 3 2 1 1 1
90 5: 3 1 1 1 1
91 6: 3 1 1 2 1
92 5: 3 1 2 1
93 6: 3 1 3 1
94 11: 3 1 3 1
95 1: 3 1 2 1
96 2: 4 1 2 1
97 1: 4 1 1 1
98 3: 4 1 1
99 4: 3 1 1
100 7: 3 1 1 1
This time, restricting output to only occasions when N is a power of two requires no peculiar bit-counting function. Just change the interpretation loop to <lang Fortran> DO I = 1,MS !Here we go!
IT = FRACTRAN(LF) !Do it!
IF (ALL(NPPOW(2:LP).EQ.0)) CALL SHOWN(I,IT) !Show it!
IF (IT.LE.0) EXIT !Quit it?
END DO !The next step.</lang>
Output:
Interpreter for J. H. Conway's FRACTRAN language.
14 fractions, as follow:
17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1
Start with N = 2, step limit 6666666
Require 10 primes only, up to Prime(10) = 29
N as powers of prime factors
Step F#: 2 3 5 7 11 13 17 19 23 29
0 : 1
19 9: 2
69 9: 3
280 9: 5
707 9: 7
2363 9: 11
3876 9: 13
8068 9: 17
11319 9: 19
19201 9: 23
36866 9: 29
45551 9: 31
75224 9: 37
101112 9: 41
117831 9: 43
152025 9: 47
215384 9: 53
293375 9: 59
327020 9: 61
428553 9: 67
507519 9: 71
555694 9: 73
700063 9: 79
808331 9: 83
989526 9: 89
1273490 9: 97
1434366 9: 101
1530213 9: 103
1710923 9: 107
1818254 9: 109
2019962 9: 113
2833089 9: 127
3104685 9: 131
3546320 9: 137
3720785 9: 139
4549718 9: 149
4755581 9: 151
5329874 9: 157
5958403 9: 163
6400897 9: 167
Execution took about two seconds.
Add and Multiply
Examples taken from Wikipaedia...
Interpreter for J. H. Conway's FRACTRAN language.
1 fraction, as follows:
3/2
Start with N = 72, step limit 66
Require 2 primes only, up to Prime(2) = 3
N as powers of prime factors
Step F#: 2 3
0 : 3 2
1 1: 2 3
2 1: 1 4
3 1: 5
4 : 5
The initial value is 72 = 23x32 so that "register" two holds 3 and register three holds 2. On completion, register three holds 5, the sum of 2 and 3.
Interpreter for J. H. Conway's FRACTRAN language.
6 fractions, as follow:
455/33, 11/13, 1/11, 3/7, 11/2, 1/3
Start with N = 72, step limit 66
Require 6 primes only, up to Prime(6) = 13
N as powers of prime factors
Step F#: 2 3 5 7 11 13
0 : 3 2
1 5: 2 2 1
2 1: 2 1 1 1 1
3 2: 2 1 1 1 1
4 1: 2 2 2 1
5 2: 2 2 2 1
6 3: 2 2 2
7 4: 2 1 2 1
8 4: 2 2 2
9 5: 1 2 2 1
10 1: 1 1 3 1 1
11 2: 1 1 3 1 1
12 1: 1 4 2 1
13 2: 1 4 2 1
14 3: 1 4 2
15 4: 1 1 4 1
16 4: 1 2 4
17 5: 2 4 1
18 1: 1 5 1 1
19 2: 1 5 1 1
20 1: 6 2 1
21 2: 6 2 1
22 3: 6 2
23 4: 1 6 1
24 4: 2 6
25 6: 1 6
26 6: 6
27 : 6
Here, register two holds 3 and register three holds 2. Their product appears in register five.
