如何判断牛顿法是否失败 [英] How to tell if Newtons-Method Fails

问题描述

我正在为无约束的优化问题创建一个基本的牛顿法算法，但是我从算法中得到的结果与我的预期不同。它是一个简单的目标函数，因此很明显该算法应收敛于（1,1）。这是由我之前在这里创建的梯度下降算法所证实的：

I am creating a basic Newton-method algorithm for an unconstrained optimization problem, and my results from the algorithm are not what I expected. It is a simple objective function so it is clear that the algorithm should converge on (1,1). This is confirmed by a gradient descent algorithm I created previously, here:

def grad_descent(x, t, count, magnitude):
    xvalues.append(x)
    gradvalues.append(np.array([dfx1(x), dfx2(x)]))
    fvalues.append(f(x))   
    temp=x-t*dfx(x)
    x = temp
    magnitude = mag(dfx(x))    
    count+=1

    return xvalues, gradvalues, fvalues, count

我为牛顿法创建算法的尝试是在这里：

My attempt at creating an algorithm for Newtons-Method is here:

def newton(x, t, count, magnitude):
  xvalues=[]
  gradvalues=[]
  fvalues=[]
  temp=x-f(x)/dfx(x)

  while count < 10:
    xvalues.append(x)
    gradvalues.append(dfx(x))
    fvalues.append(f(x))  

    temp=x-t*f(x)/dfx(x)
    x = temp
    magnitude = mag(dfx(x))    
    count+=1
    if count > 100:
      break
  return xvalues, gradvalues, fvalues, count

目标函数和梯度函数：

f = lambda x: 100*np.square(x[1]-np.square(x[0])) + np.square((1-x[0]))
dfx = lambda x: np.array([-400*x[0]*x[1]+400*np.power(x[0],3)+2*x[0]-2, 200*(x[1]-np.square(x[0]))])

以下是初始条件。请注意，牛顿方法未使用alpha和beta。

Here are the initial conditions. Note that alpha and beta are not used in the newton method.

x0, t0, alpha, beta, count = np.array([-1.1, 1.1]), 1, .15, .7, 1
magnitude = mag(np.array([dfx1(x0), dfx2(x0)]))

调用函数：

xvalues, gradvalues, fvalues, iterations = newton(x0, t0, count, magnitude)

这会产生非常奇怪的结果。以下是相应x输入的x值，梯度值和函数解的前10个迭代：

This produce very strange results. Here are the first 10 iterations of the xvalues, gradient values, and function solution for its respective x input:

[array([-1.1,  1.1]), array([-0.99315589,  1.35545455]), array([-1.11651296,  1.11709035]), array([-1.01732476,  1.35478987]), array([-1.13070578,  1.13125051]), array([-1.03603697,  1.35903467]), array([-1.14368874,  1.14364506]), array([-1.05188162,  1.36561528]), array([-1.15600558,  1.15480705]), array([-1.06599492,  1.37360245])]
[array([-52.6, -22. ]), array([142.64160215,  73.81918332]), array([-62.07323963, -25.90216846]), array([126.11789251,  63.96803995]), array([-70.85773749, -29.44900758]), array([114.31050737,  57.13241151]), array([-79.48668009, -32.87577304]), array([104.93863096,  51.83206539]), array([-88.25737032, -36.308371  ]), array([97.03403558, 47.45145765])]
[5.620000000000003, 17.59584998020613, 6.156932949106968, 14.29937453260906, 6.7080172227439725, 12.305727666787176, 7.297442528545537, 10.926625703722639, 7.944104584786208, 9.89743708419569]  

这是最终输出：

final_value = print('Final set of x values: ', xvalues[-1])
final_grad = print('Final gradient values: ', gradvalues[-1])
final_f = print('Final value of the object function with optimized inputs: ', fvalues[-1])
final_grad_mag = print('Final magnitude of the gradient with optimized inputs: ', mag(np.array([dfx1(xvalues[-1]), dfx2(xvalues[-1])])))
total_iterations = print('Total iterations: ', iterations)

