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cpuEvaluate.py
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cpuEvaluate.py
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# Python3 program to find the next optimal move for a player
# This function returns true if there are moves
# remaining on the board. It returns false if
# there are no moves left to play.
def isMovesLeft(board) :
for i in range(3) :
for j in range(3) :
if (board[i][j] == ' ') :
return True
return False
# This is the evaluation function as discussed
# in this source link ( http://goo.gl/sJgv68 )
def evaluate(b, player) :
if player == 'X':
opponent = 'O'
else:
opponent = 'X'
# Checking for Rows for X or O victory.
for row in range(3) :
if (b[row][0] == b[row][1] and b[row][1] == b[row][2]) :
if (b[row][0] == player) :
return 10
elif (b[row][0] == opponent) :
return -10
# Checking for Columns for X or O victory.
for col in range(3) :
if (b[0][col] == b[1][col] and b[1][col] == b[2][col]) :
if (b[0][col] == player) :
return 10
elif (b[0][col] == opponent) :
return -10
# Checking for Diagonals for X or O victory.
if (b[0][0] == b[1][1] and b[1][1] == b[2][2]) :
if (b[0][0] == player) :
return 10
elif (b[0][0] == opponent) :
return -10
if (b[0][2] == b[1][1] and b[1][1] == b[2][0]) :
if (b[0][2] == player) :
return 10
elif (b[0][2] == opponent) :
return -10
# Else if none of them have won then return 0
return 0
# This is the minimax function. It considers all
# the possible ways the game can go and returns
# the value of the board
def minimax(board, depth, isMax, player) :
if player == 'X':
opponent = 'O'
else:
opponent = 'X'
score = evaluate(board, player)
# If Maximizer has won the game return his/her
# evaluated score
if (score == 10) :
return score
# If Minimizer has won the game return his/her
# evaluated score
if (score == -10) :
return score
# If there are no more moves and no winner then
# it is a tie
if (isMovesLeft(board) == False) :
return 0
# If this maximizer's move
if (isMax) :
best = -1000
# Traverse all cells
for i in range(3) :
for j in range(3) :
# Check if cell is empty
if (board[i][j]==' ') :
# Make the move
board[i][j] = player
# Call minimax recursively and choose
# the maximum value
best = max( best, minimax(board,
depth + 1,
not isMax, player) )
# Undo the move
board[i][j] = ' '
return best
# If this minimizer's move
else :
best = 1000
# Traverse all cells
for i in range(3) :
for j in range(3) :
# Check if cell is empty
if (board[i][j] == ' ') :
# Make the move
board[i][j] = opponent
# Call minimax recursively and choose
# the minimum value
best = min(best, minimax(board, depth + 1, not isMax, player))
# Undo the move
board[i][j] = ' '
return best
# This will return the best possible move for the player
def findBestMove(board, player) :
if player == 'X':
opponent = 'O'
else:
opponent = 'X'
bestVal = -1000
bestMove = (-1, -1)
# Traverse all cells, evaluate minimax function for
# all empty cells. And return the cell with optimal
# value.
for i in range(3) :
for j in range(3) :
# Check if cell is empty
if (board[i][j] == ' ') :
# Make the move
board[i][j] = player
# compute evaluation function for this
# move.
moveVal = minimax(board, 0, False, player)
# Undo the move
board[i][j] = ' '
# If the value of the current move is
# more than the best value, then update
# best/
if (moveVal > bestVal) :
bestMove = (i, j)
bestVal = moveVal
print("The value of the best Move is :", bestVal)
print(bestMove)
return bestMove
# Driver code
# this code is contributed by divyesh072019