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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