python_scripts_rcsmit/sudoku.py at master · rcsmit/python_scripts_rcsmit · GitHub

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# -*- coding: utf-8 -*-
"""
Created on Thu Oct 29 21:39:12 2020

@author: rcxsm
"""

#https://towardsdatascience.com/solve-sudokus-automatically-4032b2203b64

sudoku =    [
            [8, 1, 0, 0, 3, 0, 0, 2, 7],
            [0, 6, 2, 0, 5, 0, 0, 9, 0],
            [0, 7, 0, 0, 0, 0, 0, 0, 0],
            [0, 9, 0, 6, 0, 0, 1, 0, 0],
            [1, 0, 0, 0, 2, 0, 0, 0, 4],
            [0, 0, 8, 0, 0, 5, 0, 7, 0],
            [0, 0, 0, 0, 0, 0, 0, 8, 0],
            [0, 2, 0, 0, 1, 0, 7, 5, 0],
            [3, 8, 0, 0, 7, 0, 0, 4, 2]]

def printsudoku():
    print("\n\n\n\n\n")
    for i in range(len(sudoku)):
        line = ""
        if i == 3 or i == 6:
            print("---------------------")
        for j in range(len(sudoku[i])):
            if j == 3 or j == 6:
                line += "| "
            line += str(sudoku[i][j])+" "
        print(line)

def findNextCellToFill(sudoku):
    for x in range(9):
        for y in range(9):
            if sudoku[x][y] == 0:
                return x, y
    return -1, -1

def isValid(sudoku, i, j, e):
    rowOk = all([e != sudoku[i][x] for x in range(9)])
    if rowOk:
        columnOk = all([e != sudoku[x][j] for x in range(9)])
        if columnOk:
            secTopX, secTopY = 3*(i//3), 3*(j//3)
            for x in range(secTopX, secTopX+3):
                for y in range(secTopY, secTopY+3):
                    if sudoku[x][y] == e:
                        return False
            return True
    return False

def solveSudoku(sudoku, i=0, j=0):
    i, j = findNextCellToFill(sudoku)
    if i == -1:
        return True
    for e in range(1, 10):
        if isValid(sudoku, i, j, e):
            sudoku[i][j] = e
            if solveSudoku(sudoku, i, j):
                return True
            sudoku[i][j] = 0
    return False

printsudoku()

solveSudoku(sudoku)
printsudoku()