Python实现AI的俄罗斯方块小游戏
python教程
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·
2021-09-11 02:45
前言
代码效果展示
开发工具
环境搭建
原理简介
(1)游戏制作
(2)AI源码实现
# 简单的AI算法
for d_now in current_direction_range:
x_now_min, x_now_max, y_now_min, y_now_max = self.inner_board.current_tetris.getRelativeBoundary(d_now)
for x_now in range(-x_now_min, self.inner_board.width - x_now_max):
board = self.getFinalBoardData(d_now, x_now)
for d_next in next_direction_range:
x_next_min, x_next_max, y_next_min, y_next_max = self.inner_board.next_tetris.getRelativeBoundary(d_next)
distances = self.getDropDistances(board, d_next, range(-x_next_min, self.inner_board.width-x_next_max))
for x_next in range(-x_next_min, self.inner_board.width-x_next_max):
score = self.calcScore(copy.deepcopy(board), d_next, x_next, distances)
if not action or action[2] < score:
action = [d_now, x_now, score]
return action
可消除的行数;
堆积后的俄罗斯方块内的虚洞数量;
堆积后的俄罗斯方块内的小方块数量;
堆积后的俄罗斯方块的最高点;
堆积后的俄罗斯方块的高度(每一列都有一个高度)标准差;
堆积后的俄罗斯方块的高度一阶前向差分;
堆积后的俄罗斯方块的高度一阶前向差分的标准差;
堆积后的俄罗斯方块的最高点和最低点之差。
代码实现如下:
# 空位统计
hole_statistic_0 = [0] * width
hole_statistic_1 = [0] * width
# 方块数量
num_blocks = 0
# 空位数量
num_holes = 0
# 每个x位置堆积俄罗斯方块的最高点
roof_y = [0] * width
for y in range(height-1, -1, -1):
# 是否有空位
has_hole = False
# 是否有方块
has_block = False
for x in range(width):
if board[x + y * width] == tetrisShape().shape_empty:
has_hole = True
hole_statistic_0[x] += 1
else:
has_block = True
roof_y[x] = height - y
if hole_statistic_0[x] > 0:
hole_statistic_1[x] += hole_statistic_0[x]
hole_statistic_0[x] = 0
if hole_statistic_1[x] > 0:
num_blocks += 1
if not has_block:
break
if not has_hole and has_block:
removed_lines += 1
# 数据^0.7之和
num_holes = sum([i ** .7 for i in hole_statistic_1])
# 最高点
max_height = max(roof_y) - removed_lines
# roof_y做差分运算
roof_dy = [roof_y[i]-roof_y[i+1] for i in range(len(roof_y)-1)]
# 计算标准差E(x^2) - E(x)^2
if len(roof_y) <= 0:
roof_y_std = 0
else:
roof_y_std = math.sqrt(sum([y**2 for y in roof_y]) / len(roof_y) - (sum(roof_y) / len(roof_y)) ** 2)
if len(roof_dy) <= 0:
roof_dy_std = 0
else:
roof_dy_std = math.sqrt(sum([dy**2 for dy in roof_dy]) / len(roof_dy) - (sum(roof_dy) / len(roof_dy)) ** 2)
# roof_dy绝对值之和
abs_dy = sum([abs(dy) for dy in roof_dy])
# 最大值与最小值之差
max_dy = max(roof_y) - min(roof_y)
# 计算得分
score = removed_lines * 1.8 - num_holes * 1.0 - num_blocks * 0.5 - max_height ** 1.5 * 0.02 - roof_y_std * 1e-5 - roof_dy_std * 0.01 - abs_dy * 0.2 - max_dy * 0.3
return score
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