FreeBASIC
Added a compiler condition to make the program work with the old GMP.bi header file <lang FreeBasic>' version 06-07-2015 ' compile with: fbc -s console ' uses gmp
- Include Once "gmp.bi"
' in case the two #define's are missing from 'gmp.bi' define them now
- Ifndef mpq_numref
#Define mpq_numref(Q) (@(Q)->_mp_num) #Define mpq_denref(Q) (@(Q)->_mp_den)
- EndIf
Dim As String prog(0 To ...) = {"17/91", "78/85", "19/51", "23/38", "29/33",_ "77/29", "95/23", "77/19", "1/17", "11/13", "13/11", "15/14", "15/2", "55/1"}
Dim As UInteger i, j, c, max = UBound(prog) Dim As Integer scanbit
Dim As ZString Ptr gmp_str : gmp_str = Allocate(10000) Dim As Mpq_ptr in_, out_ in_ = Allocate(Len(__mpq_struct)) : Mpq_init(in_) out_ = Allocate(Len(__mpq_struct)) : Mpq_init(out_) Dim As mpz_ptr num, den num = Allocate(Len(__mpz_struct)) : Mpz_init(num) den = Allocate(Len(__mpz_struct)) : Mpz_init(den)
Dim As mpq_ptr instruction(max) For i = 0 To max
instruction(i) = Allocate(Len(__mpq_struct)) mpq_init(instruction(i)) mpq_set_str(instruction(i), prog(i), 10 )
Next
mpq_set_str(in_ ,"2",10) i = 0 : j = 0 Print "2"; Do
mpq_mul(out_, instruction(i), in_)
i = i + 1
den = mpq_denref(out_)
If mpz_cmp_ui(den, 1) = 0 Then
Mpq_get_str(gmp_str, 10, out_)
Print ", ";*gmp_str;
mpq_swap(in_, out_)
i = 0
j = j + 1
End If
Loop Until j > 14
' this one only display if the integer is 2^p, p being prime mpq_set_str(in_ ,"2",10) i = 0 : j = 0 : c = 0 Print : Print : Print Print "count iterations prime 2^prime"
Do
mpq_mul(out_, instruction(i), in_)
i = i + 1
j = j + 1
den = mpq_denref(out_)
If mpz_cmp_ui(den, 1) = 0 Then
num = mpq_numref(out_)
scanbit = mpz_scan1(num, 0)
' if scanbit = 0 then number is odd
If scanbit > 0 Then
' return from mpz_scan1(num, scanbit+1) is -1 for power of 2
If mpz_scan1(num, scanbit +1) = -1 Then
If c <= 20 Then Mpq_get_str(gmp_str, 10, out_) Else *gmp_str = ""
c = c + 1
Print Using "##### ################### ######## "; c; j; scanbit;
Print *gmp_str
If InKey <> "" Then Exit Do
End If
End If
mpq_swap(in_, out_)
i = 0
End If
Loop
' Loop Until scanbit > 300 ' Loop Until InKey <> "" ' Loop Until scanbit > 300 Or InKey <> "" ' stopping conditions will slow down the hole loop ' loop will check for key if it's printing a result
' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132, 116
count iterations prime 2^prime
1 129 2 4
2 425 3 8
3 1563 5 32
4 3735 7 128
5 11674 11 2048
6 18811 13 8192
7 38010 17 131072
8 52854 19 524288
9 88134 23 8388608
10 166070 29 536870912
11 204575 31 2147483648
12 333931 37 137438953472
13 446506 41 2199023255552
14 519556 43 8796093022208
15 667496 47 140737488355328
16 940183 53 9007199254740992
17 1274660 59 576460752303423488
18 1419935 61 2305843009213693952
19 1853979 67 147573952589676412928
20 2191673 71 2361183241434822606848
shorten output file
42 34533967 181
43 40326168 191
Go
Basic task: This compiles to produce a program that reads the limit, starting number n, and list of fractions as command line arguments, with the list of fractions as a single argument. <lang go>package main
import (
"fmt" "log" "math/big" "os" "strconv" "strings"
)
func compile(src string) ([]big.Rat, bool) {
s := strings.Fields(src)
r := make([]big.Rat, len(s))
for i, s1 := range s {
if _, ok := r[i].SetString(s1); !ok {
return nil, false
}
}
return r, true
}
func exec(p []big.Rat, n *big.Int, limit int) {
var q, r big.Int
rule:
for i := 0; i < limit; i++ {
fmt.Printf("%d ", n)
for j := range p {
q.QuoRem(n, p[j].Denom(), &r)
if r.BitLen() == 0 {
n.Mul(&q, p[j].Num())
continue rule
}
}
break
}
fmt.Println()
}
func usage() {
log.Fatal("usage: ft <limit> <n> <prog>")
}
func main() {
if len(os.Args) != 4 {
usage()
}
limit, err := strconv.Atoi(os.Args[1])
if err != nil {
usage()
}
var n big.Int
_, ok := n.SetString(os.Args[2], 10)
if !ok {
usage()
}
p, ok := compile(os.Args[3])
if !ok {
usage()
}
exec(p, &n, limit)
}</lang>
- Command line usage, with program compiled as "ft":
> ft 15 2 "17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1" 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132
Extra credit: This invokes above program with appropriate arguments, and processes the output to obtain the 20 primes. <lang go>package main
import (
"fmt" "log" "math/big" "os" "os/exec"
)
func main() {
c := exec.Command("ft", "1000000", "2", `17/91 78/85 19/51 23/38
29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1`)
c.Stderr = os.Stderr
r, err := c.StdoutPipe()
if err != nil {
log.Fatal(err)
}
if err = c.Start(); err != nil {
log.Fatal(err)
}
var n big.Int
for primes := 0; primes < 20; {
if _, err = fmt.Fscan(r, &n); err != nil {
log.Fatal(err)
}
l := n.BitLen() - 1
n.SetBit(&n, l, 0)
if n.BitLen() == 0 && l > 1 {
