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35-liner: Backtracking Algo solution in Speedy category for Sudoku Solver by Stensen
def is_empty(board):
for i in range(len(board)):
for j in range(len(board[0])):
if board[i][j] == 0: return (i, j) # Empty Coordinates: col, row
return None
def is_valid(board, number, position):
#=> Check Rows <=#
for i in range(len(board)):
if board[position[0]][i] == number and position[1] != i: return False
#=> Check Columns <=#
for i in range(len(board)):
if board[i][position[1]] == number and position[0] != i: return False
#=> Check Boxes <=#
xbox = position[1] // 3
ybox = position[0] // 3
for i in range(ybox * 3, ybox * 3 + 3):
for j in range(xbox * 3, xbox * 3 + 3):
if board[i][j] == number and (i, j) != position: return False
return True
def backtrack_solve(board):
empty = is_empty(board)
if not empty: return True
else: row, col = empty
for i in range(1, 10):
if is_valid(board, i, (row, col)):
board[row][col] = i
if backtrack_solve(board): return True
board[row][col] = 0
return False
def checkio(board):
backtrack_solve(board)
return board
Oct. 23, 2020
