[深度学习]实现一个博弈型的AI,从五子棋开始(1) – xerwin

好久没有写过博客了,多久,大概8年???最近重新把写作这事儿捡起来……最近在折腾AI,写个AI相关的给团队的小伙伴们看吧。

 

搞了这么多年的机器学习,从分类到聚类,从朴素贝叶斯到SVM,从神经网络到深度学习,各种神秘的项目里用了无数次,但是感觉干的各种事情离我们生活还是太远了。最近AlphaGo Zero的发布,深度学习又火了一把,小伙伴们按捺不住内心的躁动,要搞一个游戏AI,好吧,那就从规则简单、老少皆宜的五子棋开始讲起。

 

好了,废话就说这么多,下面进入第一讲,实现一个五子棋。

 

小伙伴:此处省去吐槽一万字,说好的讲深度学习,怎么开始扯实现一个五子棋程序了,大哥你不按套路出牌啊……

我:工欲善其事必先利其器,要实现五子棋的AI,连棋都没有,AI个锤子!

老罗:什么事?

……

 

五子棋分为有禁手和无禁手,我们先实现一个普通版本的无禁手版本作为例子,因为这个不影响我们实现一个AI。补充说明一下,无禁手黑棋必胜,经过比赛和各种研究,人们逐渐知道了这个事实就开始想办法来限制黑棋先手优势。于是出现了有禁手规则,规定黑棋不能下三三,四四和长连。但随着比赛的结果的研究的继续进行,发现其实即使是对黑棋有禁手限制,还是不能阻止黑棋开局必胜的事实,像直指开局中花月,山月,云月,溪月,寒星等,斜指开局中的名月,浦月,恒星,峡月,岚月都是黑棋必胜。于是日本人继续提出了交换和换打的思想,到了后来发展成了国际比赛中三手交换和五手二打规则,防止执黑者下出必胜开局或者在第五手下出必胜打。所以结论是,在不正规的比赛规则或者无禁手情况下,黑棋必胜是存在的。

 

(1)五子棋下棋逻辑实现

这里用Python来实现,因为之后的机器学习库也是Python的,方便一点。

界面和逻辑要分开,解耦合,这个是毋庸置疑的,并且之后还要训练AI,分离这是必须的。所以我们先来实现一个五子棋的逻辑。

我们先来考虑五子棋是一个15*15的棋盘,棋盘上的每一个交叉点(或格子)上一共会有3种状态:空白、黑棋、白棋,所以先建个文件 consts.py

做如下定义:

from enum import Enum

N = 15

class ChessboardState(Enum):
    EMPTY = 0
    BLACK = 1
    WHITE = 2

 

 

棋盘的状态,我们先用一个15*15的二维数组chessMap来表示,建一个类 gobang.py

currentI、currentJ、currentState 分别表示当前这步着棋的坐标和颜色,再定义一个get和set函数,最基本的框架就出来了,代码如下:

from enum import Enum
from consts import *

class GoBang(object):
    def __init__(self):
        self.__chessMap = [[ChessboardState.EMPTY for j in range(N)] for i in range(N)]
        self.__currentI = -1
        self.__currentJ = -1
        self.__currentState = ChessboardState.EMPTY

    def get_chessMap(self):
        return self.__chessMap

    def get_chessboard_state(self, i, j):
        return self.__chessMap[i][j]

    def set_chessboard_state(self, i, j, state):
        self.__chessMap[i][j] = state
        self.__currentI = i
        self.__currentJ = j
        self.__currentState = state

 

 

这样界面端可以调用get函数来获取各个格子的状态来决定是否绘制棋子,以及绘制什么样的棋子;每次下棋的时候呢,在对应的格子上,通过坐标来设置棋盘Map的状态。

所以最基本的展示和下棋,上面的逻辑就够了,接下来干什么呢,得考虑每次下棋之后,set了对应格子的状态,是不是需要判断当前有没有获胜。所以还需要再加两个函数来干这个事情,思路就是从当前位置从东、南、西、北、东南、西南、西北、东北8个方向,4根轴,看是否有连续的大于5颗相同颜色的棋子出现。假设我们目前落子在棋盘正中,需要判断的位置如下图所示的米字形。

 

 