fmt.Printf("%d ", l)
primes++
}
}
fmt.Println()
}</lang>
- Output:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Haskell
Running the program
<lang haskell>import Data.List (find) import Data.Ratio (Ratio, (%), denominator)
fractran :: (Integral a) => [Ratio a] -> a -> [a] fractran fracts n = n :
case find (\f -> n `mod` denominator f == 0) fracts of Nothing -> [] Just f -> fractran fracts $ truncate (fromIntegral n * f)</lang>
Example:
λ> let prog = [17 % 91,78 % 85,19 % 51,23 % 38,29 % 33,77 % 29,95 % 23,77 % 19,1 % 17,11 % 13,13 % 11,15 % 14,15 % 2,55 % 1] λ> take 15 $ fractran prog 2 [2,15,825,725,1925,2275,425,390,330,290,770,910,170,156,132]
Reading the program
Additional import <lang Haskell>import Data.List.Split (splitOn)</lang> <lang Haskell>readProgram :: String -> [Ratio a] readProgram = map (toFrac . splitOn "/") . splitOn ","
where toFrac [n,d] = read n % read d</lang>
Example of running the program:
λ> let prog = readProgram "17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23 , 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1" λ> take 15 $ fractran prog 2 [2,15,825,725,1925,2275,425,390,330,290,770,910,170,156,132]
Generation of primes
Additional import <lang Haskell>import Data.Maybe (mapMaybe)</lang> <lang Haskell>primes = catMaybes $ map log2 $ fractran prog 2
where
prog = [17 % 91, 78 % 85, 19 % 51, 23 % 38, 29 % 33
,77 % 29, 95 % 23, 77 % 19, 1 % 17, 11 % 13
,13 % 11, 15 % 14, 15 % 2, 55 % 1]
log2 = fmap (+ 1) . findIndex (== 2) . takeWhile even . iterate (`div` 2)</lang>
λ> take 20 primes [1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67]
Icon and Unicon
Works in both languages:
<lang unicon>record fract(n,d)
procedure main(A)
fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2)
end
procedure fractran(s, n, limit)
execute(parse(s),n, limit)
end
procedure parse(s)
f := []
s ? while not pos(0) do {
tab(upto(' ')|0) ? put(f,fract(tab(upto('/')), (move(1),tab(0))))
move(1)
}
return f
end
procedure execute(f,d,limit)
/limit := 15
every !limit do {
if d := (d%f[i := !*f].d == 0, (writes(" ",d)/f[i].d)*f[i].n) then {}
else break write()
}
write()
end</lang>
- Output:
->fractan 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 ->
J
Solution: <lang j>toFrac=: '/r' 0&".@charsub ] NB. read fractions from string fractran15=: ({~ (= <.) i. 1:)@(toFrac@[ * ]) ^:(<15) NB. return first 15 Fractran results</lang>
Example: <lang j> taskstr=: '17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1'
taskstr fractran15 2
2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132</lang>
Java
<lang java>import java.util.Vector; import java.util.regex.Matcher; import java.util.regex.Pattern;
public class Fractran{
public static void main(String []args){
new Fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2);
}
final int limit = 15;
Vector<Integer> num = new Vector<>();
Vector<Integer> den = new Vector<>();
public Fractran(String prog, Integer val){
compile(prog);
dump();
exec(2);
}
void compile(String prog){
Pattern regexp = Pattern.compile("\\s*(\\d*)\\s*\\/\\s*(\\d*)\\s*(.*)");
Matcher matcher = regexp.matcher(prog);
while(matcher.find()){
num.add(Integer.parseInt(matcher.group(1)));
den.add(Integer.parseInt(matcher.group(2)));
matcher = regexp.matcher(matcher.group(3));
}
}
void exec(Integer val){
int n = 0;
while(val != null && n<limit){
System.out.println(n+": "+val);
val = step(val);
n++;
}
}
Integer step(int val){
int i=0;
while(i<den.size() && val%den.get(i) != 0) i++;
if(i<den.size())
return num.get(i)*val/den.get(i);
return null;
}
void dump(){
for(int i=0; i<den.size(); i++)
System.out.print(num.get(i)+"/"+den.get(i)+" ");
System.out.println();
}
}</lang>
JavaScript
<lang javascript> var num = new Array(); var den = new Array(); var val ;
function compile(prog){
var regex = /\s*(\d*)\s*\/\s*(\d*)\s*(.*)/m;
while(regex.test(prog)){
num.push(regex.exec(prog)[1]);
den.push(regex.exec(prog)[2]);
prog = regex.exec(prog)[3];
}
}
function dump(prog){
for(var i=0; i<num.length; i++) document.body.innerHTML += num[i]+"/"+den[i]+" "; document.body.innerHTML += "
";
}
function step(val){
var i=0; while(i<den.length && val%den[i] != 0) i++; return num[i]*val/den[i];
}
function exec(val){
var i = 0;
while(val && i<limit){
document.body.innerHTML += i+": "+val+"
";
val = step(val);
i ++;
}
}
// Main compile("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1"); dump(); var limit = 15; exec(2); </lang>
Mathematica / Wolfram Language
This isn't as efficient as possible for long lists of fractions, since it doesn't stop doing n*listelements once it finds an integer. Instead, it computes "is integer?" for n*{all list elements}. For short lists that's probably not a big deal.
