I am looking for a workflow that would clean (and possibly straighten) old and badly scanned images of musical scores (like the one below).
I've tried to use denoise, hough filters, imagemagick geometry filters, and I am struggling to identify the series of filters that remove the scanner noise/bias.
Just some quick ideas:
Remove grayscale noise: Do a low pass filter (darks), since the music is darker than a lot of the noise. Remaining noise is mostly vertical lines.
Rotate image: Sum grayscale values for each column of the image. You'll get a vector with the total pixel lightness in that column. Use gradient descent or search on the rotation of the image (within some bounds like +/-15deg rotation) to maximize the variance of that vector. Idea here is that the vertical noise lines indicate vertical alignment, and so we want the columns of the image to align with these noise lines (= maximized variance).
Remove vertical line noise: After rotation, take median value of each column. The greater the distance (squared difference) a pixel is from that median darkness, the more confident we are it is its true color (e.g. a pure white or black pixel when vertical noise was gray). Since noise is non-white, you could try blending this distance by the whiteness of the median for an alternative confidence metric. Ideally, I think here you'd train some 7x7x2 convolution filter (2 channels being pixel value and distance from median) to estimate true value of the pixel. That would be the most minimal machine learning approach, not using some full-fledged NN. However, given your lack of training data, we'll have to come up with our own heuristic for what the true pixel value is. You likely will need to play around with it, but here's what I think might work:
Set some threshold of confidence; above that threshold we take the value as is. Below the threshold, set to white (the binary expected pixel value for the entire page).
For all values below threshold, take the max confidence value within a +/-2 pixels L1 distance (e.g. 5x5 convolution) as that pixel's value. Seems like features are separated by at least 2 pixels, but for lower resolutions that window size may need to be adjusted. Since white pixels may end up being more confident overall, you could experiment with prioritizing darker pixels (increase their confidence somehow).
Clamp the image contrast and maybe run another low pass filter.
I have an array of data from a grayscale image that I have segmented sets of contiguous points of a certain intensity value from.
Currently I am doing a naive bounding box routine where I find the minimum and maximum (x,y) [row, col] points. This obviously does not provide the smallest possible box that contains the set of points which is demonstrable by simply rotating a rectangle so the longest axis is no longer aligned with a principal axis.
What I wish to do is find the minimum sized oriented bounding box. This seems to be possible using an algorithm known as rotating calipers, however the implementations of this algorithm seem to rely on the idea that you have a set of vertices to begin with. Some details on this algorithm: https://www.geometrictools.com/Documentation/MinimumAreaRectangle.pdf
My main issue is in finding the vertices within the data that I currently have. I believe I need to at least find candidate vertices in order to reduce the amount of iterations I am performing, since the amount of points is relatively large and treating the interior points as if they are vertices is unnecessary if I can figure out a way to not include them.
Here is some example data that I am working with:
Here's the segmented scene using the naive algorithm, where it segments out the central objects relatively well due to the objects mostly being aligned with the image axes:
.
In red, you can see the current bounding boxes that I am drawing utilizing 2 vertices: top-left and bottom-right corners of the groups of points I have found.
The rotation part is where my current approach fails, as I am only defining the bounding box using two points, anything that is rotated and not axis-aligned will occupy much more area than necessary to encapsulate the points.
Here's an example with rotated objects in the scene:
Here's the current naive segmentation's performance on that scene, which is drawing larger than necessary boxes around the rotated objects:
Ideally the result would be bounding boxes aligned with the longest axis of the points that are being segmented, which is what I am having trouble implementing.
Here's an image roughly showing what I am really looking to accomplish:
You can also notice unnecessary segmentation done in the image around the borders as well as some small segments, which should be removed with some further heuristics that I have yet to develop. I would also be open to alternative segmentation algorithm suggestions that provide a more robust detection of the objects I am interested in.
I am not sure if this question will be completely clear, therefore I will try my best to clarify if it is not obvious what I am asking.
It's late, but that might still help. This is what you need to do:
expand pixels to make small segments connect larger bodies
find connected bodies
select a sample of pixels from each body
find the MBR ([oriented] minimum bounding rectangle) for selected set
For first step you can perform dilation. It's somehow like DBSCAN clustering. For step 3 you can simply select random pixels from a uniform distribution. Obviously the more pixels you keep, the more accurate the MBR will be. I tested this in MATLAB:
% import image as a matrix of 0s and 1s
oI = ~im2bw(rgb2gray(imread('vSb2r.png'))); % original image
% expand pixels
dI = imdilate(oI,strel('disk',4)); % dilated
% find connected bodies of pixels
CC = bwconncomp(dI);
L = labelmatrix(CC) .* uint8(oI); % labeled
% mark some random pixels
rI = rand(size(oI))<0.3;
sI = L.* uint8(rI) .* uint8(oI); % sampled
% find MBR for a set of connected pixels
for i=1:CC.NumObjects
[Y,X] = find(sI == i);
mbr(i) = getMBR( X, Y );
end
You can also remove some ineffective pixels using some more processing and morphological operations:
remove holes
find boundaries
find skeleton
In MATLAB:
I = imfill(I, 'holes');
I = bwmorph(I,'remove');
I = bwmorph(I,'skel');
I was taking a look at Convolutional Neural Network from CS231n Convolutional Neural Networks for Visual Recognition. In Convolutional Neural Network, the neurons are arranged in 3 dimensions(height, width, depth). I am having trouble with the depth of the CNN. I can't visualize what it is.
