Fundamentals of Digital Image and Video Processing - Week 12 Solutions

Hi Coursera people,

This is a follower submitted code. Thanks to Mokam for this code.

Solution to question 8 of Week 12.

b = [-2,-6,-9,1,8,10,1,-9,-4,-3]';
S = 3;
for i = 1:10
for j = 1:10
A(i,j) = sin(i+j);
if (i==j)
A(i,j) = A(i,j) +1;
end
end
end
for i = 1: 10
Anorm(:,i) = A(:,i)/norm(A(:,i));
end
A = Anorm;
x = zeros(10,1);
r = b;
omg = [];
A_omg = [];
while nnz(x)<3 nbsp="" p="">i = 0;
maxnorm = 0;
for j=1:10
if(any(j==omg))
else
x_j = norm(transpose(A(:,j)) * r);
if(maxnormmaxnorm = x_j;
i = j;
end
end
end
omg = [omg, i];
A_omg = [A_omg, A(:,i)];
z_omega_star = pinv(A_omg' * A_omg) * A_omg' * b;
r = b - A_omg * z_omega_star;


x = zeros(10,1);
for j=1:length(omg)
index = omg(1,j);
x(index,1) = z_omega_star(j,1);
end
end

Answer is the matrix omg, in ascending order.

-Cheers,
Vijay.

Fundamentals of Digital Image and Video Processing - Week 11 Solutions

Hi Coursera people,

Week 11 was not very mathematically intensive, except for the fact that I couldn't grasp the equations. I have to admit that I did understand all of the concepts of segmentation that were spoken about in the class. I was very keen to try the quiz, as this week's material was pretty challenging! The question on programming was so skillfully worded that it could baffle and deter a normal soul! And that feeling when you see a full 11 point score on the first attempt is so intriguing; you just can't express through words :)

Here is the question number 9 of week 11 :

In this problem, you will use Accumulative Difference Image (ADI) to calculate the motion of an object. The object is a bright rectangle moving with a constant speed in a dark background. Your task is to find out the speed of the object in the horizontal direction (x direction) and in the vertical direction (y direction), as well as the total space this object occupied while moving. The total space is defined as the total number of pixels that this object occupies at least once during its movement. Download the MATLAB code "motion_ADI.m" from here. The code has detailed comments regarding each functioning part. Basically, the code generates the reference frame and 10 consecutive frames containing the moving object. All you need to do is to decide on the appropriate threshold T in line 23 in the code and implement the three equations for ADI in the video lectures regarding motion-based segmentation. Starting your code flowing line 37 and finish it before the end of the for-loop. The rest of the code will calculate the speed of the moving object and the total space it occupies for you. Enter the values of speed_X_Direction, speed_Y_Direction, and total_space_occupied in the box below.

Here is my attempt at the code : (thanks to the discussions in the forums by fellow learners)

clear all
close all


A = zeros(256,256); % initialize a 256*256 image

% initialize absolute ADI, positive ADI and Negative ADI
% all initialized to zero
% Note that all ADIs are of the same size with the image
% DO NOT change the name of the ADIs as they will be used later
ADI_abs = zeros(256,256);
ADI_pos = zeros(256,256);
ADI_neg = zeros(256,256);

% initialize the starting position of the moving object
% the moving object is a rectangle similar to the example in the lecture
% slides
start1 = 100;
start2 = 150;
start3 = 40;
start4 = 110;

%threshold T as in euations in the lecture slides regarding ADI
T = 0.1;

%initialize the reference frame R
A(start1:start2, start3:start4) = 1;

%visualize the object and in the reference frame R
figure,imshow(A,[], 'border','tight');

j = 0;
for i = 5: 5 :50
        j = j + 12;
        A2 = zeros(256,256);
        A2(start1 + i: start2 + i, start3 + j: start4 + j) = 1;
        ADI_abs(abs(A-A2) > T) = ADI_abs(abs(A-A2) > T) +1;
        ADI_pos((A-A2) > T) = ADI_pos((A-A2) > T) +1;
        ADI_neg((A-A2) < -T) = ADI_neg((A-A2) < -T) +1;
     
        % You need to code up the follwing part that calculate the ADIs
        % Namely, the absolute ADI, the positive ADI and the negative ADI
        % Equations can be found in lecture slides regarding ADIs
        % You need to decide on the appropriate threshold T for this case
        % at line 23
end

% The following part will calculate the moving speed
% and the total space(in pixel number) occupied by the moving object
[row, col] = find(ADI_neg > 0);
speed_X_Direction = (max(col) - start4) / 10
speed_Y_Direction = (max(row) - start2) / 10
total_space_occupied = sum(sum(ADI_abs > 0))

% The following part helps you to visualize the ADIs you compute
% compare them with the example shown in lecture
% You should be getting someting very similar
figure,imshow(ADI_abs,[], 'border','tight');
figure,imshow(ADI_pos,[], 'border','tight');
figure,imshow(ADI_neg,[], 'border','tight');

-Cheers,
Vijay.

