20
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Inspired (with the explanation stolen from) this

Background

Say you have two lists A = [a_1, a_2, ..., a_n] and B = [b_1, b_2, ..., b_n] of integers. We say A is potentially-divisible by B if there is a permutation of B that makes a_i divisible by b_i for all i. The problem is then: is it possible to reorder (i.e. permute) B so that a_i is divisible by b_i for all i? For example, if you have

A = [6, 12, 8]
B = [3, 4, 6]

Then the answer would be True, as B can be reordered to be B = [3, 6, 4] and then we would have that a_1 / b_1 = 2, a_2 / b_2 = 2, and a_3 / b_3 = 2, all of which are integers, so A is potentially-divisible by B.

As an example which should output False, we could have:

A = [10, 12, 6, 5, 21, 25]
B = [2, 7, 5, 3, 12, 3]

The reason this is False is that we can't reorder B as 25 and 5 are in A, but the only divisor in B would be 5, so one would be left out.

Your task

Your task is, obviously, to determine whether two lists (given as input) are potentially divisible. You may take input in any accepted manner, as with output.

Duplicates in the lists are a possibility, and the only size restrictions on the integers are your language. All integers in both lists will be larger than 0, and both lists will be of equal size.

As with all s the output values must be 2 distinct values that represent true and false.

This is a so shortest code wins!

Test cases

Input, input => output

[6, 12, 8], [3, 4, 6] => True
[10, 5, 7], [1, 5, 100] => False
[14, 10053, 6, 9] [1,1,1,1] => True
[12] [7] => False
[0, 6, 19, 1, 3] [2, 3, 4, 5, 6] => undefined
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  • 3
    \$\begingroup\$ @Shaggy from the question: both lists will be of equal size \$\endgroup\$ – caird coinheringaahing Aug 27 '17 at 20:16
  • 2
    \$\begingroup\$ Why is the last test case undefined? \$\endgroup\$ – Dennis Aug 27 '17 at 21:29
  • 1
    \$\begingroup\$ @Dennis one of the lists has a 0 in it \$\endgroup\$ – caird coinheringaahing Aug 27 '17 at 21:31
  • 1
    \$\begingroup\$ Right. (Not sure why, 0 is divisible by all integers.) Do the two outputs have to be truthy and falsy, or just consistent? \$\endgroup\$ – Dennis Aug 27 '17 at 21:41
  • \$\begingroup\$ @Dennis 1) it's in case 0 is in the second list, to avoid 0 division errors 2) just consistent \$\endgroup\$ – caird coinheringaahing Aug 27 '17 at 21:55

23 Answers 23

10
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Jelly, 5 bytes

Œ!%ḄẠ

Returns 0 for True, 1 for False.

Try it online!

How it works

Œ!%ḄẠ  Main link. Arguments: A, B (arrays)

Œ!     Generate all permutations of A.
  %    Take each permutation modulo B (element-wise).
   Ḅ   Convert all resulting arrays from binary to integer.
       This yields 0 iff the permutation is divisible by B.
    Ạ  All; yield 0 if the result contains a 0, 1 otherwise.
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9
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Husk, 7 6 5 bytes

Saved 2 bytes thanks to @Zgarb

▼▲‡¦P

Takes argents in reverse order and returns 1 for True and 0 for False.

Try it online!

Explanation

    P     -- Permutations of the first argument
  ‡       -- Deep zip (vectorises function) with second argument
   ¦      --   Does x divide y
 ▲        -- Get the maximum of that list, returns [1,1...1] if present
▼         -- Get the minimum of that list, will return 0 unless the list is all 1s
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  • \$\begingroup\$ VΠMz¦P should work for 6 bytes. \$\endgroup\$ – Zgarb Aug 27 '17 at 19:46
  • \$\begingroup\$ Are those considered "two distinct values"? \$\endgroup\$ – geokavel Aug 27 '17 at 19:49
  • \$\begingroup\$ Oh, and Mz can be . \$\endgroup\$ – Zgarb Aug 27 '17 at 20:06
  • \$\begingroup\$ I think you need ▼▲ instead of ▲▼. Nice idea in any case! \$\endgroup\$ – Zgarb Aug 27 '17 at 20:24
5
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05AB1E, 7 bytes

Input: takes lists B and A (reversed order)
Output: 1 when true, 0 otherwise

œvIyÖPM

Try it online!

Explanations:

œvIyÖPM    Complete program
œ          Pushes all permutations of B as a list
 v         For each permutation
  I        Pushes last input on top of the stack
   yÖ      Computes a % b == 0 for each element of A and B
     P     Pushes the total product of the list
      M    Pushes the largest number on top of the stack
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5
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MATL, 8 7 6 bytes

1 byte off using an idea from Dennis' Jelly answer

Y@\!aA

Inputs are B, then A. Output is 0 if divisible or 1 if not.

