Budge

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The Budge programming language

Budge (bʌdʒ, b’dzh) is an esoteric programming language. It uses Gödel numbering to represent registers and their values by relying on the Fundamental Theorem of Arithmetic (prime factorization). The language uses similar constructs as FRACTRAN, however, it provides a more convenient way to construct loops and uses integers rather than fractions to denote instructions. It also abstracts prime numbers in the code, allowing for direct register access. A negative integer will decrease a register’s value, while a positive integer will increase a register’s value.

Syntax and semantics

Syntax:

<posn>  ::= "1" | "2" | ...               <negn> ::= "-1" | "-2" | ...
<stmt>  ::= <posn> | <negn> | "("<posn>","<stmts>")"
<stmts> ::= <stmt>","<stmts> | <stmt>     <code> ::= "("<stmts>")"

The code is represented with lists. Let p(n) denote the n-th prime number. Let sign(n) = 1 if n is positive, and -1 otherwise; this will determine whether we need to multiply or divide. We can now define the semantics of the language, i.e., <code> is one of:

  • Code without loops: A sequence of numbers, e.g. (1, 2, 3). The semantics are such that we iterate through all numbers in the sequence, one by one until the end of the sequence is reached. In each iteration of a number n, calculate i' = p(|n|)^{sign(n)}. For the given input i, if i * i' is a natural number, set i the value of i * i'. Otherwise, do not change i and skip to the next instruction. The first pair of parenthesis is the program’s entry point.
  • Code with loops: A sequence of numbers mixed with other sequences, e.g. (1, 2, (1, -1, 2, 3), 4). The instructions n_k, n_{k+1}, ... within (n_0, n_1, ..., (x, n_k, n_{k+1}, ...), ...) are repeated in a cycle until p(|x|) no longer divides i. The check is done only at the start of the iteration before calculating n_k. The calculation halts whenever i * p(|x|)^{-1} is no longer a natural number, i.e., the value of the register x (within i) is 1.

That is, the outermost sequence (main program) will not loop, and any sequence contained within it will loop based on the condition.

For the general audience: Imagine several jars with marbles. The command -n removes a marble (if any) from the jar number n, and n adds a marble to the jar number n. Loops can be interpreted as a way to repeat some commands while a jar n is not empty.

Budge addition visualization

Example program: Addition

Input: 2^x * 3^y. Output: 2^n.

The following code adds x and y: ((2, -2, 1))

Here is how it will be evaluated for input: i = 2^3 * 3^3 = 216. We iterate until i/p(2) is no longer a natural number, i.e., i/3:

  1. Initially, i = 216, so 216/3 = 72 which is a natural number, proceed with calculation.
  2. Calculate p(|n|)^{sign(n)} for n = -2: p(2)^{-1} = 3^{-1} = 1/3. Since 216 * 1/3 is a natural number, set i to 216 * 1/3 = 72.
  3. Calculate p(|n|)^{sign(n)} for n = 1: p(1)^{1} = 2^{1}. Since 72 * 2 is a natural number, set i to 72 * 2 = 144.
  4. At this point, we go back and check the condition if 144/3 is a natural number - proceed with the calculation.
  5. Calculate p(|n|)^{sign(n)} for n = -2: p(2)^{-1} = 3^{-1} = 1/3. Since 144 * 1/3 is a natural number, set i to 144 * 1/3 = 48.
  6. Calculate p(|n|)^{sign(n)} for n = 1: p(1)^{1} = 2^{1}. Since 48 * 2 is a natural number, set i to 48 * 2 = 96.
  7. At this point, we go back and check the condition if 96/3 is a natural number - proceed with the calculation.
  8. Calculate p(|n|)^{sign(n)} for n = -2: p(2)^{-1} = 3^{-1} = 1/3. Since 96 * 1/3 is a natural number, set i to 96 * 1/3 = 32.
  9. Calculate p(|n|)^{sign(n)} for n = 1: p(1)^{1} = 2^{1}. Since 32 * 2 is a natural number, set i to 32 * 2 = 64.
  10. Now the condition that 32/3 is a natural number is not fulfilled, so the calculation halts.

The value of i is now equal to 64, which is 2^6, i.e. the sum of 3 and 3.

Turing completeness

We claim Budge is Turing-complete. Since Budge can multiply, divide, run a primality test for numbers, and perform code repeatedly, it can simulate FRACTRAN. Since FRACTRAN is proven to be Turing complete, so is Budge.

External resources