<lang Mathematica>fractionlist = {17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1}; n = 2; steplimit = 20; j = 0; break = False; While[break == False && j <= steplimit,
newlist = n fractionlist; isintegerlist = IntegerQ[#] & /@ newlist; truepositions = Position[isintegerlist, True]; If[Length[truepositions] == 0, break = True, Print[ToString[j] <> ": " <> ToString[n]]; n = newlist[[truepositions1, 1]]; j++; ] ]</lang>
- Output:
0: 2 1: 15 2: 825 3: 725 4: 1925 5: 2275 6: 425 7: 390 8: 330 9: 290 10: 770 11: 910 12: 170 13: 156 14: 132 15: 116 16: 308 17: 364 18: 68 19: 4 20: 30
PARI/GP
In this version ideas were borrowed from C++, Java and JavaScript.
<lang parigp> \\ FRACTRAN \\ 4/27/16 aev fractran(val,ft,lim)={ my(ftn=#ft,fti,di,L=List(),j=0); while(val&&j<lim, listput(L,val);
for(i=1,ftn, fti=ft[i]; di=denominator(fti);
if(val%di==0, break));\\fend i
val= numerator(fti)*val/di; j++);\\wend j
return(Vec(L)); }
{\\ Executing: my(v=[17/91,78/85,19/51,23/38,29/33,77/29,95/23,77/19,1/17,11/13,13/11,15/14,15/2,55/1]); print(fractran(2,v,15)); } </lang>
- Output:
[2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132]
Perl
Instead of printing all steps, I chose to only print those steps which were a power of two. This makes the fact that it's a prime-number-generating program much clearer.
<lang perl>use strict; use warnings; use Math::BigRat;
my ($n, @P) = map Math::BigRat->new($_), qw{ 2 17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1 };
$|=1; MAIN: for( 1 .. 5000 ) { print " " if $_ > 1; my ($pow, $rest) = (0, $n->copy); until( $rest->is_odd ) { ++$pow; $rest->bdiv(2); } if( $rest->is_one ) { print "2**$pow"; } else { #print $n; } for my $f_i (@P) { my $nf_i = $n * $f_i; next unless $nf_i->is_int; $n = $nf_i; next MAIN; } last; }
print "\n"; </lang>
If you uncomment the
#print $n
, it will print all the steps.
Perl 6
A Fractran program potentially returns an infinite list, and infinite lists are a common data structure in Perl 6. The limit is therefore enforced only by slicing the infinite list. <lang perl6>sub fractran(@program) {
2, { +first Int, map (* * $_).narrow, @program } ... 0
} say fractran(<17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11
15/14 15/2 55/1>)[^100];</lang>
- Output:
(2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116 308 364 68 4 30 225 12375 10875 28875 25375 67375 79625 14875 13650 2550 2340 1980 1740 4620 4060 10780 12740 2380 2184 408 152 92 380 230 950 575 2375 9625 11375 2125 1950 1650 1450 3850 4550 850 780 660 580 1540 1820 340 312 264 232 616 728 136 8 60 450 3375 185625 163125 433125 380625 1010625 888125 2358125 2786875 520625 477750 89250 81900 15300 14040 11880 10440 27720 24360 64680 56840 150920 178360 33320 30576 5712 2128 1288)
Extra credit:
We can weed out all the powers of two into another infinite constant list based on the first list. In this case the sequence is limited only by our patience, and a ^C from the terminal. The .msb method finds the most significant bit of an integer, which conveniently is the base-2 log of the power-of-two in question.
