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。