Try it online!

Explanation

Y@     % Implicit input: row vector B. Matrix of all permutations, each on a row
\      % Implicit input: row vector A. Modulo, element-wise with broadcast. Gives
       % a matrix in which each row contains the moduli of each permutation of B
       % with respect to A
!a     % True for rows that contain at least a nonzero value
A      % True if all values are true. Implicit display
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3
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Mathematica, 52 bytes

Cases[Permutations@#2,p_/;And@@IntegerQ/@(#/p)]!={}& 

thanks @ngenisis for -5 bytes

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  • 2
    \$\begingroup\$ Cases is generally shorter: Cases[Permutations@#2,p_/;And@@IntegerQ/@(#/p)]!={}& \$\endgroup\$ – ngenisis Aug 27 '17 at 20:40
3
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JavaScript (ES6), 67 63 bytes

Returns a boolean.

f=([x,...a],b)=>!x||b.some((y,i)=>x%y?0:f(a,c=[...b],c[i]=1/0))

Test cases

f=([x,...a],b)=>!x||b.some((y,i)=>x%y?0:f(a,c=[...b],c[i]=1/0))

console.log(f([6, 12, 8],[3, 4, 6]))        // true
console.log(f([10, 5, 7],[1, 5, 100]))      // false
console.log(f([14, 10053, 6, 9],[1,1,1,1])) // true
console.log(f([12],[7]))                    // false

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3
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Haskell, 79 74 68 62 61 bytes

import Data.List
f a=any((<1).sum.zipWith rem a).permutations

Try it online!

Saved 1 byte thanks to @nimi

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  • 1
    \$\begingroup\$ 61 bytes: f a=any((<1).sum.zipWith rem a).permutations. \$\endgroup\$ – nimi Aug 27 '17 at 21:05
3
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R + combinat, 69 66 58 bytes

-3 bytes thanks to Jarko Dubbeldam

another -8 bytes thanks to Jarko

function(a,b)any(combinat::permn(b,function(x)all(!a%%x)))

oddly, R doesn't have a builtin for generating all permutations. Returns a boolean.

Additionally, with Jarko's second improvement, any coerces the list to a vector of logical with a warning.

Try it online! (r-fiddle)

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  • 1
    \$\begingroup\$ All(x<1) is longer than any(!x) and you should be able to replace sum with any \$\endgroup\$ – JAD Aug 28 '17 at 15:17
  • \$\begingroup\$ @JarkoDubbeldam good call. thank you. \$\endgroup\$ – Giuseppe Aug 28 '17 at 15:20
  • \$\begingroup\$ Oh, and you can omit unlist, yay for implicit coercion. \$\endgroup\$ – JAD Aug 28 '17 at 15:35
  • \$\begingroup\$ @JarkoDubbeldam excellent. \$\endgroup\$ – Giuseppe Aug 28 '17 at 15:39
2
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Mathematica, 42 bytes

MemberQ[Permutations@#2,a_/;And@@(a∣#)]&
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1
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Jelly, 7 bytes

Œ!ọẠ€1e

Try it online!

Factorial complexity in the length of the list.

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  • \$\begingroup\$ Does Jelly really not have an any builtin? TIL \$\endgroup\$ – caird coinheringaahing Aug 27 '17 at 19:19
1
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Pyth - 11 bytes

sm!s%Vdvz.p

Test Suite.

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  • \$\begingroup\$ You beat me to it :((( - Was just about to post something similar \$\endgroup\$ – Mr. Xcoder Aug 27 '17 at 19:02
1
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J, 27 bytes

0=[:*/(A.~i.@!@#)@]+/@:|"1[

Try it online!

Takes the first list as the left argument and the second list as the right.

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  • 1
    \$\begingroup\$ 21 bytes (|"1~e.~0*[)i.@!@#A.] \$\endgroup\$ – miles Aug 28 '17 at 19:10
1
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CJam, 20 17 bytes

:A;e!{A\.%:+!}#W>

Test Version

Function that takes array B as the first argument and array A as the second argument. Note that in the test version I switch the order to A then B.

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1
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JavaScript (ES6), 100 bytes

f=(a,b)=>!a[0]||a.some((c,i)=>b.some((d,j)=>c%d<1&f(e=[...a],d=[...b],e.splice(i,1),d.splice(j,1))))

Somewhat inefficient; an extra & would speed it up.

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1
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PHP, 112 180 178 bytes

I was thinking too short.

function($a,$b){for($p=array_keys($b);++$i<count($b);){foreach($b as$k=>$x)$f|=$a[$k]%$x;if($f=!$f)return 1;$p[$i]?[$b[$j],$b[$i],$i]=[$b[$i],$b[$j=$i%2*--$p[$i]],0]:$p[$i]=$i;}}

anonymous function takes two arrays, returns NULL for falsy and 1 for truthy.
Throws an error if second array contains 0.

