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 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]) 

My own implementation solves Sudokus the same way I solve them in my head but how does this cryptic algorithm work?

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

Answer 1

Well, you can make things a little easier by fixing up the syntax:

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]) 

Cleaning up a little:

from sys import exit, argv def r(a):   i = a.find('0')   if i == -1:     exit(a)   for m in '%d' % 5**18:     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:])  r(argv[1]) 

Okay, so this script expects a command-line argument, and calls the function r on it. If there are no zeros in that string, r exits and prints out its argument.

(If another type of object is passed, None is equivalent to passing zero, and any other object is printed to sys.stderr and results in an exit code of 1. In particular, sys.exit(‘some error message’) is a quick way to exit a program when an error occurs. See http://www.python.org/doc/2.5.2/lib/module-sys.html)

I guess this means that zeros correspond to open spaces, and a puzzle with no zeros is solved. Then there’s that nasty recursive expression.

The loop is interesting: for m in'%d'%5**18

Why 5**18? It turns out that '%d'%5**18 evaluates to '3814697265625'. This is a string that has each digit 1-9 at least once, so maybe it’s trying to place each of them. In fact, it looks like this is what r(a[:i]+m+a[i+1:]) is doing: recursively calling r, with the first blank filled in by a digit from that string. But this only happens if the earlier expression is false. Let’s look at that:

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)]

So the placement is done only if m is not in that monster list. Each element is either a number (if the first expression is nonzero) or a character (if the first expression is zero). m is ruled out as a possible substitution if it appears as a character, which can only happen if the first expression is zero. When is the expression zero?

It has three parts that are multiplied:

• (i-j)%9 which is zero if i and j are a multiple of 9 apart, i.e. the same column.
• (i/9^j/9) which is zero if i/9 == j/9, i.e. the same row.
• (i/27^j/27|i%9/3^j%9/3) which is zero if both of these are zero:
• • i/27^j^27 which is zero if i/27 == j/27, i.e. the same block of three rows
• • i%9/3^j%9/3 which is zero if i%9/3 == j%9/3, i.e. the same block of three columns

If any of these three parts is zero, the entire expression is zero. In other words, if i and j share a row, column, or 3×3 block, then the value of j can’t be used as a candidate for the blank at i. Aha!

from sys import exit, argv def r(a):   i = a.find('0')   if i == -1:     exit(a)   for m in '3814697265625':     okay = True     for j in range(81):       if (i-j)%9 == 0 or (i/9 == j/9) or (i/27 == j/27 and i%9/3 == j%9/3):         if a[j] == m:           okay = False           break     if okay:       # At this point, m is not excluded by any row, column, or block, so let's place it and recurse       r(a[:i]+m+a[i+1:])  r(argv[1]) 

Note that if none of the placements work out, r will return and back up to the point where something else can be chosen, so it’s a basic depth first algorithm.

Not using any heuristics, it’s not particularly efficient. I took this puzzle from Wikipedia (http://en.wikipedia.org/wiki/Sudoku):

$ time python sudoku.py 530070000600195000098000060800060003400803001700020006060000280000419005000080079 534678912672195348198342567859761423426853791713924856961537284287419635345286179  real    0m47.881s user    0m47.223s sys 0m0.137s 

Addendum: How I would rewrite it as a maintenance programmer (this version has about a 93x speedup 🙂

import sys  def same_row(i,j): return (i/9 == j/9) def same_col(i,j): return (i-j) % 9 == 0 def same_block(i,j): return (i/27 == j/27 and i%9/3 == j%9/3)  def r(a):   i = a.find('0')   if i == -1:     sys.exit(a)    excluded_numbers = set()   for j in range(81):     if same_row(i,j) or same_col(i,j) or same_block(i,j):       excluded_numbers.add(a[j])    for m in '123456789':     if m not in excluded_numbers:       # At this point, m is not excluded by any row, column, or block, so let's place it and recurse       r(a[:i]+m+a[i+1:])  if __name__ == '__main__':   if len(sys.argv) == 2 and len(sys.argv[1]) == 81:     r(sys.argv[1])   else:     print 'Usage: python sudoku.py puzzle'     print '  where puzzle is an 81 character string representing the puzzle read left-to-right, top-to-bottom, and 0 is a blank'