Fundamentals of Digital Image and Video Processing - Week 10 Solutions

Hi Coursera people,

Well, for a pretty lecture intensive week, the solution to the problem posted was rather simple!

Just run the code without changing any parameters, and it should work fine!

PS : Remember to change the path of the images in the downloaded program!

-Cheers,
Vijay.

Fundamentals of Digital Image and Video Processing - Week 9 Solutions

Hi Coursera people,

I have to admit that the initial part of this week was very highly mathematical and I couldn't understand the videos. I was expecting the assignment to be equally tough. But to my surprise, the assignment was a straightforward implementation of a built-in Matlab function!

Here goes the Question number 7 of Week 9 :

In this problem you will get hands-on experience in JPEG image compression. Follow the instructions below to complete this problem. (1) Download the original 8-bit grayscale image here, and load it into a MATLAB array. (2) Perform JPEG compression by using the MATLAB function "imwrite". For the purpose of this problem, you need to specify 5 input arguments. The first argument is the MATLAB array containing the input image; the second argument is a string specifying the output file name; the third argument is 'jpg' (including the single quotes); the fourth argument is the string 'quality' (including the single quotes); and the last argument is a number that specifies the quality level used for compression. The quality level is an integer between 0 and 100. For this step, set the quality level to be 75 (the defaut value). After the function "imwrite" is invoked, a new JPEG image will be created in the location that was specified by you. (3) Load
the newly created JPEG image into a MATLAB array. Compute the PSNR between the JPEG compressed image and the original image. Note that the image loaded into MATLAB is of type 'uint8' (i.e., 8-bit integer). In order to compute the PSNR, you need to convert these arrays into 'double'. (4) Repeat steps (2) and (3) with the quality level set at 10. Enter the PSNR values corresponding to the JPEG images at quality level 75 and 10, respectively. Enter the numbers to 2 decimal points.

Here is my attempt at the code :

Original = imread('D:\private\MS related\~Coursera courses\Image and Video processing\Week 9\Cameraman256.bmp');
Original_double = im2double(Original);
imwrite(Original,'D:\private\MS related\~Coursera courses\Image and Video processing\Week 9\Converted.jpg','jpg','quality',75);
Converted = imread('D:\private\MS related\~Coursera courses\Image and Video processing\Week 9\Converted.jpg');
Converted_double = im2double(Converted);
MSE1= mean(mean((Original_double - Converted_double).^2,2));
MaxI=1;
PSNR1=10*log10((MaxI^2)/MSE1);
imwrite(Original,'D:\private\MS related\~Coursera courses\Image and Video processing\Week 9\ConvertedLowQual.jpg','jpg','quality',10);
ConvertedLowQual = imread('D:\private\MS related\~Coursera courses\Image and Video processing\Week 9\ConvertedLowQual.jpg');
ConvertedLowQual_double = im2double(ConvertedLowQual);
MSE2 = mean(mean((Original_double - ConvertedLowQual_double).^2,2)); % get the MSE
PSNR2=10*log10((MaxI^2)/MSE2);
PSNR1
PSNR2

-Cheers,
Vijay.

Fundamentals of Digital Image and Video Processing - Week 8 Solutions

Hi Coursera people,

I never knew this week would be so much interesting! I could understand almost every bit of information spoken in the class! :)

Here is the question number 7 of week 8.

In this problem, you will write a MATLAB program to compute the entropy of a given gray-scale image. Follow the instructions below to finish this problem. (1) Download the input image from here. The input is a gray-scale image with pixel values in the range [0,255]. Treat the pixel intensities in this image as symbols emitted from a DMS. (2) Build a probability model (i.e., an alphabet with associated probabilities) corresponding to this input image. Specifically, this alphabet consists of symbols {0,1,2,⋯,255}. In order to find the probabilities associated with each symbol, you will need to scan over all the pixels in this image, and for each pixel, adjust the probability associated with that pixel's intensity value accordingly, or in other words find the histogram of the image. Make sure you normalize the probability model correctly such that each probability is a real-valued number in [0,1]. (3) Compute the entropy using the formula that you have learned in class. Enter the result below to at least 2 decimal points.

Here is my attempt at the code.

