Gradient Checking

In this Notebook:

  • I will implement and use gradient checking.

We are part of a team working to make mobile payments available globally, and are asked to build a deep learning model to detect fraud–whenever someone makes a payment, you want to see if the payment might be fraudulent, such as if the user’s account has been taken over by a hacker.

But backpropagation is quite challenging to implement, and sometimes has bugs. Because this is a mission-critical application, our company’s CEO wants to be really certain that our implementation of backpropagation is correct. Our CEO says, “Give me a proof that our backpropagation is actually working!” To give this reassurance, we are going to use “gradient checking”.

Let’s do it!

# Packages
import numpy as np
from testCases import *
from gc_utils import sigmoid, relu, dictionary_to_vector, vector_to_dictionary, gradients_to_vector

1) How does gradient checking work?

Backpropagation computes the gradients $\frac{\partial J}{\partial \theta}$, where $\theta$ denotes the parameters of the model. $J$ is computed using forward propagation and our loss function.

Because forward propagation is relatively easy to implement, we’re confident we got that right, and so we’re almost 100% sure that we’re computing the cost $J$ correctly. Thus, we can use our code for computing $J$ to verify the code for computing $\frac{\partial J}{\partial \theta}$.

Let’s look back at the definition of a derivative (or gradient):

If not familiar with the “$\displaystyle \lim_{\varepsilon \to 0}$” notation, it’s just a way of saying “when $\varepsilon$ is really really small.”

We know the following:

  • $\frac{\partial J}{\partial \theta}$ is what we want to make sure we’re computing correctly.
  • We can compute $J(\theta + \varepsilon)$ and $J(\theta - \varepsilon)$ (in the case that $\theta$ is a real number), since we’re confident our implementation for $J$ is correct.

Lets use equation (1) and a small value for $\varepsilon$ to convince our CEO that our code for computing $\frac{\partial J}{\partial \theta}$ is correct!

2) 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.

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

Figure 1 : 1D linear model

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”).

Let’s 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.

def forward_propagation(x, theta):
    Implement the linear forward propagation (compute J) presented in Figure 1 (J(theta) = theta * x)

    x -- a real-valued input
    theta -- our parameter, a real number as well

    J -- the value of function J, computed using the formula J(theta) = theta * x

    J = theta*x

    return J
x, theta = 2, 4
J = forward_propagation(x, theta)
print ("J = " + str(J))
J = 8

Now let’s 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 us from doing the calculus, we should get $dtheta = \frac { \partial J }{ \partial \theta} = x$.

def backward_propagation(x, theta):
    Computes the derivative of J with respect to theta (see Figure 1).

    x -- a real-valued input
    theta -- our parameter, a real number as well

    dtheta -- the gradient of the cost with respect to theta

    dtheta = x

    return dtheta
x, theta = 2, 4
dtheta = backward_propagation(x, theta)
print ("dtheta = " + str(dtheta))
dtheta = 2

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

The below cell will implement the following:

  • 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: We 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 $10^{-7}$), we can be quite confident that we have computed our gradient correctly. Otherwise, there may be a mistake in the gradient computation.
def gradient_check(x, theta, epsilon = 1e-7):
    Implement the backward propagation presented in Figure 1.

    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)

    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.
    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

    # Check if gradapprox is close enough to the output of backward_propagation()
    grad = backward_propagation(x, theta)

    numerator = np.linalg.norm(grad-gradapprox)    # Step 1'
    denominator = np.linalg.norm(grad)+np.linalg.norm(gradapprox)  # Step 2'
    difference = numerator/denominator             # Step 3'

    if difference < 1e-7:
        print ("The gradient is correct!")
        print ("The gradient is wrong!")

    return difference
x, theta = 2, 4
difference = gradient_check(x, theta)
print("difference = " + str(difference))
The gradient is correct!
difference = 2.91933588329e-10

Yay!, the difference is smaller than the $10^{-7}$ threshold. So we can have high confidence that we’ve correctly computed the gradient in backward_propagation().

