archieve: homework8
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"""
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连续型——Hopfield神经网络求解TSP
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1、初始化权值(A, D, U0)
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2、计算N个城市的距离矩阵dxy
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3、初始化神经网络的输入Uxi和输出Vxi
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4、利用动力微分方程计算:dUxi / dt
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5、由一阶欧拉方法更新计算:Uxi(t + 1) = Uxi(t) + dUxi / dt * step
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6、由非线性函数sigmoid更新计算:Vxi(t) = 0.5 * (1 + th(Uxi / U0))
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7、计算能量函数E
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8、检查路径是否合法
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"""
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import numpy as np
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from matplotlib import pyplot as plt
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# 代价函数(具有三角不等式性质)
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def price_cn(vec1, vec2):
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# 元素的平方和再开根号
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return np.linalg.norm(np.array(vec1) - np.array(vec2))
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# 计算该方案下,总的路径长度
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def calc_distance(path):
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dis = 0.0
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for i in range(len(path) - 1):
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dis += distance[path[i]][path[i + 1]]
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return dis
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# 得到城市之间的距离矩阵
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def get_distance(citys):
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N = len(citys)
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# 构造一个N*N的零矩阵
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distance = np.zeros((N, N))
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for i, curr_point in enumerate(citys):
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line = []
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# 计算不同城市之间的距离
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[line.append(price_cn(curr_point, other_point)) if i != j else line.append(0.0) for j, other_point in
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enumerate(citys)]
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# 把距离添加到矩阵相应的位置上
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distance[i] = line
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return distance
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# 动态方程计算微分方程du
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def calc_du(V, distance):
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a = np.sum(V, axis=0) - 1 # 按列相加 - 1
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b = np.sum(V, axis=1) - 1 # 按行相加 - 1
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t1 = np.zeros((N, N))
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t2 = np.zeros((N, N))
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for i in range(N):
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for j in range(N):
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t1[i, j] = a[j]
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for i in range(N):
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for j in range(N):
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t2[j, i] = b[j]
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# 将第一列移动到最后一列
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c_1 = V[:, 1:N]
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# 构造一个N行1列的零举证
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c_0 = np.zeros((N, 1))
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c_0[:, 0] = V[:, 0]
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# 把c_1和c_0在行方向上连接
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c = np.concatenate((c_1, c_0), axis=1)
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c = np.dot(distance, c)
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return -A * (t1 + t2) - D * c
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# 更新神经网络的输入U
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# Uxi(t+1) = Uxi(t) + dUxi/dt * step
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def calc_U(U, du, step):
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return U + du * step
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# 更新神经网络的输出V
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# Vxi(t) = 0.5 * (1 + th(Uxi/U0))
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def calc_V(U, U0):
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return 1 / 2 * (1 + np.tanh(U / U0))
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# 计算当前网络的能量
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def calc_energy(V, distance):
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t1 = np.sum(np.power(np.sum(V, axis=0) - 1, 2))
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t2 = np.sum(np.power(np.sum(V, axis=1) - 1, 2))
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idx = [i for i in range(1, N)]
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idx = idx + [0]
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Vt = V[:, idx]
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t3 = distance * Vt
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t3 = np.sum(np.sum(np.multiply(V, t3)))
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e = 0.5 * (A * (t1 + t2) + D * t3)
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return e
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# 检查路径的正确性
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def check_path(V):
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newV = np.zeros([N, N])
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route = []
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for i in range(N):
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mm = np.max(V[:, i])
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for j in range(N):
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if V[j, i] == mm:
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newV[j, i] = 1
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route += [j]
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break
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return route, newV
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# 可视化画出哈密顿回路和能量趋势
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def draw_H_and_E(citys, H_path, energys):
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fig = plt.figure()
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# 绘制哈密顿回路
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ax1 = fig.add_subplot(121)
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# 设置x轴的数值显示范围
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plt.xlim(0, 7)
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# 设置y轴的数值显示范围
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plt.ylim(0, 7)
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for (from_, to_) in H_path:
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# 绘制城市点,大小为0.2,颜色为红色
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p1 = plt.Circle(citys[from_], 0.2, color='red')
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p2 = plt.Circle(citys[to_], 0.2, color='red')
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ax1.add_patch(p1)
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ax1.add_patch(p2)
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ax1.plot((citys[from_][0], citys[to_][0]), (citys[from_][1], citys[to_][1]), color='red')
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ax1.annotate(text=chr(97 + to_), xy=citys[to_], xytext=(-8, -4), textcoords='offset points', fontsize=20)
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ax1.axis('equal')
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ax1.grid()
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# 绘制能量趋势图
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ax2 = fig.add_subplot(122)
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ax2.plot(np.arange(0, len(energys), 1), energys, color='red')
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plt.show()
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# 定义城市坐标
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citys = np.array([[2, 6], [2, 4], [1, 3], [4, 6], [5, 5], [4, 4], [6, 4], [3, 2], [7, 7], [9, 3]])
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# 定义城市与城市之间的距离矩阵
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distance = get_distance(citys)
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# 计算城市个数
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N = len(citys)
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# 初始化参数A和D
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A = N * N
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D = N / 2
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U0 = 0.0009 # 初始输入
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step = 0.0001 # 步长
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num_iter = 10000 # 迭代次数
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# 初始化神经网络的输入状态(电路的输入U)
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U = 1 / 2 * U0 * np.log(N - 1) + (2 * (np.random.random((N, N))) - 1)
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# 初始化神经网络的输出状态(电路的输出V)
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V = calc_V(U, U0)
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energys = np.array([0.0 for x in range(num_iter)]) # 每次迭代的能量
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best_distance = np.inf # 最优距离
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best_route = [] # 最优路线
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H_path = [] # 哈密顿回路
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# 开始迭代训练网络
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for n in range(num_iter):
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# 利用动态方程计算du
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du = calc_du(V, distance)
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# 由一阶欧拉法更新下一个时间的输入状态(电路的输入U)
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U = calc_U(U, du, step)
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# 由sigmoid函数更新下一个时间的输出状态(电路的输出V)
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V = calc_V(U, U0)
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# 计算当前网络的能量E
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energys[n] = calc_energy(V, distance)
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# 检查路径的合法性
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route, newV = check_path(V)
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if len(np.unique(route)) == N:
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route.append(route[0])
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dis = calc_distance(route)
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if dis < best_distance: # 如果dis小于现有最好的best_distance则把best_distance替换为dis
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H_path = []
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# 更新dis
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best_distance = dis
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# 跟新route
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best_route = route
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[H_path.append((route[i], route[i + 1])) for i in range(len(route) - 1)]
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print('第{}次迭代找到的次优解距离为:{},能量为:{},路径为:'.format(n, best_distance, energys[n]))
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[print(chr(97 + v), end=',' if i < len(best_route) - 1 else '\n') for i, v in enumerate(best_route)]
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if len(H_path) > 0:
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draw_H_and_E(citys, H_path, energys)
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else:
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print('没有找到最优解')
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