Set consolidation

See also:

Given two sets of items then if any item is common to any set then the result of applying consolidation to those sets is:

The two input sets if no common item exists between those two sets of items.
Or one set of all the items from both sets if they share a common item.

Given N sets of items then the result is the same as repeatedly replacing all combinations of two sets by their consolidation until no further consolidation between set pairs is possible.

Example 1:

   Given the two sets {A,B} and {C,D} then there is no common
   element between the sets and the result is the same as the
   input.

Example 2:

   Given the two sets {A,B} and {B,D} then there is a common
   element B between the sets and the result is the set {B,D,A}.
   
   (Note that order of items in a set is immaterial - {A,B,D} is
   the same as {B,D,A} and {D,A,B} etc).

Example 3:

   Given the three sets {A,B} and {C,D} and {D,B} then there is no
   common element between the sets {A,B} and {C,D} but the sets
   {A,B} and and  {D,B} do share a common element that
   consolidates to produce the result {B,D,A}.
   On examining this result with the remaining set {C,D} they
   share a common element and so consolidate to the final output
   of the single set {A,B,C,D}

Example 4:

   The consolidation of the five sets:
       {H,I,K}, {A,B}, {C,D}, {D,B}, and {F,G,H}
   Is the two sets:
       {'A', 'C', 'B', 'D'}, and {'G', 'F', 'I', 'H', 'K'}

Python

The docstring contains solutions to all the examples as well as a check to show the order-independence of the sets given to the consolidate function. <lang python>def consolidate(sets):

   
   >>> # Define some variables
   >>> A,B,C,D,E,F,G,H,I,J,K = 'A,B,C,D,E,F,G,H,I,J,K'.split(',')
   >>> # Consolidate some lists of sets
   >>> consolidate([{A,B}, {C,D}])
   [{'A', 'B'}, {'C', 'D'}]
   >>> consolidate([{A,B}, {B,D}])
   [{'A', 'B', 'D'}]
   >>> consolidate([{A,B}, {C,D}, {D,B}])
   [{'A', 'C', 'B', 'D'}]
   >>> consolidate([{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}])
   [{'A', 'C', 'B', 'D'}, {'G', 'F', 'I', 'H', 'K'}]
   >>> consolidate([{A,H}, {H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}])
   [{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}]
   >>> consolidate([{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}, {A,H}])
   [{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}]
   >>> # Confirm order-independence
   >>> import itertools
   >>> sets = [{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}, {A,H}]
   >>> answer = consolidate(sets)
   >>> for perm in itertools.permutations(sets):
           assert consolidate(perm) == answer

   >>> len(list(itertools.permutations(sets)))
   720
   >>>     
   
   setlist = [s for s in sets if s]
   for i, s1 in enumerate(setlist):
       if s1:
           for j, s2 in enumerate(setlist[i+1:], i+1):
               intersection = s1.intersection(s2)
               if intersection:
                   s2.update(s1)
                   s1.clear()
                   s1 = s2
   return [s for s in setlist if s]

</lang>