How Neural Network Works
Let’s directly discuss the core of neural work. That said, the backpropagation algorithm:
- Input x: Set the corresponding activation \(a^{1}\) for the input layer.
- Feedforward: For each l=2,3,…,L compute \(z^{l} = w^l a^{l-1}+b^l\) and \(a^{l} = \sigma(z^{l})\)
- Output error \(\delta^{L}\): Compute the vector \(\delta^{L} = \nabla_a C \odot \sigma'(z^L)\)
- Backpropagate the error: For each l=L−1,L−2,…,2, compute \(\delta^{l} = ((w^{l+1})^T \delta^{l+1}) \odot \sigma'(z^{l})\)
- Output: The gradient of the cost function is given by \(\frac{\partial C}{\partial w^l_{jk}} = a^{l-1}_k \delta^l_j\) and \(\frac{\partial C}{\partial b^l_j} = \delta^l_j\)
This is copied from Michael Nelson’s book Neural Networks and Deep Learning. Is it overwhelming? It might be at the first time you see it. But it is not after one month of studying around it. Let me explain.
Input
There are 5 phases. The first phase is Input. Here we use handwritten digits as input. Our task is to recognize them. One handwritten digit has 784 pixels, which is 28*28. In every pixel, there is a grayscale value which is range from 0 to 255. So the activation means that we use some function to activate it, to change its original value to a new value for the convenience of processing.
Say, we have now 1000 pictures of 784 pixels. We now train it to recognize what digit they show. We have now 100 pictures to test that learning effect. If the program can recognize 97 pictures’ digits, we say its accuracy is 97%.
So we would loop over the 1000 pictures, to train out the weights and biases. We make weights and biases more correct every time we give it a new picture to learn.
One batch training result is to be reflected in 10 neurons. Here, the 10 neurons represent from 0 to 9 and its value is range from 0 to 1 to indicate how their confidence about its accuracy.
And the input is 784 neurons. How can we reduce 784 neurons to 10 neurons? Here is the thing. Let’s suppose we have two layers. What does the layer mean? That is the first layer, we have 784 neurons. In the second layer, we have 10 neurons.
We give each neuron in the 784 neurons a weight, say,
\[w_1, w_2, w_3, w_4, ... , w_{784}\]And give the first layer, a bias, that is, \(b_1\).
And so for the first neuron in the second layer, its value is:
\[w_1*a_1 + w_2*a_2+...+ w_{784}*a_{784}+b_1\]But these weights and a bias are for \(neuron^2_{1}\)(the first one in the second layer). To the \(neuron^2_{2}\), we need another set of weights and a bias.
How about the sigmoid function? We use the sigmoid function to map the value of the above from 0 to 1.
\[\begin{eqnarray} \sigma(z) \equiv \frac{1}{1+e^{-z}} \end{eqnarray}\] \[\begin{eqnarray} \frac{1}{1+\exp(-\sum_j w_j x_j-b)} \end{eqnarray}\]We also use the sigmoid function to activate the first layer. That said, we change that grayscale value to the range from 0 to 1. So now, every neuron in every layer has a value from 0 to 1.
So now for our two-layer network, the first layer has 784 neurons, and the second layer has 10 neurons. We train it to get the weights and biases.
We have 784 * 10 weights and 10 biases. In the second layer, for every neuron, we will use 784 weights and 1 biases to calculate its value. The code here like,
def __init__(self, sizes):
self.num_layers = len(sizes)
self.sizes = sizes
self.biases = [np.random.randn(y, 1) for y in sizes[1:]]
self.weights = [np.random.randn(y, x)
for x, y in zip(sizes[:-1], sizes[1:])]
Feedforward
Feedforward: For each l=2,3,…,L compute \(z^{l} = w^l a^{l-1}+b^l\) and \(a^{l} = \sigma(z^{l})\)
Notice here, we use the value of the last layer, that is \(a^{l-1}\) and the current layer’s weight, \(w^l\) and its bias \(b^l\) to do the sigmoid to get the value of the current layer, \(a^{l}\).
