intro neuron and neural network

intro to neuron and neural-network
https://victorzhou.com/blog/intro-to-neural-networks/

Machine Learning for Beginners: An Introduction to Neural Networks

A simple explanation of how they work and how to implement one from scratch in Python.

Here’s something that might surprise you: neural networks aren’t that complicated! The term “neural network” gets used as a buzzword a lot, but in reality they’re often much simpler than people imagine.

This post is intended for complete beginners and assumes ZERO prior knowledge of machine learning. We’ll understand how neural networks work while implementing one from scratch in Python.

Let’s get started!

1. Building Blocks: Neurons

First, we have to talk about neurons, the basic unit of a neural network. A neuron takes inputs, does some math with them, and produces one output. Here’s what a 2-input neuron looks like:

3 things are happening here. First, each input is multiplied by a weight:

x1→x1∗w1x_1 \rightarrow x_1 * w_1x1​→x1​∗w1​ x2→x2∗w2x_2 \rightarrow x_2 * w_2x2​→x2​∗w2​

Next, all the weighted inputs are added together with a bias $b$:

(x1∗w1)+(x2∗w2)+b(x_1 * w_1) + (x_2 * w_2) + b(x1​∗w1​)+(x2​∗w2​)+b

Finally, the sum is passed through an activation function:

y=f(x1∗w1+x2∗w2+b)y = f(x_1 * w_1 + x_2 * w_2 + b)y=f(x1​∗w1​+x2​∗w2​+b)

The activation function is used to turn an unbounded input into an output that has a nice, predictable form. A commonly used activation function is the sigmoid function:

The sigmoid function only outputs numbers in the range $(0,1)$. You can think of it as compressing $(-\infty ,+\infty )$ to $(0,1)$ - big negative numbers become ~$0$, and big positive numbers become ~$1$.

A Simple Example

Assume we have a 2-input neuron that uses the sigmoid activation function and has the following parameters:

$$w=[0,1]$$ $$b=4$$

$w=[0,1]$ is just a way of writing w1=0,w2=1w_1 = 0, w_2 = 1w1​=0,w2​=1 in vector form. Now, let’s give the neuron an input of $x=[2,3]$. We’ll use the dot product to write things more concisely:

(w⋅x)+b=((w1∗x1)+(w2∗x2))+b=0∗2+1∗3+4=7\begin{aligned} (w \cdot x) + b &= ((w_1 * x_1) + (w_2 * x_2)) + b \\ &= 0 * 2 + 1 * 3 + 4 \\ &= 7 \\ \end{aligned}(w⋅x)+b​=((w1​∗x1​)+(w2​∗x2​))+b=0∗2+1∗3+4=7​ y=f(w⋅x+b)=f(7)=0.999y = f(w \cdot x + b) = f(7) = \boxed{0.999}y=f(w⋅x+b)=f(7)=0.999​

The neuron outputs $0.999$ given the inputs $x=[2,3]$. That’s it! This process of passing inputs forward to get an output is known as feedforward.

Coding a Neuron

Time to implement a neuron! We’ll use NumPy, a popular and powerful computing library for Python, to help us do math:

import numpy as np

def sigmoid(x):
# Our activation function: f(x) = 1 / (1 + e^(-x))
return 1 / (1 + np.exp(-x))

class Neuron:
def init(self, weights, bias):
self.weights = weights
self.bias = bias

def feedforward(self, inputs):
# Weight inputs, add bias, then use the activation function
total = np.dot(self.weights, inputs) + self.bias
return sigmoid(total)

weights = np.array([0, 1]) # w1 = 0, w2 = 1
bias = 4 # b = 4
n = Neuron(weights, bias)

x = np.array([2, 3]) # x1 = 2, x2 = 3
print(n.feedforward(x)) # 0.9990889488055994

Recognize those numbers? That’s the example we just did! We get the same answer of $0.999$.

2. Combining Neurons into a Neural Network

A neural network is nothing more than a bunch of neurons connected together. Here’s what a simple neural network might look like:

This network has 2 inputs, a hidden layer with 2 neurons (h1h_1h1​ and h2h_2h2​), and an output layer with 1 neuron (o1o_1o1​). Notice that the inputs for o1o_1o1​ are the outputs from h1h_1h1​ and h2h_2h2​ - that’s what makes this a network.

