Hi there pythonistas! We all know that Sudoku is a great game. Some of us even bet on this game but did you know that you can use python to make a Sudoku solver ? In this post I am going to share with you a Sudoku solver written in python. From now on you will win all Sudoku challenges. However let me tell you that I am not the original writer of this script. I stumbled over this script on stack-overflow. So without wasting any time let me share the script with you:
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: r(a[:i]+m+a[i+1:]) if __name__ == '__main__': if len(sys.argv) == 2 and len(sys.argv) == 81: r(sys.argv) 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'
Hey there wait! let me share with you a shorter obfuscated version of the same Sudoku solving script. However this short version of Sudoku solver is a lot slower but I think that I should share it with you just to show you that even in python obfuscated code can be written. So here is the shorter obfuscated version:
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)
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