Well am referring the following paper and trying to implement the algorithm as given in matlab
The only problem is how do i find a noisy pixel i.e Pixel with impulse noise?
X seems to be the impulse pixel in an image which i have to calculate
_
____________________________________________
Input – Noisy Image h
_______________________________________________
Step 1: Compute X
for every pixel repeat steps from 2 to 7
Step 2: Initialize w = 3
Step 3: If X(i,j) ≠ Impulse pixel
goto step 7
Step 4: ∆i,j = { h(i1,j1) | i-(w-1)/2 ≤ i1 ≤ i+(w-1)/2,
j-(w-1)/2 ≤ j1 ≤ j+(w-1)/2}
b=no. of black pixels in the window
w=no. of white pixels in the window
Step 5: If ∆i,j ≠ NULL
p(i,j) = mean(∆i,j
)
d(i,j) = | h(i,j) – p(i,j) |
else if (w < wmax)
w=w+2
goto step 4
else
if (b>w)
h(i,j)=0
else
h(i,j)=255
Step 7: Goto next pixel
Step 8: Calculate threshold t, from detailed coefficient
matrix d
for every pixel
Step 9: If (d(i,j)>t)
h(i,j)=p(i,j)
____________________________
Edit: To implement the PSM or the median filter method we
need to set some parameters and a threshold value. This
threshold value is dependent on the image and the noise
density. So, to restore different images we need to check for
a range of threshold values and find out the best one. So, in
our proposed algorithm we removed the need to define a threshold value. The algorithm is intelligent and determines
the threshold automatically.
The article you are trying to implement is obviously badly written...
For instance in the algorithm w means 2 things: the size of the window, and the number of white pixels!!!
Both the step 1 and 7, are refering to the same loop.
Anyway, to me, the "impulse pixels" are all the pixels a which are either equal to 0 or 255.
Basically, they are the pixels which are part of the "salt and pepper" noise.
So basically, you can find them by doing:
[impulsepixelsY,impulasPixelX]=find((im==0)|(im==255));
From the paper it seems that the "impulse pixels" are just the noisy pixels, in the case of salt & pepper noise. Furthermore, it also seems that the algorithm provides an "intelligent" mechanism to calculate the denoised value of a noisy pixel if its value is above a threshold (which it calculates adaptively).
So, what about "If X(i,j) ≠ Impulse pixel " ? Well, apparently, the authors assume to know (!) which pixels are noisy (!!), which makes the whole thing rather ridiculous, since this info is almost impossible to know.
I might also add that the rather stunning results presented in the paper are most probably due to this fact.
P.S. Regarding the argument that <"impulse pixels" are all the pixels a which are either equal to 0 or 255>, it is wrong. The set of pixels that have either 0 or 255 intensity value, includes the noisy pixels as well as proper pixels that just happen to have such a value. In this case, the algorithm will most probably collapse since it will denoise healthy pixels as well.
Related
I have two sparse matrices "Matrix1" and "Matrix2" of the same size p x n.
By sparse matrix I mean that it contains a lot of exactly zero elements.
I want to show the two matrices under the same colormap and a unique colorbar. Doing this in MATLAB is straightforward:
bottom = min(min(min(Matrix1)),min(min(Matrix2)));
top = max(max(max(Matrix1)),max(max(Matrix2)));
subplot(1,2,1)
imagesc(Matrix1)
colormap(gray)
caxis manual
caxis([bottom top]);
subplot(1,2,2)
imagesc(Matrix2)
colormap(gray)
caxis manual
caxis([bottom top]);
colorbar;
My problem:
In fact, when I show the matrix using imagesc(Matrix), it can ignore the noises (or backgrounds) that always appear with using imagesc(10*log10(Matrix)).
That is why, I want to show the 10*log10 of the matrices. But in this case, the minimum value will be -Inf since the matrices are sparse. In this case caxis will give an error because bottom is equal to -Inf.
What do you suggest me? How can I modify the above code?
Any help will be very appreciated!
A very important point is that the minimum value in your matrix will always be 0. Leveraging this, a very simple way to address your problem is to add 1 inside the log operation so that values that map to 0 in the original matrix also map to 0 in the log operation. This avoids the -Inf error that you're encountering. In fact, this is a very common way of visualizing the Fourier Transform if you will. Adding 1 to the logarithm ensures that the transform has no negative values in the output, yet the derivative or its rate of change remains intact as the effect is simply a translation of the curve by 1 unit to the left.
Therefore, simply do imagesc(10*log10(1 + Matrix));, then the minimum is always bounded at 0 while the maximum is unbounded but subject to the largest value that is seen in Matrix.
it is about a non-parameteric density estimation.
