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技术资源汇总(杭电支持版)/4.人工智能/4.3.4.1程序示例.md
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341
技术资源汇总(杭电支持版)/4.人工智能/4.3.4.1程序示例.md
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# 程序示例
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::: tip
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阅读程序并运行
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完成习题
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:::
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::: tip 📥
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本节附件下载 <Download url="https://cdn.xyxsw.site/code/4-Lecture.zip"/>
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:::
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## Hospital(局部搜索)
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```python
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import random
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class Space():
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def __init__(self, height, width, num_hospitals):
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"""创建一个具有给定维度的新状态空间"""
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self.height = height # 高度
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self.width = width # 宽度
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self.num_hospitals = num_hospitals # 医院数量
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self.houses = set() # 住房位置集合
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self.hospitals = set() # 医院位置集合
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def add_house(self, row, col):
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"""在状态空间中的特定位置添加住房"""
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self.houses.add((row, col))
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def available_spaces(self):
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"""返回住房或医院当前未使用的所有单元格"""
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# 考虑所有可能的单元格
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candidates = set(
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(row, col)
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for row in range(self.height)
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for col in range(self.width)
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)
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# 排除所有住房和医院
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for house in self.houses:
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candidates.remove(house)
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for hospital in self.hospitals:
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candidates.remove(hospital)
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return candidates
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def hill_climb(self, maximum=None, image_prefix=None, log=False):
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"""执行爬山算法找到解决方案"""
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count = 0
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# 从随机初始化的医院位置开始
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self.hospitals = set()
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for i in range(self.num_hospitals):
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self.hospitals.add(random.choice(list(self.available_spaces())))
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...
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# 执行算法，直到达到最大迭代次数
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while maximum is None or count < maximum:
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count += 1
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best_neighbors = []
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best_neighbor_cost = None
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# 考虑所有医院移动
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for hospital in self.hospitals:
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# 考虑一下那家医院的所有邻居
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for replacement in self.get_neighbors(*hospital):
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# 生成一组相邻的医院
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neighbor = self.hospitals.copy()
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neighbor.remove(hospital)
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neighbor.add(replacement)
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# 检查邻居是否是迄今为止最好的
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cost = self.get_cost(neighbor)
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if best_neighbor_cost is None or cost < best_neighbor_cost:
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best_neighbor_cost = cost
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best_neighbors = [neighbor]
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elif best_neighbor_cost == cost:
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best_neighbors.append(neighbor)
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# 没有一个邻居比目前的状态更好
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if best_neighbor_cost >= self.get_cost(self.hospitals):
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return self.hospitals
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# 移动到价值最高的邻居
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else:
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...
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self.hospitals = random.choice(best_neighbors)
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...
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def random_restart(self, maximum, image_prefix=None, log=False):
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"""多次重复爬山算法"""
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best_hospitals = None
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best_cost = None
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# 重复爬山算法的固定次数
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for i in range(maximum):
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hospitals = self.hill_climb()
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cost = self.get_cost(hospitals)
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if best_cost is None or cost < best_cost:
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best_cost = cost
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best_hospitals = hospitals
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...
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else:
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...
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...
