When performing any of these arithmatic operations on the images, we do it pixel by pixel. i.e. when we add image to image
we are adding the gray level at pixel
image
to the gray level at pixel
in image
. In all these case I assume the images are of the same size.
(a) : It is possible to find histogram of
in terms of
under any one of these conditions:
1. (i.e two images are same size and same gray level)
Steps to find histogram for this condition:
For each peak in at gray level
move to the right to gray level
2. is a binary image (2 gray levels only), and
is the binary inverse image of
(i.e. if pixel
is
in image
then the same pixel
is
in image
)
Steps to find histogram for this condition:
Since there are only 2 gray levels in or
, then take switch the positions of the peaks.
3. where
is some integer. i.e. the images differ from each others only by the intensity level. i.e
is a shifted to the right version of
and the peaks in
are spread out. (Both images are the same size)
Steps to find histogram for this condition:
Starting from left to right in , number each peak. Call this number
where
is the total number of peaks.Both
will have the same
but will be located at different gray levels.
Then For each peak in
, with gray level
build a new peak in
with same hight (frequency) as peak
, but at gray level shifted to the right to new gray level of
(b) This case is the same as case (a), since we can let
and then consider
, where
is the negative of image
(c) It is possible to find histogram of
in terms of
under any one of these conditions:
1. (i.e two images are same size and same gray level).
To build the histogram do:
Starting from left to right in , number each peak. Call this number
where
is the number of peaks.Both
will have the same
since the same image.
Then For each peak in
, with gray level
build a new peak in
with same hight (frequency) as peak
, but at gray level shifted to the right to new gray level of
2. is a binary image (2 gray levels only), and
is the binary inverse image of
(i.e. if pixel
is
in image
then the same pixel
is
in image
) .
Then histogram will be one peak at gray level 0 (all black image)
(d) It is possible to find histogram of
in terms of
under any one of these conditions:
1. (i.e two images are same size and same gray level).
The histogram in this case will be all at one gray level 1. (All black) (Assuming black is at gray level 1.) if Black is at gray level 0 and white at 255, then it is not possible to divide any 2 images with each others since we will get a divide by zero error.
2. Can not divide binary images (assuming we assign 0 and 1 for the gray level)
Created by Mathematica (October 23, 2004)