shifting a part of image (not rectangular shape)

I have an rectangular shape image. In which there is an inclined line (not vertical or horizontal) separating that image into 2 parts. I want to shift one part of the image down by a fixed amount of pixels. how can do that

Answer 1

[Walter Roberson]

The question is not clear. Are you pulling one part of the image "away" from the dividing line, or are you pushing it "into" the dividing line?

If you are pulling it "away" from the dividing line, then what should go in to the new space between the dividing line and the new location, and what should happen with the part of the image that would protrude beneath the old bottom of the image?

If you are pushing it "into" the dividing line, should the pixels that would go above the line just be discarded? And what should go in to the empty space left by moving the image?

Is the inclined line a straight line defined by a start and end point, or do its boundaries need to be found by examining the image to detect some distinctive property of the boundary?

[bes]

I know the starting and ending point of the straight line.

it is a 500*500 image. one part should shifted so that it should start at row 5 and finishes row 505. need not to display the pixel between 500-505. pixel between 0-5 can be empty white space. (this is a colour image. containing colour blocks)

[Walter Roberson]

So the original pixels 495-500 can be thrown away if they would go beyond the edge?

Based upon your original description, I thought perhaps you wanted to shift a cut rectangular block -- rectangular until it hit the dividing line. But your answer has left me thinking that perhaps you meant something else?

Does the straight line only go through part of the overall image? When the rows are shifted, should the portion of the image between the left or right edge of the line and the edge of the image be left as-is?

An example might help.

http://www.mathworks.com/matlabcentral/answers/7924-where-can-i-upload-images-and-files-for-use-on-matlab-answers

[Walter Roberson]

That last should have been written in terms of the upper and lower part of the line edge, not the left and right part.

Answer 2

[bes]

 im2=ones(500,500);
im2(1:100, :)=1;
 im2(100:200,:)=2;
 im2(200:300,:)=3;
 im2(300:400,:)=4;
 im2(400:500,:)=5;
 imagesc(im2);

this is the input image

[x, y] = getline; % get the input line to separate the image into 2 parts

In output the righthand side part of the image should shifted down

Answer 3

[bes]

http://www.uploadhouse.com/viewfile.php?id=14792283

this is the input image

[bes]

http://rapidshare.com/files/1810672008/input.jpg

[Walter Roberson]

I will take that as an implicit confirmation that the line always goes the entire height of the image.

Does the line sometimes lean the other direction, going "upward" bottom left to top right?

[bes]

it can go to upward bottom left to top right

[bes]

ya it always goes the entire height

Answer 4

[bes]

 I have solved this problem using this code
I have downloaded imTransD function from <http://www.csse.uwa.edu.au/~pk/research/matlabfns/Projective/imTransD.m>

 [xCor, yCor] = getline;   % get the line input from user
  cor = [xCor(1), xCor(2), yCor(1), yCor(2)];
  [rows, cols] = size(im);
  [x y] = meshgrid(1:cols, 1:rows);
        
        % Eqn of line joining end points (x1 y1) and (x2 y2) can be   
        %    parameterised by
        %    x*(y1-y2) + y*(x2-x1) + y2*x1 - y1*x2 = 0
        
        
  mask = (x*(yCor(1)-yCor(2))+ y*(xCor(2)-xCor(1))+ yCor(2)*xCor(1)- yCor(1)*xCor(2)) > 0;
  m = (yCor(2)-yCor(1))/(xCor(2)-xCor(1));
  dy = d;
  dx = dy/m;
  T = [1 0 dx; 0 1 dy; 0 0 1];  
      % Transformation matrix ( shift along x  direction is d and in y direction is (d/m)
  im2 = imTransD(im, T, size(im));
  im = im.*(mask) + removenan(im2.*double(~mask), 0);
  figure, imagesc(im);