sudoku.py

# Define the board
board = [
    [5, 3, 0, 0, 7, 0, 0, 0, 0],
    [6, 0, 0, 1, 9, 5, 0, 0, 0],
    [0, 9, 8, 0, 0, 0, 0, 6, 0],
    [8, 0, 0, 0, 6, 0, 0, 0, 3],
    [4, 0, 0, 8, 0, 3, 0, 0, 1],
    [7, 0, 0, 0, 2, 0, 0, 0, 6],
    [0, 6, 0, 0, 0, 0, 2, 8, 0],
    [0, 0, 0, 4, 1, 9, 0, 0, 5],
    [0, 0, 0, 0, 8, 0, 0, 7, 9]
]

def is_valid(board, row, col, num):
    # Constraint 1: Each number must appear only once in each row.
    for x in range(9):
        if board[row][x] == num:
            return False
            
    # Constraint 2: Each number must appear only once in each column.
    for x in range(9):
        if board[x][col] == num:
            return False
            
    # Constraint 3: Each number must appear only once in each 3x3 box.
    startRow = row - row % 3
    startCol = col - col % 3
    for i in range(3):
        for j in range(3):
            if board[i + startRow][j + startCol] == num:
                return False
    return True

def solve_sudoku(board):
    for i in range(9):
        for j in range(9):
            # Find an unfilled cell
            if board[i][j] == 0:
                for num in range(1,10):
                    # Check if placing 'num' in the unfilled cell violates any constraint
                    if is_valid(board, i, j, num):
                        # Assign 'num' to the cell as it doesn't violate any constraint
                        board[i][j] = num
                        # Recursively attempt to fill the rest of the grid
                        if solve_sudoku(board):
                            return True
                        # If no valid arrangement could be found, undo the current cell assignment
                        board[i][j] = 0
                # If no number can be assigned to this cell, backtrack
                return False 
    # All cells are filled successfully
    return True 

if __name__ == "__main__":
    if (solve_sudoku(board)):
        print(board)
    else:
        print("No solution exists")