那代码怎么写呢,最最笨的办法,按照字面意思来翻译咯,比如横轴,先看当前位置左边有多少颗连续同色的,再看右边有多少颗连续同色的,左边加右边,就是当前横轴上的连续数,如果大于5,则胜利。

    def have_five(self, current_i, current_j):
        #四个方向计数 竖 横 左斜 右斜
        hcount = 1

        temp = ChessboardState.EMPTY

        #H-左
        for j in range(current_j - 1, -1, -1):  #横向往左 from (current_j - 1) to 0
            temp = self.__chessMap[current_i][j]
            if temp == ChessboardState.EMPTY or temp != self.__currentState:
                break
            hcount = hcount + 1
#H-右 for j in range(current_j + 1, N): #横向往右 from (current_j + 1) to N temp = self.__chessMap[current_i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break hcount = hcount + 1
#H-结果 if hcount >= 5: return True

 

 

以此类推,再看竖轴、再看左斜、再看右斜。于是,have_five函数变成这样了:

    def have_five(self, current_i, current_j):
        #四个方向计数 横 竖 左斜 右斜
        hcount = 1
        vcount = 1
        lbhcount = 1
        rbhcount = 1

        temp = ChessboardState.EMPTY

        #H-左
        for j in range(current_j - 1, -1, -1):  #横向往左 from (current_j - 1) to 0
            temp = self.__chessMap[current_i][j]
            if temp == ChessboardState.EMPTY or temp != self.__currentState:
                break
            hcount = hcount + 1
        #H-右
        for j in range(current_j + 1, N):  #横向往右 from (current_j + 1) to N
            temp = self.__chessMap[current_i][j]
            if temp == ChessboardState.EMPTY or temp != self.__currentState:
                break
            hcount = hcount + 1
        #H-结果
        if hcount >= 5:
            return True
#V-上 for i in range(current_i - 1, -1, -1): # from (current_i - 1) to 0 temp = self.__chessMap[i][current_j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break vcount = vcount + 1 #V-下 for i in range(current_i + 1, N): # from (current_i + 1) to N temp = self.__chessMap[i][current_j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break vcount = vcount + 1 #V-结果 if vcount >= 5: return True
#LB-上 for i, j in zip(range(current_i - 1, -1, -1), range(current_j - 1, -1, -1)): temp = self.__chessMap[i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break lbhcount = lbhcount + 1 #LB-下 for i, j in zip(range(current_i + 1, N), range(current_j + 1, N)): temp = self.__chessMap[i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break lbhcount = lbhcount + 1 #LB-结果 if lbhcount >= 5: return True
#RB-上 for i, j in zip(range(current_i - 1, -1, -1), range(current_j + 1, N)): temp = self.__chessMap[i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break rbhcount = rbhcount + 1 #RB-下 for i, j in zip(range(current_i + 1, N), range(current_j - 1, -1, -1)): temp = self.__chessMap[i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break rbhcount = rbhcount + 1 #LB-结果 if rbhcount >= 5: return True

 

 

这样是不是就写完了,五子棋的逻辑全部实现~ 

NO,别高兴得太早,我想说,我好恶心,上面那个代码,简直丑爆了,再看一眼,重复的写了这么多for,这么多if,这么多重复的代码块,让我先去吐会儿……

好了,想想办法怎么改,至少分了4根轴,是重复的对不对,然后每根轴分别从正负两个方向去统计,最后加起来,两个方向,也是重复的对不对。

于是我们能不能只写一个方向的代码,分别调2次,然后4根轴,分别再调4次,2*4=8,一共8行代码搞定试试。

因为有45°和135°这两根斜轴的存在,所以方向上应该分别从x和y两个轴来控制正负,于是可以这样,先写一个函数,按照方向来统计:

xdirection=0,ydirection=1       表示从y轴正向数;

xdirection=0,ydirection=-1     表示从y轴负向数;

xdirection=1,ydirection=1       表示从45°斜轴正向数;

……

不一一列举了,再加上边界条件的判断,于是有了以下函数:

    def count_on_direction(self, i, j, xdirection, ydirection, color):
        count = 0
        for step in range(1, 5): #除当前位置外,朝对应方向再看4步
            if xdirection != 0 and (j + xdirection * step < 0 or j + xdirection * step >= N):
                break
            if ydirection != 0 and (i + ydirection * step < 0 or i + ydirection * step >= N):
                break
            if self.__chessMap[i + ydirection * step][j + xdirection * step] == color:
                count += 1
            else:
                break
        return count

 

 

