CNNVR: CNN

机器学习

Input volume of size $[W1 \times H1 \times D1]$, using $K$ neurons with receptive fields $F \times F$ and applying them at strides of $S$ gives output volumn $[W2, H2, D2]$, where $W2 = \frac{W1 - F}{S} - 1$, $H2 = \frac{H1 - F}{S} - 1$, and $D2 = K$. In practice is common to zero pad the border with $\frac{F-1}{2}$ $0$s.

The number weight will be $H2 \times W2 \times K \times F \times F \times D1$, and is too big. Weight sharing reduce this number to $K \times F \times F \times D1$. Note: Sometimes globally weight sharing is not a good idea.

Size tricks

• Start with image that has power-of-2 size.
• Use stride $1$ filter size $3 \times 3$ pad input with a border of $0$s.
• Use pool size $2 \times 2$.