显示3d图此处
代码：

a 3d plot is shown here code:

x = np.array([i[0] for i in xvalues])
y = np.array([i[1] for i in xvalues])
z = np.array(fvalues)
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.scatter(x, y, z, label='Newton Method')
ax.legend()

是这样做的原因是，最初的猜测非常接近最佳点，或者我的算法中是否存在一些我无法捕捉到的错误？任何建议将不胜感激。看来解决方案甚至可能会出现问题，但很难分辨

Is the reasoning for this because the initial guess is so close to the optimal point, or is there some error in my algorithm that I am not catching? Any advice would be greatly appreciated. It looks like the solution may even be oscillating, but it is difficult to tell

推荐答案

所以我终于弄清楚了到底发生了什么有了这个。一切都是关于Python将我的变量存储为什么数据结构。因此，我将所有值都设置为 float32，并初始化了要迭代的变量。工作代码在这里：

So I finally figured out what was going on with this. It was all about what data structures Python was storing my variables as. As such, I set all my values to 'float32' and initialized the variables being iterated. Working code is here:

f = lambda x: 100*np.square(x[1]-np.square(x[0])) + np.square((1-x[0]))
dfx = lambda x: np.array([-400*x[0]*x[1]+400*np.power(x[0],3)+2*x[0]-2, 200*(x[1]-np.square(x[0]))], dtype='float32')
dfx11 = lambda x: -400*(x[1])+1200*np.square(x[0])+2
dfx12 = lambda x: -400*x[0]
dfx21 = lambda x: -400*x[0]
dfx22 = lambda x: 200
hessian = lambda x: np.array([[dfx11(x), dfx12(x)], [dfx21(x), dfx22(x)]], dtype='float32')
inv_hessian = lambda x: inv(hessian(x))
mag = lambda x: math.sqrt(sum(i**2 for i in x))

def newton(x, t, count, magnitude):
  xvalues=[]
  gradvalues=[]
  fvalues=[]
  temp = np.zeros((2,1))

  while magnitude > .000005:
    xvalues.append(x)
    gradvalues.append(dfx(x))
    fvalues.append(f(x))      

    deltaX = np.array(np.dot(-inv_hessian(x), dfx(x)))
    temp = np.array(x+t*deltaX)
    x = temp
    magnitude = mag(deltaX)    
    count+=1
  return xvalues, gradvalues, fvalues, count

x0, t0, alpha, beta, count = np.array([[-1.1], [1.1]]), 1, .15, .7, 1
xvalues, gradvalues, fvalues, iterations = newton(x0, t0, count, magnitude)

final_value = print('Final set of x values: ', xvalues[-1])
final_grad = print('Final gradient values: ', gradvalues[-1])
final_f = print('Final value of the object function with optimized inputs: ', fvalues[-1])
final_grad_mag = print('Final magnitude of the gradient with optimized inputs: ', mag(np.array([dfx1(xvalues[-1]), dfx2(xvalues[-1])])))
total_iterations = print('Total iterations: ', iterations
print(xvalues)

输出：

Final set of x values:  [[0.99999995]
 [0.99999987]]
Final gradient values:  [[ 9.1299416e-06]
 [-4.6193604e-06]]
Final value of the object function with optimized inputs:  [5.63044182e-14]
Final magnitude of the gradient with optimized inputs:  1.02320249276675e-05
Total iterations:  9
[array([[-1.1],
       [ 1.1]]), array([[-1.00869558],
       [ 1.00913081]]), array([[-0.25557778],
       [-0.50186648]]), array([[-0.24460602],
       [ 0.05971173]]), array([[ 0.97073805],
       [-0.53472879]]), array([[0.97083687],
       [0.94252417]]), array([[0.99999957],
       [0.99914868]]), array([[0.99999995],
       [0.99999987]])]

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