so by google translating i figured out
Вход means input
Слой means layer.
Свертка means convolution (this
must be the number of filters?)
Шаг means step (this must be stride?)
субдискр means subdiskr (i guess this is pooling?)
Now my question is how would a
size 22x256 image result in 6x256 with a 5 filters?
The filter size (kernel) that i found out results in 6x256 is [17,1] with 1 filter. From layer 1 to layer a kernel size of [1,8] and stride [1,8] is what i found to work. This just does not look like anything on this graph though.
In the paper they wrote this about the layer between 1 and 2
"The second layer allows to reduce the dimensionality of the signal in time, producing a weighted average of the signal over 16 values"
Heres a clear explanation how the sizes of the inputs vary with proceeding among the layers.
In the input the dimensions that you are giving are 28 wide and 28 height and depth as 1. For filters in layer1 the depth dimension of filter must be equal to the depth of the input. so the dimension of the filter will be 5x5x1, applying one filter the dimension is reduced (due to strides)to produce 14x14x1 dimension activation map, so applying 32 such filters will give you 32 activations maps. Combining all of these 14x14x32 is output of the layer 1 and input to your second layer. Again in second layer you need to apply a filter of dimension 5(width)x5(height)x32(depth) on the layer to produce one activation map of 14x14x1 , stacking all the 64 activation maps give you output dimension of the second layer as 14x14x64 and so on.
In the figure that you posted looks very different in representation. Check the standard ones in your language.
I asked the authors:
They told me that they used 1 dimensional CNN.
This Means that the first number is the depth and the second number is the Width:
depth # width.
Related
I just wanted to confirm whether all the images in a mini-batch get the same dropout mask or not.
For clarification: Suppose a 1000 sized vector is passed through the Dropout layer with a mini batch size of 100.
Now, the 21st and 31st elements of the 1st image's vector gets dropped out.
Is it necessary that the 21st and 31st elements will get dropped for all the remaining 99 images in the batch? Or is it the case that every image gets a separate mask?
No, each image in a batch gets an independent, completely random mask.
Actually, the Dropout layer doesn't even care about the shape of its input: It counts the number of elements in the bottom Blob (i.e. for a batch size of 10 images, 3 channels and 224x224, it would be 10 * 3 * 224 * 224 = 1505280), and generates the same number of independent random numbers.
I'm trying to implement the proposed model in a CVPR paper (Deep Interactive Object Selection) in which the data set contains 5 channels for each input sample:
1.Red
2.Blue
3.Green
4.Euclidean distance map associated to positive clicks
5.Euclidean distance map associated to negative clicks (as follows):
To do so, I should fine tune the FCN-32s network using "object binary masks" as labels:
As you see, in the first conv layer I have 2 extra channels, so I did net surgery to use pretrained parameters for the first 3 channels and Xavier initialization for 2 extras.
For the rest of the FCN architecture, I have these questions:
Should I freeze all the layers before "fc6" (except the first conv layer)? If yes, how the extra channels of the first conv will be learned? Are the gradients strong enough to reach the first conv layer during training process?
What should be the kernel size of the "fc6"? should I keep 7? I saw in "Caffe net_surgery" notebook that it depends on the output size of the last layer ("pool5").
The main problem is the number of outputs of the "score_fr" and "upscore" layers, since I'm not doing class segmentation (to use 21 for 20 classes and the background), how should I change it? What about 2? (one for object and the other for the non-object (background) area)?
Should I change "crop" layer "offset" to 32 to have center crops?
In case of changing each of these layers, what is the best initialization strategy for them? "bilinear" for "upscore" and "Xavier" for the rest?
Should I convert my binary label matrix values into zero-centered ( {-0.5,0.5} ) status, or it is OK to use them with the values in {0,1} ?
Any useful idea will be appreciated.
PS:
I'm using Euclidean loss, while I'm using "1" as the number of outputs for "score_fr" and "upscore" layers. If I use 2 for that, I guess it should be softmax.
I can answer some of your questions.
The gradients will reach the first layer so it should be possible to learn the weights even if you freeze the other layers.
Change the num_output to 2 and finetune. You should get a good output.
