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Building Your First Neural Network From Scratch: A Step-by-Step Guide with the MNIST Dataset

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    Qi Wang


Table of Contents


Hey everyone, welcome back! This post is one of my first posts that actually serves as a tutorial instead of introducing the concept, so bear with me if the format is not as fine tuned. Also for the first time, there is a substantial prerequisite to this blog. Most importantly, I am assuming the knowledge of how neural networks work from a fundamental level. If you don't know, that's completely fine! I recommend watching this video series by 3Blue1Brown to get started. Feel free to return to this post after! Another prerequisite, though less important, is the knowledge of calculus. It'll be fine if you aren't too confident, I will show most of the derivations in this blog. Anyways, let's get started building our neural network.

What We Are Looking For

Before diving into the specifics, let's go over what we ideally want to see as an end product. Obviously, we want the performance of this model to be high. Note: a multi-layer neural network is not the best way of doing image recognition, CNNs are much more fit for the task. However, for the MNIST dataset we should still be able to get a relatively high accuracy with a simple neural network. Lastly, we want a neural network that can be trained quickly. Who wants to wait hours?!?!


Vectorization is the process of doing operations on entire arrays instead of applying them individually. You can imagine it to be almost parallelization. Essentially, in one forward pass of the neural network, we want to go through everything in the training sample, and in one backward pass, we want to backpropagate through every parameter update. This often creates a neural network that is 10x faster and potentially even higher with larger datasets. To do vectorization, we will be making use of the Numpy library which supports many types of array operations and is much faster to begin with internally as it works with C.

Let's Talk Some Matrix Multiplications!

As you may already know, a neural network is made of a bunch of parameters wi,biw_i, b_i. One neuron's value in the network is often calculated like such:

σ(i=0n(wixi)+b)\sigma (\sum_{i=0}^{n} (w_i x_i) + b)

The σ\sigma represents the activation function. Following our goal of a efficient neural network, we will be vectorizing this process! So we want to somehow get results for all of the neurons in a given layer... Matrix Multiplications! Let's say every layer has cc neurons with nn inputs to each. We will create a c×nc \times n matrix with all the weight parameters in that layer, and a c×1c \times 1 matrix for all bias parameters. You can observe that with the input being n×mn \times mnn being the size of the input, and mm being the amount of data in the training dataset — we can multiply the weights matrix with this matrix and add the bias matrix on top of it to produce a c×xc \times x matrix that contains the values for each neuron for each entry in the dataset before the activation function is applied.

You can see that if we keep chaining the input matrix to the weights in the next layers, we are eventually left with a matrix with values for the final layer. All of these is with simple matrix operations and dramatically reduces the runtime compared to using for loops in python.

Architecture For MNIST Dataset

To tackle this problem, we will use a neural network with 2 hidden layers of 16 neurons (with a tanh activation) connected to the input of 784 (the image given is 28×2828 \times 28 which will be flattened and normalized) and lastly to an output of 10 neurons with a softmax activation to calculate the probabilities of each class.

Below we will write a few functions to initialize these parameters of our neural network. It is important to initilize the weight parameters to be random small numbers so it is compatible with a gradient descent approach for learning.

def initialize_parameters(n_in, n_outs):
    Three Layer Neural Network (Hard Coded)
    # Layer 1 parameters
    W1 = np.random.randn(n_outs[0], n_in) * 0.1
    b1 = np.zeros((n_outs[0], 1))
    # Layer 2
    W2 = np.random.randn(n_outs[1], n_outs[0]) * 0.1
    b2 = np.zeros((n_outs[1], 1))
    # Layer 3
    W3 = np.random.randn(n_outs[2], n_outs[1]) * 0.1
    b3 = np.zeros((n_outs[2], 1))
    parameters = {
        'W1': W1,
        'b1': b1,
        'W2': W2,
        'b2': b2,
        'W3': W3,
        'b3': b3
    return parameters

For this blog post, I've hardcoded the three layers. I tend to think this is more intuitive to understand rather than sticking layers in a loop. Afterwards, it could be an exercise to reimplement this to work for any amount of layers with any amount of neurons in each.

