第二课第一周-Gradient Checking

1-dimensional gradient checking

Consider a 1D linear function $J(\theta) = \theta x$. The model contains only a single real-valued parameter $\theta$, and takes $x$ as input.

You will implement code to compute $J(.)$ and its derivative $\frac{\partial J}{\partial \theta}$. You will then use gradient checking to make sure your derivative computation for $J$ is correct.

The diagram above shows the key computation steps: First start with $x$, then evaluate the function $J(x)$ (“forward propagation”). Then compute the derivative $\frac{\partial J}{\partial \theta}$ (“backward propagation”).

Exercise: implement “forward propagation” and “backward propagation” for this simple function. I.e., compute both $J(.)$ (“forward propagation”) and its derivative with respect to $\theta$ (“backward propagation”), in two separate functions.

# GRADED FUNCTION: forward_propagation

def 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 J

Exercise: Now, implement the backward propagation step (derivative computation) of Figure 1. That is, compute the derivative of $J(\theta) = \theta x$ with respect to $\theta$. To save you from doing the calculus, you should get $dtheta = \frac { \partial J }{ \partial \theta} = x$.

# GRADED FUNCTION: backward_propagation

def 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 dtheta

Exercise: To show that the backward_propagation() function is correctly computing the gradient $\frac{\partial J}{\partial \theta}$, let’s implement gradient checking.

Instructions:

• First compute “gradapprox” using the formula above (1) and a small value of $\varepsilon$. Here are the Steps to follow:
  1. $\theta^{+} = \theta + \varepsilon$
  2. $\theta^{-} = \theta - \varepsilon$
  3. $J^{+} = J(\theta^{+})$
  4. $J^{-} = J(\theta^{-})$
  5. $gradapprox = \frac{J^{+} - J^{-}}{2 \varepsilon}$
• Then compute the gradient using backward propagation, and store the result in a variable “grad”
• Finally, compute the relative difference between “gradapprox” and the “grad” using the following formula:
  $$ difference = \frac {\mid\mid grad - gradapprox \mid\mid_2}{\mid\mid grad \mid\mid_2 + \mid\mid gradapprox \mid\mid_2} \tag{2}$$
  You will need 3 Steps to compute this formula:
  • 1’. compute the numerator using np.linalg.norm(…)
  • 2’. compute the denominator. You will need to call np.linalg.norm(…) twice.
  • 3’. divide them.
• If this difference is small (say less than $10^{-7}$), you can be quite confident that you have computed your gradient correctly. Otherwise, there may be a mistake in the gradient computation.

# GRADED FUNCTION: gradient_check

def 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

---

N-dimensional gradient checking

The following figure describes the forward and backward propagation of your fraud detection model.

Let’s look at your implementations for forward propagation and backward propagation.

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

backward propagation:

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) * 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 = 4./m * np.sum(dZ1, axis=1, keepdims = True)
    
    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

How does gradient checking work?.

As in 1) and 2), you want to compare “gradapprox” to the gradient computed by backpropagation. The formula is still:

$$ \frac{\partial J}{\partial \theta} = \lim_{\varepsilon \to 0} \frac{J(\theta + \varepsilon) - J(\theta - \varepsilon)}{2 \varepsilon} \tag{1}$$

However, $\theta$ is not a scalar anymore. It is a dictionary called “parameters”. We implemented a function “dictionary_to_vector()“ for you. It converts the “parameters” dictionary into a vector called “values”, obtained by reshaping all parameters (W1, b1, W2, b2, W3, b3) into vectors and concatenating them.

The inverse function is “vector_to_dictionary“ which outputs back the “parameters” dictionary.

Exercise: Implement gradient_check_n().

Instructions: Here is pseudo-code that will help you implement the gradient check.

For each i in num_parameters:

• To compute J_plus[i]:
  1. Set $\theta^{+}$ to np.copy(parameters_values)
  2. Set $\theta^{+}_i$ to $\theta^{+}_i + \varepsilon$
  3. Calculate $J^{+}_i$ using to forward_propagation_n(x, y, vector_to_dictionary($\theta^{+}$ )).
• To compute J_minus[i]: do the same thing with $\theta^{-}$
• Compute $gradapprox[i] = \frac{J^{+}_i - J^{-}_i}{2 \varepsilon}$

Thus, you get a vector gradapprox, where gradapprox[i] is an approximation of the gradient with respect to parameter_values[i]. You can now compare this gradapprox vector to the gradients vector from backpropagation. Just like for the 1D case (Steps 1’, 2’, 3’), compute:
$$ difference = \frac {| grad - gradapprox |_2}{| grad |_2 + | gradapprox |_2 } \tag{3}$$

# GRADED FUNCTION: gradient_check_n

def 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)
    num_parameters = parameters_values.shape[0]
    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 ###

    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

Note

• Gradient Checking is slow! Approximating the gradient with $\frac{\partial J}{\partial \theta} \approx \frac{J(\theta + \varepsilon) - J(\theta - \varepsilon)}{2 \varepsilon}$ is computationally costly. For this reason, we don’t run gradient checking at every iteration during training. Just a few times to check if the gradient is correct.
• Gradient Checking, at least as we’ve presented it, doesn’t work with dropout. You would usually run the gradient check algorithm without dropout to make sure your backprop is correct, then add dropout.

What you should remember from this notebook:

• Gradient checking verifies closeness between the gradients from backpropagation and the numerical approximation of the gradient (computed using forward propagation).
• Gradient checking is slow, so we don’t run it in every iteration of training. You would usually run it only to make sure your code is correct, then turn it off and use backprop for the actual learning process.