Python sudoku shortest code ever

def r(a):i=a.find('0');~i or exit(a);[m
in[(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)]or r(a[:i]+m+a[i+1:])for m in'%d'%5**18]

from sys import*;r(argv[1]) 

Src:

http://www.scottkirkwood.com/2006/07/shortest-sudoku-solver-in-python.html

Original Source:

  Mark Byer's site has some more Sudoku solvers    (can be found on web archive as web site is down)

Expalanation

https://jakevdp.github.io/blog/2013/04/15/code-golf-in-python-sudoku/

NOTE

"A=0 #False

B=20#True

print(A or B) #B

A=1 #True

B=0#False

print(A or B) #A

you might expect the result to be either True or False. Instead, Python does something a bit clever. If the result is False, it returns A (which, naturally, evaluates to False). 

If the result is True, it returns A if A evaluates to True, and B otherwise. 

We can use this fact to remove the if statement completely from the set comprehension