## Tuesday, 16 July 2013

### Information - How much do we need?

The good thing about doing research is that there is a possibility that you can come across a new technique/method everyday. Recently I have been working with a lot of information from different datasets and trying to extract the useful information content, which can be eventually used to describe the major trend in the dataset.

A major hurdle in this is to quantify that how much information is actually informative. One possible method which is widely used to achieve this is called Singular Value Decomposition (SVD). As this is a widely used technique, the details of this method can be easily found across the web. This blog post will visualize the content information and try to comprehend how much is required for computer vision applications. Here I am using one of the applications of SVD, which is to compress the content of an image. Although there exist better approaches to achieve this, the content used can be quantified easily using SVD.

Below is a sample grayscale image that I have used for this blog post. This is the original image, without any compression using SVD. Notice that there are a lot of regions in the image that look similar.

When the above image is decomposed using singular value decomposition, it gives a total of 480 singular values. Compression in the image is determined by using lesser number of these decomposed singular values for reconstruction. More detail about this can be found in literature, however the aim of this post is to visualize how the information looks at different levels of compression. Below is the visualization of compression using different number of singular values.

From the above image, it is clear that most of the information is stored in the first 55 singular values yielding to a massive 88.5% compression with little loss of visual quality.

To quantify the errors, I have compared the output from compression process with the original image. The results support the observation made above. These results are summarized in the graph below. It can be seen the errors decrease in an exponential trend with the increase in singular values used.

I hope the above example helps you visualize the change in data and think about how much information is really required. There is no definitive boundary on this, however one thing that can be done see the graph is to choose a value which gives an adequate balance between data compression and square error.

The code for the above method is given below:

``` close allclear allclc %reading and converting the imageinImage=imread('fruits.jpg');inImage=rgb2gray(inImage);inImageD=double(inImage); % decomposing the image using singular value decomposition[U,S,V]=svd(inImageD); % Using different number of singular values (diagonal of S) to compress and% reconstruct the imagedispEr = [];numSVals = [];for N=5:25:300    % store the singular values in a temporary var    C = S;        % discard the diagonal values not required for compression    C(N+1:end,:)=0;    C(:,N+1:end)=0;        % Construct an Image using the selected singular values    D=U*C*V';            % display and compute error    figure;    buffer = sprintf('Image output using %d singular values', N)    imshow(uint8(D));    title(buffer);    total = double(size(C, 1));    cur = double(N);        pCompress = ((total-cur)/total)*100    buffer = sprintf('%d singular values', N)    text(10,460, [buffer], 'Color', 'w', 'FontSize', 15);        buffer = sprintf('%0.01f', pCompress);            text(270,460, [buffer '% Compressed'], 'Color', 'w', 'FontSize', 20);        im = getframe();    buffer = sprintf('%d.png', N);    imwrite(im.cdata, buffer);        error=sum(sum((inImageD-D).^2));        % store vals for display    dispEr = [dispEr; error];    numSVals = [numSVals; N];end % dislay the error graphfigure; title('Error in compression');plot(numSVals, dispEr);grid onxlabel('Number of Singular Values used');ylabel('Error between compress and original image');```

#### 1 comment:

1. I saw your post on stackoverflow ,but even when i am saving the image the min size i can achive is the actual size .I used matlab please help me in this.