DeepLearing学习笔记-改善深层神经网络(第一周作业-3-梯度校验)

1-背景：

在神经网络计算过程中，对后向传播的梯度进行校验，确保其计算无误。至于，前向传播，由于相对简单，所以，一般不会出错，在前向传播的基础上利用计算出来的代价J我们可以进行后向梯度的校验。

公式原理如下：

∂J∂θ=limε→0J(θ+ε)−J(θ−ε)2ε(1)

在前向传播的基础上，我们可以获取到J，也就可以上公式1的计算，将计算的结果值和反向传播的梯度值按照一定的方式做比较即可。

2- 一维梯度校验

当 J(θ)=θx，线性模型如下:

该模型的前向传播和后向传播的代码如下：

# GRADED FUNCTION: forward_propagationdef forward_propagation(x, theta):    """    Implement the linear forward propagation (compute J) presented in Figure 1 (J(theta) = theta * x)    Arguments:    x -- a real-valued input    theta -- our parameter, a real number as well    Returns:    J -- the value of function J, computed using the formula J(theta) = theta * x    """    ### START CODE HERE ### (approx. 1 line)    J = theta * x    ### END CODE HERE ###    return Jx, theta = 2, 4J = forward_propagation(x, theta)print ("J = " + str(J))

输出结果：
J = 8

由于J(θ)=θx 所以dtheta=∂J∂θ=x

# GRADED FUNCTION: backward_propagationdef backward_propagation(x, theta):    """    Computes the derivative of J with respect to theta (see Figure 1).    Arguments:    x -- a real-valued input    theta -- our parameter, a real number as well    Returns:    dtheta -- the gradient of the cost with respect to theta    """    ### START CODE HERE ### (approx. 1 line)    dtheta = x    ### END CODE HERE ###    return dthetax, theta = 2, 4dtheta = backward_propagation(x, theta)print ("dtheta = " + str(dtheta))

输出结果：
dtheta = 2

2-1 一维梯度校验

步骤如下：

• 计算梯度近似值 “gradapprox”:

  1. θ+=θ+ε
  2. θ−=θ−ε
  3. J+=J(θ+)
  4. J−=J(θ−)
  5. gradapprox=J+−J−2ε

• 进行反向传播，获取梯度值 “grad”

• 按照如下公式计算两者difference :
  
  difference=∣∣grad−gradapprox∣∣2∣∣grad∣∣2+∣∣gradapprox∣∣2(2)

范数的计算可以用np.linalg.norm(...)
当difference 足够小(<10−7)，则可以视为梯度校验通过。

2-2 梯度校验代码：

# GRADED FUNCTION: gradient_checkdef gradient_check(x, theta, epsilon = 1e-7):    """    Implement the backward propagation presented in Figure 1.    Arguments:    x -- a real-valued input    theta -- our parameter, a real number as well    epsilon -- tiny shift to the input to compute approximated gradient with formula(1)    Returns:    difference -- difference (2) between the approximated gradient and the backward propagation gradient    """    # Compute gradapprox using left side of formula (1). epsilon is small enough, you don't need to worry about the limit.    ### START CODE HERE ### (approx. 5 lines)    thetaplus = theta + epsilon                              # Step 1    thetaminus = theta - epsilon                             # Step 2    J_plus = forward_propagation(x, thetaplus)                                  # Step 3    J_minus = forward_propagation(x, thetaminus)                                # Step 4    gradapprox = (J_plus-J_minus)/(2*epsilon)                              # Step 5    ### END CODE HERE ###    # Check if gradapprox is close enough to the output of backward_propagation()    ### START CODE HERE ### (approx. 1 line)    grad = backward_propagation(x, theta)    ### END CODE HERE ###    ### START CODE HERE ### (approx. 1 line)    numerator = np.linalg.norm(grad - gradapprox)                               # Step 1'    denominator = np.linalg.norm(grad)+np.linalg.norm(gradapprox)               # Step 2'    difference = numerator/denominator                              # Step 3'    ### END CODE HERE ###    if difference < 1e-7:        print ("The gradient is correct!")    else:        print ("The gradient is wrong!")    return difference

测试：

x, theta = 2, 4difference = gradient_check(x, theta)print("difference = " + str(difference))

