XXIIVV

Fractran is insanely difficult to program in, but based on one of the most bizarrely elegant concepts of computation.

A Fractran program is an ordered list of positive fractions together with an initial positive integer input. The program is run by updating the accumulator.

 
—The Book of Numbers, John Conway

Any number that can't be divided by any other number, apart from itself and one, is prime. Since primes can't be divided, we can think of them as the DNA of other numbers. In Fractran, each prime is a register and their exponent is their value.

The Accumulator

The state of the accumulator is held as a single number, whose prime factorization holds these registers(2, 3, 5, 7, 11, 13, 17, ..). If the state of the accumulator is 1008(2⁴ × 3² × 7), r2 has the value 4, r3 has the value 2, r7 has the value 1, and all other registers are unassigned.

accumulatorregisters
r2r3r5r7
611
1812
1008421
54022501234

The Operators

A fractran operation is a positive fraction, each fraction represents an instruction that tests one or more registers, represented by the prime factors of its denominator. The Fractran computer goes through each fraction in order, in terms of our current accumulator value.

18(21 × 32) 2/3 = 8(23) addition r2+r3->r2

To run the adder operation(2/3), we will take the state of the accumulator. If multiplying it by this fraction will give us an integer, we will do so and start again at the beginning of the program. Otherwise, we will stop and consider the program complete. We will do this repeatedly until we can no longer produce an integer with this method.

stepsstateregisters
r2r3
1181218 × 2/3 = 12/1INT, RESTART
2122112 × 2/3 = 8/1INT, RESTART
3838 × 2/3 = 16/3NOT INT, END

To add the values 1 and 2, we will store the values in registers 2 and 3, our starting state is therefore 18(21 × 32).

For each step of the program, we will multiply our state with the program(18 × 2/3 = 12, 12 × 2/3 = 8, ..) until our our working value cannot be reduced to a whole number(16/3), we have exhausted the program. Alternatively, the program 3/2 will do the same operation but store the result in the register 3.

576(26 × 32) 1/6 = 16(24) subtraction r2-r3->r2

Operations become more readable when broken down into their primes. We can think of every prime number as having a register which can take on non-negative integer values. Each fraction is an instruction that operates on some of the registers.

2/3 15/256 21/20
(21)/(31) (31 × 51)/(26) (31 × 71)/(22 × 51)
if(r3 >= 1){ 
	r3 -= 1;
	r2 += 1;
	return;
}
if(r2 >= 6){ 
	r2 -= 6;
	r3 += 1;
	r5 += 1;
	return;
}
if(r2 >= 2 && r5 >= 1){ 
	r2 -= 2; 
	r5 -= 1; 
	r3 += 1; 
	r7 += 1;
	return;
}

You can interpret a fraction as saying if the current value of each register is greater than or equal to the the value specified by the denominator, you subtract from the registers all of the values in the denominator, add all the values specified in the numerator, and then jump back to the first instruction. Otherwise, if any register is less than the value specified in the denominator, continue to the next fraction.

The Programs

A Fractran program is a list of fractions together with an initial positive integer input n. The program is run by updating the integer n as follows:

Let's put together an adder program similar from the one above(2/3) but which writes to a third register. The following program first moves the content in r2 to r3, and then the content of r3 to r5.

18(21 × 32) 3/2 5/3 = 125(53) addition r2+r3->r5(9 steps)

Alternatively, a faster way to do this would be to directly move powers of 2 over to 5, then powers of 3.