<lang perl6>for fractran <17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11
15/14 15/2 55/1> {
say $++, "\t", .msb, "\t", $_ if .log %% log(2);
}</lang>
- Output:
0 1 2 1 2 4 2 3 8 3 5 32 4 7 128 5 11 2048 6 13 8192 7 17 131072 8 19 524288 9 23 8388608 ^C
Python
Python: Generate series from a fractran program
<lang python>from fractions import Fraction
def fractran(n, fstring='17 / 91, 78 / 85, 19 / 51, 23 / 38, 29 / 33,'
'77 / 29, 95 / 23, 77 / 19, 1 / 17, 11 / 13,'
'13 / 11, 15 / 14, 15 / 2, 55 / 1'):
flist = [Fraction(f) for f in fstring.replace(' ', ).split(',')]
n = Fraction(n)
while True:
yield n.numerator
for f in flist:
if (n * f).denominator == 1:
break
else:
break
n *= f
if __name__ == '__main__':
n, m = 2, 15
print('First %i members of fractran(%i):\n ' % (m, n) +
', '.join(str(f) for f,i in zip(fractran(n), range(m))))</lang>
- Output:
First 15 members of fractran(2): 2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132
Python: Generate primes
Use fractran above as a module imported into the following program.
<lang python> from fractran import fractran
def fractran_primes():
for i, fr in enumerate(fractran(2), 1):
binstr = bin(fr)[2:]
if binstr.count('1') == 1:
prime = binstr.count('0')
if prime > 1: # Skip 2**0 and 2**1
yield prime, i
if __name__ == '__main__':
for (prime, i), j in zip(fractran_primes(), range(15)):
print("Generated prime %2i from the %6i'th member of the fractran series" % (prime, i))</lang>
- Output:
Generated prime 2 from the 20'th member of the fractran series Generated prime 3 from the 70'th member of the fractran series Generated prime 5 from the 281'th member of the fractran series Generated prime 7 from the 708'th member of the fractran series Generated prime 11 from the 2364'th member of the fractran series Generated prime 13 from the 3877'th member of the fractran series Generated prime 17 from the 8069'th member of the fractran series Generated prime 19 from the 11320'th member of the fractran series Generated prime 23 from the 19202'th member of the fractran series Generated prime 29 from the 36867'th member of the fractran series Generated prime 31 from the 45552'th member of the fractran series Generated prime 37 from the 75225'th member of the fractran series Generated prime 41 from the 101113'th member of the fractran series Generated prime 43 from the 117832'th member of the fractran series Generated prime 47 from the 152026'th member of the fractran series
Racket
Simple version, without sequences.
<lang Racket>#lang racket
(define (displaysp x)
(display x) (display " "))
(define (read-string-list str)
(map string->number
(string-split (string-replace str " " "") ",")))
(define (eval-fractran n list)
(for/or ([e (in-list list)])
(let ([en (* e n)])
(and (integer? en) en))))
(define (show-fractran fr n s)
(printf "First ~a members of fractran(~a):\n" s n)
(displaysp n)
(for/fold ([n n]) ([i (in-range (- s 1))])
(let ([new-n (eval-fractran n fr)])
(displaysp new-n)
new-n))
(void))
(define fractran
(read-string-list
(string-append "17 / 91, 78 / 85, 19 / 51, 23 / 38, 29 / 33,"
"77 / 29, 95 / 23, 77 / 19, 1 / 17, 11 / 13,"
"13 / 11, 15 / 14, 15 / 2, 55 / 1")))
(show-fractran fractran 2 15)</lang>
- Output:
First 15 members of fractran(2): 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132
REXX
Programming note: extra blanks can be inserted in the fractions before and/or after the solidus [/].
