200 行代码实现简易版 2048 游戏,2002048,state 存储当前状态


创建游戏文件 2048.py

首先导入需要的包:

Python
import curses
from random import randrange, choice
from collections import defaultdict

主逻辑

用户行为

所有的有效输入都可以转换为”上,下,左,右,游戏重置,退出”这六种行为,用 actions 表示

Python
actions = ['Up', 'Left', 'Down', 'Right', 'Restart', 'Exit']

有效输入键是最常见的 W(上),A(左),S(下),D(右),R(重置),Q(退出),这里要考虑到大写键开启的情况,获得有效键值列表:

Python
letter_codes = [ord(ch) for ch in 'WASDRQwasdrq']

将输入与行为进行关联:

Python
actions_dict = dict(zip(letter_codes, actions * 2))

状态机

处理游戏主逻辑的时候我们会用到一种十分常用的技术:状态机,或者更准确的说是有限状态机(FSM)

你会发现 2048 游戏很容易就能分解成几种状态的转换。

state 存储当前状态, state_actions 这个词典变量作为状态转换的规则,它的 key 是状态,value 是返回下一个状态的函数:

  • Init: init()
    • Game
  • Game: game()
    • Game
    • Win
    • GameOver
    • Exit
  • Win: lambda: not_game(‘Win’)
    • Init
    • Exit
  • Gameover: lambda: not_game(‘Gameover’)
    • Init
    • Exit
  • Exit: 退出循环

状态机会不断循环,直到达到 Exit 终结状态结束程序。

下面是经过提取的主逻辑的代码,会在后面进行补全:

Python
def main(stdscr):
 
    def init():
        #重置游戏棋盘
        return 'Game'
 
    def not_game(state):
        #画出 GameOver 或者 Win 的界面
        #读取用户输入得到action,判断是重启游戏还是结束游戏
        responses = defaultdict(lambda: state) #默认是当前状态,没有行为就会一直在当前界面循环
        responses['Restart'], responses['Exit'] = 'Init', 'Exit' #对应不同的行为转换到不同的状态
        return responses[action]
 
    def game():
        #画出当前棋盘状态
        #读取用户输入得到action
        if action == 'Restart':
            return 'Init'
        if action == 'Exit':
            return 'Exit'
        #if 成功移动了一步:
            if 游戏胜利了:
                return 'Win'
            if 游戏失败了:
                return 'Gameover'
        return 'Game'
 
 
    state_actions = {
            'Init': init,
            'Win': lambda: not_game('Win'),
            'Gameover': lambda: not_game('Gameover'),
            'Game': game
        }
 
    state = 'Init'
 
    #状态机开始循环
    while state != 'Exit':
        state = state_actions[state]()

用户输入处理

阻塞+循环,直到获得用户有效输入才返回对应行为:

Python
def get_user_action(keyboard):
    char = "N"
    while char not in actions_dict:
        char = keyboard.getch()
    return actions_dict[char]

矩阵转置与矩阵逆转

加入这两个操作可以大大节省我们的代码量,减少重复劳动,看到后面就知道了。

矩阵转置:

Python
def transpose(field):
    return [list(row) for row in zip(*field)]

矩阵逆转(不是逆矩阵):

Python
def invert(field):
    return [row[::-1] for row in field]

创建棋盘

初始化棋盘的参数,可以指定棋盘的高和宽以及游戏胜利条件,默认是最经典的 4×4~2048。

Python
class GameField(object):
def __init__(self, height=4, width=4, win=2048):
    self.height = height       #高
    self.width = width         #宽
    self.win_value = 2048      #过关分数
    self.score = 0             #当前分数
    self.highscore = 0         #最高分
    self.reset()               #棋盘重置

棋盘操作

随机生成一个 2 或者 4

Python
def spawn(self):
        new_element = 4 if randrange(100) > 89 else 2
        (i,j) = choice([(i,j) for i in range(self.width) for j in range(self.height) if self.field[i][j] == 0])
        self.field[i][j] = new_element

重置棋盘

Python
def reset(self):
    if self.score > self.highscore:
        self.highscore = self.score
    self.score = 0
    self.field = [[0 for i in range(self.width)] for j in range(self.height)]
    self.spawn()
    self.spawn()

一行向左合并

(注:这一操作是在 move 内定义的,拆出来是为了方便阅读)

Python
def move_row_left(row):
    def tighten(row): # 把零散的非零单元挤到一块
        new_row = [i for i in row if i != 0]
        new_row += [0 for i in range(len(row) - len(new_row))]
        return new_row
 
    def merge(row): # 对邻近元素进行合并
        pair = False
        new_row = []
        for i in range(len(row)):
            if pair:
                new_row.append(2 * row[i])
                self.score += 2 * row[i]
                pair = False
            else:
                if i + 1 < len(row) and row[i] == row[i + 1]:
                    pair = True
                    new_row.append(0)
                else:
                    new_row.append(row[i])
        assert len(new_row) == len(row)
        return new_row
    #先挤到一块再合并再挤到一块
    return tighten(merge(tighten(row)))