18(21 × 32) 5/2 5/3 = 125(53) addition r2+r3->r5(7 steps)

Each of the 7 steps of this last program looks like:

18 5/2 5/3           [18]  r2=01 r3=02
------------------   -----------------
18 × 5/2 = 45/1      [45]  r3=02 r5=01
45 × 5/2 = 225/2 
45 × 5/3 = 75/1      [75]  r3=01 r5=02
75 × 5/2 = 375/2 
75 × 5/3 = 125/1     [125] r5=03
125 × 5/2 = 625/2 
125 × 5/3 = 625/3    [125] r5=03

Both of these programs are destructive, meaning that they drain the registers of their original values. We can make (2/3) less destructive with (10/3) by storing a copy of r3 in r5. And we can create a non-destructive adder but this requires coming in with the program with the flag r7 set:

126(21 × 32 × 71) 7/11 715/14 935/21 1/7 2/13 3/17 = 2250(21 × 32 × 53)

As an extra demonstration, let us consider the following programs representing all the logic gates:

Program07142142
AND Gate5/42 1/21 1/14 1/71115
OR Gate5/42 5/21 5/14 1/71555
XOR Gate1/42 5/21 5/14 1/71551
NAND Gate1/42 5/21 5/14 5/75551
NOR Gate1/42 1/21 1/14 5/75111
XNOR Gate5/42 1/21 1/14 5/75115
— Submit an edit to fractran_guide.htm(137 lines)

Interpreter

A simple Fractran interpreter, written in ANSI C, showing the value in the registers as it steps through the program.

#include <stdio.h>

/* 
Copyright (c) 2020 Devine Lu Linvega

Permission to use, copy, modify, and distribute this software for any
purpose with or without fee is hereby granted, provided that the above
copyright notice and this permission notice appear in all copies.

THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES
WITH REGARD TO THIS SOFTWARE.
*/

typedef struct Fraction {
	unsigned int num, den;
} Fraction;

typedef struct Machine {
	int len;
	Fraction acc, program[256];
} Machine;

static int
gcd(int a, int b)
{
	if(b == 0)
		return a;
	return gcd(b, a % b);
}

static Fraction
Frac(unsigned int num, unsigned int den)
{
	Fraction f;
	unsigned int d = gcd(num, den);
	f.num = num / d;
	f.den = den / d;
	return f;
}

static void
printstate(Machine *m)
{
	unsigned int fac = 2, num = m->acc.num;
	printf("[%d] ", num);
	while(num > 1) {
		if(num % fac == 0) {
			unsigned int pow = 1;
			printf("r%02u=", fac);
			num /= fac;
			while(!(num % fac)) {
				num /= fac;
				pow++;
			}
			printf("%02u", pow);
			if(num != 1)
				putchar(' ');
		} else
			fac++;
	}
	putchar('\n');
}

static void
run(Machine *m)
{
	int i = 0, steps = 0;
	while(i < m->len && m->acc.num) {
		Fraction res, *f = &m->program[i++];
		res = Frac(m->acc.num * f->num, m->acc.den * f->den);
		printf("%u × %u/%u = %u/%u \n",
			m->acc.num,
			f->num,
			f->den,
			res.num,
			res.den);
		if(res.den == 1) {
			m->acc = res;
			printstate(m);
			i = 0;
		}
		steps++;
	}
	if(steps) {
		printstate(m);
		printf("Completed in %d steps.\n", steps);
	}
}

static void
push(Machine *m, char *w)
{
	Fraction f;
	if(!m->acc.den) {
		if(sscanf(w, "%u", &m->acc.num) > 0)
			m->acc.den = 1;
		return;
	}
	if(sscanf(w, "%u/%u", &f.num, &f.den) > 0)
		m->program[m->len++] = f;
}

static Machine m;

int
main(void)
{
	int len = 0;
	char c, word[64];
	while((c = fgetc(stdin)) != EOF) {
		if(c == ' ' || c == '\n') {
			word[len] = '\0';
			len = 0;
			push(&m, word);
		} else
			word[len++] = c;
		if(c == '\n')
			break;
	}
	printstate(&m);
	run(&m);
	return 0;
}
— Submit an edit to fractran.c.txt(123 lines)
A common man marvels at uncommon things; a wise man marvels at the commonplace.
—Confucius

incoming(1): firth

Last update on 14Y10, edited 5 times. +18/37fh ----|-