You are given a set of positive integers. You must arrange them into pairs such that:

  • Each pair contains 2 numbers, one of which is a multiple of another. For example, 8 is a multiple of 4, and 9 is a multiple of 9.
  • If the same number occurs many times in the initial set, it can be used that many times in the pairs; a number can even be paired with another occurence of the same number
  • The maximum possible number of pairs is obtained.

Output must be the number of pairs. Shortest code wins.

Sample data

2,3,4,8,9,18 -> 3

7,14,28,42,56 -> 2

7,1,9,9,4,9,9,1,3,9,8,5 -> 6

8,88,888,8888,88888,888888 -> 3

2,6,7,17,16,35,15,9,83,7 -> 2

  • 3
    \$\begingroup\$ Anyone know whether this problem is NP-complete? I think the smallest "hard" set is 2,3,4,8,9,18. (Each number in that list is a factor and/or multiple of at least two other numbers in the list, but it has only one solution.) \$\endgroup\$ – Neil Nov 26 '15 at 22:55

Haskell, 109 107 76 70 bytes

Thanks to nimi for saving 33 bytes and teaching me some more Haskell. :)
Thanks to xnor for saving another 6 bytes.

import Data.List
f l=maximum$0:[1+f t|a:b:t<-permutations l,a`mod`b<1]

Yay, my first Haskell golf. It works the same as all the answers so far (well, not quite: it only counts the length of the longest prefix of valid pairs in each permutation, but that's equivalent and is actually what my original CJam code did).

For extra golfitude it's also extra inefficient by recursively generating all permutations of the suffix each time the first two elements of a permutation are a valid pair.

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  • \$\begingroup\$ Is the f= necessary? \$\endgroup\$ – Alex A. Nov 25 '15 at 17:22
  • \$\begingroup\$ @AlexA. I'm not sure what the standard policy on PPCG for unnamed functions in Haskell is, but I've checked a few other Haskell answers and they used named functions. Also, you'd technically have to use parentheses around the function if you wanted to use it as an unnamed function, so that would be the same byte count anyway I guess. \$\endgroup\$ – Martin Ender Nov 25 '15 at 17:26
  • \$\begingroup\$ @nimi Thanks for letting me know. :) Do you see anything else that could be shortened? The import for chunksOf is painful. I don't really know Haskell's standard library to be able to tell if there is a shorter equivalent function. I tried implementing it myself, but it came out two or three bytes longer than the import. \$\endgroup\$ – Martin Ender Nov 26 '15 at 10:35
  • \$\begingroup\$ ohhh, catching both [] and [_] at the same time by putting g x=[] second is really clever. I'll give that a try. Thanks :) \$\endgroup\$ – Martin Ender Nov 26 '15 at 10:56
  • \$\begingroup\$ Looks a bit shorter to define the whole function recursively: f l=maximum$0:[1+f t|(a:b:t)<-permutations l,a`mod`b<1]. \$\endgroup\$ – xnor Nov 26 '15 at 11:19

CJam, 22 18 bytes


Try it online.

Expects input in the form of a CJam-style list.

This is a bit inefficient for larger lists (and Java will probably run out of memory unless you give it more).


q~     e# Read and evaluate input.
e!     e# Get all distinct permutations.
{      e# Map this block onto each permutation...
  2/   e#   Split the list into (consecutive) pairs. There may be a single element at the
       e#   end, which doesn't participate in any pair.
  ::%  e#   Fold modulo onto each chunk. If it's a pair, this computes the modulo, which
       e#   yields 0 if the first element is a multiple of the second. If the list has only
       e#   one element, it will simply return that element, which we know is positive.
  0e=  e#   Count the number of zeroes (valid pairs).
:e>    e# Find the maximum of the list by folding max() onto it.
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  • \$\begingroup\$ It does not give output for [1 2 3 4 5 6 7 8 9 10] However [7 1 9 9 4 9 9 1 3 9 8 1] which is a longer list, works properly. Why is that? \$\endgroup\$ – ghosts_in_the_code Nov 25 '15 at 11:25
  • \$\begingroup\$ @ghosts_in_the_code Because the former has more distinct permutations. 10! = 3628800, but 12! / 5! / 3! = 665280. So it runs out of memory for the first case. If you ran it from the console with the Java interpreter, you could tell Java to use more memory and the first case would work as well (although it might take a while, don't know). \$\endgroup\$ – Martin Ender Nov 25 '15 at 12:29

Pyth, 13 bytes


The time and storage complexity is really terrible. The first thing I do is to create a list with all permutations of the originally list. This takes n*n! storage. Input lists with length 9 already take quite a long time.

Try it online: Demonstration or Test Suite


            Q   read the list of integer
          .p    create the list of all permutations
  m             map each permutation d to:
      cd2          split d into lists of length 2
    %M             apply modulo to each of this lists
   /     Z         count the zeros (=number of pairs with the first 
                   item divisible by the second)
 S              sort these values
e               and print the last one (=maximum)
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Mathematica, 95 93 87 83 79 60 58 bytes


Takes a few seconds for the larger examples.

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Matlab (120+114=234)

  function w=t(y,z),w=0;for i=1:size(z,1),w=max(w,1+t([y,z(i,:)],feval(@(d)z(d(:,1)&d(:,2),:),~ismember(z,z(i,:)))));end


  a=input('');h=bsxfun(@mod,a,a');v=[];for i=1:size(h,1) b=find(~h(i,:));v=[v;[(2:nnz(b))*0+i;b(b~=i)]'];end;t([],v)

  • the topper function is called by the main part.

  • the input is in the form [. . .]

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  j=@(e,x)['b(:,' num2str(e(x)) ')'];r=@(e,y)arrayfun(@(t)['((mod(' j(e,1) ',' j(e,t) ')==0|mod(' j(e,t) ',' j(e,1) ')==0)&',(y<4)*49,[cell2mat(strcat(r(e(setdiff(2:y,t)),y-2),'|')) '0'],')'],2:y,'UniformOutput',0);a=input('');i=nnz(a);i=i-mod(i,2);q=0;while(~q)b=nchoosek(a,i);q=[cell2mat(strcat((r(1:i,i)),'|')) '0'];q=nnz(b(eval(q(q~=0)),:));i=i-2;end;fix((i+2)/2)

  • This apparently is longer, but oneliner and executive, and I managed to escape perms function because it takes forever.

  • This function takes many reprises to run quiet well due to anonymous functions, I m open to suggestions here :)

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