A hidden layer is any layer between the input (first) layer and output (last) layer. There can be multiple hidden layers!

An Example: Feedforward

Let’s use the network pictured above and assume all neurons have the same weights $w=[0,1]$, the same bias $b=0$, and the same sigmoid activation function. Let h1,h2,o1h_1, h_2, o_1h1​,h2​,o1​ denote the outputs of the neurons they represent.

What happens if we pass in the input $x=[2,3]$?

h1=h2=f(w⋅x+b)=f((0∗2)+(1∗3)+0)=f(3)=0.9526\begin{aligned} h_1 = h_2 &= f(w \cdot x + b) \\ &= f((0 * 2) + (1 * 3) + 0) \\ &= f(3) \\ &= 0.9526 \\ \end{aligned}h1​=h2​​=f(w⋅x+b)=f((0∗2)+(1∗3)+0)=f(3)=0.9526​ o1=f(w⋅[h1,h2]+b)=f((0∗h1)+(1∗h2)+0)=f(0.9526)=0.7216\begin{aligned} o_1 &= f(w \cdot [h_1, h_2] + b) \\ &= f((0 * h_1) + (1 * h_2) + 0) \\ &= f(0.9526) \\ &= \boxed{0.7216} \\ \end{aligned}o1​​=f(w⋅[h1​,h2​]+b)=f((0∗h1​)+(1∗h2​)+0)=f(0.9526)=0.7216​​

The output of the neural network for input $x=[2,3]$ is $0.7216$. Pretty simple, right?

A neural network can have any number of layers with any number of neurons in those layers. The basic idea stays the same: feed the input(s) forward through the neurons in the network to get the output(s) at the end. For simplicity, we’ll keep using the network pictured above for the rest of this post.

Coding a Neural Network: Feedforward

Let’s implement feedforward for our neural network. Here’s the image of the network again for reference:

import numpy as np

# … code from previous section here

class OurNeuralNetwork:
‘’’
A neural network with:
- 2 inputs
- a hidden layer with 2 neurons (h1, h2)
- an output layer with 1 neuron (o1)
Each neuron has the same weights and bias:
- w = [0, 1]
- b = 0
‘’’
def init(self):
weights = np.array([0, 1])
bias = 0

# The Neuron class here is from the previous section
self.h1 = Neuron(weights, bias)
self.h2 = Neuron(weights, bias)
self.o1 = Neuron(weights, bias)

def feedforward(self, x):
out_h1 = self.h1.feedforward(x)
out_h2 = self.h2.feedforward(x)

# The inputs for o1 are the outputs from h1 and h2
out_o1 = self.o1.feedforward(np.array([out_h1, out_h2]))

return out_o1

network = OurNeuralNetwork()
x = np.array([2, 3])
print(network.feedforward(x)) # 0.7216325609518421

We got $0.7216$ again! Looks like it works.

3. Training a Neural Network, Part 1

Say we have the following measurements:

Name	Weight (lb)	Height (in)	Gender
Alice	133	65	F
Bob	160	72	M
Charlie	152	70	M
Diana	120	60	F

Let’s train our network to predict someone’s gender given their weight and height:

We’ll represent Male with a $0$ and Female with a $1$, and we’ll also shift the data to make it easier to use:

Name	Weight (minus 135)	Height (minus 66)	Gender
Alice	-2	-1	1
Bob	25	6	0
Charlie	17	4	0
Diana	-15	-6	1

I arbitrarily chose the shift amounts ($135$ and $66$) to make the numbers look nice. Normally, you’d shift by the mean.

Loss

Before we train our network, we first need a way to quantify how “good” it’s doing so that it can try to do “better”. That’s what the loss is.

We’ll use the mean squared error (MSE) loss:

MSE=1n∑i=1n(ytrue−ypred)2\text{MSE} = \frac{1}{n} \sum_{i=1}^n (y_{true} - y_{pred})^2MSE=n1​i=1∑n​(ytrue​−ypred​)2

Let’s break this down:

• $n$ is the number of samples, which is $4$ (Alice, Bob, Charlie, Diana).
• $y$ represents the variable being predicted, which is Gender.
• ytruey_{true}ytrue​ is the true value of the variable (the “correct answer”). For example, ytruey_{true}ytrue​ for Alice would be $1$ (Female).
• ypredy_{pred}ypred​ is the predicted value of the variable. It’s whatever our network outputs.

