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Golf a FRACTRAN interpreter

+6
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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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General comments (4 comments)

6 answers

+6
−0

APL (Dyalog Unicode), 33 bytes

{×x←⊃(2⌷⍵)(÷⍨(/⍨)0=|)⍺×1⌷⍵:x∇⍵⋄⍺}

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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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General comments (2 comments)
+4
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Brachylog, 16 bytes

g₂;.z∋zbᵗ×ᵐ/ℤ↰|w

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This is a function submission, which inputs n from the left and A from the right (as a list of two-element lists). Output is to standard output.

Explanation

g₂;.z∋zbᵗ×ᵐ/ℤ↰|w
g₂                  place two list wrappers around the left input
  ;.                append the right input
    z               cycling zip; because the left input is length 1,
                      this pairs it with each element of the right input
     ∋              find {the first} element for which no assertions fail
[at this point, the element looks like [[n], [num, denom]]
 for some pair [num, denom] in A]
      z             rearrange to [[n, num], [n, denom]]
       bᵗ           remove first of last pair ([[n, num], [denom]])
         ×ᵐ         take product of each inner list ([n×num, denom])
           /ℤ       divide; assert result (n×num÷denom) is an integer
             ↰      recurse (loop back to start of program)
              |     if all else fails (i.e. ∋ found no elements)
               w    output the current value to standard output
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+3
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Python 3, 76 71 bytes

Saved 5 bytes thanks to user

def f(p,n):l=[n*p//q for(p,q)in p if n%q<1];return f(p,l[0])if l else n

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

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General comments (1 comment)
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JavaScript, 44 bytes

a=>g=n=>a.every(([p,q])=>(x=n*p/q)%1)?n:g(x)

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Japt v2.0a0, 18 bytes

The can be removed if we can take an array of floats instead.

T=V£*Xr÷Ãæv1)?ßT:U

Try it

T=V£*Xr÷Ãæv1)?ßT:U     :Implicit input of integer U and 2D-array V
T=                     :Assign to variable T
  V£                   :  Map each X in V
    *                  :    Multiply U by
     Xr÷               :    X reduced by division
        Ã              :  End map
         æ             :  Get the first element that
          v1           :    Is divisible by 1
            )          :End assignment
             ?         :If truthy (not undefined)
              ßT       :Recursive call with arguments U=T and V unchanged
                :U     :Else return U
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+1
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Ruby, 51 50 bytes

-1 from Razetime

->p,n{while i=p.find{(_1*n).to_f%1==0}
n*=i
end
n}

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-1 changing to `n*=i` (1 comment)

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