In the link they said The CONV layer's parameters consist of a set of learnable filters. Every filter is small spatially (along width and height), but extends through the full depth of the input volume.
For example loook at this picture. Sorry if the image is too crappy.
I can grasp the idea that we take a small area off the image, then compare it with the "Filters". So the filters will be collection of small images? Also they said We will connect each neuron to only a local region of the input volume. The spatial extent of this connectivity is a hyperparameter called the receptive field of the neuron. So is the receptive field has the same dimension as the filters? Also what will be the depth here? And what do we signify using the depth of a CNN?
So, my question mainly is, if i take an image having dimension of [32*32*3] (Lets say i have 50000 of these images, making the dataset [50000*32*32*3]), what shall i choose as its depth and what would it mean by the depth. Also what will be the dimension of the filters?
Also it will be much helpful if anyone can provide some link that gives some intuition on this.
EDIT:
So in one part of the tutorial(Real-world example part), it says The Krizhevsky et al. architecture that won the ImageNet challenge in 2012 accepted images of size [227x227x3]. On the first Convolutional Layer, it used neurons with receptive field size F=11, stride S=4 and no zero padding P=0. Since (227 - 11)/4 + 1 = 55, and since the Conv layer had a depth of K=96, the Conv layer output volume had size [55x55x96].
Here we see the depth is 96. So is depth something that i choose arbitrarily? or something i compute? Also in the example above(Krizhevsky et al) they had 96 depths. So what does it mean by its 96 depths? Also the tutorial stated Every filter is small spatially (along width and height), but extends through the full depth of the input volume.
So that means the depth will be like this? If so then can i assume Depth = Number of Filters?
In Deep Neural Networks the depth refers to how deep the network is but in this context, the depth is used for visual recognition and it translates to the 3rd dimension of an image.
In this case you have an image, and the size of this input is 32x32x3 which is (width, height, depth). The neural network should be able to learn based on this parameters as depth translates to the different channels of the training images.
UPDATE:
In each layer of your CNN it learns regularities about training images. In the very first layers, the regularities are curves and edges, then when you go deeper along the layers you start learning higher levels of regularities such as colors, shapes, objects etc. This is the basic idea, but there lots of technical details. Before going any further give this a shot : http://www.datarobot.com/blog/a-primer-on-deep-learning/
UPDATE 2:
Have a look at the first figure in the link you provided. It says 'In this example, the red input layer holds the image, so its width and height would be the dimensions of the image, and the depth would be 3 (Red, Green, Blue channels).' It means that a ConvNet neuron transforms the input image by arranging its neurons in three dimeonsion.
As an answer to your question, depth corresponds to the different color channels of an image.
Moreover, about the filter depth. The tutorial states this.
Every filter is small spatially (along width and height), but extends through the full depth of the input volume.
Which basically means that a filter is a smaller part of an image that moves around the depth of the image in order to learn the regularities in the image.
UPDATE 3:
For the real world example I just browsed the original paper and it says this : The first convolutional layer filters the 224×224×3 input image with 96 kernels of size 11×11×3 with a stride of 4 pixels.
In the tutorial it refers the depth as the channel, but in real world you can design whatever dimension you like. After all that is your design
The tutorial aims to give you a glimpse of how ConvNets work in theory, but if I design a ConvNet nobody can stop me proposing one with a different depth.
Does this make any sense?
Depth of CONV layer is number of filters it is using.
Depth of a filter is equal to depth of image it is using as input.
For Example: Let's say you are using an image of 227*227*3.
Now suppose you are using a filter of size of 11*11(spatial size).
This 11*11 square will be slided along whole image to produce a single 2 dimensional array as a response. But in order to do so, it must cover every aspect inside of 11*11 area. Therefore depth of filter will be depth of image = 3.
Now suppose we have 96 such filter each producing different response. This will be depth of Convolutional layer. It is simply number of filters used.
I'm not sure why this is skimped over so heavily. I also had trouble understanding it at first, and very few outside of Andrej Karpathy (thanks d00d) have explained it. Although, in his writeup (http://cs231n.github.io/convolutional-networks/), he calculates the depth of the output volume using a different example than in the animation.
