Python 井字游戏
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【中文标题】Python 井字游戏【英文标题】：Python tic tac toe game 【发布时间】：2014-04-12 17:02:36 【问题描述】：我不确定是否需要所有代码，所以我会发布它：
# Tic-Tac-Toe
# Plays the game of tic-tac-toe against a human opponent
# global constants
X = "X"
O = "O"
EMPTY = " "
TIE = "TIE"
NUM_SQUARES = 9
def display_instruct():
"""Display game instructions."""
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 yourself, human. The ultimate battle is about to begin. \n
"""
)
def ask_yes_no(question):
"""Ask a yes or no question."""
response = None
while response not in ("y", "n"):
response = input(question).lower()
return response
def ask_number(question, low, high):
"""Ask for a number within a range."""
response = None
while response not in range(low, high):
response = int(input(question))
return response
def pieces():
"""Determine if player or computer goes first."""
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():
"""Create new game board."""
board = []
for square in range(NUM_SQUARES):
board.append(EMPTY)
return board
def display_board(board):
"""Display game board on screen."""
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])
def legal_moves(board):
"""Create list of legal moves."""
moves = []
for square in range(NUM_SQUARES):
if board[square] == EMPTY:
moves.append(square)
return moves
def winner(board):
"""Determine the game winner."""
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[:]
# the best positions to have, in order
BEST_MOVES = (4, 0, 2, 6, 8, 1, 3, 5, 7)
print("I shall take square number,", end="")
# if computer can win, take that move
for move in legal_moves(board):
board[move] = computer
if winner(board) == computer:
print(move)
return move
# done checking this move, undo it
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 checkin this move, undo it
board[move] = EMPTY
# since no one can 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):
"""Switch turns."""
if turn == X:
return O
else:
return X
def congrat_winner(the_winner, computer, human):
"""Congratulate the winner."""
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.")
elif the_winner == human:
print("No, no! It cannot be! Somehow you tricked me, human. \n" \
"But never again! I, the computer, so swear it!")
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.")
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)
# start the program
main()
input("\n\nPress the enter key to quit.")
这是我正在阅读的书中的一个例子，我还没有完全理解，我认为直到：
for row in WAYS_TO_WIN:
if board[row[0]] == board[row[1]] == board[row[2]] != EMPTY:
winner = board[row[0]]
return winner
谁能解释一下这个函数的作用，更具体地说是什么条件
if board[row[0]] == board[row[1]] == board[row[2]] != EMPTY: 正在测试？
【问题讨论】：
【参考方案1】：列表项
def tic_tac_toe(): 板 = [1, 2, 3, 4, 5, 6, 7, 8, 9] 结束 = 假 win_commbinations = ((0, 1, 2), (3, 4, 5), (6, 7, 8), (0, 3, 6), (1, 4, 7), (2, 5, 8) , (0, 4, 8), (2, 4, 6))
定义绘制（）： 打印（板[0]，板[1]，板[2]） 打印（板[3]，板[4]，板[5]） 打印（板[6]，板[7]，板[8]） 打印（）
定义 p1(): n = 选择号码（） 如果板[n] == "X" 或板[n] == "O"： print("\n你不能去那里。再试一次") p1() 别的： 板[n] = "X"
定义 p2(): n = 选择号码（） 如果板[n] == "X" 或板[n] == "O"： print("\n你不能去那里。再试一次") p2() 别的： 板[n] =“O”
def 选择号码（）： 而真： 而真： 一=输入（） 尝试： a = int(a) 一个 -= 1 如果 a 在范围 (0, 9) 内： 返回一个 别的： print("\n那不在板上。再试一次") 继续 除了ValueError： print("\n这不是一个数字。再试一次") 继续
def check_board(): 计数 = 0 对于 win_commbinations 中的一个： 如果板[a[0]] == 板[a[1]] == 板[a[2]] == "X"： print("玩家 1 获胜！\n") print("恭喜！\n") 返回真
if board[a[0]] == board[a[1]] == board[a[2]] == "O":
print("Player 2 Wins!\n")
print("Congratulations!\n")
return True
for a in range(9):
if board[a] == "X" or board[a] == "O":
count += 1
if count == 9:
print("The game ends in a Tie\n")
return True