(ytrue−ypred)2(y_{true} - y_{pred})^2(ytrue​−ypred​)2 is known as the squared error. Our loss function is simply taking the average over all squared errors (hence the name mean squared error). The better our predictions are, the lower our loss will be!

Better predictions = Lower loss.

Training a network = trying to minimize its loss.

An Example Loss Calculation

Let’s say our network always outputs $0$ - in other words, it’s confident all humans are Male . What would our loss be?

Name	ytruey_{true}ytrue​	ypredy_{pred}ypred​	(ytrue−ypred)2(y_{true} - y_{pred})^2(ytrue​−ypred​)2
Alice	1	0	1
Bob	0	0	0
Charlie	0	0	0
Diana	1	0	1

MSE=14(1+0+0+1)=0.5\text{MSE} = \frac{1}{4} (1 + 0 + 0 + 1) = \boxed{0.5}MSE=41​(1+0+0+1)=0.5​

Code: MSE Loss

Here’s some code to calculate loss for us:

import numpy as np

def mse_loss(y_true, y_pred):
# y_true and y_pred are numpy arrays of the same length.
return ((y_true - y_pred) ** 2).mean()

y_true = np.array([1, 0, 0, 1])
y_pred = np.array([0, 0, 0, 0])

print(mse_loss(y_true, y_pred)) # 0.5

If you don't understand why this code works, read the NumPy quickstart on array operations.

Nice. Onwards!

4. Training a Neural Network, Part 2

We now have a clear goal: minimize the loss of the neural network. We know we can change the network’s weights and biases to influence its predictions, but how do we do so in a way that decreases loss?

This section uses a bit of multivariable calculus. If you’re not comfortable with calculus, feel free to skip over the math parts.

For simplicity, let’s pretend we only have Alice in our dataset:

Name	Weight (minus 135)	Height (minus 66)	Gender
Alice	-2	-1	1

Then the mean squared error loss is just Alice’s squared error:

MSE=11∑i=11(ytrue−ypred)2=(ytrue−ypred)2=(1−ypred)2\begin{aligned} \text{MSE} &= \frac{1}{1} \sum_{i=1}^1 (y_{true} - y_{pred})^2 \\ &= (y_{true} - y_{pred})^2 \\ &= (1 - y_{pred})^2 \\ \end{aligned}MSE​=11​i=1∑1​(ytrue​−ypred​)2=(ytrue​−ypred​)2=(1−ypred​)2​

Another way to think about loss is as a function of weights and biases. Let’s label each weight and bias in our network:

Then, we can write loss as a multivariable function:

L(w1,w2,w3,w4,w5,w6,b1,b2,b3)L(w_1, w_2, w_3, w_4, w_5, w_6, b_1, b_2, b_3)L(w1​,w2​,w3​,w4​,w5​,w6​,b1​,b2​,b3​)

Imagine we wanted to tweak w1w_1w1​. How would loss $L$ change if we changed w1w_1w1​? That’s a question the partial derivative ∂L∂w1\frac{\partial L}{\partial w_1}∂w1​∂L​ can answer. How do we calculate it?

Here’s where the math starts to get more complex. Don’t be discouraged! I recommend getting a pen and paper to follow along - it’ll help you understand.

To start, let’s rewrite the partial derivative in terms of ∂ypred∂w1\frac{\partial y_{pred}}{\partial w_1}∂w1​∂ypred​​ instead:

∂L∂w1=∂L∂ypred∗∂ypred∂w1\frac{\partial L}{\partial w_1} = \frac{\partial L}{\partial y_{pred}} * \frac{\partial y_{pred}}{\partial w_1}∂w1​∂L​=∂ypred​∂L​∗∂w1​∂ypred​​ This works because of the Chain Rule.

