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.