Question

matlab

Answer 1

@ikram002p Do you know much about pixel shifting?

Answer 2

nah sorry :'( as i don't have the program.

Answer 3

Okay no worries, thanks.

Answer 4

@jim_thompson5910

Answer 5

Should I prompt the user or not? I am not sure really how to set up the for loop without it.

Answer 6

I am suppose to use a nested for loop @jim_thompson5910

Answer 7

I got the image to shift without the loop

Answer 8

do I have the general idea? I got the prompts to work, however can't get the new image to display (simg)

Answer 9

why are you translating pixel by pixel? the function `imtranslate` shifts EVERY pixel of the image by the same amount

http://www.mathworks.com/help/images/ref/imtranslate.html?searchHighlight=imtranslate

Answer 10

I'm guessing your teacher wants you to do every pixel?

Answer 11

oh I see now, the prof wants you to use nested loops

Answer 12

yes and I am suppose to shift pixel by pixel?

Answer 13

working on a small if else statement for original image.

Answer 14

let me think

Answer 15

clear, clc, close all; % clears all windows

% loads image provided by instructor into matlab
img = imread('dostoevsky.tif');

% x increases from 0 to 450, while
% y decreses from 600 to 0 from bottom to top
% displays image

% asking use if wants to see original image
original = input('Do you want to see original image? ', 's');
if (original == 'y' || original == 'Y')
figure(1);
imshow(img);
axis on;
elseif (original == 'n' || original == 'N');
fprintf('You said no. \n');
end

% asking user to input x and y to transverse image
x = input('Enter a X value, (must be a positive number): ');
y = input('Enter a Y value, (must be a positive number): ');

% for loops
for k = 1:100
if (x > 0)
% shifts image in x-positive direction
simg(k) = imtranslate(img, [x, 0], 'outputview', 'full');

elseif (x < 0)
fprintf('nothing will happen.');
end

% second for loop
for i = 1:100
if (y > 0)
% shifts image y-positive direction
simg(i) = imtranslate(img, [0, y], 'outputview', 'full');
elseif (y < 0)
fprintf('nothing will happen.');
imshow(simg)
end
end
end

Answer 16

ok trying something out

Answer 17

okay

Answer 18

hmm I'm getting weird results, one moment

Answer 19

okay. This is what I get

Do you want to see original image? y
Enter a X value, (must be a positive number): 100
Enter a Y value, (must be a positive number): 200
Subscripted assignment dimension mismatch.

Answer 20

I think I got it, but I have this weird blue/green portion

Answer 21

okay lol

Answer 22

what do you mean by blue/green portion?

Answer 23

ok what I did was hard code x = 50 and y = 100 just so I didn't have to type in x and y each time I tested this script

Answer 24

here's the m file and what the figure window looks like (2 attachments)

Answer 25

reading it now

Answer 26

% get the size of matrix img
[m,n] = size(img);


% set up matrix of numbering but 0s
% this blank matrix is set to the size of matrix img
simg = uint8(zeros(size(img)));
% this matrix will later be filled by
% the values found in matrix img

Answer 27

What does the top command do?

Answer 28

this command `[m,n] = size(img);` ?

Answer 29

yeah and don't worry about the color.

Answer 30

Does that command just get the size of the matrix?

Answer 31

simg = uint8(zeros(size(img)));

and then this command sets a zero matrix up

Answer 32

`Does that command just get the size of the matrix?`

yes the command "[m,n] = size(img);" gets the size of the matrix

m rows
n columns

Answer 33

and then what about simg = uint8(zeros(size(img)));

Answer 34

sets up a matrix of zeros

it's the same size as matrix img

Answer 35

okay let me read on some more

Answer 36

What about this bit of code?

for j = 1:1:m
% 'for' loop designed to go through each column
for k = 1:1:n
% check to see if in bounds
if( (j+y <= m) && (k+x <= n) )
% this line of code will assign the pixel value
% of matrix img to matrix simg
% with the shifts in mind
simg(j+y,k+x) = img(j,k);

Answer 37

why did you go j = 1:1:m and k = 1:1:n

Answer 38

so j+y has to less or equal to m, m being the rows and k+x has to be less or equal to n because n is the columns

Answer 39

basically we can't go beyond the size of the matrix both in row and column

Answer 40

one question at a time please

Answer 41

so you figured out the question when you asked "why did you go j = 1:1:m and k = 1:1:n" ?

Answer 42

sorry I will ask you one question at a time


Not really to the last question
so you figured out the question when you asked "why did you go j = 1:1:m and k = 1:1:n" ?

Answer 43

j starts at 1, increments by 1, and keeps increment til it gets to m

m = number of rows

Answer 44

k starts at 1, increments by 1, stops increasing til it equals n

n = number of columns

Answer 45

j and k are indices to help go through the 2 'for' loops

Answer 46

okay

Answer 47

What about this part?

% check to see if in bounds
if( (j+y <= m) && (k+x <= n) )
% this line of code will assign the pixel value
% of matrix img to matrix simg
% with the shifts in mind
simg(j+y,k+x) = img(j,k);

Answer 48

`if( (j+y <= m) && (k+x <= n) ) `
is to make sure that I didn't go past the boundaries

I don't want to write in values into the simg matrix that shouldn't be there

Answer 49

`simg(j+y,k+x) = img(j,k);`
is the part of code that does the actual shifting

though for some reason, the blue part is there and I still don't know why

Answer 50

okay makes a lot of sense. I thought that was the case but just wanted to make sure my logic was correct.

Answer 51

so img is the original and simg is the new image with the shift constraints?

Answer 52

yes
img = original
simg = shifted

Answer 53

I guess

img = image
simg = shifted+image

Answer 54

okay that makes more sense.

Answer 55

uint8

^^what does this command do

Answer 56

simg = uint8(zeros(size(img)));
^^^^
This

Answer 57

it converts the values to uint8 data types I think

Answer 58

uint = unsigned integer
uint8 = 8 bit unsigned integer

just a guess really

Answer 59

How did you know that we were not to exceed the mxn dimensions?

Answer 60

I knew the for loop needed a specific constraint but didn't know what it was suppose to be.

Answer 61

so the image shifts properly. Otherwise, the simg would be bigger than img

Answer 62

when you say 'bigger' you mean the size of the matrix?

Answer 63

IF I didn't do those bounds checks, then simg would be bigger

let's say
x = 50
y = 100
m = 200
n = 300

img would be a 200x300 matrix

simg would be a 250x400 matrix

that is if I didn't check to see if you went out of bounds or not

Answer 64

okay okay I see what you are saying.

Answer 65

Because if we didn't have the bounds there is a possibility the image would be 'hidden'. It would have been displayed somewhere not visible to the user?