In [1]:
import itertools

def overlap(a, b, min_length=3):
    """ Return length of longest suffix of 'a' matching
        a prefix of 'b' that is at least 'min_length'
        characters long.  If no such overlap exists,
        return 0. """
    start = 0  # start all the way at the left
    while True:
        start = a.find(b[:min_length], start)  # look for b's suffx in a
        if start == -1:  # no more occurrences to right
            return 0
        # found occurrence; check for full suffix/prefix match
        if b.startswith(a[start:]):
            return len(a)-start
        start += 1  # move just past previous match

def scs(ss):
    """ Returns shortest common superstring of given
        strings, which must be the same length """
    shortest_sup = None
    for ssperm in itertools.permutations(ss):
        sup = ssperm[0]  # superstring starts as first string
        for i in range(len(ss)-1):
            # overlap adjacent strings A and B in the permutation
            olen = overlap(ssperm[i], ssperm[i+1], min_length=1)
            # add non-overlapping portion of B to superstring
            sup += ssperm[i+1][olen:]
        if shortest_sup is None or len(sup) < len(shortest_sup):
            shortest_sup = sup  # found shorter superstring
    return shortest_sup  # return shortest

In [2]:
scs(['BAA', 'AAB', 'BBA', 'ABA', 'ABB', 'BBB', 'AAA', 'BAB'])


In [3]:
scs(['ABCD', 'CDBC', 'BCDA'])