Problems for player play order in Tic-tac-toe

Question

I have trouble with the job players(), because I can’t increment the variable n, the function is only returning "X". I know it sounds pretty simple, but I can’t identify the problem. Can anyone help me?

def players():
    n = 0
    if n % 2 == 0:
        n = n + 1
        return 'X'
    else:
        n = n + 1
        return 'O'

def play():
    gon = 0
    while gon != 9:
        pos = input('Insira a posição que deseja jogar: ')
        if pos == '1':
            paper[0][0] = players()
            gon += 1
        elif pos == '2':
            paper[1][0] = players()
            gon += 1
        elif pos == '3':
            paper[2][0] = players()
            gon += 1
        elif pos == '4':
            paper[3][0] = players()
            gon += 1
        elif pos == '5':
            paper[4][0] = players()
            gon += 1
        elif pos == '6':
            paper[5][0] = players()
            gon += 1
        elif pos == '7':
            paper[6][0] = players()
            gon += 1
        elif pos == '8':
            paper[7][0] = players()
            gon += 1
        elif pos == '9':
            paper[8][0] = players()
            gon += 1
        else:
            print('Dígito inválido! Tente novamente.')

        print(' %s | %s | %s ' % (paper[0][0],paper[1][0],paper[2][0]))
        print('-----------')
        print(' %s | %s | %s ' % (paper[3][0],paper[4][0],paper[5][0]))
        print('-----------')
        print(' %s | %s | %s ' % (paper[6][0],paper[7][0],paper[8][0]))
    if gon == 9:
        print('\n Empate!')

paper = [[' '],[' '],[' '],
         [' '],[' '],[' '],
         [' '],[' '],[' ']]

print('Bem-vindo ao Jogo da Velha!')
print('Cliente é X;')
print('Servidor é O.')
play()

Answer 1

The problem can be verified by making a simple table test in function players. Basically, it starts with n = 0, I mean, every time you call players(), the first thing she will do is set the value zero in the variable n. At all times, every time it’s called.

So you’ll always be in if. If you want the behavior to vary according to the value of n, then probably what you want is for the function to receive the value as parameter:

def players(n):
    if n % 2 == 0:
        return 'X'
    else:
        return 'O'

You can still simplify and improve some things in the code. For example, the function play:

def play():
    paper = [ ' ' ] * 9
    gon = 0
    n = 0
    while gon != 9:
        try:
            pos = int(input('Insira a posição que deseja jogar: '))
            if 1 <= pos <= 9:
                paper[pos - 1] = players(n)

                n = 1 - n # se n for 1, vira zero, se for zero, vira 1
                gon += 1

                for i in range(0, len(paper), 3):
                    print(' {} | {} | {} '.format(paper[i],paper[i + 1],paper[i + 2]))
                    if i < 6:
                        print('-----------')
            else:
                print('Valor deve estar entre 1 e 9')
        except ValueError:
            print('Valor inválido! Tente novamente.')

    if gon == 9:
        print('\n Empate!')

I convert the result of input to number using int (and if it’s not a number, it launches a ValueError, that I capture with except). And I only update the board if the position value is valid.

If the board is always the same, you can create it within the same function (I saw no sense in creating it outside, unless it could have varied size, then it could be a function parameter, for example).

I also made a "board" as a list with 9 positions. Of course being a list of lists (a "matrix") would be more "faithful", but leaving it as a simple list simplifies that lot of if's.

When printing the board, I used a range of 3 in 3, so it is possible to print each "line" of the "matrix" at once.

---

Of course you can improve more, because the code does not check if a position has already been occupied, does not make a fact check if there was a winner, etc. An example:

def players(n):
    if n % 2 == 0:
        return 'X'
    else:
        return 'O'

def jogo_terminou(tabuleiro):
    posicoes_vencedoras = [
        [0, 1, 2], [3, 4, 5], [6, 7, 8], # horizontal
        [0, 3, 6], [1, 4, 7], [2, 5, 8], # vertical
        [0, 4, 8], [2, 4, 6] # diagonal
    ]
    # verifica todas as combinações de posições vencedoras
    for pos1, pos2, pos3 in posicoes_vencedoras:
        if tabuleiro[pos1] == tabuleiro[pos2] == tabuleiro[pos3] and tabuleiro[pos1] in ('X', 'O'):
            print(f'Temos um vencedor: {tabuleiro[pos1]}! Parabéns!') # mudei a mensagem para mostrar quem ganhou
            # se já encontrei um vencedor, posso retornar direto (não precisa continuar verificando as outras combinações)
            return True

    # não tem vencedor, verifica se todas as posições estão ocupadas (não preciso mais do contador de jogadas)
    if all(posicao in ('X', 'O') for posicao in tabuleiro):
        print('Deu velha! Ninguém ganhou.')
        return True

    # jogo ainda não terminou
    return False

def play():
    paper = [ ' ' ] * 9
    n = 0
    while True:
        try:
            pos = int(input('Insira a posição que deseja jogar: '))
            if 1 <= pos <= 9:
                if paper[pos - 1] == ' ': # verifica se posição não está ocupada
                    paper[pos - 1] = players(n)

                    n = 1 - n # se n for 1, vira zero, se for zero, vira 1

                    for i in range(0, len(paper), 3):
                        print(' {} | {} | {} '.format(paper[i],paper[i + 1],paper[i + 2]))
                        if i < 6:
                            print('-----------')
                else:
                    print('posição já ocupada')
            else:
                print('Valor deve estar entre 1 e 9')
        except ValueError:
            print('Valor inválido! Tente novamente.')

        if jogo_terminou(paper):
            break

Answer 2

Dear jmb2001, without wanting to modify much your program but seeking to simplify it I did it in the following way:

import os

def players(flag):

    if flag:
        return 'X'
    else:
        return 'O'

def play():
    flag = True
    gon = 0
    while gon != 9:
        os.system('clear')

        print(' %s | %s | %s ' % tuple(paper[0:3]))
        print('-----------')
        print(' %s | %s | %s ' % tuple(paper[3:6]))
        print('-----------')
        print(' %s | %s | %s ' % tuple(paper[6:9]))
        print()

        pos = input('Insira a posição que deseja jogar: ')
        if pos in 'qQ':
            exit(0)

        elif pos in '123456789':
                paper[int(pos)-1] = players(flag)
                gon += 1
                flag = not flag

        else:
            print('Dígito inválido! Tente novamente.')

    if gon == 9:
        print('\n Empate!')


paper = [' '] * 9 # copie do hkotsubo

print('Bem-vindo ao Jogo da Velha!')
print('Cliente é X;')
print('Servidor é O.')

play()

I thought it best to change the name of your control variable n to flag, which means flag in English because that’s what the flag variable does.

It works as a flip-flop changing its state each time the line ( flag = not flag ) runs.

I added the instruction os.system('clear') so that the screen is cleaned and the print’s always exit in the same positions of the terminal.

Also includes an option allowing the user to terminate the program by typing q or Q.

Finally I agree with hkotsubo about the fact that it remains to verify whether the position is already occupied or not and other things more.

But you’re on the right track, keep it up...

And be Happy