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tictactoe shows off using a background image, the ImageButton widget and
background handlers.
Francois Granger sent the email below which you can adapt to the tictactoe
sample if you want a better computer opponent than Mr. Gumby.
-----Original Message-----
From: Francois Granger [mailto:fgranger@altern.org]
Sent: Tuesday, January 01, 2002 1:50 PM
To: Kevin Altis
Subject: TicTacToe
You published a TicTacToe game developped with PythonCard recently
with some questions about a winning algorythm. Since I worked on the
same subject for some time, I got a working brute force one. It needs
cleaning. See below.
seq is the previous sequence of turns in the form [(i, j), (i, j), ...]
turn is the current turn numbered from 0
======================================
class ComputerAnal(Player):
"""
The computer analyse the board
"""
def move(self, seq, turn, warn = ""):
"""
Analytical seek for solution
seq is a sequence of tuples representing previous moves.
return a tuple i, j representing the next move.
"""
prompt = warn + "Player " + str(self.token) + " : "
print prompt
if turn == 0:
i,j = 1,1
elif turn == 1 and seq[0] != (1,1):
i,j = 1,1
elif turn == 1:
i, j = 0,0
else:
b = [[0,0,0],[0,0,0],[0,0,0]]
x = 0
for move in seq:
i, j = move
if x == self.play:
b[i][j] = 1
else:
b[i][j] = -1
x = not x
row = [0,0,0]
col = [0,0,0]
diag =[0,0]
"""
find any line where there are two similar
tokens and third is empty.
first for me so I play and win
then for the other so I play and he can't win
other wise at random.
"""
for i in range(3):
row[i] = b[i][0] + b[i][1] + b[i][2]
for j in range(3):
col[j] = b[0][j] + b[1][j] + b[2][j]
diag[0] = b[0][0] + b[1][1] + b[2][2]
diag[1] = b[0][2] + b[1][1] + b[2][0]
pass
if diag[0] == 2:
for i in range(3):
if not b[i][i]:
return i,i
if diag[1] == 2:
for i in range(3):
if not b[i][2-i]:
return i,2-i
for i in range(3):
if row[i] == 2:
for j in range(3):
if not b[i][j]:
return i,j
for j in range(3):
if col[j] == 2:
for i in range(3):
if not b[i][j]:
return i,j
if diag[0] == -2:
for i in range(3):
if not b[i][i]:
return i,i
if diag[1] == -2:
for i in range(3):
if not b[i][2-i]:
return i,2-i
for i in range(3):
if row[i] == -2:
for j in range(3):
if not b[i][j]:
return i,j
for j in range(3):
if col[j] == -2:
for i in range(3):
if not b[i][j]:
return i,j
pass
i,j = rand.randint(0, 2), rand.randint(0, 2)
print i,j
return i, j
======================================
|