运行结果如下：

The gradient is correct!difference = 2.91933588329e-10

difference 明显小于阈值10−7，校验通过。

3- N微梯度校验

本文采用3层神经网络做说明，模型：LINEAR -> RELU -> LINEAR -> RELU -> LINEAR -> SIGMOID

3-1 前向传播和后向传播：

前向传播代码：

def forward_propagation_n(X, Y, parameters):    """    Implements the forward propagation (and computes the cost) presented in Figure 3.    Arguments:    X -- training set for m examples    Y -- labels for m examples     parameters -- python dictionary containing your parameters "W1", "b1", "W2", "b2", "W3", "b3":                    W1 -- weight matrix of shape (5, 4)                    b1 -- bias vector of shape (5, 1)                    W2 -- weight matrix of shape (3, 5)                    b2 -- bias vector of shape (3, 1)                    W3 -- weight matrix of shape (1, 3)                    b3 -- bias vector of shape (1, 1)    Returns:    cost -- the cost function (logistic cost for one example)    """    # retrieve parameters    m = X.shape[1]    W1 = parameters["W1"]    b1 = parameters["b1"]    W2 = parameters["W2"]    b2 = parameters["b2"]    W3 = parameters["W3"]    b3 = parameters["b3"]    # LINEAR -> RELU -> LINEAR -> RELU -> LINEAR -> SIGMOID    Z1 = np.dot(W1, X) + b1    A1 = relu(Z1)    Z2 = np.dot(W2, A1) + b2    A2 = relu(Z2)    Z3 = np.dot(W3, A2) + b3    A3 = sigmoid(Z3)    # Cost    logprobs = np.multiply(-np.log(A3),Y) + np.multiply(-np.log(1 - A3), 1 - Y)    cost = 1./m * np.sum(logprobs)    cache = (Z1, A1, W1, b1, Z2, A2, W2, b2, Z3, A3, W3, b3)    return cost, cache

这里的后向传播，故意在dW2和db1这里写错：

def backward_propagation_n(X, Y, cache):    """    Implement the backward propagation presented in figure 2.    Arguments:    X -- input datapoint, of shape (input size, 1)    Y -- true "label"    cache -- cache output from forward_propagation_n()    Returns:    gradients -- A dictionary with the gradients of the cost with respect to each parameter, activation and pre-activation variables.    """    m = X.shape[1]    (Z1, A1, W1, b1, Z2, A2, W2, b2, Z3, A3, W3, b3) = cache    dZ3 = A3 - Y    dW3 = 1./m * np.dot(dZ3, A2.T)    db3 = 1./m * np.sum(dZ3, axis=1, keepdims = True)    dA2 = np.dot(W3.T, dZ3)    dZ2 = np.multiply(dA2, np.int64(A2 > 0))    dW2 = 1./m * np.dot(dZ2, A1.T)#不用乘以2    db2 = 1./m * np.sum(dZ2, axis=1, keepdims = True)    dA1 = np.dot(W2.T, dZ2)    dZ1 = np.multiply(dA1, np.int64(A1 > 0))    dW1 = 1./m * np.dot(dZ1, X.T)    db1 = 1./m * np.sum(dZ1, axis=1, keepdims = True)#分子是1，不是4    gradients = {"dZ3": dZ3, "dW3": dW3, "db3": db3,                 "dA2": dA2, "dZ2": dZ2, "dW2": dW2, "db2": db2,                 "dA1": dA1, "dZ1": dZ1, "dW1": dW1, "db1": db1}    return gradients

3-2 梯度校验

对每个参数执行如下操作:

• 计算J_plus[i]:
  
  1. θ+= np.copy(parameters_values)
  2. θ+i = θ+i+ε
  3. 采用forward_propagation_n(x, y, vector_to_dictionary(θ+ ))计算 J+i
• 对于θ−同理计算 J_minus[i]
• 计算gradapprox[i]=J+i−J−i2ε

gradapprox中的 gradapprox[i] 对应的是参数parameter_values[i]的梯度近似值 。gradapprox 向量和后向传播的梯度的相似按照如下公式估算.:

difference=∥grad−gradapprox∥2∥grad∥2+∥gradapprox∥2(3)

