I am trying to implement Hough transform in Matlab to find circles in a picture.
In the accumulator matrix, the global maximum is 105 at A(32,31,24). So I'm able to get this: max circle
The problem is, how can i find the local maxima to find the rest of the circles?
I wrote this to find A(i,j,k) which is bigger than the 26 adjacent points (26-Connected voxel neighborhood):
[i j k]=find(A~=0) ;
f=0;
for s=1:size(i)
if(i(s)~=100&&j(s)~=100&&k(s)~=141&&i(s)~=1&&j(s)~=1&&k(s)~=1)
if (A(i(s),j(s),k(s))>A(i(s)-1,j(s),k(s))&&A(i(s),j(s),k(s))>A(i(s)+1,j(s),k(s))&&A(i(s),j(s),k(s))>A(i(s),j(s)-1,k(s))&&A(i(s),j(s),k(s))>A(i(s),j(s)+1,k(s))&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)-1,k(s))&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)+1,k(s))&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)+1,k(s))&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)-1,k(s))&&A(i(s),j(s),k(s))>A(i(s),j(s),k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s),k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s),k(s)+1)&&A(i(s),j(s),k(s))>A(i(s),j(s)-1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s),j(s)+1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)-1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)+1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)+1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)-1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s),j(s),k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s),k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s),k(s)-1)&&A(i(s),j(s),k(s))>A(i(s),j(s)-1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s),j(s)+1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)-1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)+1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)+1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)-1,k(s)-1))
f=f+1
i(s)
j(s)
k(s)
end
end
end
Why i never got these correct i,j,k and f is always 0? I think i should at least find (32,31,24) and f=1?
Can anyone help me ?
Thank you so much!
The complete code is here:
im=imread('C:\Users\dell\Desktop\tp-complet\Hough\four.png');
sigma = 0.3;
gausFilter = fspecial('gaussian',[5 5],sigma);
sobelFilter=fspecial('sobel');
img=imfilter(im,gausFilter,'replicate');
ims=edge(img,'sobel');
rmax=size(im,1);
cmax=size(im,2);
radmax=round(sqrt(rmax^2+cmax^2));
for i=1:rmax
for j=1:cmax
for k=1:radmax
A(i,j,k)=0;
end
end
end
[r c]=find(ims==1);
length=size(r);
for k=1:length
for l=1:rmax
for m=1:cmax
if((l~=r(k))&&(m~=c(k)))
x=sqrt((l-r(k))^2+(m-c(k))^2);
x=round(x);
A(l,m,x)=A(l,m,x)+1;
end
end
end
end
[i j k]=find(A~=0) ;
f=0;
for s=1:size(i)
if(i(s)~=100&&j(s)~=100&&k(s)~=141&&i(s)~=1&&j(s)~=1&&k(s)~=1)
if (A(i(s),j(s),k(s))>A(i(s)-1,j(s),k(s))&&A(i(s),j(s),k(s))>A(i(s)+1,j(s),k(s))&&A(i(s),j(s),k(s))>A(i(s),j(s)-1,k(s))&&A(i(s),j(s),k(s))>A(i(s),j(s)+1,k(s))&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)-1,k(s))&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)+1,k(s))&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)+1,k(s))&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)-1,k(s))&&A(i(s),j(s),k(s))>A(i(s),j(s),k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s),k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s),k(s)+1)&&A(i(s),j(s),k(s))>A(i(s),j(s)-1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s),j(s)+1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)-1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)+1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)+1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)-1,k(s)+1)&&A(i(s),j(s),k(s))>A(i(s),j(s),k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s),k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s),k(s)-1)&&A(i(s),j(s),k(s))>A(i(s),j(s)-1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s),j(s)+1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)-1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)-1,j(s)+1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)+1,k(s)-1)&&A(i(s),j(s),k(s))>A(i(s)+1,j(s)-1,k(s)-1))
f=f+1
i(s)
j(s)
k(s)
end
end
end
I don't have the image processing toolbox, so cannot readily run your code. However, a quick review indicates to me that you're using the find function incorrectly. When provided with 3 outputs (as you have it), find doesn't return the 3D indices, but rather row, column, and a vector of the values. That means that k is just a giant vector of 1's, and so your for loop's if k(s)~=1 statement is never satisfied. You should do something like [i,j,k]=ind2sub(size(A),find(A~=0)) if you instead want the 3D indices.
Also, FYI, this:
for i=1:rmax
for j=1:cmax
for k=1:radmax
A(i,j,k)=0;
end
end
end
Can be replaced with A=zeros(rmax,cmax,radmax);.
I have recently been tasked with using a first derivative filter on an image of myself. The instructor said that I should first fix the value of y and preform f(x+1) - f(x) on the rows and then fix the new "X" values and preform f(y+1)-f(y) on the columns.
Note: I have been asked to do this task manually, not using filter2() or any other programmed function, so please do not suggest that I use filter2() or similar. Thanks!
I tried calling up all the pixels and subtracting each successive one by doing
fid = fopen('image.raw')
myimage = fread(fid,[512 683], '*int8')
fclose(fid)
imsz = size(myimage)
x = imsz(1)
for I = 1:512
for J = 1:683
X(I) - X(I-1) = XX
But it doesnt seem to work, and I dont quite understand why. If you could help me, or point me in the right direction, I would be very appreciative.
First of all, your code is syntatically incorrect:
There is no end statement to any of your loops, and besides, you don't even need loops here.
You seem to read your image into the variable myimage, but you're using an undefined variable X when attempting to calculate the derivative.
The order of your assignment statements is reversed. The variable you wish to assign to should be written in the left hand part of the assignement.
I strongly suggest that you read online tutorials and get yourself familiar with MATLAB basics before taking on more complicated tasks.
As to your specific problem:
MATLAB encourages vectorized operations, i.e operations on entire arrays (vectors or matrices) at once. To subtract adjacent values in an array, what you're basically doing is subtracting two arrays, shifted by one element with respect to each other. For one dimensional arrays, that would translate in MATLAB to:
a(2:end) - a(1:end-1)
where a is your array. The end keyword specifies the last index in the array.
To compute the derivative of an image (a 2-D matrix), you need to decide along which axis you want to perform that operation. To approximate the derivate along the y-axis, do this:
X(2:end, :) - X(1:end-1, :)
You can verify that this gives you the same result as diff(X, 1) (or simply diff(X)). To compute the approximate derivative along the x-axis, which is equivalent to diff(X, 2), do this:
X(:, 2:end) - X(:, 1:end-1)
The colon (:) is the same as writing 1:end as the array subscript for the corresponding dimension.
If your filtered image is div then
for Y = 1:682
for X = 1:511
div(X, Y) = myimage(X + 1, Y + 1) - myimage(X,Y);
end
end
Remember the last row and the last column are not filtered!