Code:
nabla_b = [np.zeros(b.shape) for b in self.biases]
nabla_w = [np.zeros(w.shape) for w in self.weights]
# feedforward
activation = x
activations = [x]
zs = []
for b, w in zip(self.biases, self.weights):
z = np.dot(w, activation)+b
zs.append(z)
activation = sigmoid(z)
activations.append(activation)
Output error
Output error \(\delta^{L}\): Compute the vector \(\delta^{L} = \nabla_a C \odot \sigma'(z^L)\)
Let’s see what the \(\nabla\) mean.
\[\begin{eqnarray} w_k & \rightarrow & w_k' = w_k-\eta \frac{\partial C}{\partial w_k} \\ b_l & \rightarrow & b_l' = b_l-\eta \frac{\partial C}{\partial b_l} \end{eqnarray}\]Del, or nabla, is an operator used in mathematics (particularly in vector calculus) as a vector differential operator, usually represented by the nabla symbol ∇.
Here \(\eta\) is the learning rate. We use the derivative that C is respective to the weights and the bias, that is the rate change between them. That is sigmoid_prime in the below.
Code:
delta = self.cost_derivative(activations[-1], y) * \
sigmoid_prime(zs[-1])
nabla_b[-1] = delta
nabla_w[-1] = np.dot(delta, activations[-2].transpose())
def cost_derivative(self, output_activations, y):
return (output_activations-y)
Backpropagate the error
Backpropagate the error: For each l=L−1,L−2,…,2, compute \(\delta^{l} = ((w^{l+1})^T \delta^{l+1}) \odot \sigma'(z^{l})\)
for l in range(2, self.num_layers):
z = zs[-l]
sp = sigmoid_prime(z)
delta = np.dot(self.weights[-l+1].transpose(), delta) * sp
nabla_b[-l] = delta
nabla_w[-l] = np.dot(delta, activations[-l-1].transpose())
return (nabla_b, nabla_w)
Output
Output: The gradient of the cost function is given by \(\frac{\partial C}{\partial w^l_{jk}} = a^{l-1}_k \delta^l_j\) and \(\frac{\partial C}{\partial b^l_j} = \delta^l_j\)
def update_mini_batch(self, mini_batch, eta):
nabla_b = [np.zeros(b.shape) for b in self.biases]
nabla_w = [np.zeros(w.shape) for w in self.weights]
for x, y in mini_batch:
delta_nabla_b, delta_nabla_w = self.backprop(x, y)
nabla_b = [nb+dnb for nb, dnb in zip(nabla_b, delta_nabla_b)]
nabla_w = [nw+dnw for nw, dnw in zip(nabla_w, delta_nabla_w)]
self.weights = [w-(eta/len(mini_batch))*nw
for w, nw in zip(self.weights, nabla_w)]
self.biases = [b-(eta/len(mini_batch))*nb
for b, nb in zip(self.biases, nabla_b)]
Final
It is a short article. And in the most part, it just shows the code and math formula. But it is fine to me. Before writing it, I don’t understand clearly. After writing or just copying snippets from code and book, I understand most of it. After gaining confidence from the teacher Yin Wang, reading about 30% of the book Neural Networks and Deep Learning, listening to the Andrej Karpathy’s Standford lectures and Andrew Ng’s courses, discussing with my friend Qi, and tweaking with Anaconda, numpy, and Theano libraries to make the code years ago work, I now understand it.
One of the key points is the dimensions. We should know the dimensions of every symbol and variable. And it just does the differentiable computation. Let’s end with Yin Wang’s quotes:
Machine learning is really useful, one might even say beautiful theory, because it is simply calculus after a makeover! It is the old and great theory of Newton, Leibniz, in a simpler, elegant and powerful form. Machine learning is basically the use of calculus to derive and fit some functions, and deep learning is the fitting of more complex functions.
There is no ‘intelligence’ in artificial intelligence, no ‘neural’ in neural network, no ‘learning’ in machine learning, and no ‘depth’ in deep learning. There is no ‘depth’ in deep learning. What really works in this field is called ‘calculus’. So I prefer to call this field ‘differentiable computing’, and the process of building models is called ‘differentiable programming’.