美好的一天。我目前正在阅读Michael Dawson的书“ Python for Absolute Beginner ”。我有第6章的代码，当我运行代码并在主板上得到直接的“X”时，它不会返回获胜者。程序继续，直到所有空格都标有“X”或“O”。这是我的代码：
#!/usr/bin/python3
X = "X"
O = "O"
EMPTY = " "
TIE = "TIE"
NUM_SQUARES = 9
def display_instruct():
print(
"""
Welcome to the greatest intellectual challenge of all time: Tic-Tac-Toe.
This will be a showdown between your human brain and my silicon processor.
You will make your move known by entering a number, 0 - 8. The number
will correspond to the board position as illustrated:
0 | 1 | 2
-----------
3 | 4 | 5
-----------
6 | 7 | 8
Prepare your self, human. The ultimate battle is about to begin. \n
""")
def ask_yes_no(question):
response = None
while response not in ("y", "n"):
response = input(question).lower()
return response
def ask_number(question, low, high):
response = None
while response not in range(low, high):
response = int(input(question))
return response
def pieces():
go_first = ask_yes_no("Do you require the first move? (y/n): ")
if go_first == "y":
print( "\nThen take the first move. You will need it. ")
human = X
computer = O
else:
print("\nYour bravery will be your undoing... I will go first.")
computer = X
human = O
return computer, human
def new_board():
board = []
for square in range(NUM_SQUARES):
board.append(EMPTY)
return board
def display_board(board):
print("\n\t", board[0], "|", board[1], "|", board[2])
print("\t", "---------")
print("\t", board[3], "|", board[4], "|", board[5])
print("\t", "---------")
print("\t", board[6], "|", board[7], "|", board[8], "\n")
def legal_moves(board):
moves = []
for square in range(NUM_SQUARES):
if board[square] == EMPTY:
moves.append(square)
return moves
def winner(board):
WAYS_TO_WIN = ((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 row in WAYS_TO_WIN:
if board[row[0]] == board[row[1]] == board[row[2]] != EMPTY:
winner = board[row[0]]
return winner
if EMPTY not in board:
return TIE
return None
def human_move(board, human):
"""Get human move."""
legal = legal_moves(board)
move = None
while move not in legal:
move = ask_number("Where will you move? (0 - 8): ", 0, NUM_SQUARES)
if move not in legal:
print("\nThat square is already occupied, foolish human. Choose another.\n")
print("Fine....")
return move
def computer_move(board, computer, human):
"""Make computer move."""
#make a copy to work with since function will be changing list.
board = board[:]
BEST_MOVES = (4, 0, 2, 6, 8, 1, 3, 5, 7)
print("I shall take square number")
# if computer can win, take that move
for move in legal_moves(board):
board[move] = computer
if winner(board) == computer:
print(move)
return move
board[move] = EMPTY
# if human can win, block that move
for move in legal_moves(board):
board[move] = human
if winner(board) == human:
print(move)
return move
# done checking this move, undo it
board[move] = EMPTY
# since no one ca win on next move, pick best open square
for move in BEST_MOVES:
if move in legal_moves(board):
print(move)
return move
def next_turn(turn):
if turn == X:
return 0
else:
return X
def congrat_winner(the_winner, computer, human):
if the_winner != TIE:
print(the_winner, "won!\n")
else:
print("It's a tie!\n")
if the_winner == computer:
print("As I predicted, human, I am triumphant once more. \n" \
"Proof that computers are superior to humans in all regards.\n")
elif the_winner == human:
print("No, no! It cannot be! Somehow you tricked me, human. \n" \
"But never again! I, the computer, so swears it\n!")