showing all terms
<lang rexx>/*REXX program runs FRACTRAN for a given set of fractions and from a specified N. */ numeric digits 2000 /*be able to handle larger numbers. */ parse arg N terms fracs /*obtain optional arguments from the CL*/ if N== | N=="," then N=2 /*Not specified? Then use the default.*/ if terms== | terms=="," then terms=100 /* " " " " " " */ if fracs= then fracs= '17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23,',
'77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1'
/* [↑] The default for the fractions. */
f=space(fracs,0) /*remove all blanks from the FRACS list*/
do #=1 while f\==; parse var f n.# '/' d.# "," f
end /*#*/ /* [↑] parse all the fractions in list*/
- =#-1 /*the number of fractions just found. */
say # 'fractions:' fracs /*display number and actual fractions. */ say 'N is starting at ' N /*display the starting number N. */ say terms ' terms are being shown:' /*display a kind of header/title. */
do j=1 for terms /*perform the DO loop for each term. */
do k=1 for # /* " " " " " " fraction*/
if N//d.k\==0 then iterate /*Not an integer? Then ignore it. */
say right('term' j, 35) "──► " N /*display the Nth term with the N. */
N=N % d.k * n.k /*calculate next term (use %≡integer ÷)*/
iterate j /*go start calculating the next term. */
end /*k*/ /* [↑] if an integer, we found a new N*/
end /*j*/ /*stick a fork in it, we're all done. */</lang>
output using the default input:
14 fractions: 17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1
N is starting at 2
100 terms are being shown:
term 1 ──► 2
term 2 ──► 15
term 3 ──► 825
term 4 ──► 725
term 5 ──► 1925
term 6 ──► 2275
term 7 ──► 425
term 8 ──► 390
term 9 ──► 330
term 10 ──► 290
term 11 ──► 770
term 12 ──► 910
term 13 ──► 170
term 14 ──► 156
term 15 ──► 132
term 16 ──► 116
term 17 ──► 308
term 18 ──► 364
term 19 ──► 68
term 20 ──► 4
term 21 ──► 30
term 22 ──► 225
term 23 ──► 12375
term 24 ──► 10875
term 25 ──► 28875
term 26 ──► 25375
term 27 ──► 67375
term 28 ──► 79625
term 29 ──► 14875
term 30 ──► 13650
term 31 ──► 2550
term 32 ──► 2340
term 33 ──► 1980
term 34 ──► 1740
term 35 ──► 4620
term 36 ──► 4060
term 37 ──► 10780
term 38 ──► 12740
term 39 ──► 2380
term 40 ──► 2184
term 41 ──► 408
term 42 ──► 152
term 43 ──► 92
term 44 ──► 380
term 45 ──► 230
term 46 ──► 950
term 47 ──► 575
term 48 ──► 2375
term 49 ──► 9625
term 50 ──► 11375
term 51 ──► 2125
term 52 ──► 1950
term 53 ──► 1650
term 54 ──► 1450
term 55 ──► 3850
term 56 ──► 4550
term 57 ──► 850
term 58 ──► 780
term 59 ──► 660
term 60 ──► 580
term 61 ──► 1540
term 62 ──► 1820
term 63 ──► 340
term 64 ──► 312
term 65 ──► 264
term 66 ──► 232
term 67 ──► 616
term 68 ──► 728
term 69 ──► 136
term 70 ──► 8
term 71 ──► 60
term 72 ──► 450
term 73 ──► 3375
term 74 ──► 185625
term 75 ──► 163125
term 76 ──► 433125
term 77 ──► 380625
term 78 ──► 1010625
term 79 ──► 888125
term 80 ──► 2358125
term 81 ──► 2786875
term 82 ──► 520625
term 83 ──► 477750
term 84 ──► 89250
term 85 ──► 81900
term 86 ──► 15300
term 87 ──► 14040
term 88 ──► 11880
term 89 ──► 10440
term 90 ──► 27720
term 91 ──► 24360
term 92 ──► 64680
term 93 ──► 56840
term 94 ──► 150920
term 95 ──► 178360
term 96 ──► 33320
term 97 ──► 30576
term 98 ──► 5712
term 99 ──► 2128
term 100 ──► 1288
showing prime numbers
Programming note: if the number of terms specified (the 2nd argument) is negative, then only powers of two are displayed. <lang rexx>/*REXX program runs FRACTRAN for a given set of fractions and from a specified N. */ numeric digits 999; w=length(digits()) /*be able to handle gihugeic numbers. */ parse arg N terms fracs /*obtain optional arguments from the CL*/ if N== | N=="," then N=2 /*Not specified? Then use the default.*/ if terms== | terms=="," then terms=100 /* " " " " " " */ if fracs= then fracs= '17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23,',
'77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1'
/* [↑] The default for the fractions. */
f=space(fracs, 0) /*remove all blanks from the FRACS list*/
do #=1 while f\==; parse var f n.# '/' d.# "," f
end /*#*/ /* [↑] parse all the fractions in list*/
- =#-1 /*adjust the number of fractions found.*/
tell= terms>0 /*flag: show number or a power of 2.*/ !.=0; _=1 /*the default value for powers of 2. */ if \tell then do p=1 until length(_)>digits(); _=_+_; !._=1
if p==1 then @._=left(,w+9) "2**"left(p,w) ' '
else @._='(prime' right(p,w)") 2**"left(p,w) ' '
end /*p*/ /* [↑] build powers of 2 tables. */