So, we have 2 different data 220 values of "good data" and 30 values of "broke data"
we should use k-nearest-neighbour density estimation for estimate p(x |c="good data")
In case k=5 we have p(x |c=good) =(5/220)*(1/V).
If I have right understood, by k-nearest-neighbour we should determine V and then get
p(x |c=good)
If we must find out V for 5 points, then we can solve p(x|c=good)
I have a problem, how to plot und calcucale this probability.
There is picture from book http://content.foto.mail.ru/mail/zurix/_mypagephoto/h-67.jpg
What does blue curve mean on graphic of K nearest-neighbor density estimation(you can see attach)? Can this curve show boundaries of different V? If yes, where exactly boundary between is classes, each class consists of 5 points???
Thank you in advance!!
it's difficult to guess what the two curves mean without any additional information such as the figure caption or the book title.
My best guess is that the green curve is the true (one dimensional) density from which a sample of data points was drawn. The blue curves seem to be the resulting density estimation function for three different values of k .
This should illustrate the importance of choosing k properly, for k = 1, this overfits the data (high variance of the resulting density estimation function), for k = 30, this 'oversmoothes' the data (high bias of the resulting density estimation function) as it does not reproduce the bump around 0.3 .
In fact, looking at the k=1 example, it looks to me like this is not using a pure 1 / V but rather some weighting function. For a pure 1/V estimate per point, I would expect a piecewise constant function (only pieces of horizontal lines).
I read now tons of different explanations of the gaussian blur and I am really confused.
I roughly understand how the gaussian blur works.
http://en.wikipedia.org/wiki/Gaussian_blur
I understood that we choose 3*sigma as the maxium size for our mask because the values will get really small.
But my three questions are:
How do I create a gaussian mask with the sigma only?
If I understood it correctly, the mask gives me the weights, then
I place the mask on the top left pixel. I multiply the weights for
each value of the pixels in the mask. Then I move the mask to the
next pixel. I do this for all pixels. Is this correct?
I also know that 1D masks are faster. So I create a mask for x and a
mask for y. Lets say my mask would look like this. (3x3)
1 2 1
2 4 2
1 2 1
How would my x and y mask look like?
1- A solution to create a gaussian mask is to setup an N by N matrix, with N=3*sigma (or less if you want a coarser solution), and fill each entry (i,j) with exp(-((i-N/2)^2 + (j-N/2)^2)/(2*sigma^2)). As a comment mentioned, taking N=3*sigma just means that you truncate your gaussian at a "sufficiently small" threshold.
2- yes - you understood correctly. A small detail is that you'll need to normalize by the sum of your weights (ie., divide the result of what you said by the sum of all the elements of your matrix). The other option is that you can build your matrix already normalized, so that you don't need to perform this normalization at the end (the normalized gaussian formula becomes exp(-((i-N/2)^2 + (j-N/2)^2)/(2*sigma^2))/(2*pi*sigma))
3- In your specific case, the 1D version is [1 2 1] (ie, both your x and y masks) since you can obtain the matrix you gave with the multiplication transpose([1 2 1]) * [1 2 1]. In general, you can directly build these 1D gaussians using the 1D gaussian formula which is similar as the one above : exp(-((i-N/2)^2)/(2*sigma^2)) (or the normalized version exp(-((i-N/2)^2)/(2*sigma^2)) / (sigma*sqrt(2*pi)))
I'm writing a low level image processing algorithm which needs to do alot of 8-connectivity checks for pixels. For every pixel I often need to check the pixels above it, below it and on its sides and diagonals.
On the edges of the image there are special cases where there are only 5 or 3 neighbors instead of 8 neighbors for a pixels. The naive way to do it is for every access to check if the coordinates are in the right range and if not, return some default value.
I'm looking for a way to avoid all these checks since they introduce a large overhead to the algorithm. Are there any tricks to avoid it altogether?
For performance-critical code you might do something like the following:
process row 0
for r = 1 to M - 2
process row r, pixel 0
for c = 1 to N - 2
process row r, pixel c
process row r, pixel N - 1
process row M - 1
The bulk of the operations are in the inner loop and are branchless. You just handle the first/last row, and the first/last pixel of each row, as special cases. It makes the code bulkier, but that's the nature of the beast when it comes to optimisation.
If there are suitable values for your algorithm, add a 1-pixel border and only check your original pixel data.
Or, to turn #Roger's suggestion around, ignore the outermost pixel along each border, if this works for you. In other words, only process the interior pixels of your image.
