lab6343
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Arizona State University *
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Course
325
Subject
Mathematics
Date
Feb 20, 2024
Type
Pages
6
Uploaded by LieutenantMetalPartridge38
MAT 343 Lab 5 - Abhikya Reddy Ananth A=imread(
'gauss.jpg'
); %reading the image B=double(A(:,:,1)); %convert to double precision
B=B/255; %scale the values of B
[U S V] = svd(B); %compute the SVD decomposition of B
Problem 1
Compute the dimensions of U, S and V
%finding dimensions
disp(
"U="
)
U=
size(U)
%displays size of U ans = 1×2
393 393
disp(
"S="
)
S=
size(S)
%displays size of S ans = 1×2
393 613
disp(
"V="
)
V=
size(V)
%displays size of V
ans = 1×2
613 613
Problem 2
Compute the best rank-1 approximation and store it in
rank1
k = 1; % defining ran to 1 rank1 = U(:,1:k)*S(1:k, 1:k)*V(:,1:k)'; %this is to find rank 1 for B sing U, S, V Visualize rank1 by performing steps 3 -6
C = zeros(size(A)); % visualizing rank C(:,:,1) = rank1; C(:,:,2) = rank1; C(:,:,3) = rank1; C = max(0,min(1,C));
image(C), axis image %helps to view the image 1
Problem 3
Create and view a rank-10 approximation to the original picture
k=10;
%defining rank to 10 rank1 = U(:,1:k)*S(1:k, 1:k)*V(:,1:k)';
C = zeros(size(A)); % visualizing rank C(:,:,1) = rank1; C(:,:,2) = rank1; C(:,:,3) = rank1; C = max(0,min(1,C));
image(C), axis image %helps to view the image 2
Problem 4
Experiment with different ranks until you found one that gives, in your opinion, an acceptable approximation.
for k=[20 30 40] %repeating with ranks 20, 30, 40
rank1 = U(:,1:k)*S(1:k, 1:k)*V(:,1:k)';
C = zeros(size(A)); % visualizing rank C(:,:,1) = rank1; C(:,:,2) = rank1; C(:,:,3) = rank1; C = max(0,min(1,C));
figure(k);
image(C), axis image %helps to view the image end
3
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4
%In my opinion rank-40 gives an aceptable range Problem 5
What rank-r approximation exactly reproduces the original picture? Explain,
Answer: It would be rank of B , which is 255, The image is made up of different levels of grey on a 255 bit scale Problem 6
(i)
How much data is needed to represent a rank-k approximation? Explain.
Answer: After experimenting in problem 4. I found out that rank-40 gives an appropriate approximation therefore
k is 40 this data can be evaluated by using the formula k+(k*m)+(k*n) where m is 613 and n is 393 . The result
of the formula is 40280. Therefore we can say the data needed to represent rank-40 approximation is 40280.
(ii)
Find the compression rate for the value of the rank you determined in problem 4. Explain.
Answer: Compression rate = (data of rank-40 approximation)/(data used for the original image) 40280 / ( 613*393) = 0.16720006309
Thus after calculation we find that rank-40 has a compression rate of 0.16720006309 ( 16.72%)
What does the compression rate represent? Explain.
5
Answer:The compression rate represents ratio of the amount of data used in rank-approximation to the amount
of data used in the original image.
Problem 7
Find the smallest value of k such that the rank-k approximation uses the same or more amount of data as the
original picture. Explain how you obtained the answer.
Answer: In order to find the smallest value of k , we can set the compression rate to 1 , or 100% and solve for k
with the formula
k = (m*n)/(2*(m+n)) k = ( 613*393)/(2*(613+393)) = 240909/2012 = 119.736083499
Thus, the smallest value of k would be 119.736083499.
6
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