Challenges

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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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# 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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# 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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# 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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#### 1 comment thread

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

T=V£*Xr÷Ãæv1)?ßT:U


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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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# 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 comment thread

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

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