L=length(N) /*length in decimal digits of integer N*/ say # 'fractions:' fracs /*display number and actual fractions. */ say 'N is starting at ' N /*display the starting number N. */ if tell then say terms ' terms are being shown:' /*display hdr.*/
else say 'only powers of two are being shown:' /* " " */
q='(max digits used:' /*a literal used in the SAY below. */
do j=1 for abs(terms) /*perform DO loop once for each term. */
do k=1 for # /* " " " " " " fraction*/
if N//d.k\==0 then iterate /*Not an integer? Then ignore it. */
if tell then say right('term' j, 35) "──► " N /*display Nth term and N.*/
else if !.N then say right('term' j,15) "──►" @.N q right(L,w)") " N
N=N % d.k * n.k /*calculate next term (use %≡integer ÷)*/
L=max(L, length(N)) /*the maximum number of decimal digits.*/
iterate j /*go start calculating the next term. */
end /*k*/ /* [↑] if an integer, we found a new N*/
end /*j*/ /*stick a fork in it, we're done. */</lang>
output using the input of: , -50000000
(negative fifty million)
14 fractions: 17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1
N is starting at 2
only powers of two are being shown:
term 1 2**1 (max digits used: 1) 2
term 20 ──► (prime 2) 2**2 (max digits used: 4) 4
term 70 ──► (prime 3) 2**3 (max digits used: 5) 8
term 281 ──► (prime 5) 2**5 (max digits used: 8) 32
term 708 ──► (prime 7) 2**7 (max digits used: 12) 128
term 2364 ──► (prime 11) 2**11 (max digits used: 18) 2048
term 3877 ──► (prime 13) 2**13 (max digits used: 21) 8192
term 8069 ──► (prime 17) 2**17 (max digits used: 27) 131072
term 11320 ──► (prime 19) 2**19 (max digits used: 30) 524288
term 19202 ──► (prime 23) 2**23 (max digits used: 36) 8388608
term 36867 ──► (prime 29) 2**29 (max digits used: 46) 536870912
term 45552 ──► (prime 31) 2**31 (max digits used: 49) 2147483648
term 75225 ──► (prime 37) 2**37 (max digits used: 58) 137438953472
...
(some output elided.)
...
term 193455490 ──► (prime 523) 2**523 (max digits used: 808) 27459190640522438859927603196325572869077741200573221637577853836742172733590624208490238562645818219909185245565923432148487951998866575250296113164460228608
Output note: There are intermediary numbers (that aren't powers of two) that are hundreds of digits long.
Ruby
<lang ruby>str ="17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1" FractalProgram = str.split(',').map(&:to_r) #=> array of rationals
Runner = Enumerator.new do |y|
num = 2
loop{ y << num *= FractalProgram.detect{|f| (num*f).denominator == 1} }
end
prime_generator = Enumerator.new do |y|
Runner.each do |num| l = Math.log2(num) y << l.to_i if l.floor == l end
end
- demo
p Runner.take(20) p prime_generator.take(20) </lang>
- Output:
[(15/1), (825/1), (725/1), (1925/1), (2275/1), (425/1), (390/1), (330/1), (290/1), (770/1), (910/1), (170/1), (156/1), (132/1), (116/1), (308/1), (364/1), (68/1), (4/1), (30/1)] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
Scala
<lang scala>class TestFractran extends FunSuite {
val program = Fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1")
val expect = List(2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132)
test("find first fifteen fractran figures") {
assert((program .execute(2) take 15 toList) === expect)
}
}
object Fractran {
val pattern = """\s*(\d+)\s*/\s*(\d+)\s*""".r def parse(m: Match) = ((m group 1).toInt, (m group 2).toInt) def apply(program: String) = new Fractran( pattern.findAllMatchIn(program).map(parse).toList)
}
class Fractran(val numDem: List[(Int,Int)]) {
def execute(value: Int) = unfold(value) { v =>
numDem indexWhere(v % _._2 == 0) match {
case i if i > -1 => Some(v, numDem(i)._1 * v / numDem(i)._2)
case _ => None
}
}
}</lang>
Seed7
<lang seed7>$ include "seed7_05.s7i";
include "rational.s7i";
const func array integer: fractran (in integer: limit, in var integer: number, in array rational: program) is func
result
var array integer: output is 0 times 0;
local
var integer: index is 1;
var rational: newNumber is 0/1;
begin
output := [] (number);
while index <= length(program) and length(output) <= limit do
newNumber := rat(number) * program[index];
if newNumber = rat(trunc(newNumber)) then
number := trunc(newNumber);
output &:= number;
index := 1;
else
incr(index);
end if;
end while;
end func;
const proc: main is func
local
const array rational: program is []
(17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1);
var array integer: output is 0 times 0;
var integer: number is 0;
begin
output := fractran(15, 2, program);
for number range output do
write(number <& " ");
end for;
writeln;
end func;</lang>
- Output:
2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116
Program to compute prime numbers with fractran (The program has no limit, use CTRL-C to terminate it): <lang seed7>$ include "seed7_05.s7i";
include "bigrat.s7i";
const proc: fractran (in var bigInteger: number, in array bigRational: program) is func
local
var integer: index is 1;
var bigRational: newNumber is 0_/1_;
begin
while index <= length(program) do
newNumber := rat(number) * program[index];
if newNumber = rat(trunc(newNumber)) then
number := trunc(newNumber);
if 2_ ** ord(log2(number)) = number then
writeln(log2(number));
end if;
index := 1;
else
incr(index);
end if;
end while;
end func;
const proc: main is func
local
const array bigRational: program is []
(17_/91_, 78_/85_, 19_/51_, 23_/38_, 29_/33_, 77_/29_, 95_/23_, 77_/19_, 1_/17_, 11_/13_, 13_/11_, 15_/14_, 15_/2_, 55_/1_);
begin
fractran(2_, program);
end func;</lang>
- Output:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
Sidef
<lang ruby>var str ="17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1" const FractalProgram = str.split(',').map{.to_r} #=> array of rationals
func runner(n, callback) {
var num = 2.rat
n.times {
callback(num *= FractalProgram.find { |f| (f*num).de.is_one })
}
}
func prime_generator(n, callback) {
var x = 0;
runner(Math.inf, { |num|
var l = num.to_f.log2
if (l.floor == l) {
callback(l.to_i)
++x == n && return
}
})
}
STDOUT.autoflush(true)
runner(20, {|n| print (n, ' ') }) print "\n"
prime_generator(20, {|n| print (n, ' ') }) print "\n"</lang>
- Output:
15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116 308 364 68 4 30 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Tcl
<lang tcl>package require Tcl 8.6
oo::class create Fractran {
variable fracs nco
constructor {fractions} {
set fracs {} foreach frac $fractions { if {[regexp {^(\d+)/(\d+),?$} $frac -> num denom]} { lappend fracs $num $denom } else { return -code error "$frac is not a supported fraction" } } if {![llength $fracs]} { return -code error "need at least one fraction" }
}
method execute {n {steps 15}} {
set co [coroutine [incr nco] my Generate $n] for {set i 0} {$i < $steps} {incr i} { lappend result [$co] } catch {rename $co ""} return $result
}
method Step {n} {
foreach {num den} $fracs { if {$n % $den} continue return [expr {$n * $num / $den}] } return -code break
}
method Generate {n} {
yield [info coroutine] while 1 { yield $n set n [my Step $n] } return -code break
}
}
set ft [Fractran new {
17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1
}] puts [$ft execute 2]</lang>
- Output:
2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132
You can just collect powers of 2 by monkey-patching in something like this: <lang tcl>oo::objdefine $ft method pow2 {n} {
set co [coroutine [incr nco] my Generate 2]
set pows {}
while {[llength $pows] < $n} {
set item [$co] if {($item & ($item-1)) == 0} { lappend pows $item }
} return $pows
} puts [$ft pow2 10]</lang> Which will then produce this additional output:
2 4 8 32 128 2048 8192 131072 524288 8388608
zkl
<lang zkl>var fracs="17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17,"
"11/13, 13/11, 15/14, 15/2, 55/1";
fcn fractranW(n,fracsAsOneBigString){ //-->Walker (iterator)
fracs:=(fracsAsOneBigString-" ").split(",").apply(
fcn(frac){ frac.split("/").apply("toInt") }); //( (n,d), (n,d), ...)
Walker(fcn(rn,fracs){
n:=rn.value;
foreach a,b in (fracs){
if(n*a%b == 0){ rn.set(n*a/b); return(n); }
} }.fp(Ref(n),fracs))
}</lang> <lang zkl>fractranW(2,fracs).walk(20).println();</lang>
- Output:
L(2,15,825,725,1925,2275,425,390,330,290,770,910,170,156,132,116,308,364,68,4)
<lang zkl>var [const] BN=Import("zklBigNum"); // libGMP fcn fractranPrimes{
foreach n,fr in ([1..].zip(fractranW(BN(2),fracs))){
if(fr.num1s==1){
p:=(fr.toString(2) - "1").len(); // count zeros if(p>1) println("Prime %3d from the nth Fractran(%8d): %d".fmt(p,n,fr));
} }
} fractranPrimes();</lang>
- Output:
Prime 2 from the nth Fractran( 20): 4 Prime 3 from the nth Fractran( 70): 8 Prime 5 from the nth Fractran( 281): 32 ... Prime 227 from the nth Fractran(15956646): 215679573337205118357336120696157045389097155380324579848828881993728 Prime 229 from the nth Fractran(16429799): 862718293348820473429344482784628181556388621521298319395315527974912 Prime 233 from the nth Fractran(17293373): 13803492693581127574869511724554050904902217944340773110325048447598592 Prime 239 from the nth Fractran(18633402): 883423532389192164791648750371459257913741948437809479060803100646309888 ...