Start by reading the section titled 'Numpy examples'
Here, we go through iteratively.
In this case we have an 11x11x4. (why we start with 4 is kind of peculiar, as it would be easier to grasp with a depth of 3)
Really pay attention to this line:
A depth column (or a fibre) at position (x,y) would be the activations
X[x,y,:].
A depth slice, or equivalently an activation map at depth d
would be the activations X[:,:,d].
V[0,0,0] = np.sum(X[:5,:5,:] * W0) + b0
V is your output volume. The zero'th index v[0] is your column - in this case V[0] = 0 this is the first column in your output volume.
V[1] = 0 this is the first row in your output volume. V[3]= 0 is the depth. This is the first output layer.
Now, here's where people get confused (at least I did). The input depth has absolutely nothing to do with your output depth. The input depth only has control of the filter depth. W in Andrej's example.
Aside: A lot of people wonder why 3 is the standard input depth. For color input images, this will always be 3 for plain ole images.
np.sum(X[:5,:5,:] * W0) + b0 (convolution 1)
Here, we are calculating elementwise between a weight vector W0 which is 5x5x4. 5x5 is an arbitrary choice. 4 is the depth since we need to match our input depth. The weight vector is your filter, kernel, receptive field or whatever obfuscated name people decide to call it down the road.
if you come at this from a non python background, that's maybe why there's more confusion since array slicing notation is non-intuitive. The calculation is a dot product of your first convolution size (5x5x4) of your image with the weight vector. The output is a single scalar value which takes the position of your first filter output matrix. Imagine a 4 x 4 matrix representing the sum product of each of these convolution operations across the entire input. Now stack them for each filter. That shall give you your output volume. In Andrej's writeup, he starts moving along the x axis. The y axis remains the same.
Here's an example of what V[:,:,0] would look like in terms of convolutions. Remember here, the third value of our index is the depth of your output layer
[result of convolution 1, result of convolution 2, ..., ...]
[..., ..., ..., ..., ...]
[..., ..., ..., ..., ...]
[..., ..., ..., result of convolution n]
The animation is best for understanding this, but Andrej decided to swap it with an example that doesn't match the calculation above.
This took me a while. Partly because numpy doesn't index the way Andrej does in his example, at least it didn't I played around with it. Also, there's some assumptions that the sum product operation is clear. That's the key to understand how your output layer is created, what each value represents and what the depth is.
Hopefully that helps!
Since the input volume when we are doing an image classification problem is N x N x 3. At the beginning it is not difficult to imagine what the depth will mean - just the number of channels - Red, Green, Blue. Ok, so the meaning for the first layer is clear. But what about the next ones? Here is how I try to visualize the idea.
On each layer we apply a set of filters which convolve around the input. Lets imagine that currently we are at the first layer and we convolve around a volume V of size N x N x 3. As #Semih Yagcioglu mentioned at the very beginning we are looking for some rough features: curves, edges etc... Lets say we apply N filters of equal size (3x3) with stride 1. Then each of these filters is looking for a different curve or edge while convolving around V. Of course, the filter has the same depth, we want to supply the whole information not just the grayscale representation.
Now, if M filters will look for M different curves or edges. And each of these filters will produce a feature map consisting of scalars (the meaning of the scalar is the filter saying: The probability of having this curve here is X%). When we convolve with the same filter around the Volume we obtain this map of scalars telling us where where exactly we saw the curve.
Then comes feature map stacking. Imagine stacking as the following thing. We have information about where each filter detected a certain curve. Nice, then when we stack them we obtain information about what curves / edges are available at each small part of our input volume. And this is the output of our first convolutional layer.
It is easy to grasp the idea behind non-linearity when taking into account 3. When we apply the ReLU function on some feature map, we say: Remove all negative probabilities for curves or edges at this location. And this certainly makes sense.
Then the input for the next layer will be a Volume $V_1$ carrying info about different curves and edges at different spatial locations (Remember: Each layer Carries info about 1 curve or edge).
This means that the next layer will be able to extract information about more sophisticated shapes by combining these curves and edges. To combine them, again, the filters should have the same depth as the input volume.
From time to time we apply Pooling. The meaning is exactly to shrink the volume. Since when we use strides = 1, we usually look at a pixel (neuron) too many times for the same feature.
Hope this makes sense. Look at the amazing graphs provided by the famous CS231 course to check how exactly the probability for each feature at a certain location is computed.
In simple terms, it can explain as below,
Let's say you have 10 filters where each filter is the size of 5x5x3. What does this mean? the depth of this layer is 10 which is equal to the number of filters. Size of each filter can be defined as we want e.g., 5x5x3 in this case where 3 is the depth of the previous layer. To be precise, depth of each filer in the next layer should be 10 ( nxnx10) where n can be defined as you want like 5 or something else. Hope will make everything clear.