Activation Functions

Now let's implement the activation functions that we will be using for our neural network. If you do want to try other functions, that is completely fine, just remeber that the back propagation process will be a little different because each activation's derivative is different. For our neural network, we will be using the tanh activation and the softmax activation. Essentially, the tanh activation squishes the outputs of the neurons into a value between -1 and 1, and the softmax activation transforms the outputs of the ten neurons in the final layer to probabilities that add up to 1. The formulas for these functions are avaliable online, below are their implementations:

def tanh(x):
    return (np.exp(x) - np.exp(-x)) / (np.exp(x) + np.exp(-x))

def softmax(x):
    # remove largest value
    expx = np.exp(x - np.max(x))
    return expx / expx.sum(axis=0, keepdims=True)

One technique used to avoid overflowing with np.exp() is to subtract the largest element in the array from every element, the probabilities will stay the same.

Forward Propagation

As described above, we are essentially just doing a few matrix multiplications and chaining the outputs of the previous layer into the next until we are left with the final result.

def forward_propagation(X, parameters):
    Forward propagates through the network to get the resulting output node
    # Getting parameters
    W1 = parameters['W1']
    W2 = parameters['W2']
    W3 = parameters['W3']
    b1 = parameters['b1']
    b2 = parameters['b2']
    b3 = parameters['b3']
    Z1 =, X) + b1
    A1 = tanh(Z1)
    Z2 =, A1) + b2
    A2 = tanh(Z2)
    Z3 =, A2) + b3
    A3 = softmax(Z3)
    cache = {
        'Z1': Z1,
        'A1': A1,
        'Z2': Z2,
        'A2': A2,
        'Z3': Z3,
        'A3': A3
    return A3, cache

You might be wondering why we are using for matrix multiplications. Specified in the Numpy documentation, is actually matrix multiplications if both arrays are 2D. The reason why we are storing every layer's output in the cache dictionary is because during backpropagation, we need these values to calculate the gradients for the parameters. If you are not yet familiar with backpropagation, I suggest watching the 3Blue1Brown videos linked above. Next, I will go through the process of backpropagating through these layers.


Just a little recap on backpropagation. The performance of the neural network is measured by a metric called the cost function, usually, the bigger the cost value, the worse the model is. Therefore, we want our weights to move to the global minimum of this cost (or loss) function. The gradient of a multivariable function points in the direction of greatest change, so if we want to minimize our cost function we need to update our parameters like so:

wi=wiLwiαw_i = w_i - \frac{\partial L}{\partial w_i} \cdot \alpha

LL is refering to our loss function (the cost function is just the average of the loss across all the training samples). Since we are building a multiclass classifer, we will be using Categorical Cross Entropy Loss:

L=k=0c(yklog(y^k))L = -\sum_{k=0}^{c} (y_k \cdot log( \hat{y}_k ))

cc is the number of classes we are predicting in our model. To get gradients for our weights in the final layer, we must use the chain rule:

LW[L]=LA[L]A[L]Z[L]Z[L]W[L]\frac{\partial L}{\partial W^{[L]}} = \frac{\partial L}{\partial A^{[L]}} \frac{\partial A^{[L]}}{\partial Z^{[L]}} \frac{\partial Z^{[L]}}{\partial W^{[L]}}

The [L][L] is refering to which layer we are currently talking about, with L being the last layer.

