python实现翻转棋游戏（othello）

'''
作者：hhh5460
时间：2017年7月1日
'''

class Board(object):
 def __init__(self):
  self.empty = '.'
  self._board = [[self.empty for _ in range(8)] for _ in range(8)] # 规格：8*8
  self._board[3][4], self._board[4][3] = 'X', 'X'
  self._board[3][3], self._board[4][4] = 'O', 'O'
  
 # 增加 Board[][] 索引语法
 def __getitem__(self, index):
  return self._board[index]
 
 # 打印棋盘
 def print_b(self):
  board = self._board
  print(' ', ' '.join(list('ABCDEFGH')))
  for i in range(8):
   print(str(i+1),' '.join(board[i]))
   
 # 棋局终止
 def teminate(self):
  list1 = list(self.get_legal_actions('X'))
  list2 = list(self.get_legal_actions('O'))
  return [False, True][len(list1) == 0 and len(list2) == 0]
  
 # 判断赢家
 def get_winner(self):
  s1, s2 = 0, 0
  for i in range(8):
   for j in range(8):
    if self._board[i][j] == 'X':
     s1 += 1
    if self._board[i][j] == 'O':
     s2 += 1
  if s1 > s2:
   return 0 # 黑胜
  elif s1 < s2:
   return 1 # 白胜
  elif s1 == s2:
   return 2 # 平局
 # 落子
 def _move(self, action, color):
  x,y = action
  self._board[x][y] = color
  
  return self._flip(action, color)
  
 
  
 
 # 翻子（返回list）
 def _flip(self, action, color):
  flipped_pos = []
  
  for line in self._get_lines(action):
   for i,p in enumerate(line):
    if self._board[p[0]][p[1]] == self.empty: 
     break
    elif self._board[p[0]][p[1]] == color:
     flipped_pos.extend(line[:i])
     break
  
  for p in flipped_pos:
   self._board[p[0]][p[1]] = color
   
  return flipped_pos
  
 # 撤销
 def _unmove(self, action, flipped_pos, color):
  self._board[action[0]][action[1]] = self.empty
  
  uncolor = ['X', 'O'][color=='X']
  for p in flipped_pos:
   self._board[p[0]][p[1]] = uncolor
   
 # 生成8个方向的下标数组，方便后续操作
 def _get_lines(self, action):
  '''说明：刚开始我是用一维棋盘来考虑的，后来改为二维棋盘。偷懒，不想推倒重来，简单地修改了一下'''
  board_coord = [(i,j) for i in range(8) for j in range(8)] # 棋盘坐标
  
  r,c = action
  ix = r*8 + c
  r, c = ix//8, ix%8
  left = board_coord[r*8:ix] # 要反转
  right = board_coord[ix+1:(r+1)*8]
  top = board_coord[c:ix:8] # 要反转
  bottom = board_coord[ix+8:8*8:8]
  
  if r <= c:
   lefttop = board_coord[c-r:ix:9] # 要反转
   rightbottom = board_coord[ix+9:(7-(c-r))*8+7+1:9]
  else:
   lefttop = board_coord[(r-c)*8:ix:9] # 要反转
   rightbottom = board_coord[ix+9:7*8+(7-(c-r))+1:9]
  
  if r+c<=7:
   leftbottom = board_coord[ix+7:(r+c)*8:7]
   righttop = board_coord[r+c:ix:7] # 要反转
  else:
   leftbottom = board_coord[ix+7:7*8+(r+c)-7+1:7]
   righttop = board_coord[((r+c)-7)*8+7:ix:7] # 要反转
  
  # 有四个要反转，方便判断
  left.reverse()
  top.reverse()
  lefttop.reverse()
  righttop.reverse()
  lines = [left, top, lefttop, righttop, right, bottom, leftbottom, rightbottom]
  return lines
  
 # 检测，位置是否有子可翻
 def _can_fliped(self, action, color):
  flipped_pos = []
  
  for line in self._get_lines(action):
   for i,p in enumerate(line):
    if self._board[p[0]][p[1]] == self.empty: 
     break
    elif self._board[p[0]][p[1]] == color:
     flipped_pos.extend(line[:i])
     break
  return [False, True][len(flipped_pos) > 0]
  
 # 合法走法
 def get_legal_actions(self, color):
  uncolor = ['X', 'O'][color=='X']
  uncolor_near_points = [] # 反色邻近的空位
  
  board = self._board
  for i in range(8):
   for j in range(8):
    if board[i][j] == uncolor:
     for dx,dy in [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1)]:
      x, y = i+dx, j+dy
      if 0 <= x <=7 and 0 <= y <=7 and board[x][y] == self.empty and (x, y) not in uncolor_near_points:
       uncolor_near_points.append((x, y))
  for p in uncolor_near_points:
   if self._can_fliped(p, color):
    yield p

# 测试
if __name__ == '__main__':
 board = Board()
 board.print_b()
 print(list(board.get_legal_actions('X')))