Question

what I want to do is to create an algorithm, able to find all the possible bijections between two sets of objects.

A simple example, let’s suppose we have two arrays {1,2,3} {4,5,6}.

The algorithm should give me 3!= 3*2*1 =6 bijections which are the following:

1-4 2-5 3-6 \
1-4 2-6 3-5\
1-5 2-4 3-6\
1-5 2-6 3-4\
1-6 2-5 3-4 \
1-6 2-4 3-5\

Even though it seems simple at first place, I am quite stuck. Is there any standard algorithm in the theory of combinatorics, bijections or permutations to solve this problem?
Thank you in advance.

Christina

Answer 1

You should do it recursively, “choose” one variable from each – and add it to the solution – do it for all possibilities, and narrow your possible choices at each recursive call.

Pseudo code should be something like [assuming |S1| == |S2|]:

getAllBijections(S1,S2,sol):
   if (S1 is empty):
       print sol
   else:
       first <- S1.first
       S1 <- S1.deleteFirst()
       for each e in S2:
           S2.remove(e)
           sol.append(first,e)
           getAllBijections(S1,S2,sol) // note we are invoking with modified S1,S2,sol
           sol.removeLast() //remove the last sol
           S2.add(e) //return e to S2
       end for
   end if

Note that it indeed generate n! possibilities, because for each iteration you have one less element to choose from, resulting in total of n * (n-1) * ... * 1 = n! possibilities, as expected..