elif the_winner == TIE:
print("You were most lucky, human, and somehow managed to tie me. \n" \
"Celebrate today... for this is the best you will ever achieve.\n")
def main():
display_instruct()
computer, human = pieces()
turn = X
board = new_board()
display_board(board)
while not winner(board):
if turn == human:
move = human_move(board, human)
board[move] = human
else:
move = computer_move(board, computer, human)
board[move] = computer
display_board(board)
turn = next_turn(turn)
the_winner = winner(board)
congrat_winner(the_winner, computer, human)
main()
input("Press enter to exit")
答案 0 :(得分：3)
检查胜利或平局的功能不正确。目前你有这个：
def winner(board):
WAYS_TO_WIN = ((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 row in WAYS_TO_WIN:
if board[row[0]] == board[row[1]] == board[row[2]] != EMPTY:
winner = board[row[0]]
return winner
if EMPTY not in board:
return TIE
return None
请注意，return None位于循环内部，因此当第一行无法匹配时返回None。除非电路板已满，否则您将正确返回TIE，除非您在尝试匹配任何行之前检查了完整的电路板。
我们可以通过重新定位return None来使函数正确，以便在循环之后执行它。此外，在循环之前移动TIE检查是有意义的。
def winner(board):
if EMPTY not in board:
return TIE
WAYS_TO_WIN = ((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 row in WAYS_TO_WIN:
if board[row[0]] == board[row[1]] == board[row[2]] != EMPTY:
winner = board[row[0]]
return winner
return None
让我们回顾一下这个功能的作用。我们首先检查一个完整的董事会。然后我们继续考虑获胜的方法。如果它们都不匹配，我们会返回None。
您的代码中还有一个问题：
def next_turn(turn):
if turn == X:
return 0
else:
return X
为了让以后的测试if turn == human:在所有情况下都能正常运行，turn的值必须是X或O，而不是X或0。这是快速修复：
def next_turn(turn):
if turn == X:
return O
else:
return X
解决此问题的更好方法是避免使用名为O的变量。
答案 1 :(得分：0)
您还在def(congrat_winner)
正确的方法是：
def congrat_winner(the_winner, computer, human):
if the_winner != TIE:
print(the_winner, "won!\n")
else:
print("It's a tie!\n")
if the_winner == computer:
print("As I predicted, human, I am triumphant once more. \n" \
"Proof that computers are superior to humans in all regards.\n")
elif the_winner == human:
print("No, no! It cannot be! Somehow you tricked me, human. \n" \
"But never again! I, the computer, so swears it\n!")
elif the_winner == TIE:
print("You were most lucky, human, and somehow managed to tie me. \n" \
"Celebrate today... for this is the best you will ever achieve.\n")
此外，最后elif应替换为else：
def congrat_winner(the_winner, computer, human):
if the_winner != TIE:
print(the_winner, "won!\n")
else:
print("It's a tie!\n")
if the_winner == computer:
print("As I predicted, human, I am triumphant once more. \n" \
"Proof that computers are superior to humans in all regards.\n")
elif the_winner == human:
print("No, no! It cannot be! Somehow you tricked me, human. \n" \
"But never again! I, the computer, so swears it\n!")
else:
print("You were most lucky, human, and somehow managed to tie me. \n" \
"Celebrate today... for this is the best you will ever achieve.\n")
答案 2 :(得分：0)
明白你已经有了答案，但提出了一些建议 - 尝试单独测试你的功能。
＆＃xA;＆＃xA;在你的情况下，你已经有了预感您的 winner 函数存在问题。我直接从你的问题中提到了这一点，你提到“它不会让胜利者回归”。
有很多方法可以测试，但让我们保持简单并获取该功能，将其复制到一个新的python程序，扔一些游戏板，看看会发生什么。我的意思是“看”字面意思 - 在测试时你最好的朋友是 print 。假设我们像这样修改你的功能：
def winner（board）：＆＃xA; print（“赢得功能输入”）＆＃xA; WAYS_TO_WIN =（（0,1,2），＆＃xA;（3,4,5），＆＃xA;（6,7,8），＆＃xA;（0,3,6），＆＃xA ;（1,4,7），＆＃xA;（2,5,8），＆＃xA;（0,4,8），＆＃xA;（2,4,6））＆＃xA;＆ #xA;对于WAYS_TO_WIN中的行：＆＃xA; print（“for loop entered”）＆＃xA;如果board [row [0]] == board [row [1]] == board [row [2]]！= EMPTY：＆＃xA;获胜者=董事会[row [0]]＆＃xA;返回获胜者＆＃xA;＆＃xA;如果EMPTY不在船上：＆＃xA;返回TIE＆＃xA;返回无＆＃xA; ＆＃xA;＆＃xA; 这似乎太容易了，但我们立即注意到，在所有情况下，for循环只输入一次。只需要关注一小部分代码，在找到缩进错误之前不会太久。
＆＃xA;