于是乎,前面的have_five稍微长的好看了一点,可以变成这样:

def have_five(self, i, j, color):
        #四个方向计数 横 竖 左斜 右斜
        hcount = 1
        vcount = 1
        lbhcount = 1
        rbhcount = 1

        hcount += self.count_on_direction(i, j, -1, 0, color)
        hcount += self.count_on_direction(i, j, 1, 0, color)
        if hcount >= 5:
            return True

        vcount += self.count_on_direction(i, j, 0, -1, color)
        vcount += self.count_on_direction(i, j, 0, 1, color)
        if vcount >= 5:
            return True

        lbhcount += self.count_on_direction(i, j, -1, 1, color)
        lbhcount += self.count_on_direction(i, j, 1, -1, color)
        if lbhcount >= 5:
            return True

        rbhcount += self.count_on_direction(i, j, -1, -1, color)
        rbhcount += self.count_on_direction(i, j, 1, 1, color)
        if rbhcount >= 5:
            return True

 

 

还是一大排重复的代码呀,我还是觉得它丑啊,我真的不是处女座,但是这个函数是真丑啊,能不能让它再帅一点,当然可以,4个重复块再收成一个函数,循环调4次,是不是可以,好,就这么干,于是have_five就又漂亮了一点点:

    def have_five(self, i, j, color):
        #四个方向计数 横 竖 左斜 右斜
        directions = [[(-1, 0), (1, 0)], \
                      [(0, -1), (0, 1)], \
                      [(-1, 1), (1, -1)], \
                      [(-1, -1), (1, 1)]]

        for axis in directions:
            axis_count = 1
            for (xdirection, ydirection) in axis:
                axis_count += self.count_on_direction(i, j, xdirection, ydirection, color)
                if axis_count >= 5:
                    return True

        return False

 

 

嗯,感觉好多了,这下判断是否有5颗相同颜色棋子的逻辑也有了,再加一个函数来给界面层返回结果,逻辑部分的代码就差不多了:

    def get_chess_result(self):
        if self.have_five(self.__currentI, self.__currentJ, self.__currentState):
            return self.__currentState
        else:
            return ChessboardState.EMPTY

 

 

于是,五子棋逻辑代码就写完了,完整代码 gobang.py 如下:

#coding:utf-8

from enum import Enum
from consts import *

class GoBang(object):
    def __init__(self):
        self.__chessMap = [[ChessboardState.EMPTY for j in range(N)] for i in range(N)]
        self.__currentI = -1
        self.__currentJ = -1
        self.__currentState = ChessboardState.EMPTY

    def get_chessMap(self):
        return self.__chessMap

    def get_chessboard_state(self, i, j):
        return self.__chessMap[i][j]

    def set_chessboard_state(self, i, j, state):
        self.__chessMap[i][j] = state
        self.__currentI = i
        self.__currentJ = j
        self.__currentState = state

    def get_chess_result(self):
        if self.have_five(self.__currentI, self.__currentJ, self.__currentState):
            return self.__currentState
        else:
            return ChessboardState.EMPTY

    def count_on_direction(self, i, j, xdirection, ydirection, color):
        count = 0
        for step in range(1, 5): #除当前位置外,朝对应方向再看4步
            if xdirection != 0 and (j + xdirection * step < 0 or j + xdirection * step >= N):
                break
            if ydirection != 0 and (i + ydirection * step < 0 or i + ydirection * step >= N):
                break
            if self.__chessMap[i + ydirection * step][j + xdirection * step] == color:
                count += 1
            else:
                break
        return count

    def have_five(self, i, j, color):
        #四个方向计数 横 竖 左斜 右斜
        directions = [[(-1, 0), (1, 0)], \
                      [(0, -1), (0, 1)], \
                      [(-1, 1), (1, -1)], \
                      [(-1, -1), (1, 1)]]

        for axis in directions:
            axis_count = 1
            for (xdirection, ydirection) in axis:
                axis_count += self.count_on_direction(i, j, xdirection, ydirection, color)
                if axis_count >= 5:
                    return True

        return False

 

小伙伴:大哥,憋了半天,就憋出这么不到60行代码?

我:代码不在多,实现则灵……

 

明天来给它加个render,前端界面就有了,就是一个简单的完整游戏了,至于AI,别急嘛。

好吧,就这样…

 

UI部分在这里:

[深度学习]实现一个博弈型的AI,从五子棋开始(2)



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