Try it online.

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  • \$\begingroup\$ Prints the wrong result for $f([6,5],[3,5]). \$\endgroup\$ – nwellnhof Aug 28 '17 at 10:16
  • \$\begingroup\$ @nwellnhof fixed. thanks for noticing. \$\endgroup\$ – Titus Aug 28 '17 at 14:04
1
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C (gcc), 191 bytes

#define F(v)for(i=0;i<v;++i){
#define X if(f(s,n,a,b))return 1
j;f(s,n,a,b,i)int*a,*b;{if(--n){F(n)X;j=i*(n%2);b[j]^=b[n];b[n]^=b[j];b[j]^=b[n];}X;}else{F(s)if(a[i]%b[i])return 0;}return 1;}}

Try it online!

Usage: f(int size, int size, int *a, int *b)

returns 1 if divisable, 0 otherwise. See example usage on TIO.

(Gotta do permutations the hard way in C, so this is hardly competitive)

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1
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Perl 6, 38 bytes

Actually @nwellnhof's answer seems to be too much readable, so I set out to follow the fine Perl tradition of write-only code :—).

1 byte saved thanks to @nwellnhof.

{min max (@^a,) XZ%% @^b.permutations}

Try it online!

What does it do: It's an anonymous function that takes two list arguments. When we say @^a, we mean the first one, when @^b, it's the second one.

(@^a,) is a list containing the list @^a. @^b.permutations is the list of all the permutations of @^b. The "XZ%%" operator makes all possible pairs of that one list on the left and all the permutations on the right, and uses the operator "Z%%" on them, which is the standard "zip" operation using the divisibility operator %%.

The max operator gives the largest element of the list (in this case, it's the list that has the most True's in it). We then reduce it using the logical AND operator to see if all elements of that "most true" list are true, and that's the result. It's nearly exact copy of what @nwellnhof wrote, just using obscure operators to shave off bytes.

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  • \$\begingroup\$ It says permutations, it is clearly far too readable ;) \$\endgroup\$ – caird coinheringaahing Aug 28 '17 at 18:06
  • \$\begingroup\$ Well, Perl 6 has a really powerful introspection model. Perhaps I could study it in order to obscure that call? :D \$\endgroup\$ – Ramillies Aug 28 '17 at 18:08
  • \$\begingroup\$ Replace [&&] with min to save another byte. \$\endgroup\$ – nwellnhof Aug 28 '17 at 19:31
  • \$\begingroup\$ You can remove the spaces around the XZ%% \$\endgroup\$ – Jo King Mar 4 at 0:23
  • \$\begingroup\$ I wish something like {all (@^a,)Z%%@^b.permutations.any} were possible \$\endgroup\$ – Jo King Mar 4 at 0:28
1
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Brachylog, 6 bytes

pᵐz%ᵛ0

Try it online!

The predicate succeeds if the two lists are potentially divisible and fails if they are not.

pᵐ        For some pair of permutations of the two input lists,
  z       for each pair of corresponding elements
   %ᵛ0    the first mod the second is always zero.
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0
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Python 2, 92 bytes

lambda a,b:any(all(x%y<1for x,y in zip(a,p))for p in permutations(b))
from itertools import*

Try it online!

Yer basic implementation.

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0
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Python 2, 90 bytes

lambda a,b:any(sum(map(int.__mod__,a,p))<1for p in permutations(b))
from itertools import*

Try it online!

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0
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Ruby, 56 bytes

->x,y{x.permutation.any?{|p|p.zip(y).all?{|a,b|a%b==0}}}

Try it online!

Fairly straightforward, exploits the fact that permutation exists.

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0
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Scala, 60 Bytes

Golfed:

a=>b=>b.permutations exists(a zip _ forall(p=>p._1%p._2==0))

Ungolfed:

a=>b=>         // Function literal taking 2 lists of integers, a and b.
b.permutations // All permutations of b.
exists(        // Whether the given function is true for any element.
a zip _        // Zips a and the current permutation of b into a list of pairs.
forall(        // Whether the given function is true for all elements.
p=>            // Function literal taking a pair of integers.
p._1%p._2==0)) // If the remainder of integer division between the members of the pair is 0.
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0
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Japt, 12 11 bytes

Outputs true or false.

Vá de@gY vX

Test it


Explanation

Implicit input of arrays U & V (A & B, respectively)

Generate an array of all permutations of V.

d

Check if any of the elements (sub-arrays) return true.

e@

Check if every element in the current sub-array returns true when passed through the following function, with X being the current element and Y the current index.

gY

Get the element in U at index Y.

vX

Check if it's divisible by X.

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