棋盘走一步

通过对矩阵进行转置与逆转,可以直接从左移得到其余三个方向的移动操作

Python
def move(self, direction):
    def move_row_left(row):
        #一行向左合并
 
    moves = {}
    moves['Left']  = lambda field: [move_row_left(row) for row in field]
    moves['Right'] = lambda field: invert(moves['Left'](invert(field)))
    moves['Up']    = lambda field: transpose(moves['Left'](transpose(field)))
    moves['Down']  = lambda field: transpose(moves['Right'](transpose(field)))
 
    if direction in moves:
        if self.move_is_possible(direction):
            self.field = moves[direction](self.field)
            self.spawn()
            return True
        else:
            return False

判断输赢

Python
def is_win(self):
    return any(any(i >= self.win_value for i in row) for row in self.field)
 
def is_gameover(self):
    return not any(self.move_is_possible(move) for move in actions)

判断能否移动

Python
def move_is_possible(self, direction):
    defrow_is_left_movable(row):
        def change(i):
            if row[i] == 0 and row[i + 1] != 0: # 可以移动
                return True
            if row[i] != 0 and row[i + 1] == row[i]: # 可以合并
                return True
            return False
        return any(change(i) for i in range(len(row) - 1))
 
    check = {}
    check['Left']  = lambda field: any(row_is_left_movable(row) for row in field)
 
    check['Right'] = lambda field: check['Left'](invert(field))
 
    check['Up']    = lambda field: check['Left'](transpose(field))
 
    check['Down']  = lambda field: check['Right'](transpose(field))
 
    if direction in check:
        return check[direction](self.field)
    else:
        return False

绘制游戏界面

Python
def draw(self, screen):
    help_string1 = '(W)Up (S)Down (A)Left (D)Right'
    help_string2 = '     (R)Restart (Q)Exit'
    gameover_string = '           GAME OVER'
    win_string = '          YOU WIN!'
    def cast(string):
        screen.addstr(string + 'n')
 
    #绘制水平分割线
    def draw_hor_separator():
        line = '+' + ('+------' * self.width + '+')[1:]
        separator = defaultdict(lambda: line)
        if not hasattr(draw_hor_separator, "counter"):
            draw_hor_separator.counter = 0
        cast(separator[draw_hor_separator.counter])
        draw_hor_separator.counter += 1
 
    def draw_row(row):
        cast(''.join('|{: ^5} '.format(num) if num > 0 else '|      ' for num in row) + '|')
 
    screen.clear()
 
    cast('SCORE: ' + str(self.score))
    if 0 != self.highscore:
        cast('HGHSCORE: ' + str(self.highscore))
 
    for row in self.field:
        draw_hor_separator()
        draw_row(row)
 
    draw_hor_separator()
 
    if self.is_win():
        cast(win_string)
    else:
        if self.is_gameover():
            cast(gameover_string)
        else:
            cast(help_string1)
    cast(help_string2)

完成主逻辑

完成以上工作后,我们就可以补完主逻辑了!

Python
def main(stdscr):
    def init():
        #重置游戏棋盘
        game_field.reset()
        return 'Game'
 
    def not_game(state):
        #画出 GameOver 或者 Win 的界面
        game_field.draw(stdscr)
        #读取用户输入得到action,判断是重启游戏还是结束游戏
        action = get_user_action(stdscr)
        responses = defaultdict(lambda: state) #默认是当前状态,没有行为就会一直在当前界面循环
        responses['Restart'], responses['Exit'] = 'Init', 'Exit' #对应不同的行为转换到不同的状态
        return responses[action]
 
    def game():
        #画出当前棋盘状态
        game_field.draw(stdscr)
        #读取用户输入得到action
        action = get_user_action(stdscr)
 
        if action == 'Restart':
            return 'Init'
        if action == 'Exit':
            return 'Exit'
        if game_field.move(action): # move successful
            if game_field.is_win():
                return 'Win'
            if game_field.is_gameover():
                return 'Gameover'
        return 'Game'
 
 
    state_actions = {
            'Init': init,
            'Win': lambda: not_game('Win'),
            'Gameover': lambda: not_game('Gameover'),
            'Game': game
        }
 
    curses.use_default_colors()
    game_field = GameField(win=32)
 
 
    state = 'Init'
 
    #状态机开始循环
    while state != 'Exit':
        state = state_actions[state]()

运行

填上最后一行代码:

Python
curses.wrapper(main)

屏幕快照 2016-04-13 下午4.48.12

完整版代码地址

https://github.com/JLUNeverMore/easy_2048-in-200-lines

评论关闭