A = imread('C:\~Coursera courses\Image and Video processing\Week 8\Cameraman256.bmp');
for i = 1:256
DMS(i,1) = i-1;
DMS(i,2) = 0;
end
for i = 1:256
for j = 1:256
for k = 1:256
if A(i,j) == DMS(k,1)
DMS(k,2) = DMS(k,2)+1;
end
end
end
end
sum = 0;
for i = 1:256
sum = sum + DMS(i,2);
end
for i = 1:256
prob(i) = DMS(i,2)/sum;
end
ans=0;
for i = 1:256
entropy(i) = -1 * prob(i) * log2(prob(i));
ans = ans + entropy(i);
end
ans

-Cheers,
Vijay.

Fundamentals of Digital Image and Video Processing - Week 7 Solutions

Hi Coursera people,

After hours of struggle, I had found the solution to question number 7 of week 7! There is absolutely no material available online regarding calculation of frequency response of the CLS filter! Finally, the answer sprung up from the discussions forum of the course itself :) I must thank my fellow Courserans for helping me out with the code. Here goes the question number 7 of week 7.

In this problem, you will implement the Constrained Least Squares (CLS) filter and examine its performance when the regularization parameter is set at different values. You will be provided with the original image and a set of MATLAB files. Follow the instructions below to finish this problem. (1) Download the original image and the MATLAB code from here. Place the original image and all the provided MATLAB files in the same directory. (2) The file "wrapper.m" is the entry or the "main" code. It loads the original image, applies a motion blur to it, and degrades the image by adding noise. The 17th line in "wrapper.m" sets the value of the regularization parameter "alpha". (3) The MATLAB file "cls_restoration.m" has an incomplete implementation of the CLS filter. You need to un-comment line 24 in "cls_restoration.m" and complete the implementation of the CLS filter. (4) After you complete the implementation of the CLS filter, you should run "wrapper.m" with different values of alpha. Specifically, we ask you to try the following values of alpha: {0.0001, 0.001, 0.01, 0.1, 1, 10, 100}. For each value of alpha, we ask you to compute the Improvement in SNR (ISNR). Note that the computation of ISNR involves there images: the original image, the blurred and noisy image, and the restored image. After you obtain the ISNR values, enter in the box below the largest ISNR value. Enter the number with at least two decimal points.

Here is my attempt at the code :

wrapper.m

clear all
close all

%% Simulate 1-D blur and noise
image_original = im2double(imread('C:\Image and Video processing\Week 7\downloaded codes\Cameraman256.bmp', 'bmp'));
[H, W] = size(image_original);
blur_impulse = fspecial('motion', 7, 0);
image_blurred = imfilter(image_original, blur_impulse, 'conv', 'circular');
noise_power = 1e-4;
randn('seed', 1);
noise = sqrt(noise_power) * randn(H, W);
image_noisy = image_blurred + noise;

figure; imshow(image_original, 'border', 'tight');
figure; imshow(image_blurred, 'border', 'tight');
figure; imshow(image_noisy, 'border', 'tight');

%% CLS restoration
alpha = 0.0001;  % you should try different values of alpha
image_cls_restored = cls_restoration(image_noisy, blur_impulse, alpha);
figure; imshow(image_cls_restored, 'border', 'tight');

%% computation of ISNR

e1=image_original-image_noisy;
e2=image_original-image_cls_restored;
E1=mean2(e1.*e1);
E2=mean2(e2.*e2);
result=10*log(E1/E2)/log(10)


cls_restoration.m

function image_restored = cls_restoration(image_noisy, psf, alpha)

%% find proper dimension for frequency-domain processing
[image_height, image_width] = size(image_noisy);
[psf_height, psf_width] = size(psf);
dim = max([image_width, image_height, psf_width, psf_height]);
dim = next2pow(dim);

%% frequency-domain representation of degradation
psf = padarray(psf, [dim - psf_height, dim - psf_width], 'post');
psf = circshift(psf, [-(psf_height - 1) / 2, -(psf_width - 1) / 2]);
H = fft2(psf, dim, dim);

%% frequency-domain representation of Laplace operator
Laplace = [0, -0.25, 0; -0.25, 1, -0.25; 0, -0.25, 0];
Laplace = padarray(Laplace, [dim - 3, dim - 3], 'post');
Laplace = circshift(Laplace, [-1, -1]);
C = fft2(Laplace, dim, dim);

%% Frequency response of the CLS filter
% Refer to the lecture for frequency response of CLS filter
% Complete the implementation of the CLS filter by uncommenting the
% following line and adding appropriate content

R = conj(H)./(abs((H.*H))+(alpha*abs((C.*C))));

