“Venn diagram” partitioning

Paddy3118 wrote about partitioning elements in the same way a Venn diagram does. So, if we have sets A, B and C, the partitions are

These partitions can be organised in a table according to whether each of A, B and C is included or excluded:

A	B	C	partition
1	0	0	A-(B∪C)
0	1	0	B-(A∪C)
1	1	0	(A∩B)-C
0	0	1	C-(A∪B)
1	0	1	(A∩C)-B
0	1	1	(B∩C)-A
1	1	1	A∩B∩C

and the question is how to write a function that will return this sequence of partitions. Paddy3118 writes it like this:

def venndiv(*sets):
    bins = tuple(set() for n in range(2**len(sets)))
    for i, si in enumerate(sets): # Over every individual item of each set
        for item in si:
            binindex = 0
            for j, sj in enumerate(sets): # Test for membership of each set
                if item in sj:
                    binindex += 2**j # Binary count binning
            bins[binindex].add(item)
    return bins[1:] # bins[0] - an item in no set is impossible.

which is perfectly good: the explicit manipulation of index numbers is a little inelegant, perhaps. This post describes a different, pleasingly simple, way to express the venndiv function.

The first thing to notice is that there is a systematic way to build up the partitions by starting with A∪B∪C and taking intersections and differences with A, B and C. Where there is a 1 in the table, take an intersection with that set, and where there is a 0 subtract that set. You can do these steps in any order. Here is the table again, with systematic expressions included:

A	B	C	partition	systematic
0	0	0	∅	(((A∪B∪C) – C) – B) – A
1	0	0	A-(B∪C)	(((A∪B∪C) – C) – B) ∩ A
0	1	0	B-(A∪C)	(((A∪B∪C) – C) ∩ B) – A
1	1	0	(A∩B)-C	(((A∪B∪C) – C) ∩ B) ∩ A
0	0	1	C-(A∪B)	(((A∪B∪C) ∩ C) – B) – A
1	0	1	(A∩C)-B	(((A∪B∪C) ∩ C) – B) ∩ A
0	1	1	(B∩C)-A	(((A∪B∪C) ∩ C) ∩ B) – A
1	1	1	A∩B∩C	(((A∪B∪C) ∩ C) ∩ B) ∩ A

So we can build the partitions starting with the C’s, then the B’s and then the A’s, like this:

The bottom row has all the partitions, in the right order. So our algorithm is: make a recursive call to fetch the second-to-last row, then split each bin into (bin – A) and (bin ∩ A). In Python:

def venndiv(*sets):
    def vd(x, *xs):
        bins = vd(*xs) if xs else (set.union(*sets),)
        return sum(tuple( (bin - x, bin & x) for bin in bins ), ())
    return vd(*sets)[1:]

Or maybe it’s clearer as a loop, rather than a recursive function:

def venndiv(*sets):
    bins = ( set.union(*sets), )
    for x in reversed(sets):
        bins = sum(tuple( (bin - x, bin & x) for bin in bins ), ())
    return bins[1:]