## Please explain difference in image variance algorithms ... E-10 for all tests

### Eric Diaz (view profile)

on 2 May 2014
Latest activity Commented on by Eric Diaz

### Eric Diaz (view profile)

on 3 May 2014
Can someone please explain why these different algorithms are giving slightly different results for the variance of an image?
%References:
% Load images and get basic parameters
% You may need to load your own image or one of Matlab's demo images, like cameraman
nrows = size(img,1);
ncols = size(img,2);
nzs = size(img,3);
h = ones(7);
n = sum(h(:));
n1 = n - 1;
%Preinitialize
M2a = zeros(nrows,ncols,nzs);
M1a = zeros(nrows,ncols,nzs);
varImA = zeros(nrows,ncols,nzs);
M2b = zeros(nrows,ncols,nzs);
M1b = zeros(nrows,ncols,nzs);
varImB = zeros(nrows,ncols,nzs);
M2c = zeros(nrows,ncols,nzs);
M1c = zeros(nrows,ncols,nzs);
varImC = zeros(nrows,ncols,nzs);
imgSq = img.^2;
%Algorithm #1
M2a = imfilter(imgSq, h/n1 , 'symmetric');
M1a = imfilter(img, h, 'symmetric').^2 / (n*n1);
varImA = (max((M2a - M1a),0));
%Algorithm #2
M2b = imfilter(imgSq, h/n1 , 'symmetric');
M1b = imfilter(img, h/n, 'symmetric');
M1S = (M1b.^2)*n/n1;
varImB = max((M2b - M1S),0);
%Algorithm #3
for i=1:1:nzs;
M2c(:,:,i) = filter2(h, imgSq(:,:,i)) / n;
end
for i=1:1:nzs;
M1c(:,:,i) = filter2(h, img(:,:,i)) / n;
end
varImC = n/n1.*(M2c-M1c.^2);
% Test the difference (need to crop image borders to get relevant results)
Test1 = varImA(20:end-20,20:end-20,:) - varImB(20:end-20,20:end-20,:);
Test2 = varImA(20:end-20,20:end-20,:) - varImC(20:end-20,20:end-20,:);
Test3 = varImB(20:end-20,20:end-20,:) - varImC(20:end-20,20:end-20,:);
% Display results;
max(Test1(:))
min(Test1(:))
max(Test2(:))
min(Test2(:))
max(Test3(:))
min(Test3(:))

Jan

### Jan (view profile)

on 3 May 2014
Loading cameraman.tif replies an integer matrix, such that it sufficient as test data. There are some typos in "VarImB" and "VarImC", which both need a lowercase "v".

on 3 May 2014
Edited by Jan

### Jan (view profile)

on 3 May 2014

When I use img = rand(256, 256, 3) as input data and fix the typos, I get differences of the magnitude 6.6613e-16. This is 3*eps, which means, that the differences are very tiny. Such differences are caused by the limited precision of the representation of numbers. See FAQ: Why is 0.3-0.2-0.1 not equal to 0.

Eric Diaz

### Eric Diaz (view profile)

on 3 May 2014
That's kind of what I was thinking.
I'm wondering why I am getting differences on the order of E-10 with my image data, which is imported as UINT12 --> double.