 Convolutional Neural Network

# Convolutional Neural Network

Donny August 19, 2018
August 26, 2018
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This article is my reflection on my previous work FaceLock, a project to recognize user's face and lock the computer if the user doesn't present in a certain time. CNN is used to recognize different faces. I watch the Coursera course Convolutional Neural Networks by Andrew Ng to understand more about CNN, so it's also a learning note about it.

## One Layer of a Convolutional Network

In a non-convolutional network, we have the following formula:

$$Z = W \cdot x + b \\ a = g(Z)$$

Similarly, in the convolutional network, we can have:

$$Z = W * x + b \\ a = g(Z)$$

@ $$*$$ is a convolution operation.

@ $$x$$ is the input matrix.

@ $$W$$ is the filter. Different filter can detect different feature, e.g. vertical edge, diagonal edge, etc.

@ $$b$$ is the bias.

@ $$g$$ is a activation function.

@ $$a$$ is the output matrix, and can be fed to the next layer. ## Calculating the Number

### The Number of the Parameters

Suppose we have 10 filters which are $$3 \times 3 \times 3$$ in one layer of a neural network, how many parameters does the layer have?

In one filter, there are $$3 \times 3 \times 3 + 1 = 28$$ (1 bias) parameters. So there are $$10 \times 28 = 280$$ parameters in this layer.

### The Shape of the Output

Suppose we have a $$x \times y \times ch$$ (ch is the number of channel) input, $$n$$ numbers of $$r \times c \times ch$$ filters, the stride is set to (0, 0), not padding. what will be the output shape?

The output shape will be $$(x - r + 1) \times (y - c + 1) \times n$$.

Now change the stride to (a, b), what will be the output shape?

The output shape will be $$(\lfloor \frac{x - r}{a} \rfloor + 1) \times (\lfloor \frac{y - c}{b} \rfloor + 1) \times n$$.

Now add padding $$(p_x, p_y)$$, what will be the output shape again?

The output shape will be $$(\lfloor \frac{x + 2 p_x - r}{a} \rfloor + 1) \times (\lfloor \frac{y + 2 p_y - c}{b} \rfloor + 1) \times n$$.

## Summary of Notation

Define:

\begin{align*} f^{[l]} & = \text{filter size} \\ p^{[l]} & = \text{padding} \\ s^{[l]} & = \text{stride} \\ n^{[l]}_c & = \text{number of filters} \\ \end{align*}

Than we have:

\begin{align*} & \text{Input shape:} & n^{[l-1]}_H \times n^{[l-1]}_W \times n^{[l-1]}_c \\ & \text{Output shape:} & n^{[l]}_H \times n^{[l]}_W \times n^{[l]}_c \\ & \text{s.t.} & n^{[l]}_H = \lfloor \frac{n^{[l-1]}_H + 2 p^{[l]} - f^{[l]}}{s^{[l]}} \rfloor + 1 \\ & \quad & n^{[l]}_W = \lfloor \frac{n^{[l-1]}_W + 2 p^{[l]} - f^{[l]}}{s^{[l]}} \rfloor + 1 \\ & \text{Filter shape:} & f^{[l]} \times f^{[l]} \times n^{[l-1]}_c \\ & \text{Activations:} & a^{[l]} \rightarrow n^{[l]}_H \times n^{[l]}_W \times n^{[l]}_c \\ & \text{Weight:} & f^{[l]} \times f^{[l]} \times n^{[l-1]}_c \times n^{[l]}_c \\ & \text{bias:} & 1 \times 1 \times 1 \times n^{[l]}_c \end{align*}

In batch gradient descent, Activation is

$$A^{[l]} \rightarrow m \times n^{[l]}_H \times n^{[l]}_W \times n^{[l]}_c$$

where $$A^{[l]}$$ is a set of $$m$$ activations if there are $$m$$ examples.