Python 中最短的数独解决器-它是如何工作的？

A lot of the short sudoku solvers just recursively try every possible legal number left until they have successfully filled the cells. I haven't taken this apart, but just skimming it, it looks like that's what it does.

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)

r(a) is a recursive function which attempts to fill in a 0 in the board in each step.

i=a.find('0');~i or exit(a) is the on-success termination. If no more 0 values exist in the board, we're done.

m is the current value we will try to fill the 0 with.

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)] evaluates to truthy if it is obivously incorrect to put m in the current 0. Let's nickname it "is_bad". This is the most tricky bit. :)

is_bad or r(a[:i]+m+a[i+1:] is a conditional recursive step. It will recursively try to evaluate the next 0 in the board iff the current solution candidate appears to be sane.

for m in '%d'%5**18 enumerates all the numbers from 1 to 9 (inefficiently).

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 3x3 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'

The code doesn't actually work. You can test it yourself. Here is a sample unsolved sudoku puzzle:

807000003602080000000200900040005001000798000200100070004003000000040108300000506

You can use this website (http://www.sudokuwiki.org/sudoku.htm), click on import puzzle and simply copy the above string. The output of the python program is: 817311213622482322131224934443535441555798655266156777774663869988847188399979596

which does not correspond to the solution. In fact you can already see a contradiction, two 1s in the first row.