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December 15, 2015 06:39
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tictactoe in python for 2 players
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"""tictactoe game for 2 players | |
from blogpost: http://thebillington.co.uk/blog/posts/writing-a-tic-tac-toe-game-in-python by BILLY REBECCHI, | |
slightly improved by Horst JENS""" | |
from __future__ import print_function | |
choices = [] | |
for x in range (0, 9) : | |
choices.append(str(x + 1)) | |
playerOneTurn = True | |
winner = False | |
def printBoard() : | |
print( '\n -----') | |
print( '|' + choices[0] + '|' + choices[1] + '|' + choices[2] + '|') | |
print( ' -----') | |
print( '|' + choices[3] + '|' + choices[4] + '|' + choices[5] + '|') | |
print( ' -----') | |
print( '|' + choices[6] + '|' + choices[7] + '|' + choices[8] + '|') | |
print( ' -----\n') | |
while not winner : | |
printBoard() | |
if playerOneTurn : | |
print( "Player 1:") | |
else : | |
print( "Player 2:") | |
try: | |
choice = int(input(">> ")) | |
except: | |
print("please enter a valid field") | |
continue | |
if choices[choice - 1] == 'X' or choices [choice-1] == 'O': | |
print("illegal move, plase try again") | |
continue | |
if playerOneTurn : | |
choices[choice - 1] = 'X' | |
else : | |
choices[choice - 1] = 'O' | |
playerOneTurn = not playerOneTurn | |
for x in range (0, 3) : | |
y = x * 3 | |
if (choices[y] == choices[(y + 1)] and choices[y] == choices[(y + 2)]) : | |
winner = True | |
printBoard() | |
if (choices[x] == choices[(x + 3)] and choices[x] == choices[(x + 6)]) : | |
winner = True | |
printBoard() | |
if((choices[0] == choices[4] and choices[0] == choices[8]) or | |
(choices[2] == choices[4] and choices[4] == choices[6])) : | |
winner = True | |
printBoard() | |
print ("Player " + str(int(playerOneTurn + 1)) + " wins!\n") |
Amazing code but it doesn't have a way to show the game is draw
I kinda changed the code to solve that
link: https://gist.github.com/vivek1339/a35edf19740bc5c58bb729b1bbf690ca/eac36b2e6f90139c33332b04e7f5c26c5de2198b
Doesn't stop when game draws...
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tnx............. bro