Primes

Multiplying two numbers is the same as adding the counts of each prime factors, and division is the same as subtracting the counts. To experiment with primes, have a look at Fractran. For example, using numbers made up of the 3 first primes(2, 3, 5), 2250 is equal to 2^1 x 3^2 x 5^3.

To find the prime factorization of a number, start by dividing the number by the first prime number 2 and continue dividing by 2 until you get a decimal or remainder. Then divide by 3, 5, 7, etc. until the only numbers left are prime numbers.

Peasant Multiplication

In the first column, divide the first number by 2, dropping the remainder if any, until 1 is reached. In the second column, write the numbers obtained by successive multiplication by 2. The answer is found by adding the numbers in the doubling column with odd numbers in their first column.

Addition of 5

When adding 5 to a digit greater than 5, it is easier to first subtract 5 and then add 10.

7 + 5 = 12.
Also 7 - 5 = 2; 2 + 10 = 12.

Subtraction of 5

When subtracting 5 from a number ending with a a digit smaller than 5, it is easier to first add 5 and then subtract 10.

23 - 5 = 18.
Also 23 + 5 = 28; 28 - 10 = 18.

Division by 5

Similarly, it's often more convenient instead to multiply first by 2 and then divide by 10.

1375/5 = 2750/10 = 275.

Multiplication by 5

It's often more convenient instead of multiplying by 5 to multiply first by 10 and then divide by 2.

137×5 = 1370/2 = 685.

Division by 5

Similarly, it's often more convenient instead to multiply first by 2 and then divide by 10.

1375/5 = 2750/10 = 275.

Division/multiplication by 4

Replace either with a repeated operation by 2.

124/4 = 62/2 = 31. Also,
124×4 = 248×2 = 496.

Division/multiplication by 25

Use operations with 4 instead.

37×25 = 3700/4 = 1850/2 = 925.

Division/multiplication by 8

Replace either with a repeated operation by 2.

124×8 = 248×4 = 496×2 = 992.

Division/multiplication by 125

Use operations with 8 instead.

37×125 = 37000/8 = 18500/4 = 9250/2 = 4625.

Misc

• a = x – y$ where – means subtraction, a dyadic use of the symbol
• a = -y$ where – means negative, a monadic use of the same symbol

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.

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.

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.

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:

• For each fraction in the list for which the multiplication of the accumulator and the fraction is an integer, replace the accumulator by the result of that multiplication.
• Repeat this rule until no fraction in the list produces an integer when multiplied by the accumulator, then halt.

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:

#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;

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

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;
}

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');
}

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);
	}
}

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;
}

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;
}