The first thing you need to note is
receptive field of a neuron is 3D
ie If the receptive field is 5x5 the neuron will be connected to 5x5x(input depth) number of points. So whatever be your input depth, one layer of neurons will only develop 1 layer of output.
Now, the next thing to note is
depth of output layer = depth of conv. layer
ie The output volume is independent of the input volume, and it only depends on the number filters(depth). This should be pretty obvious from the previous point.
Note that the number of filters (depth of the cnn layer) is a hyper parameter. You can take it whatever you want, independent of image depth. Each filter has it's own set of weights enabling it to learn a different feature on the same local region covered by the filter.
The depth of the network is the number of layers in the network. In the Krizhevsky paper, the depth is 9 layers (modulo a fencepost issue with how layers are counted?).
If you are referring to the depth of the filter (I came to this question searching for that) then this diagram of LeNet is illustrating
Source http://yann.lecun.com/exdb/publis/pdf/lecun-01a.pdf
How to create such a filter; Well in python like https://github.com/alexcpn/cnn_in_python/blob/main/main.py#L19-L27
Which will give you a list of numpy arrays and length of the list is the depth
Example in the code above,but adding a depth of 3 for color (RGB), the below is the network. The first Convolutional layer is a filter of shape (5,5,3) and depth 6
Input (R,G,B)= [32.32.3] *(5.5.3)*6 == [28.28.6] * (5.5.6)*1 = [24.24.1] * (5.5.1)*16 = [20.20.16] *
FC layer 1 (20, 120, 16) * FC layer 2 (120, 1) * FC layer 3 (20, 10) * Softmax (10,) =(10,1) = Output
In Pytorch
np.set_printoptions(formatter={'float': lambda x: "{0:0.2f}".format(x)})
# Generate a random image
image_size = 32
image_depth = 3
image = np.random.rand(image_size, image_size)
# to mimic RGB channel
image = np.stack([image,image,image], axis=image_depth-1) # 0 to 2
image = np.moveaxis(image, [2, 0], [0, 2])
print("Image Shape=",image.shape)
input_tensor = torch.from_numpy(image)
m = nn.Conv2d(in_channels=3,out_channels=6,kernel_size=5,stride=1)
output = m(input_tensor.float())
print("Output Shape=",output.shape)
Image Shape= (3, 32, 32)
Output Shape= torch.Size([6, 28, 28])
According to "Lowe, David G. "Distinctive image features from scale-invariant keypoints." International journal of
computer vision 60.2 (2004): 91-110 "
"It is important to avoid all boundary affects in which the descriptor
abruptly changes as a sample shifts smoothly from being within one
histogram to another or from one orientation to another. Therefore,
trilinear interpolation is used to distribute the value of each
gradient sample into adjacent histogram bins. In other words, each
entry into a bin is multiplied by a weight of 1−d for each dimension,
where d is the distance of the sample from the central value of the
bin as measured in units of the histogram bin spacing."
I am calculating the orientation[t] and location of gradient(x,y) which will be in floating point. Currently, I was just
providing the gradient magnitude to 3d histogram values[t][x][y] ( means the lower bound of floating point values of t,x
and y). But, according to paper, I have to distribute the gradient magnitude to adjacent bins. I am not sure about how
to distribute it.
I got my answer on following link:
HOG Trilinear Interpolation of Histogram Bins
I have one problem in the second step which is to accumulate weighted votes for gradient orientation over spatial cells.
Assuming the cell is 8*8. Let me use two matrix GO[8][8]([1 9]), GM[8][8] to represent the gradient orientation and gradient magnitude respectively.
The gradient orientation ranges from 0 - 180 and there are 9 orientation bins.
According to my understanding of HOG, for every pixel in a cell, adding its gradient magnitude to its corresponding orientation bin. In this way, we can have the histogram for every cell.
But there is one sentence thats confusing me.
"To reduce aliasing, votes(gradient magnitude) are interpolated
trilinearly between the neighbouring bin centers in both orientation
and position."1
Why interpolated? How to interpolate? Can someone explains more detailed? No reducing aliasing.
Thanks in advance.
1 This sentence is in Navneet Dalal's PHD thesis, p38, line 4.
Interpolation is a standard technique for computing histograms. The idea here is that each value is not simply placed into one bin, but is distributed between two neighboring bins (assuming a 1d histogram), based on how far away it is from the center of the original bin.
The purpose of this is to deal with situations when a small error in your measurement can cause a value to be placed into a different bin. This is a very good thing to do for any type of histogram, not just for HOGs, assuming you have the CPU cycles.
There is also bi-linear and tri-linear interpolation for 2d and 3d histograms, where each value is distributed between 4 and 8 neighboring bins respectively.