Comments on Golf a FRACTRAN interpreter
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Golf a FRACTRAN interpreter
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
[APL (Dyalog Unicode)], 33 byt …
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[Brachylog], 16 bytes g …
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[Python 3], 76 71 bytes Sav …
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JavaScript, 44 bytes …
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Japt v2.0a0, 18 bytes The ` …
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[Ruby], 51 50 bytes -1 from R …
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J, 36 char ``` {.2{(}.,(=< …
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Ruby, 44 bytes ->p,n{n=I …
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Post
APL (Dyalog Unicode), 33 bytes
{×x←⊃(2⌷⍵)(÷⍨(/⍨)0=|)⍺×1⌷⍵:x∇⍵⋄⍺}
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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