井字棋游戏

1 问题

如何用python完成井字棋游戏。

2 方法

打印棋盘;
让玩家选择棋子;
随机决定谁先走；
玩家选择落子位置；
判断输赢；
是否再来一次

通过实验、实践等证明提出的方法是有效的，是能够解决开头提出的问题。

代码清单 1

import random
def drawBoard(board):
# "board"是长度为10的列表，为了方便输入，忽略第一个元素board[0]
print('\n\n\n\n')
print('\t\t\t┌─┬─┬─┐')
print('\t\t\t│'+board[7]+' │'+board[8]+' │'+board[9]+' │')
print('\t\t\t├─┼─┼─┤')
print('\t\t\t│'+board[4]+' │'+board[5]+' │'+board[6]+' │')
print('\t\t\t├─┼─┼─┤')
print('\t\t\t│'+board[1]+' │'+board[2]+' │'+board[3]+' │')
print('\t\t\t└─┴─┴─┘')
def inputPlayerLetter():
letter = ''
while not (letter == 'X' or letter == 'O'):
print('Do you want to be X or O?')
letter = input().upper()
if letter == 'X':
return ['X', 'O']
else:
return ['O', 'X']
def whoGoesFirst():
if random.randint(0, 1) == 0:
return 'computer'
else:
return 'player'
def playAgain():
print('Do you want to play again? (yes or no)')
return input().lower().startswith('y')
def makeMove(board, letter, move):
board[move] = letter
def isWinner(bo, le):
return ((bo[7] == le and bo[8] == le and bo[9] == le) or
(bo[4] == le and bo[5] == le and bo[6] == le) or
(bo[1] == le and bo[2] == le and bo[3] == le) or
(bo[7] == le and bo[4] == le and bo[1] == le) or
(bo[8] == le and bo[5] == le and bo[2] == le) or
(bo[9] == le and bo[6] == le and bo[3] == le) or
(bo[7] == le and bo[5] == le and bo[3] == le) or
(bo[9] == le and bo[5] == le and bo[1] == le))
def getBoardCopy(board):
dupeBoard = []
for i in board:
dupeBoard.append(i)
return dupeBoard
def isSpaceFree(board, move):
return board[move] == ' '
def getPlayerMove(board):
move = ' '
while move not in '1 2 3 4 5 6 7 8 9'.split() or not isSpaceFree(board, int(move)):
print('What is your next move? (1-9)')
move = input()
return int(move)
def chooseRandomMoveFromList(board, movesList):
possibleMoves = []
for i in movesList:
if isSpaceFree(board, i):
possibleMoves.append(i)
if len(possibleMoves) != 0:
return random.choice(possibleMoves)
else:
return None
def getComputerMove(board, computerLetter):
if computerLetter == 'X':
playerLetter = 'O'
else:
playerLetter = 'X'
for i in range(1, 10):
copy = getBoardCopy(board)
if isSpaceFree(copy, i):
makeMove(copy, computerLetter, i)
if isWinner(copy, computerLetter):
return i
for i in range(1, 10):
copy = getBoardCopy(board)
if isSpaceFree(copy, i):
makeMove(copy, playerLetter, i)
if isWinner(copy, playerLetter):
return i
move = chooseRandomMoveFromList(board, [1, 3, 7, 9])
if move != None:
return move
if isSpaceFree(board, 5):
return 5
return chooseRandomMoveFromList(board, [2, 4, 6, 8])
def isBoardFull(board):
for i in range(1, 10):
if isSpaceFree(board, i):
return False
return True
print('Welcome to Tic Tac Toe!')
while True:
theBoard = [' '] * 10
playerLetter, computerLetter = inputPlayerLetter()
turn = whoGoesFirst()
print('The ' + turn + ' will go first.')
gameIsPlaying = True
while gameIsPlaying:
if turn == 'player':
drawBoard(theBoard)
move = getPlayerMove(theBoard)
makeMove(theBoard, playerLetter, move)
if isWinner(theBoard, playerLetter):
drawBoard(theBoard)
print('Hooray! You have won the game!')
gameIsPlaying = False
else:
if isBoardFull(theBoard):
drawBoard(theBoard)
print('The game is a tie!')
break
else:
turn = 'computer'
else:
move = getComputerMove(theBoard, computerLetter)
makeMove(theBoard, computerLetter, move)
if isWinner(theBoard, computerLetter):
drawBoard(theBoard)
print('The computer has beaten you! You lose.')
gameIsPlaying = False
else:
if isBoardFull(theBoard):
drawBoard(theBoard)
print('The game is a tie!')
break
else:
turn = 'player'
if not playAgain():
break

3 结语

针对如何使用python完成井字棋游戏的问题，提出该方法，通过本次实验，证明该方法是有效的，该程序设计的井字棋游戏具有一定的挑战性，无法满足所有阶段玩家的游戏需求。

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