Shortest Sudoku Solver in Python - How does it work?

Question

I was playing around with my own Sudoku solver and was looking for some pointers to good and fast design when I came across this:

def r(a):i=a.find(\'0\');~i         


        

Answer 1

unobfuscating it:

def r(a):
    i = a.find('0') # returns -1 on fail, index otherwise
    ~i or exit(a) # ~(-1) == 0, anthing else is not 0
                  # thus: if i == -1: exit(a)
    inner_lexp = [ (i-j)%9*(i/9 ^ j/9)*(i/27 ^ j/27 | i%9/3 ^ j%9/3) or a[j] 
                   for j in range(81)]  # r appears to be a string of 81 
                                        # characters with 0 for empty and 1-9 
                                        # otherwise
    [m in inner_lexp or r(a[:i]+m+a[i+1:]) for m in'%d'%5**18] # recurse
                            # trying all possible digits for that empty field
                            # if m is not in the inner lexp

from sys import *
r(argv[1]) # thus, a is some string

So, we just need to work out the inner list expression. I know it collects the digits set in the line -- otherwise, the code around it makes no sense. However, I have no real clue how it does that (and Im too tired to work out that binary fancyness right now, sorry)