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return best_hospitals
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def get_cost(self, hospitals):
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"""计算从住房到最近医院的距离总和"""
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cost = 0
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for house in self.houses:
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cost += min(
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abs(house[0] - hospital[0]) + abs(house[1] - hospital[1])
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for hospital in hospitals
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)
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return cost
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def get_neighbors(self, row, col):
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"""返回尚未包含住房或医院的邻居"""
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candidates = [
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(row - 1, col),
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(row + 1, col),
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(row, col - 1),
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(row, col + 1)
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]
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neighbors = []
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for r, c in candidates:
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if (r, c) in self.houses or (r, c) in self.hospitals:
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continue
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if 0 <= r < self.height and 0 <= c < self.width:
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neighbors.append((r, c))
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return neighbors
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def output_image(self, filename):
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"""生成所有房屋和医院的图像（不作要求）"""
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from PIL import Image, ImageDraw, ImageFont
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cell_size = 100
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cell_border = 2
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cost_size = 40
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padding = 10
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# Create a blank canvas
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img = Image.new(
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"RGBA",
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(self.width * cell_size,
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self.height * cell_size + cost_size + padding * 2),
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"white"
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)
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house = Image.open("assets/images/House.png").resize(
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(cell_size, cell_size)
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)
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hospital = Image.open("assets/images/Hospital.png").resize(
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(cell_size, cell_size)
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)
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font = ImageFont.truetype("assets/fonts/OpenSans-Regular.ttf", 30)
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draw = ImageDraw.Draw(img)
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for i in range(self.height):
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for j in range(self.width):
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# Draw cell
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rect = [
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(j * cell_size + cell_border,
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i * cell_size + cell_border),
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((j + 1) * cell_size - cell_border,
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(i + 1) * cell_size - cell_border)
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]
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draw.rectangle(rect, fill="black")
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if (i, j) in self.houses:
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img.paste(house, rect[0], house)
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if (i, j) in self.hospitals:
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img.paste(hospital, rect[0], hospital)
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# Add cost
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draw.rectangle(
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(0, self.height * cell_size, self.width * cell_size,
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self.height * cell_size + cost_size + padding * 2),
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"black"
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)
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draw.text(
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(padding, self.height * cell_size + padding),
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f"Cost: {self.get_cost(self.hospitals)}",
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fill="white",
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font=font
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)
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img.save(filename)
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# 创建一个状态空间并随机添加住房
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s = Space(height=10, width=20, num_hospitals=3)
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for i in range(15):
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s.add_house(random.randrange(s.height), random.randrange(s.width))
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# 使用局部搜索来确定医院位置
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hospitals = s.random_restart(maximum=100, image_prefix="hospitals", log=True)
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```
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## Production(线性规划)
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```python
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import scipy.optimize
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# Objective Function: 50x_1 + 80x_2
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# Constraint 1: 5x_1 + 2x_2 <= 20
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# Constraint 2: -10x_1 + -12x_2 <= -90
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result = scipy.optimize.linprog(
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[50, 80], # Cost function: 50x_1 + 80x_2
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A_ub=[[5, 2], [-10, -12]], # Coefficients for inequalities
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b_ub=[20, -90], # Constraints for inequalities: 20 and -90
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)
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if result.success:
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print(f"X1: {round(result.x[0], 2)} hours")
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print(f"X2: {round(result.x[1], 2)} hours")
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else:
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print("No solution")
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```
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## Schedule(约束满足)
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```python
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"""没有任何启发式或推理的自然回溯搜索"""
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VARIABLES = ["A", "B", "C", "D", "E", "F", "G"]
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CONSTRAINTS = [
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("A", "B"),
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("A", "C"),
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("B", "C"),
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("B", "D"),
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("B", "E"),