Now, in the more general case, our cost function $J$ has more than a single 1D input. When we are training a neural network, $\theta$ actually consists of multiple matrices $W^{[l]}$ and biases $b^{[l]}$! It is important to know how to do a gradient check with higher-dimensional inputs. Let’s do it!

3) N-dimensional gradient checking

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

Figure 2 : deep neural network

Let’s look at our 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.

    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)

    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"]

    Z1 =, X) + b1
    A1 = relu(Z1)
    Z2 =, A1) + b2
    A2 = relu(Z2)
    Z3 =, 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

Now, run backward propagation.

def backward_propagation_n(X, Y, cache):
    Implement the backward propagation presented in figure 2.

    X -- input datapoint, of shape (input size, 1)
    Y -- true "label"
    cache -- cache output from forward_propagation_n()

    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 *, A2.T)
    db3 = 1./m * np.sum(dZ3, axis=1, keepdims = True)

    dA2 =, dZ3)
    dZ2 = np.multiply(dA2, np.int64(A2 > 0))
    dW2 = 1./m *, A1.T) * 2
    db2 = 1./m * np.sum(dZ2, axis=1, keepdims = True)

    dA1 =, dZ2)
    dZ1 = np.multiply(dA1, np.int64(A1 > 0))
    dW1 = 1./m *, 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

We obtained some results on the fraud detection test set but we are not 100% sure of our model. Nobody’s perfect! Let’s implement gradient checking to verify if our gradients are correct.

How does gradient checking work?.

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

However, $\theta$ is not a scalar anymore. It is a dictionary called “parameters”. “dictionary_to_vector()” was implemented for us. 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.

Figure 2 : dictionary_to_vector() and vector_to_dictionary()
You will need these functions in gradient_check_n()

The “gradients” dictionary was also converted into a vector “grad” using gradients_to_vector(). We don’t need to worry about that.

Now let’s implement gradient_check_n().

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

The below cell will implement the following:

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, we get a vector gradapprox, where gradapprox[i] is an approximation of the gradient with respect to parameter_values[i]. We can now compare this gradapprox vector to the gradients vector from backpropagation. Just like for the 1D case (Steps 1’, 2’, 3’), compute:

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

    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)

    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
        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

        # Compute J_minus[i]. Inputs: "parameters_values, epsilon". Output = "J_minus[i]".
        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

        # Compute gradapprox[i]
        gradapprox[i] = (J_plus[i]-J_minus[i])/(2*epsilon)

    # Compare gradapprox to backward propagation gradients by computing difference.
    numerator = np.linalg.norm(grad-gradapprox)                 # Step 1'
    denominator = np.linalg.norm(grad)+np.linalg.norm(gradapprox)       # Step 2'
    difference = numerator/denominator                          # Step 3'

    if difference > 2e-7:
        print ("\033[93m" + "There is a mistake in the backward propagation! difference = " + str(difference) + "\033[0m")
        print ("\033[92m" + "Your backward propagation works perfectly fine! difference = " + str(difference) + "\033[0m")

    return difference
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)
There is a mistake in the backward propagation! difference = 0.285093156654

It seems that there were errors in the backward_propagation_n code in the code! Good that we’ve implemented the gradient check.

Things to try:

Go back to backward_propagation and try to find/correct the errors (Hint: check dW2 and db1). Rerun the gradient check when you think you’ve fixed it. Remember you’ll need to re-execute the cell defining backward_propagation_n() if you modify the code.

Can you get gradient check to declare your derivative computation correct? Even though this part of the assignment isn’t graded, we strongly urge you to try to find the bug and re-run gradient check until you’re convinced backprop is now correctly implemented.


  • 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.

Congrats, you can be confident that your deep learning model for fraud detection is working correctly! You can even use this to convince your CEO. :)

Key points for this section are:

  • 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.

last edited: 31/05/19

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