I have a simple program, at it's heart is a two dimensional array of floats, supposedly representing gas concentrations, I have been trying to come up with a simple algorithm that will model the gas expanding outwards, like a cloud, eventually ending up with the same concentration of the gas everywhere across the grid.
For example a given state progression could be:
(using ints for simplicity)
starting state
00000
00000
00900
00000
00000
state after 1 pass of algorithm
00000
01110
01110
01110
00000
one more pas should give a 5x5 grid all containing the value 0.36 (9/25).
I've tried it out on paper but no matter how I try, I cant get my head around the algorithm to do this.
So my question is, how should I set about trying to code this algorithm? I've tried a few things, applying a convolution, trying to take each grid cell in turn and distributing it to its neighbours, but they all end up having undesirable effects, such as ending up eventually with less gas than I originally started with, or all of gas movement being in one direction instead of expanding outwards from the centre. I really can't get my head around it at all and would appreciate any help at all.
It's either a diffusion problem if you ignore convection or a fluid dynamics/mass transfer problem if you don't. You would start with equations for conservation of mass and momentum for an Eulerian (fixed control volume) viewpoint if you were solving from scratch.
It's a transient problem, so you need to perform an integration to advance the state from time t(n) to t(n+1). You show a grid, but nothing about how you're solving in time. What integration scheme have you tried? Explicit? Implicit? Crank-Nicholson? If you don't know, you're not approaching the problem correctly.
One book that I really liked on this subject was S.W. Patankar's "Numerical Heat Transfer and Fluid Flow". It's a little dated now, but I liked the treatment. It's still good after 29 years, but there might be better texts since I was reading on the subject. I think it's approachable for somebody looking into it for the first time.
In the example you give, your second stage has a core of 1's. Usually diffusion requires a concentration gradient, so most diffusion related techniques won't change the 1 in the middle on the next iteration (nor would they have got to that state after the first one, but it's a bit easier to see once you've got a block of equal values). But as the commenters on your post say, that's not likely to be the cause of a net movement. Reducing the gas may be edge effects, but can also be a question of rounding errors - set the cpu to round half even, and total the gas and apply a correction now and again.
It looks like you're trying to implement a finite difference solver for the heat equation with Neumann boundary conditions (insulation at the edges). There's a lot of literature on this kind of thing. The Wikipedia page on finite difference method describes a simple but stable method, but for Dirichlet boundary conditions (constant density at edges). Modifying the handling of the boundary conditions shouldn't be too difficult.
It looks like what you want is something like a smoothing algorithm, often used in programs like Photoshop, or old school demo effects, like this simple Flame Effect.
Whatever algorithm you use, it will probably help you to double buffer your array.
A typical smoothing effect will be something like:
begin loop forever
For every x and y
{
b2[x,y] = (b1[x,y] + (b1[x+1,y]+b1[x-1,y]+b1[x,y+1]+b1[x,y-1])/8) / 2
}
swap b1 and b2
end loop forever
See Tom Forsyth's Game Programming Gems article. Looks like it fulfils your requirements, but if not then it should at least give you some ideas.
Here's a solution in 1D for simplicity:
The initial setup is with a concentration of 9 at the origin (), and 0 at all other positive and negative coordinates.
initial state:
0 0 0 0 (9) 0 0 0 0
The algorithm to find next iteration values is to start at the origin and average current concentrations with adjacent neighbors. The origin value is a boundary case and the average is done considering the origin value, and its two neighbors simultaneously, i.e. average among 3 values. All other values are effectively averaged among 2 values.
after iteration 1:
0 0 0 3 (3) 3 0 0 0
after iteration 2:
0 0 1.5 1.5 (3) 1.5 1.5 0 0
after iteration 3:
0 .75 .75 2 (2) 2 .75 .75 0
after iteration 4:
.375 .375 1.375 1.375 (2) 1.375 1.375 .375 .375
You do these iterations in a loop. Outputting the state every n number of iterations. You may introduce a time constant to control how many iterations represent one second of clock-on-the-wall time. This is also a function of what length units the integer coordinates represent. For a given H/W system, you can tune this value empirically. You may also introduce a steady state tolerance value to control when the program says " all neighbor values are within this tolerance" or "no value changed between iterations by more than this tolerance" and so the algorithm has reached a steady state solution.
The concentration for each iteration given a starting concentration can be obtained by the equation:
concentration = startingConcentration/(2*iter + 1)**2
iter is the time iteration. So for your example.
startingConcentration = 9
iter = 0
concentration = 9/(2*0 + 1)**2 = 9
iter = 1
concentration = 9/(2*1 + 1)**2 = 1
iter = 2
concentration = 9/(2*2 + 1)**2 = 9/25 = .35
you can set the value of the array after each "time step"