I think you'll need to experiment with each of the options and see how the accuracy is.
You can use the values 0,1.
I'm looking for a bit of guidance on using CONVN to calculate moving averages in one dimension on a 3d matrix. I'm getting a little caught up on the flipping of the kernel under the hood and am hoping someone might be able to clarify the behaviour for me.
A similar post that still has me a bit confused is here:
CONVN example about flipping
The Problem:
I have daily river and weather flow data for a watershed at different source locations.
So the matrix is as so,
dim 1 (the rows) represent each site
dim 2 (the columns) represent the date
dim 3 (the pages) represent the different type of measurement (river height, flow, rainfall, etc.)
The goal is to try and use CONVN to take a 21 day moving average at each site, for each observation point for each variable.
As I understand it, I should just be able to use a a kernel such as:
ker = ones(1,21) ./ 21.;
mat = randn(150,365*10,4);
avgmat = convn(mat,ker,'valid');
I tried playing around and created another kernel which should also work (I think) and set ker2 as:
ker2 = [zeros(1,21); ker; zeros(1,21)];
avgmat2 = convn(mat,ker2,'valid');
The question:
The results don't quite match and I'm wondering if I have the dimensions incorrect here for the kernel. Any guidance is greatly appreciated.
Judging from the context of your question, you have a 3D matrix and you want to find the moving average of each row independently over all 3D slices. The code above should work (the first case). However, the valid flag returns a matrix whose size is valid in terms of the boundaries of the convolution. Take a look at the first point of the post that you linked to for more details.
Specifically, the first 21 entries for each row will be missing due to the valid flag. It's only when you get to the 22nd entry of each row does the convolution kernel become completely contained inside a row of the matrix and it's from that point where you get valid results (no pun intended). If you'd like to see these entries at the boundaries, then you'll need to use the 'same' flag if you want to maintain the same size matrix as the input or the 'full' flag (which is default) which gives you the size of the output starting from the most extreme outer edges, but bear in mind that the moving average will be done with a bunch of zeroes and so the first 21 entries wouldn't be what you expect anyway.
However, if I'm interpreting what you are asking, then the valid flag is what you want, but bear in mind that you will have 21 entries missing to accommodate for the edge cases. All in all, your code should work, but be careful on how you interpret the results.
BTW, you have a symmetric kernel, and so flipping should have no effect on the convolution output. What you have specified is a standard moving averaging kernel, and so convolution should work in finding the moving average as you expect.
Good luck!
How to use Morton Order in range search?
From the wiki, In the paragraph "Use with one-dimensional data structures for range searching",
it says
"the range being queried (x = 2, ..., 3, y = 2, ..., 6) is indicated
by the dotted rectangle. Its highest Z-value (MAX) is 45. In this
example, the value F = 19 is encountered when searching a data
structure in increasing Z-value direction. ......BIGMIN (36 in the
example).....only search in the interval between BIGMIN and MAX...."
My questions are:
1) why the F is 19? Why the F should not be 16?
2) How to get the BIGMIN?
3) Are there any web blogs demonstrate how to do the range search?
EDIT: The AWS Database Blog now has a detailed introduction to this subject.
This blog post does a reasonable job of illustrating the process.
When searching the rectangular space x=[2,3], y=[2,6]:
The minimum Z Value (12) is found by interleaving the bits of the lowest x and y values: 2 and 2, respectively.
The maximum Z value (45) is found by interleaving the bits of the highest x and y values: 3 and 6, respectively.
Having found the min and max Z values (12 and 45), we now have a linear range that we can iterate across that is guaranteed to contain all of the entries inside of our rectangular space. The data within the linear range is going to be a superset of the data we actually care about: the data in the rectangular space. If we simply iterate across the entire range, we are going to find all of the data we care about and then some. You can test each value you visit to see if it's relevant or not.
An obvious optimization is to try to minimize the amount of superfluous data that you must traverse. This is largely a function of the number of 'seams' that you cross in the data -- places where the 'Z' curve has to make large jumps to continue its path (e.g. from Z value 31 to 32 below).