Softmax Derivative

First, let's start with the derivation for A[L]Z[L]\frac{\partial A^{[L]}}{\partial Z^{[L]}}. aia_i will represent the activation value at neuron ii and zjz_j will represent the value at neuron jj before the activation.

aizj=zj(ezik=0cezk)\frac{\partial a_i}{\partial z_j} = \frac{\partial}{\partial z_j}(\frac{e^{z_i}}{\sum_{k=0}^{c} e^{z_k}})

To derive this, we will use the quotient rule. Let f(x)=ezif(x) = e^{z_i} and g(x)=k=0cezkg(x) = \sum_{k=0}^{c} e^{z_k}. By taking their derivatives with respect to zjz_j, we get:

zj(f(x))={eziif i=j0if ij\frac{\partial}{\partial z_j} (f(x)) = \begin{cases} e^{z_i} &\text{if } i = j \\ 0 &\text{if } i \neq j \end{cases}
g(x)=ez0+ez1++ezj++ezczj(g(x))=ezjg(x) = e^{z_0} + e^{z_1} + \cdots + e^{z_j} + \cdots + e^{z_c} \\ \frac{\partial}{\partial z_j} (g(x)) = e^{z_j}

A quick reminder on the quotient rule:

h(x)=f(x)g(x)f(x)g(x)(g(x))2h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}

For i=ji = j, we get:

aizj=ezig(x)eziezj(g(x))2=aiezig(x)ezjg(x)=aiaiaj=ai(1aj)\begin{darray}{rcl} \frac{\partial a_i}{\partial z_j} &=& \frac{e^{z_i}g(x) - e^{z_i}e^{z_j}}{(g(x))^2} \\[10pt] &=& a_i - \frac{e^{z_i}}{g(x)} \cdot \frac{e^{z_j}}{g(x)} \\[10pt] &=& a_i - a_i a_j \\[5pt] &=& a_i \cdot (1 - a_j) \end{darray}

For iji \neq j, we get:

aizj=0eziezj(g(x))2=aiaj\begin{darray}{rcl} \frac{\partial a_i}{\partial z_j} &=& \frac{0 - e^{z_i}e^{z_j}}{(g(x))^2} \\[10pt] &=& -a_ia_j \end{darray}

Loss Derivative

So we now have aizj\frac{\partial a_i}{z_j} and we will combine this with Lak\frac{\partial L}{\partial a_k} to get Lzi\frac{\partial L}{\partial z_i}.

I've pasted the loss function over again for convience:

L=k=0c(yklog(y^k))L = -\sum_{k=0}^{c} (y_k \cdot log( \hat{y}_k ))

Let's do some calculus now, notice y^k=ak\hat{y}_k = a_k because the activation layer feeds into the loss function. Also a side note, the loglog down below are natural logs.

Lzi=k=0cyklog(ak)akakzi=k=0cykakakzi=(yi(1ai)+k=0,kicykak(akai))=(yiyiaik=0,kicykakakai)=(yik=0cykai)=(yiai)=aiyi\begin{darray}{rcl} \frac{\partial L}{\partial z_i} &=& -\sum_{k=0}^{c} y_k \frac{\partial log(a_k)}{\partial a_k} \frac{\partial a_k}{\partial z_i} \\[10pt] &=& -\sum_{k=0}^{c} \frac{y_k}{a_k} \frac{\partial a_k}{\partial z_i} \\[10pt] &=& -(y_i(1-a_i) + \sum_{k=0, k \neq i}^{c} \frac{y_k}{a_k} \cdot (-a_ka_i)) \\[10pt] &=& -(y_i - y_ia_i - \sum_{k=0, k \neq i}^{c} \frac{y_k}{a_k} \cdot a_ka_i) \\[10pt] &=& -(y_i - \sum_{k=0}^{c}y_ka_i) \\[15pt] &=& -(y_i - a_i) \\[5pt] &=& a_i - y_i \end{darray}

The reason I was able to reduce k=0cykai\sum_{k=0}^{c}y_ka_i down to aia_i is because the sum of probabilities across all classes is one for our case.

Now, we know we can find Lzi\frac{\partial L}{\partial z_i} with aiyia_i-y_i, so to vectorize this, we simply subtract the two matrices: AYA - Y.

