Stanford Winter Quarter 2016 class: CS231n: Convolutional Neural Networks for Visual Recognition. Lecture 7. Get in touch on Twitter @cs231n, or on Reddit /r/cs231n.
At 24:30 in Lecture #7 a student asks about why filter size F is odd. I believe this is because odd length filters have an integer center. e.g. 2x2 filter has center at 0.5,0.5 but 3x3 filter has center at 1,1 (assuming indices starting at 0).
I think it does a lot to do with EDGE DETECTION in image processing, we use different filter to find the property in different directions of a image, there are 6 filters often being used list below: Roberts filter(2x2), Sobel filter(3x3), Prewitt filter(3x3), Laplacian filter(3x3), Gauss-Laplacian filter(5x5), Krisch filter(3x3)... i thought this would helpful to understand: yangloong.github.io/posts/image_processing/edge_detection/edge_detection.html
4:10 the filter does a dot product across the depth of the image, so the image initially had a depth of 3 but after convolving the filter we get an activation map with depth 1
Has there been any study on the performance (e.g. metrics like accuracy or training time) of convolutional networks that are trained on source data that is in RGB layers vs something like hue, saturation, intensity?
I love the way you make everything so easy. I have been studying and practising conv nets for roughly 2 years now but still just looking at your videos every time bring a new intuition and understanding. Also, can u make videos covering the latest convnet trends?
1) Sir at 15:32 I used my own predicted formula 7*7 input and 3*3 filter with applied stride 2 has (7-2^2)*(7-2^2) => (7-4)*(7-4) => 3*3 output matrix. Similarly for N*N input and F*F filter with applied stride "S" gives [N-(F-1)^S] * [N-(F-1)^S] output matrix . Sir, please correct me if I'm wrong...! 2)Why we're using square matrices everywhere and why not Rectangular matrices..?? Is there any complex computational problem occurs here ...!!
At 27:24 he says we always work with squares. It is not true anymore. Modern architectures such as ResNet can work with inputs of any size. It all depends on how you process data at the end of the network. If you use global average pooling per feature layer neat the end of the net, then you do not care about spatial dimensions of input image.
So the "dot product" of a section of image and a filter is actually the dot product of 2 n-dimensional VECTORS not matrices correct? because the dot product of 2 3D matrices is completely different than 2 N-dimensional vectors.
Correct, I don't know why it wasn't explicitly stated here. x is the flattened vector representation of the 5x5x3 of that section of the input. I realize you asked this question a year or more ago, but someone else might also be curious.
Actually I'm pretty sure he meant dot product of two matrices. If we use a 5x5x3 (75) filter on a 32x32x3 input volume, we get 28x28 (784) unique inputs to the filter. If we are using a set of 6 filters, that means we need to calculate the dot product of two matrices with shapes (784, 75) and (75, 6). The result has the shape (784, 6) which we then reshape to (28, 28, 6). I realize you answered this question a year or more ago, but someone else might also be curious.
The lecture is so good that no one was even bothered by how the matrix was glitching out... or maybe they all know about the matrix?????? *Dramatic Pause*
Minute 59:27, I try to recalculate the total memory by summing all the red numbers, and I end up with around 15M, not 24M. Unless I count the fully connected layers with 4M instead of 4K for 4096 and so on. Which wrong?
why can't the other professors make diagram like these??!! instead I got 2d diagrams stacked on top of each other, and half of the other stuff is not even there. smh. It doesn't take that much time, and they can reuse them every year. JUST WHY?!!!
Thank you for this good lecture. I'm new to Conv and I'm not sure how did you compute the number of neurones in the fully connected layers? for example, in Alex's case, how did he get FC6 = 4096 neurons?
if you're confused by the example at 19:50, it is very simple but can be confusing because in the slides the rule for the output size was defined as (N-F) / stride + 1. the lack of parenthesis in that definition causes the confusion as it really should be ((N-F) / stride) + 1 which will then give you the correct answer. wrong: (32 + 2 + 2 - 5) / 1+1 = 31/2? correct: ((32+2+2-5)/1) + 1 = (31/1)+1 = 32
As far as I understand, filters are initializing randomly. So, how do we know that two seperate filters in the same layer doesn't look for the same thing after reducing the loss function?
I think you could get answer if you will think about distribution you'll use for weights initialization. So let say we will use uniform dist... It is possible that features will look for similar patterns but definitely not for the same ones. On the other side if you would do bad initialization of random generator I mean with bad or same seeds it is possible to get in the problem you're describing...
I guess i heard of him saying depth is a hyper parameter we need to choose and fine tune , perhaps start with recommended one and change based on experiment ...
Teacher, can you increase the English subtitles so that I can understand what the teacher taught, because I am a foreign student, my English listening is weak.
The dimension of the input(image) is 32X32X3. I guess 3 is for RGB. So it's actually 3 matrices one for each among R, G and B stacked together. Also same with the filters. Can anyone explain how the final volume has 3rd dimension as 1?
Which color you want to detect? The reason it is 3 layer, is because the original image have been split into 3 different channel: RGB. THerefore, its corresponding output will then only have RGB layer, and if u want any other colour, you have to combine them from individual filter (RGB).
The network will tune itself to create a filter that recognizes colors, if that is an important "feature" to distinguish images. First CONV layer takes dot product with each of the 3 colors (3 input depth). So colors get encoded into the weights this way. They are no longer held segregated into layers of their own now, they are encoded into the output of each filter. Each filter encodes colors (and spacial features) independently from each other. So each "depth" of the output contains a different mix of values indicating how important color is to it (for each pixel location). In this way color importance gets remixed and distributed. Then weights for each filter adjust themselves at backprop, taking into account color, location, and brightness. These become the "features" each filter looks for in any subspace of the image. Filter depth must always match the depth of the input layer it acts upon. So First filter would always be depth of 3 if given and RGB image. If the image is Grayscale and encoded as a 2D, then Depth would be 0 (2D), and filter would also be 2D. In that case, a 1*1 would be appropriate.
+apple-sauce You 'flatten' it. It just means that you unroll the last pooling layer into a single long sausage. It doesn't matter how you do the unrolling though as long as you stay consistent because FC networks do not take into account any spacial coherence and it will just learn the correct mapping through BP.
I assume you are referring to the java script demo , if so , you have it her ... cs.stanford.edu/people/karpathy/convnetjs/demo/cifar10.html ... I type the url pausing the screen haha
+Anirudha Kumar The product of the filter and the part of the image must be summed up, so it will be just a scalar value. Then you slide a filter by stride step to the next part of the image, and do the same thing. Repeat this, until you will reach the end of the image. As a result you will have a 2D matrix consisting of this summed up products across all image.
+Dmitriy Bobrenko In my memory, the dot product of an i*j matrix and a j*k matrix is an i*k matrix, how could it obtain a single scalar value? Need these matrix transformed as vectors, and the dot product here is just the sum of corresponding elements' product?
+Jingkang Yang yes, you are absolutely correct. The dot product will be i*k matrix, then to get a scalar value, we need to sum up all this matrix elements into a single value.
+Anirudha Kumar I'm not sure that I agree with Dmitriy's replies. Someone please correct me if I'm wrong but: First of all, each filter matrix is 3D (e.g. see 28:44). So, there is no additional step to generate a scalar for a single pixel. E.g. the first layer will examine all three color channels. Thus a single filter will reduce the 3xNxN image to a 1xNxN data matrix. The dimensionality of the second layer is due to the fact that there are F filters in layer 1, resulting in a FxNxN matrix.
