Steve on Image Processing

April 28th, 2006

I've written previously about defining a spatial transform as a function, (x,y) = T{(u,v)}, that maps points from one space (input space) to another (output space). Given such a function, how do you go about spatially transforming an image?

The most obvious procedure is called forward mapping. It works like this:

Consider each input image pixel in turn. For each input image pixel:

• Determine its location in input space, (u_k,v_k).
• Map that location to output space using (x_k,y_k) = T{(u_k,v_k)}.
• Figure out which output pixel contains the location (x_k,y_k).
• Copy the input pixel value to that output pixel.

Forward mapping has two main disadvantages as a computational procedure: gaps and overlaps. Depending on specific spatial transform function, you may have some output pixels that did not receive any input image pixels; these are the gaps. You may also have some output pixels that received more than one input image pixel; these are the overlaps. In both cases, it is challenging to figure out a reasonable way to set those output pixels.

One way to overcome these problems is to map pixel rectangles in input space to output space quadrilaterals, as shown here:

With this procedure, the input pixel value is allocated to different output pixels depending on the relative fractional coverage of the quadrilateral. Although this procedure can produce good results, it is complicated to implement and it takes a long time to compute.

Because of the problems with forward mapping, many (most?) image spatial transform implementations use a different technique, called inverse mapping. I'll describe that in a future post.

05:06 UTC | Posted in Spatial transforms | Permalink |

You can follow any responses to this entry through the RSS 2.0 feed. You can skip to the end and leave a response. Pinging is currently not allowed.

6 Responses to “Spatial transformations: Forward mapping”

1. Leonardo replied on April 4th, 2008 at 17:37 UTC :
   

   Dear Steve,
   I have an input image from which I digitized grid points.
   I know the real distance between the grid points, so I can create an output array. I would like to specify a transformation using these points, so I can assign the real location to each point from the input.
   Is it possible?
   Best
   Leonardo

2. Steve replied on April 4th, 2008 at 21:49 UTC :
   

   Leonardo—Sounds like you could use cp2tform to estimate a spatial transform fitting your points.

3. Rudolph replied on August 21st, 2009 at 09:23 UTC :
   

   Hi Steve,

   I need to perform lense calibration on an image taken with specific camera. The lense coefficients are available.

   The distortions on the lense are radial distortion, decentering of the principal point and some affinity. According to me I’ll have to calculate the (subpixel units) distance which I have to move each pixel in the x and y direction. So each pixel will have it’s own resulting transformation. I went through most of your blogs and I want to create a custom tform struct which I can implement on bulk images.

   Now I can’t see how I will be able to do this with inverse mapping, true? Will imtransform deal with the overlapping and gaps for me automatically with chosen interpolation? How do I deal with a pixel that moves 0.166 pixels up for instance? How will matlab save the info on output space?

   Any guidence would be appreciated.

4. Rudolph replied on August 28th, 2009 at 08:05 UTC :
   

   I managed to overcome the idea of linear indexing which imtransform use to determine the custom spatial transform. Thanks

5. douglas replied on September 22nd, 2009 at 06:10 UTC :
   

   Could you tell how to apply the grid points (data points from the transformation grid) in the transformation ?

   [Points edited out for space.]

6. Steve replied on September 22nd, 2009 at 12:30 UTC :
   

   Douglas—Use cp2tform to infer a spatial transformation from your point set. One of the spatially varying ones might be best. Then use imtransform to warp your image.

Leave a Reply

<pre class="code">
a = magic(3);
sum(a)
</pre>

These postings are the author's and don't necessarily represent the opinions of The MathWorks.