We can calculate ∂L∂ypred\frac{\partial L}{\partial y_{pred}}∂ypred​∂L​ because we computed L=(1−ypred)2L = (1 - y_{pred})^2L=(1−ypred​)2 above:

∂L∂ypred=∂(1−ypred)2∂ypred=−2(1−ypred)\frac{\partial L}{\partial y_{pred}} = \frac{\partial (1 - y_{pred})^2}{\partial y_{pred}} = \boxed{-2(1 - y_{pred})}∂ypred​∂L​=∂ypred​∂(1−ypred​)2​=−2(1−ypred​)​

Now, let’s figure out what to do with ∂ypred∂w1\frac{\partial y_{pred}}{\partial w_1}∂w1​∂ypred​​. Just like before, let h1,h2,o1h_1, h_2, o_1h1​,h2​,o1​ be the outputs of the neurons they represent. Then

ypred=o1=f(w5h1+w6h2+b3)y_{pred} = o_1 = f(w_5h_1 + w_6h_2 + b_3)ypred​=o1​=f(w5​h1​+w6​h2​+b3​) f is the sigmoid activation function, remember?

Since w1w_1w1​ only affects h1h_1h1​ (not h2h_2h2​), we can write

∂ypred∂w1=∂ypred∂h1∗∂h1∂w1\frac{\partial y_{pred}}{\partial w_1} = \frac{\partial y_{pred}}{\partial h_1} * \frac{\partial h_1}{\partial w_1}∂w1​∂ypred​​=∂h1​∂ypred​​∗∂w1​∂h1​​ ∂ypred∂h1=w5∗f′(w5h1+w6h2+b3)\frac{\partial y_{pred}}{\partial h_1} = \boxed{w_5 * f'(w_5h_1 + w_6h_2 + b_3)}∂h1​∂ypred​​=w5​∗f′(w5​h1​+w6​h2​+b3​)​ More Chain Rule.

We do the same thing for ∂h1∂w1\frac{\partial h_1}{\partial w_1}∂w1​∂h1​​:

h1=f(w1x1+w2x2+b1)h_1 = f(w_1x_1 + w_2x_2 + b_1)h1​=f(w1​x1​+w2​x2​+b1​) ∂h1∂w1=x1∗f′(w1x1+w2x2+b1)\frac{\partial h_1}{\partial w_1} = \boxed{x_1 * f'(w_1x_1 + w_2x_2 + b_1)}∂w1​∂h1​​=x1​∗f′(w1​x1​+w2​x2​+b1​)​ You guessed it, Chain Rule.

x1x_1x1​ here is weight, and x2x_2x2​ is height. This is the second time we’ve seen f′(x)f'(x)f′(x) (the derivate of the sigmoid function) now! Let’s derive it:

f(x)=11+e−xf(x) = \frac{1}{1 + e^{-x}}f(x)=1+e−x1​ f′(x)=e−x(1+e−x)2=f(x)∗(1−f(x))f'(x) = \frac{e^{-x}}{(1 + e^{-x})^2} = f(x) * (1 - f(x))f′(x)=(1+e−x)2e−x​=f(x)∗(1−f(x))

We’ll use this nice form for f′(x)f'(x)f′(x) later.

We’re done! We’ve managed to break down ∂L∂w1\frac{\partial L}{\partial w_1}∂w1​∂L​ into several parts we can calculate:

∂L∂w1=∂L∂ypred∗∂ypred∂h1∗∂h1∂w1\boxed{\frac{\partial L}{\partial w_1} = \frac{\partial L}{\partial y_{pred}} * \frac{\partial y_{pred}}{\partial h_1} * \frac{\partial h_1}{\partial w_1}}∂w1​∂L​=∂ypred​∂L​∗∂h1​∂ypred​​∗∂w1​∂h1​​​

This system of calculating partial derivatives by working backwards is known as backpropagation, or “backprop”.

Phew. That was a lot of symbols - it’s alright if you’re still a bit confused. Let’s do an example to see this in action!

Example: Calculating the Partial Derivative

We’re going to continue pretending only Alice is in our dataset:

Name	Weight (minus 135)	Height (minus 66)	Gender
Alice	-2	-1	1

Let’s initialize all the weights to $1$ and all the biases to $0$. If we do a feedforward pass through the network, we get:

h1=f(w1x1+w2x2+b1)=f(−2+−1+0)=0.0474\begin{aligned} h_1 &= f(w_1x_1 + w_2x_2 + b_1) \\ &= f(-2 + -1 + 0) \\ &= 0.0474 \\ \end{aligned}h1​​=f(w1​x1​+w2​x2​+b1​)=f(−2+−1+0)=0.0474​ h2=f(w3x1+w4x2+b2)=0.0474h_2 = f(w_3x_1 + w_4x_2 + b_2) = 0.0474h2​=f(w3​x1​+