This can be mitigated by employing the BIGMIN and LITMAX functions to identify these seams and navigate back to the rectangle. To minimize the amount of irrelevant data we evaluate, we can:
Keep a count of the number of consecutive pieces of junk data we've visited.
Decide on a maximum allowable value (maxConsecutiveJunkData) for this count. The blog post linked at the top uses 3 for this value.
If we encounter maxConsecutiveJunkData pieces of irrelevant data in a row, we initiate BIGMIN and LITMAX. Importantly, at the point at which we've decided to use them, we're now somewhere within our linear search space (Z values 12 to 45) but outside the rectangular search space. In the Wikipedia article, they appear to have chosen a maxConsecutiveJunkData value of 4; they started at Z=12 and walked until they were 4 values outside of the rectangle (beyond 15) before deciding that it was now time to use BIGMIN. Because maxConsecutiveJunkData is left to your tastes, BIGMIN can be used on any value in the linear range (Z values 12 to 45). Somewhat confusingly, the article only shows the area from 19 on as crosshatched because that is the subrange of the search that will be optimized out when we use BIGMIN with a maxConsecutiveJunkData of 4.
When we realize that we've wandered outside of the rectangle too far, we can conclude that the rectangle in non-contiguous. BIGMIN and LITMAX are used to identify the nature of the split. BIGMIN is designed to, given any value in the linear search space (e.g. 19), find the next smallest value that will be back inside the half of the split rectangle with larger Z values (i.e. jumping us from 19 to 36). LITMAX is similar, helping us to find the largest value that will be inside the half of the split rectangle with smaller Z values. The implementations of BIGMIN and LITMAX are explained in depth in the zdivide function explanation in the linked blog post.
It appears that the quoted example in the Wikipedia article has not been edited to clarify the context and assumptions. The approach used in that example is applicable to linear data structures that only allow sequential (forward and backward) seeking; that is, it is assumed that one cannot randomly seek to a storage cell in constant time using its morton index alone.
With that constraint, one's strategy begins with a full range that is the mininum morton index (16) and the maximum morton index (45). To make optimizations, one tries to find and eliminate large swaths of subranges that are outside the query rectangle. The hatched area in the diagram refers to what would have been accessed (sequentially) if such optimization (eliminating subranges) had not been applied.
After discussing the main optimization strategy for linear sequential data structures, it goes on to talk about other data structures with better seeking capability.
I am simulating a digital filter, which is 4-stage.
Stages are:
CIC
half-band
OSR
128
Input is 4 bits and output is 24 bits. I am confused about the 24 bits output.
I use MATLAB to generate a 4 bits signed sinosoid input (using SD tool), and simulated with modelsim. So the output should be also a sinosoid. The issue is the output only contains 4 different data.
For 24 bits output, shouldn't we get a 2^24-1 different data?
What's the reason for this? Is it due to internal bit width?
I'm not familiar with Modelsim, and I don't understand the filter terminology you used, but...Are your filters linear systems? If so, an input at a given frequency will cause an output at the same frequency, though possibly different amplitude and phase. If your input signal is a single tone, sampled such that there are four values per cycle, the output will still have four values per cycle. Unless one of the stages performs sample rate conversion the system is behaving as expected. As as Donnie DeBoer pointed out, the word width of the calculation doesn't matter as long as it can represent the four values of the input.
Again, I am not familiar with the particulars of your system so if one of the stages does indeed perform sample rate conversion, this doesn't apply.
Forgive my lack of filter knowledge, but does one of the filter stages interpolate between the input values? If not, then you're only going to get a maximum of 2^4 output values (based on the input resolution), regardless of your output resolution. Just because you output to 24-bit doesn't mean you're going to have 2^24 values... imagine running a digital square wave into a D->A converter. You have all the output resolution in the world, but you still only have 2 values.
Its actually pretty simple:
Even though you have 4 bits of input, your filter coefficients may be more than 4 bits.
Every math stage you do adds bits. If you add two 4-bit values, the answer is a 5 bit number, so that adding 0xf and 0xf doesn't overflow. When you multiply two 4-bit values, you actually need 8 bits of output to hold the answer without the possibility of overflow. By the time all the math is done, your 4-bit input apparently needs 24-bits to hold the maximum possible output.