Info
This question is closed. Reopen it to edit or answer.
I need you help to convert this equation to matlab code
1 view (last 30 days)
Show older comments
clear all;
close all;
clc;
tic
%-----------------read image
o=imread('4.2.07.tiff');
[m,n]=size(o);
%----------------random pixel insertion ------
for i=1:m
for j=1:n
if j==1
I(i,j)=randi(i)
else
I(i,j)=o(i,j-1)
end
end
end
this code does'nt work
0 Comments
Answers (2)
drummer
on 23 Oct 2020
Edited: drummer
on 23 Oct 2020
Looks funny this exercise.
By assigning i to the rand function, it is creating matrices as i increases. So rand(2) creates a 2 x 2 matrix of random values , rand(3) creates a matrix of 3 x 3 and so on...
So rand(i) would be growing in dimension sizes, as i iterates. And not assigning a single random value to the ith element.
@Samiu is correct, I think, regarding the limits of j.
Well, after all, this exercise should give an image with random values only in the first column, and return the original image otherwise?
Check if it is what you're looking for.
- The attached file shows the original in the left, and a zoom in the right to show the random values assigned to the 1st column.
Cheers
clear all;
close all;
clc;
% tic
%-----------------read image
o = imread('pout.tif');
figure
imshow(o);
title('original image');
%%
[m, n] = size(o);
I = zeros(m, n);
%----------------random pixel insertion ------
for i = 1 : m
for j = 1 : n + 1
if j == 1
I(i,j) = rand;
else
I(i,j) = o(i,j-1);
end
end
end
figure
imshow(I, 'DisplayRange', []);
title('Image I')
0 Comments
Samiu Haque
on 17 Sep 2020
I think the second for loop is supposed to iterate 'n+1' times and not 'n' times. As I(i,j) has the dimension of m*(n+1)
0 Comments
This question is closed.
See Also
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!