Backpropagating Through the Layers

To get the LW[L]\frac{\partial L}{\partial W^{[L]}} with LL being the final layer, we need to find Z[L]W[L]\frac{\partial Z^{[L]}}{W^{[L]}} since LZ[L]\frac{\partial L}{\partial Z^{[L]}} is already found. Since Z[L]=W[L]A[L1]+B[L]Z^{[L]} = W^{[L]}A^{[L-1]} + B^{[L]}, Z[L]W[L]\frac{\partial Z^{[L]}}{\partial W^{[L]}} is just equal to A[L1]A^{[L-1]}. To vectorize this, we can do, A2.T). Remeber, this gradient is summed across all mm samples in your batch, so you will need to multiply the result of this by 1m\frac{1}{m} to get the average. To find LB[L]\frac{\partial L}{\partial B^{[L]}}, we do the same thing. Except, this time, Z[L]B[L]=1\frac{\partial Z^{[L]}}{\partial B^{[L]}} = 1, so it is actually just equal to Z[L]Z^{[L]}. To vectorize this, we can do np.sum(dZ3,axis=1,keepdims=True). axis=1 for our specific case is summing across the entire batch, so once again, remeber to take the average by multiplying this by 1m\frac{1}{m}.

With this, we are basically done. We just need to do one more thing, the chain rule from this layer: Z[L]Z^{[L]} to the previous layer: Z[L1]Z^{[L-1]}. The only thing in between these two is the activation function. For our neural network, we are using tanh() activations across both layers. As an exercise, feel free to derive the derivative by hand, but I have provided it down below:

1tanh(x)21 - tanh(x)^2

Therefore, applying our chain rule:

LZ[L1]=LA[L1](1(A[L1])2)\frac{\partial L}{\partial Z^{[L-1]}} = \frac{\partial L}{\partial A^{[L-1]}} \cdot (1 - (A^{[L-1]})^2)

To vectorize this, we can do dZ2 =,dZ3) * (1 - np.power(A2,2)).

You might ask, what about the rest? Well, if you take another look above, you'll realize the rest of the process is a simple repetition of these operations that you just saw. Feel free to now go implement your own backpropagation function, but I have also provided my code below:

def backward_propagation(X, y, cache, parameters):
    Backpropagates through network and gets the gradients for each parameter to perform gradient descent on
    m = X.shape[1]
    # Getting our parameters
    W1 = parameters['W1']
    W2 = parameters['W2']
    W3 = parameters['W3']
    b1 = parameters['b1']
    b2 = parameters['b2']
    b3 = parameters['b3']
    # Getting our results in each layer
    Z1 = cache['Z1']
    Z2 = cache['Z2']
    Z3 = cache['Z3']
    A1 = cache['A1']
    A2 = cache['A2']
    A3 = cache['A3']
    dZ3 = A3 - y
    dW3 = 1 / m *, A2.T)
    db3 = 1 / m * np.sum(dZ3,axis=1,keepdims=True)
    dZ2 =,dZ3) * (1 - np.power(A2,2))
    dW2 = 1 / m *, A1.T)
    db2 = 1 / m * np.sum(dZ2, axis=1, keepdims=True)
    dZ1 =, dZ2) * (1 - np.power(A1,2))
    dW1 = 1 / m *, X.T)
    db1 = 1 / m * np.sum(dZ1, axis=1, keepdims=True)
    grad = {
        'dW1': dW1,
        'db1': db1,
        'dW2': dW2,
        'db2': db2,
        'dW3': dW3,
        'db3': db3,
    return grad

Remeber, after the backpropagation step, we need to update these parameters:

def update_parameters(parameters, grads, learning_rate=0.001):
    # Getting our parameters
    W1 = parameters['W1']
    W2 = parameters['W2']
    W3 = parameters['W3']
    b1 = parameters['b1']
    b2 = parameters['b2']
    b3 = parameters['b3']
    # Getting gradients
    dW1 = grads['dW1']
    db1 = grads['db1']
    dW2 = grads['dW2']
    db2 = grads['db2']
    dW3 = grads['dW3']
    db3 = grads['db3']
    W1 = W1 - dW1 * learning_rate
    b1 = b1 - db1 * learning_rate
    W2 = W2 - dW2 * learning_rate
    b2 = b2 - db2 * learning_rate
    W3 = W3 - dW3 * learning_rate
    b3 = b3 - db3 * learning_rate
    parameters = {
        'W1': W1,
        'b1': b1,
        'W2': W2,
        'b2': b2,
        'W3': W3,
        'b3': b3
    return parameters    

