Let us begin with some layout. This is a typical grid use case so that you won't have to manually manage grid. At the lowest level, a grid structure essentially places buttons by calculating coordinates. But this gets tedious over time, that's why a grid layout is provided. Also, it was a clever way of cheking win by being symbol agnostic. A normal tic tac toe game would have used checkwin(symbol) to determine win.

On strings

To escape' within' ' you can use" instead. From

'\'S TURN.'

to

"'S TURN."

and you can also use string formatting to clear up some clutter.

"PLAYER {}'S TURN.".format(turn)

On layout

Modifying yourcreate_button function to this allows a grid structure

def create_button(x, y):    button = tk.Button(width=10, height=2, command=lambda: callback(button))    button.grid(row=x, column=y)    return button

Then we modify others since different layouts can't be mixed

label_button = ttk.Button(                        game,                         text="PLAYER {}'S TURN.".format(turn),                         style='Fun.TButton', width=20,                         state='disabled')label_button.grid(row=0, column=1)invalid = tk.Label(text='')invalid.grid(row=4, column=1)

adding the buttons can be then done as

buttons = []buttons.append(create_button(1, 0))buttons.append(create_button(1, 1))buttons.append(create_button(1, 2))buttons.append(create_button(2, 0))buttons.append(create_button(2, 1))buttons.append(create_button(2, 2))buttons.append(create_button(3, 0))buttons.append(create_button(3, 1))buttons.append(create_button(3, 2))

You can use a loop for the row anditertools.cycle for the0, 1, 2 if you want to simplify it.

Thecheck_win function

• Simplifying if

Adding a() to if statements allows you to writeor without\

if ... :    ...toif (...):    ...

thus the win_function can be simplified from

if (buttons[0]['text'] == buttons[1]['text'] == buttons[2]['text'] != '') or \(buttons[3]['text'] == buttons[4]['text'] == buttons[5]['text'] != '') or \

to

def check_win():    if (        (buttons[0]['text'] == buttons[1]['text'] == buttons[2]['text'] != '') or        (buttons[3]['text'] == buttons[4]['text'] == buttons[5]['text'] != '') or        ...        ):        return True    else:        return False

• Simplifying values write-up

This can also be further simplified by defining a function to replacebuttons[0]['text']

def btext(i):    return buttons[i]['text']

and using it

def check_win():    if (        (btext(0) == btext(1) == btext(2) != '') or        (btext(3) == btext(4) == btext(5) != '') or        (btext(6) == btext(7) == btext(8) != '') or        (btext(0) == btext(3) == btext(6) != '') or        (btext(1) == btext(4) == btext(7) != '') or        (btext(2) == btext(5) == btext(8) != '') or        (btext(2) == btext(4) == btext(6) != '') or        (btext(0) == btext(6) == btext(8) != '')        ):        return True    else:        return False

On architecture

A common pattern is the MVC (Model, View, Controller). While checking and updating gui directly works here, you might consider adding states in a structure like this:

board = [    ['', '', ''],    ['', '', ''],    ['', '', '']]

Operations are done on this and the gui is updated according to this.

From :Python Tic Tac Toe Game

    win_commbinations = ((0, 1, 2), (3, 4, 5), (6, 7, 8), (0, 3, 6), (1, 4, 7), (2, 5, 8), (0, 4, 8), (2, 4, 6))for a in win_commbinations:            if board[a[0]] == board[a[1]] == board[a[2]] == "X":                print("Player 1 Wins!\n")                print("Congratulations!\n")                return True            if board[a[0]] == board[a[1]] == board[a[2]] == "O":                print("Player 2 Wins!\n")                print("Congratulations!\n")                return True        for a in range(9):            if board[a] == "X" or board[a] == "O":                count += 1            if count == 9:                print("The game ends in a Tie\n")                return True

This alternative solution is a bit cleaner + includes a "Tie" check, which your original solution doesn't check ( just remove if you consider it irrelevant ofc )

• \$\begingroup\$You have not reviewed the code, but instead presented your own solution. While the code may improve some aspects, it does not have a tkinter GUI like the original code does.\$\endgroup\$

  spyr03

  – spyr03

  2019-08-01 16:11:52 +00:00

  CommentedAug 1, 2019 at 16:11

• python
• python-3.x
• tic-tac-toe
• tkinter