3-2 梯度校验

# GRADED FUNCTION: gradient_check_ndef gradient_check_n(parameters, gradients, X, Y, epsilon = 1e-7):    """    Checks if backward_propagation_n computes correctly the gradient of the cost output by forward_propagation_n    Arguments:    parameters -- python dictionary containing your parameters "W1", "b1", "W2", "b2", "W3", "b3":    grad -- output of backward_propagation_n, contains gradients of the cost with respect to the parameters.     x -- input datapoint, of shape (input size, 1)    y -- true "label"    epsilon -- tiny shift to the input to compute approximated gradient with formula(1)    Returns:    difference -- difference (2) between the approximated gradient and the backward propagation gradient    """    # Set-up variables    parameters_values, _ = dictionary_to_vector(parameters)    grad = gradients_to_vector(gradients)#将字典转为向量形式    num_parameters = parameters_values.shape[0]#参数个数    J_plus = np.zeros((num_parameters, 1))    J_minus = np.zeros((num_parameters, 1))    gradapprox = np.zeros((num_parameters, 1))    # Compute gradapprox    for i in range(num_parameters):        # Compute J_plus[i]. Inputs: "parameters_values, epsilon". Output = "J_plus[i]".        # "_" is used because the function you have to outputs two parameters but we only care about the first one        ### START CODE HERE ### (approx. 3 lines)        thetaplus = np.copy(parameters_values)                                      # Step 1        thetaplus[i][0] = thetaplus[i][0]+epsilon                                # Step 2        J_plus[i], _ = forward_propagation_n(X, Y, vector_to_dictionary(thetaplus))                                   # Step 3        ### END CODE HERE ###        # Compute J_minus[i]. Inputs: "parameters_values, epsilon". Output = "J_minus[i]".        ### START CODE HERE ### (approx. 3 lines)        thetaminus = np.copy(parameters_values)                                     # Step 1        thetaminus[i][0] = thetaminus[i][0]-epsilon                               # Step 2                J_minus[i], _ =  forward_propagation_n(X, Y, vector_to_dictionary(thetaminus))                                  # Step 3        ### END CODE HERE ###        # Compute gradapprox[i]        ### START CODE HERE ### (approx. 1 line)        gradapprox[i] = (J_plus[i]-J_minus[i])/(2*epsilon)        ### END CODE HERE ###    # Compare gradapprox to backward propagation gradients by computing difference.    ### START CODE HERE ### (approx. 1 line)    numerator = np.linalg.norm(grad-gradapprox)                                            # Step 1'    denominator = np.linalg.norm(grad)+np.linalg.norm(gradapprox)                                         # Step 2'    difference = numerator/denominator                                          # Step 3'    ### END CODE HERE ###    #注意这里epsilon值的问题，如果是1e-6是可以的，使得梯度校验通过，epsilon值越小，反而和导数越不一致    if difference > 1e-7:        print ("\033[93m" + "There is a mistake in the backward propagation! difference = " + str(difference) + "\033[0m")    else:        print ("\033[92m" + "Your backward propagation works perfectly fine! difference = " + str(difference) + "\033[0m")    return difference

测试代码：

X, Y, parameters = gradient_check_n_test_case()cost, cache = forward_propagation_n(X, Y, parameters)gradients = backward_propagation_n(X, Y, cache)difference = gradient_check_n(parameters, gradients, X, Y)

测试结果：
There is a mistake in the backward propagation! difference = 0.285093156654
明显存在梯度计算问题。将反向梯度计算的dW2和db1进行修改，重新运行：
There is a mistake in the backward propagation! difference = 1.18855520355e-07
虽然现实出现问题，但是different值已经很接近阈值了，此时，我们单独修改计算双边梯度之后的epsilon=1e-6,不修改判断的阈值。
输出：
Your backward propagation works perfectly fine! difference = 8.26588225515e-09
所以，要注意看difference值是否和阈值相距很大。本文为何就差那么一些，导致需要修改epsilon，可能是由于代价函数在局部存在毛刺，导致估算值和后向梯度计算结果，存在超于阈值的偏差。另外，relu的导数在0处有歧义，也可能导致此处的不够准确。