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("C", "E"),
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("C", "F"),
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("D", "E"),
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("E", "F"),
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("E", "G"),
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("F", "G")
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]
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def backtrack(assignment):
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"""运行回溯搜索以查找赋值"""
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# 检查赋值是否完成
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if len(assignment) == len(VARIABLES):
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return assignment
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# 尝试一个新变量
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var = select_unassigned_variable(assignment)
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for value in ["Monday", "Tuesday", "Wednesday"]:
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new_assignment = assignment.copy()
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new_assignment[var] = value
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if consistent(new_assignment):
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result = backtrack(new_assignment)
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if result is not None:
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return result
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return None
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def select_unassigned_variable(assignment):
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"""按顺序选择尚未赋值的变量"""
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for variable in VARIABLES:
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if variable not in assignment:
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return variable
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return None
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def consistent(assignment):
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"""检查分配是否一致"""
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for (x, y) in CONSTRAINTS:
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# 仅考虑变量赋值都已指定的弧
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if x not in assignment or y not in assignment:
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continue
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# 如果两者的值相同，则不一致
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if assignment[x] == assignment[y]:
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return False
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# 如果没有不一致的地方，那么赋值是一致的
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return True
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solution = backtrack(dict())
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print(solution)
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```
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使用命令`pip install python-constraint`安装 constraint 库
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```python
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from constraint import *
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problem = Problem()
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# 添加变量
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problem.addVariables(
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["A", "B", "C", "D", "E", "F", "G"],
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["Monday", "Tuesday", "Wednesday"]
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)
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# 添加约束
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CONSTRAINTS = [
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("A", "B"),
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("A", "C"),
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("B", "C"),
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("B", "D"),
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("B", "E"),
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("C", "E"),
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("C", "F"),
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("D", "E"),
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("E", "F"),
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("E", "G"),
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("F", "G")
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]
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for x, y in CONSTRAINTS:
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problem.addConstraint(lambda x, y: x != y, (x, y))
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# Solve problem
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for solution in problem.getSolutions():
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print(solution)
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```
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## Quiz
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1. 对于以下哪一项，即使多次重新运行算法，也会始终找到相同的解决方案？
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假设一个问题的目标是最小化成本函数，并且状态空间中的每个状态都有不同的成本。
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1. Steepest-ascent hill-climbing，每次从不同的初始状态开始
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2. Steepest-ascent hill-climbing，每次都从相同的初始状态开始
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3. Stochastic hill-climbing，每次从不同的初始状态开始
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4. Stochastic hill-climbing，每次都从相同的初始状态开始
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5. 无论是 steepest-ascent 还是 stochastic hill climbing，只要你总是从同一个初始状态开始
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6. 无论是 steepest-ascent 还是 stochastic hill climbing，只要每次都从不同的初始状态开始
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7. 没有任何版本的爬山算法能保证每次都能得到相同的解决方案
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2. 下面两个问题都会问你关于下面描述的优化问题。
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一位农民正在尝试种植两种作物，`作物 1` 和`作物 2`，并希望实现利润最大化。农民将从种植的每英亩`作物 1` 中获得 500 美元的利润，从种植的每英亩`作物 2` 中获得 400 美元的利润。然而，农民今天需要在早上 7 点到晚上 7 点之间的 12 个小时内完成所有的种植。种植一英亩的`作物 1` 需要 3 个小时，种植一英亩`作物 2` 需要 2 个小时。农民在供应方面也很有限：他有足够的供应种植 10 英亩的`作物 1`，有足够的资源种植 4 英亩的`作物 2`。假设变量 C1 表示要种植的`作物 1` 的英亩数，变量 C2 表示要种植`作物 2` 的英亩数。
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对于这个问题，什么是有效的目标函数？
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1. 10 \* C1 + 4 \* C2
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2. -3 \* C1 - 2 \* C2
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3. 500 \* 10 \* C1 + 400 \* 4 \* C2
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4. 500 \* C1 + 400 \* C2
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5. C1 + C2
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3. 这个问题的制约因素是什么？
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1. 3 \* C1 + 2 \* C2 <= 12, C1 <= 10, C2 <= 4
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2. 3 \* C1 + 2 \* C2 <= 12, C1 + C2 <= 14
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3. 3 \* C1 <= 10, 2 \* C2 <= 4
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4. C1 + C2 <= 12, C1 + C2 <= 14
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4. 下面的问题将问你以下考试安排约束满足图，其中每个节点代表一个课程。每门课程都与可能的考试日的初始域相关联（大多数课程可能在周一、周二或周三；少数课程已经被限制在一天内）。两个节点之间的边意味着这两个课程必须在不同的日子进行考试。
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在对整个问题满足弧一致性之后，变量 C、D 和 E 的结果域是什么？
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1. C 的域是\{Mon，Tue\}，D 的域是\{Wed\}，E 的域是\{Mon\}
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2. C 的域是\{Mon\}，D 的域是\{Wed\}，E 的域为\{Tue\}
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3. C 的域是\{Mon\}，D 的域是\{Tue\}，E 的域为\{Wed\}
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4. C 的域是\{Mon\}，D 的域是\{Mon，Wed\}，E 的域是\{Tue，Wed\}
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5. C 的域是\{Mon，Tue，Wed\}，D 的域是\{Mon，Wed\}，E 的域是\{Mon，Tue，Wed\}
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6. C 的域是\{Mon\}，D 的域是\{Mon，Wed\}，E 的域是\{Mon，Tue，Wed\}
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