%% CLS filtering
Y = fft2(image_noisy, dim, dim);
image_restored_frequency = R .* Y;
image_restored = ifft2(image_restored_frequency);
image_restored = image_restored(1 : image_height, 1 : image_width);


next2pow.m

function result = next2pow(input)
if input <= 0
    fprintf('Error: input must be positive!\n');
    result = -1;
else
    index = 0;
    while 2 ^ index < input
        index = index + 1;
    end
    result = 2 ^ index;
end


And don't get freaked if you get negative ISNR values as the result; it is absolutely normal. Here are the observations for the various values of alpha.
0.0001    -5.9191
0.001      -1.5292
0.01        3.4933
0.1          4.3048
1             2.1471
10           0.4866

-Cheers,
Vijay.

Fundamentals of Digital Image and Video Processing - Week 6 Solutions

Hi Coursera people,

This week's lectures were very exasperating with regards to their length! Yes, I understand that there is a time limit of 12 weeks to complete the material; but it is equally important for followers to understand what is happening right? Anyways, with what I could manage to understand, I've cooked up this week's solution. Its very surprising to learn that a few lines of code will solve the purpose...

Here is question number 6 of week 6.

This problems pertains to inverse filtering. You should review the corresponding slides in the video lectures to refresh your memory before attempting this problem. To help you understand how inverse filter is implemented and applied, we have provided you with a MATLAB script here. Download the script and the original image, and open the script using MATLAB. Once you open the script, you will see on Line 8 the statement "T = 1e-1". This defines the threshold value used in the inverse filter. The script simulates the blur due to motion and applies inverse filtering for its removal. We encourage you to try different values of the threshold and see how it affects the performance of the inverse filter. We ask you to enter the ISNR value below when the threshold is set to 0.5. Make sure you enter the number with at least 2 decimal points.

Here's the code that was given as a part of the question. I've added the code that gives the solution at the end.

% inverse filter with thresholding

clear all
close all
clc

% specify the threshold T
T = 0.5;

%% read in the original, sharp and noise-free image
original = im2double(rgb2gray((imread('C:\Image and Video processing\Week 6\original_cameraman.jpg'))));
[H, W] = size(original);

%% generate the blurred and noise-corrupted image for experiment
motion_kernel = ones(1, 9) / 9;  % 1-D motion blur
motion_freq = fft2(motion_kernel, 1024, 1024);  % frequency response of motion blur
original_freq = fft2(original, 1024, 1024);
blurred_freq = original_freq .* motion_freq;  % spectrum of blurred image
blurred = ifft2(blurred_freq);
blurred = blurred(1 : H, 1 : W);
blurred(blurred < 0) = 0;
blurred(blurred > 1) = 1;
noisy = imnoise(blurred, 'gaussian', 0, 1e-4);


%% Restoration from blurred and noise-corrupted image
% generate restoration filter in the frequency domain
inverse_freq = zeros(size(motion_freq));
inverse_freq(abs(motion_freq) < T) = 0;
inverse_freq(abs(motion_freq) >= T) = 1 ./ motion_freq(abs(motion_freq) >= T);
% spectrum of blurred and noisy-corrupted image (the input to restoration)
noisy_freq = fft2(noisy, 1024, 1024);
% restoration
restored_freq = noisy_freq .* inverse_freq;
restored = ifft2(restored_freq);
restored = restored(1 : H, 1 : W);
restored(restored < 0) = 0;
restored(restored > 1) = 1;

%% analysis of result
noisy_psnr = 10 * log10(1 / (norm(original - noisy, 'fro') ^ 2 / H / W));
restored_psnr = 10 * log10(1 / (norm(original - restored, 'fro') ^ 2 / H / W));


%% visualization
figure; imshow(original, 'border', 'tight');
figure; imshow(blurred, 'border', 'tight');
figure; imshow(noisy, 'border', 'tight');
figure; imshow(restored, 'border', 'tight');
figure; plot(abs(fftshift(motion_freq(1, :)))); title('spectrum of motion blur'); xlim([0 1024]);
figure; plot(abs(fftshift(inverse_freq(1, :)))); title('spectrum of inverse filter'); xlim([0 1024]);

%% Calculation of ISNR

e1=original-noisy;
e2=original-restored;
E1=mean2(e1.*e1);
E2=mean2(e2.*e2);
result=10*log(E1/E2)/log(10)




P.S. : The answer is varying with regards to the second decimal as this code is run several times on the same machine. Don't ask me why! I'm as clueless as you are :p
Just type "2.85" in the answer area, and you get a full 3 points!

-Cheers,
Vijay.