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Just thought I might make things a bit more clear
here:

The FAQ talked about doing a reverse rotation to
avoid holes in the rotated image, which I followed.
The question is: even if you do it the reverse way,
you still end up with some pixels that fall out
of the range of the source image. How do you deal
with this - for now, I just arbitrarily assign them
the same value as the boundary pixels. Is there a
better way?

Thanks.

In article <4de7gt$j0i@charm.magnus.acs.ohio-state.edu>,
xiaoyi@bmecg.bme.ohio-state.edu (Xiaoyi Wu) writes:
> Path: magnus.acs.ohio-state.edu!bmecg.bme.ohio-state.edu!xiaoyi
> From: xiaoyi@bmecg.bme.ohio-state.edu (Xiaoyi Wu)
> Newsgroups: comp.graphics.algorithms,comp.lang.c
> Subject: 2D image rotation problem
> Date: 15 Jan 1996 18:47:57 GMT
> Organization: Biomedical Engineering OSU
> Lines: 12
> Distribution: world
> Message-ID: <4de7gt$j0i@charm.magnus.acs.ohio-state.edu>
> Reply-To: xiaoyi@mozart.bme.ohio-state.edu (Xiaoyi Wu)
> NNTP-Posting-Host: bmecg.bme.ohio-state.edu
> Xref: magnus.acs.ohio-state.edu comp.graphics.algorithms:26053
comp.lang.c:153880
> 
> I did read the FAQ and digged up Graphics Gems, but
> no source code was given. So, here it comes:
> 
> I need to rotate a 2D image by a specified angle with
> respect to a specified point within the image.
> 
> Has anyone written a program like this before? I'd
> greatly appreciate any inputs. 
> 
> Thanks.
> 
> Xiaoyi

On 15 Jan 1996 20:04:49 GMT Xiaoyi Wu (xiaoyi@bmecg.bme.ohio-state.edu) wrote:
>> Just thought I might make things a bit more clear
>> here:

>> The FAQ talked about doing a reverse rotation to
>> avoid holes in the rotated image, which I followed.
>> The question is: even if you do it the reverse way,
>> you still end up with some pixels that fall out
>> of the range of the source image. How do you deal
>> with this - for now, I just arbitrarily assign them
>> the same value as the boundary pixels. Is there a
>> better way?

It really depends on your application.

Typically, you would specify a background colour and assign this to 
the "outside" pixels.  Alternatively, you could crop the output image 
so there aren't any "outside" pixels.  The key, here, is to think about 
your application.

It really depends on your application.

Regards,
Martin
E-mail: mleese@omg.unb.ca
WWW:    http://www.omg.unb.ca/~mleese/
________________________________________________________________________
 Want to know how Ambisonics can improve the sound of your LPs and CDs?
  See the Ambisonic Surround Sound FAQ.  Details on my WWW Home page.

xiaoyi@bmecg.bme.ohio-state.edu (Xiaoyi Wu) wrote:
>Just thought I might make things a bit more clear
>here:
>
>The FAQ talked about doing a reverse rotation to
>avoid holes in the rotated image, which I followed.
>The question is: even if you do it the reverse way,
>you still end up with some pixels that fall out
>of the range of the source image. How do you deal
>with this - for now, I just arbitrarily assign them
>the same value as the boundary pixels. Is there a
>better way?

Hmm, to keep it in the range of the source image, couldn't you use AND 
with your index?  Um, this is how I use to do it - the rotation failed 
but it was never out of the range of the source file. :)  Um, this is 
assuming you're using powers of 2 for the dimensions of your image of 
course; if you're not then you might have to suffer a divide (eeek) i.e. 
use the MOD of your index into the image and the dimensions multiplied.

Um, assuming your image dimensions are 32x32, 

   ; calculate the index [si] first
   and si, (32*32)-1
   lea si, [image+si]
   movsb

I'm curious though.. could you explain how you did the reverse rotation? 
I was thinking about this and doing landscapes or flags.  It always had 
holes in it since I was calculating the X and Y values to plot on screen 
but if I had just done a loop over the X and Y values then this would not 
have happened - instead, I would have just calculated the index into the 
image array and covered the entire area.  The problem is, I don't know 
how to get the reverse function.  Is it just the inverse?

Hi Kap, actually after a night's thought about it i figured it is
impossible to do a rotation without creating holes unless the image is
infintely large - of course you can kinda fix them by duplicating the
closest pixels to the pixels that fall out of the source image. 

another option which seems to be the way you did it (at least the way i 
interprete your response) is to ignore the pixels that need to be 
reconstructed using non-existent pixels, and end up with a "rotated" 
image that has blank strips in it.

for me, both would work 'cuz the object in my images are in the center so 
i can afford to ignore the edges. but for other applications i guess i 
still need to be educated.

the reverse rotation is just replacing your theta with -theta :) then 
when you do the interpolation you will be working with pixel values you know.

thanks. 

Xiaoyi

On Tue, 16 Jan 1996, Kap Shiwram wrote:
> 
> Hmm, to keep it in the range of the source image, couldn't you use AND 
> with your index?  Um, this is how I use to do it - the rotation failed 
> but it was never out of the range of the source file. :)  Um, this is 
> assuming you're using powers of 2 for the dimensions of your image of 
> course; if you're not then you might have to suffer a divide (eeek) i.e. 
> use the MOD of your index into the image and the dimensions multiplied.
> 
> Um, assuming your image dimensions are 32x32, 
> 
>    ; calculate the index [si] first
>    and si, (32*32)-1
>    lea si, [image+si]
>    movsb
> 
> I'm curious though.. could you explain how you did the reverse rotation? 
> I was thinking about this and doing landscapes or flags.  It always had 
> holes in it since I was calculating the X and Y values to plot on screen 
> but if I had just done a loop over the X and Y values then this would not 
> have happened - instead, I would have just calculated the index into the 
> image array and covered the entire area.  The problem is, I don't know 
> how to get the reverse function.  Is it just the inverse?
> 
> 
> 

Xiaoyi

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