Convolution Kernel using a user defined function. How to deal with negative pixel values? - matlab

Question

I've declared a function that will be used to calculate the convolution of an image using an arbitrary 3x3 kernel. I also created a script that will prompt the user to select both an image as well as enter the convolution kernel of their choice. However, I do not know how to go about dealing with negative pixel values that will arise for various kernels. How would I implement a condition into my script that will deal with these negative values?
This is my function:
function y = convul(x,m,H,W)
y=zeros(H,W);
for i=2:(H-1)
for j=2:(W-1)
Z1=(x(i-1,j-1))*(m(1,1));
Z2=(x(i-1,j))*(m(1,2));
Z3=(x(i-1,j+1))*(m(1,3));
Z4=(x(i,j-1))*(m(2,1));
Z5=(x(i,j))*(m(2,2));
Z6=(x(i,j+1))*(m(2,3));
Z7=(x(i+1,j-1))*(m(3,1));
Z8=(x(i+1,j))*(m(3,2));
Z9=(x(i+1,j+1))*(m(3,3));
y(i,j)=Z1+Z2+Z3+Z4+Z5+Z6+Z7+Z8+Z9;
end
end
And this is the script that I've written that prompts the user to enter an image and select a kernel of their choice:
[file,path]=uigetfile('*.bmp');
x = imread(fullfile(path,file));
x_info=imfinfo(fullfile(path,file));
W=x_info.Width;
H=x_info.Height;
L=x_info.NumColormapEntries;
prompt='Enter a convulation kernel m: ';
m=input(prompt)/9;
y=convul(x,m,H,W);
imshow(y,[0,(L-1)]);
I've tried to use the absolute value of the convolution, as well as attempting to locate negatives in the output image, but nothing worked.
This is the original image:
This is the image I get when I use the kernel [-1 -1 -1;-1 9 -1; -1 -1 -1]:
I don't know what I'm doing wrong.

Answer 1

MATLAB is rather unique in how it handles operations between different data types. If x is uint8 (as it likely is in this case), and m is double (as it likely is in this case), then this operation:
Z1=(x(i-1,j-1))*(m(1,1));
returns a uint8 value, not a double. Arithmetic in MATLAB always takes the type of the non-double argument. (And you cannot do arithmetic between two different types unless one of them is double.)
MATLAB does integer arithmetic with saturation. That means that uint8(5) * -1 gives 0, not -5, because uint8 cannot represent a negative value.
So all your Z1..Z9 are uint8 values, negative results have been set to 0. Now you add all of these, again with saturation, leading to a value of at most 255. This value is assigned to the output (a double). So it looks like you are doing your computations correctly and outputting a double array, but you are still clamping your result in an odd way.
A Correct implementation would cast each of the values of x to double before multiplying by a potentially negative number. For example:
for i = 2:H-1
for j = 2:W-1
s = 0;
s = s + double(x(i-1,j-1))*m(1,1);
s = s + double(x(i-1,j))*m(1,2);
s = s + double(x(i-1,j+1))*m(1,3);
s = s + double(x(i,j-1))*m(2,1);
s = s + double(x(i,j))*m(2,2);
s = s + double(x(i,j+1))*m(2,3);
s = s + double(x(i+1,j-1))*m(3,1);
s = s + double(x(i+1,j))*m(3,2);
s = s + double(x(i+1,j+1))*m(3,3);
y(i,j) = s;
end
end
(Note that I removed your use of 9 different variables, I think this is cleaner, and I also removed a lot of your unnecessary brackets!)
A simpler implementation would be:
for i = 2:H-1
for j = 2:W-1
s = double(x(i-1:i+1,j-1:j+1)) .* m;
y(i,j) = sum(s(:));
end
end