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 ![FormBox[RowBox[{n, =, RowBox[{1., ., K}]}], TraditionalForm]](HTMLFiles/problem_3_7_34.gif){kind=small}
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 ![FormBox[RowBox[{n, =, RowBox[{1., ., K}]}], TraditionalForm]](HTMLFiles/problem_3_7_55.gif){kind=small}
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)