Final Model

Putting all these steps together, we get the below function:

def model(X, y, learning_rate, num_iterations = 150, print_cost=True):
    # Initialize parameters
    parameters = initialize_parameters(784, [16, 16, 10])
    W1 = parameters['W1']
    W2 = parameters['W2']
    W3 = parameters['W3']
    b1 = parameters['b1']
    b2 = parameters['b2']
    b3 = parameters['b3']
    # Loop (gradient descent)
    for i in range(0, num_iterations):
        if i > 75:
            learning_rate = 0.1
        A3, cache = forward_propagation(X, parameters)
        cost = -np.mean(y * np.log(A3 + 1e-8))
        y_hat = np.argmax(A3, axis=0)
        Y = np.argmax(y, axis=0)
        accuracy = (y_hat == Y).mean()
        grads = backward_propagation(X, y, cache, parameters)
        parameters = update_parameters(parameters, grads, learning_rate)
        if print_cost and i % 10 == 0:
            print ("Cost after iteration %i: %f Accuracy: %f" %(i, cost, accuracy * 100))
    return parameters

Let's now train our network. I will be loading the MNIST dataset with the tensorflow library. We also need to implement a one_hot_encoding function because we need a 10-dimensional vector for our loss function.

from tensorflow.keras.datasets import mnist

def one_hot(a, num_classes):
  return np.squeeze(np.eye(num_classes)[a.reshape(-1)])

# Load dataset
(train_X, train_y), (test_X, test_y) = mnist.load_data()

print('X_train: ' + str(train_X.shape))
print('Y_train: ' + str(train_y.shape))
print('X_test:  '  + str(test_X.shape))
print('Y_test:  '  + str(test_y.shape))

# Flatten images and one_hot_encode
train_X = train_X.reshape(60000, 784).T
train_X = train_X / 255
train_y = one_hot(train_y, 10).T

# Running the network
parameters = model(train_X, train_y, 1)

My final output:

Cost after iteration 0: 0.230083 Accuracy: 13.260000
Cost after iteration 10: 0.204155 Accuracy: 40.781667
Cost after iteration 20: 0.153983 Accuracy: 47.766667
Cost after iteration 30: 0.118734 Accuracy: 67.136667
Cost after iteration 40: 0.096660 Accuracy: 73.488333
Cost after iteration 50: 0.083073 Accuracy: 77.815000
Cost after iteration 60: 0.072998 Accuracy: 80.786667
Cost after iteration 70: 0.115302 Accuracy: 60.456667
Cost after iteration 80: 0.061357 Accuracy: 84.006667
Cost after iteration 90: 0.060022 Accuracy: 84.413333
Cost after iteration 100: 0.058777 Accuracy: 84.738333
Cost after iteration 110: 0.057598 Accuracy: 85.070000
Cost after iteration 120: 0.056478 Accuracy: 85.361667
Cost after iteration 130: 0.055412 Accuracy: 85.630000
Cost after iteration 140: 0.054397 Accuracy: 85.903333

Since the network is not very large, it is able to converge still with a large learning rate.

To test on one of the test images:

from matplotlib import pyplot as plt

# This is on jupyter notebook btw
test_X.reshape(10000, 784)
test_X = test_X / 255
im = test_X[0]
pixels = im.reshape((28, 28))
plt.imshow(pixels, cmap='gray') # an image of 7

A3, _ = forward_propagation(im.reshape(784, 1), parameters)

y_hat = np.argmax(A3, axis=0)
y_hat # output: array([7])


And there you have it, you just built a neural network that classifies the MNIST dataset from scratch! If it was a lot of information, make sure to try to do some of these derivations yourself on paper to solidify your foundation! Hopefully, you enjoyed this post, it definetly took me hours to fully format and create everything. Good luck on your neural network journey, and see you next time!

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