This is the content of Course 4, Week 1 (C4W1) of Deep Learning Specialization.
(C4W1L01) Computer Vision
--For an image of 64 $ \ times $ 64, the number of data will be $ 64 \ times 64 \ times 3 = 12288 $ (considering RGB) --For 1000 $ \ times $ 1000 images, it will be 3 million --Introduce convolution to solve size problem --convolution is useful (maybe) beyond computer vision
(C4W1L02) Edge Detection Example
--In the computer vision problem, recognition is performed in the order of edge → part → whole. --For Vertical Edge Detection, use the 3 $ \ times $ 3 filter below.
1 0 -1 1 0 -1 1 0 -1
conv_forward
tf.nn.conv2d
Conv2D
(C4W1L03) More edge detection
1 0 -1 1 0 -1 1 0 -1
1 1 1 0 0 0 -1 -1 -1
--Make the filter element a variable ($ w_i $) to learn a good filter
(C4W1L04) Padding
--If the input image is $ n \ times n $ and the filter is $ f \ times f $, the output image will be $ (n-f + 1) \ times (n-f + 1) $.
Disadvantage --The image becomes smaller with each convolution operation --The corner pixel is used only once (discarding information)
Padding --Pixels around the input image to make it larger (typically 0) -If $ p $ = padding, the output image will be $ (n + 2p-f + 1) \ times (n + 2p-f + 1) $
Valid and Same convolution
-$ f $ is usually odd ($ 3 \ times 3 $ is common, $ 5 \ times 5 $ or $ 7 \ times 7 $ is fine)
(C4W1L05) Strided convolution
--The number of pixels that move the filter is called stride --input image; $ n \ times n $, filter; $ f \ times f $, padding; $ p $, stride; $ s $, output image; $ \ lfloor \ frac {n + 2p-f} { s} + 1 \ rfloor \ times \ lfloor \ frac {n + 2p-f} {s} + 1 \ rfloor $ (truncate if not divisible)
--In mathematical texts, the convolution operation flips the filter over. By doing so, the associative law ($ (A \ ast B) \ ast C = A \ ast (B \ ast C) $) can be satisfied. ――But in deep learning, the filter is calculated as it is without turning it over.
(C4W1L06) Convolutions Over Volume
-Deep Learning Specialization (Coursera) Self-study record (table of contents)
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