虽然没有结束： 画（） 结束 = check_board() 如果结束 == 真： 休息 print("玩家 1 选择放置十字架的位置") p1() 打印（） 画（） 结束 = check_board() 如果结束 == 真： 休息 print("玩家 2 选择放置 nought 的位置") p2() 打印（）
if input("再次播放 (y/n)\n") == "y": 打印（） tic_tac_toe()
【讨论】：
【参考方案2】：我会简单地说。 Row 是一个变量，分配给元组 WAYS_TO_WIN 中的每个元组。 在第一次迭代中，row = (0,1,2) 它检查值是否为 0==1==2。
在第二次迭代中，row = (3,4,5) 它检查值是否为 3==4==5。
Row 转到外部元组ways_to_win 的每个内部元组，直到达到 row = (2,4,6)。 这就是程序正在做的事情。
【讨论】：
【参考方案3】：我是 Python 新手。下面的井字游戏脚本来自我的一个练习。它使用了不同的方法。
对于数据结构，我使用整数值 0 表示空白单元格，+1 表示计算机放置的单元格，-1 表示用户放置的单元格。
主要好处是我可以使用 lineValue（即一行中所有三个单元格值的总和）来跟踪每行的状态。所有 8 行值都存储在列表 lineValues 中。这可以使决策更容易。例如，当轮到我（计算机）时，如果有 lineValue==2 的线，我知道我会赢。否则，如果有 lineValue ==-2 的线，我必须阻止这些线的交叉点（如果有）。
做出决定的关键是函数findMostValuableCell。它的作用是找出哪个单元格对下一步最有价值（即，对于特定 lineValue，哪个单元格出现在最多行数中）。此脚本中没有试用测试（假设测试）。它使用了很多列表推导。
希望对你有帮助。
ttt = [0 for i in range(9)]
lines = [[0, 1, 2],[3, 4, 5],[6, 7, 8],[0, 3, 6],[1, 4, 7],[2, 5, 8],[0, 4, 8],[2, 4, 6]]
lineValues = [0 for i in range(8)]
userChar = 1: "O", -1: "X", 0: "_"
turn = -1 # defalut to user move first
#*****************************************************
def main():
global userChar, turn
if input("Do you want me to start first? (Y/N)").lower()=="y":
userChar = 1:"X",-1:"O",0:"_"
turn = 1
display()
while not hasWinner():
if 0 in ttt:
nextMove(turn)
turn *= -1
display()
else:
print("It's a tie!")
break
#*****************************************************
def hasWinner():
if max(lineValues) == 3:
print("******** I win!! ********")
return True
elif min(lineValues) == -3:
print("******** You win ********")
return True
#*****************************************************
def nextMove(turn):
if turn== -1: #User's turn
print("It's your turn now (" + userChar[-1]+"):")
while not isUserMoveSuccessful(input("Please choose your cell number:")):
print("Your choice is not valid!")
else: #Computer's turn
print("It's my turn now...")
for lineValue in [2,-2,-1,1,0]:
cell = findMostValuableCell(lineValue)
if cell>=0: #found a cell for placement
markCell(cell, turn)
print ("I chose cell", str(cell),"." )
return
#*****************************************************
def isUserMoveSuccessful(userInput):
s = list(userInput)[0]
if '012345678'.find(s)>=0 and ttt[int(s)]==0:
markCell(int(s), turn)
return True
#*****************************************************
def findMostValuableCell(lineValue):
if set(ttt)==0:
return 1
allLines = [i for i in range(8) if lineValues[i]==lineValue]
allCells =[j for line in allLines for j in lines[line] if ttt[j]==0]
cellFrequency = dict((c, allCells.count(c)) for c in set(allCells))
if len(cellFrequency)>0: # get the cell with highest frequency.
return max(cellFrequency, key=cellFrequency.get)
else:
return -1
#*****************************************************
def markCell(cell, trun):
global lineValues, ttt
ttt[cell]=turn
lineValues = [sum(cellValue) for line in lines for cellValue in [[ttt[j] for j in line]]]
#*****************************************************
def display():
print(' _ _ _\n'+''.join('|'+userChar[ttt[i]]+('|\n' if i%3==2 else '') for i in range(9)))
#*****************************************************
main()
【讨论】：
【参考方案4】：它只是检查当前棋盘以查看是否有任何获胜的单元格组合（如行数组中所列）具有 (a) 相同的值和 (b) 该值不是 EMPTY。
注意：在 Python 中，如果 a == b == c != d，则检查 a ==b AND b == c AND c != d
所以如果单元格 0、1 和 2 都有 X，那么在第一次通过循环时，它将从获胜例程返回 X。
【讨论】：
【参考方案5】：查看发生了什么的最佳方法是在运行此代码时将一些print 语句放入此代码中。
从事物名称的方式来看，您可以看出您正在寻找是否有人赢得了比赛。您从井字游戏的规则中知道，如果 X 或 O 在一行、一列或对角线上有三个，则该玩家获胜。您在board[x] == board[y] == board[z] 中看到我们可能在这里连续测试三个。那么x, y z 是什么？好吧，看看WAYS_TO_WIN。该数组中的行表示行、列或对角线中的索引。因此，我们正在测试行、列或对角线是否包含相同的字符，并且该字符不是EMPTY（即" " [空格] 字符）。
【讨论】：
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