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Golf a FRACTRAN interpreter

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Description

From the Esolangs wiki,

In Fractran,

  • a program consists of a finite list of positive, rational numbers.
  • The input to a program is a positive integer n.
  • The list is then searched in order, for a rational number $p/q$ such that $n×p/q$ is an integer.
  • Then n is replaced by $n×p/q$ and the search is restarted from the beginning of the list.
  • The program halts when no such number $p/q$ can be found, and the final n becomes the output from the program.
  • Output the final value of $n$.

Your task is to implement an interpreter for this language.

Input

You are to take two inputs:

  • $n$, an integer
  • $A$, an array of fractions, which may be taken as a list of pairs, or in the rational datatype of your language.

Output

A single integer, the final value of $n$.

Test Cases

Formatted as

program
input
output
78/55, 5/3, 1/5, 11/2, 5/7
1096135733
328842888196762472689573703

3/2
1296
6561

455/33, 11/13, 1/11, 3/7, 11/2, 1/3
72
15625
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2 comments

I don't understand how the second test case can ever terminate. Since there is a fraction 55/1 at the end and two integers produce an integer when multiplied, it would never get past that fraction. user‭ 13 days ago

yeah, I wrote that wrong. It's an infinite loop which cycles through 2^n, where n is prime. Razetime‭ 13 days ago

2 answers

+4
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APL (Dyalog Unicode), 33 bytes

{×x←⊃(2⌷⍵)(÷⍨(/⍨)0=|)⍺×1⌷⍵:x∇⍵⋄⍺}

Try it online!

The first case doesn't work because it gets really big, but the other two do. The input is taken on the left and the fractions are taken as a table on the right.

{×x←⊃(2⌷⍵)(÷⍨(/⍨)0=|)⍺×1⌷⍵:x∇⍵⋄⍺}
                      ⍺×            ⍝ n multiplied by
                        1⌷⍵         ⍝ The first row of the right argument (every p)
      (2⌷⍵)                        ⍝ Second row of right arg (all q's)
                    |              ⍝ All n×p modulo q
                  0=               ⍝ Check which ones are 0 (rational)
           ÷⍨                      ⍝ Make another vector of 'n×p÷q's
               /⍨                  ⍝ And keep the ones that were rational
    ⊃                              ⍝ Pick the first (0 if empty)
  x←                               ⍝ Assign to x
 ×                                 ⍝ Sign of x
                                 ⍺ ⍝ If sign is 0, return n
                            x∇⍵   ⍝ Otherwise, call again with x as new n
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2 comments

finding tests for this is a bit of a pain.. everything's in some weird modulo or prime factorization format aaaaa Razetime‭ 11 days ago

I'll change the first test case to your output in ⎕FR←1287⋄⎕PP←34 Razetime‭ 11 days ago

+1
−0

Python 3, 76 bytes

def f(p,n):l=[n*p//q for(p,q)in p if n%q==0];return n if l==[]else f(p,l[0])

Try it online!

This code assumes that the fractions are given as completely cancelled pairs of integers.

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