[Insight-users] Silly questions about registration

[Luis Ibanez]

Hi David,


1) The ImageRegistration1.cxx example was tunned
    for the image used in the SoftwareGuide. This
    image is found in Insight/Examples/Data.

    The behavior that you observe in your run is
    typical of a too large step length. In this
    example, the step lenght was hard wired as 4.0.

    You may want to start by reducing the steplength
    to a much smaller value until the optimizer advances
    in the desired direction. Then you can play increasing
    the step size again until gaining a reasonable speed
    for registration.

2) The transform resulting from the registration
    process is exactly the one that you need to provide
    to the itk::ResampleImageFilter in order to map
    data from the moving image space to the fixed
    image space.

    Note that the reason why the transform is actually
    the one that maps points from the fixed image
    to the moving image is that this is the only way
    in which the final image can be computed whitout
    'missing' pixels.  The Resample image filter visits
    all the pixels in the output image grid, and use
    the transform to map its positions to the moving
    image space.

    This is explained in more detail in the software
    guide on Section 5.7.1, pages 136-152 where the
    ResampleImageFilter is described. It is quite
    important to read this section before moving into
    the Image registration section.


Please let us know if you have further questions.


Thanks


  Luis


-----


David Holmes wrote:
> I'm going over the registration framework for ITK and
> running some tests so that I better understand the
> model.  I have several silly questions.   
> 
>      (1)     Pulling the code from Registration1
> example, I took two images that were the same, applied
> translations, and ran the registration.  Here is how
> it
>         progressed:
> 
>         0 = 2690.87 : [2.05794, -3.43]
>         1 = 3538.93 : [2.79055, -7.36234]
>         2 = 3945.74 : [1.41266, -11.1175]
>         3 = 4030.27 : [1.45081, -15.1173]
>         4 = 4296.43 : [0.881159, -19.0766]
>         5 = 4622.25 : [-1.29036, -22.4358]
>         6 = 4726.89 : [-3.55759, -25.7312]
>         7 = 5030.82 : [-5.44799, -29.2563]
>         8 = 5368.15 : [-8.17364, -32.1839]
>         9 = 5666.88 : [-10.8422, -35.1636]
>         .
>         .
>         .
> 
>      As you can see, it looks like it is maximizing,
> even though I am using the Means Squared metric.  So I
> explicitly set the MaximizeOff flag-Same response. 
> So,
>      for kicks, I tried MaximizeOn().  Here?s what I
> got:
> 
>          
> 
>         0 = 2690.87 : [-2.05794, 3.43]
>         1 = 898.249 : [-4.63303, 6.49087]
>         2 = 1243.98 : [-3.05864, 5.25746]
>         3 = 27.0703 : [-2.62099, 3.30593]
>         4 = 720.144 : [-2.93342, 4.25587]
>         5 = 200.952 : [-3.01993, 5.25212]
>         6 = 24.1303 : [-2.99806, 4.7526]
>         7 = 22.6493 : [-2.99862, 5.0026]
>         8 = 0.309667 : [-3.11456, 4.95589]
> 
>      Is something perhaps backwards there, or I am all
> messed up?
> 
>          
> 
>      (2)     Using the example out of the software
> guide again, this question is about what is returned. 
> Generally speak, with Analyze, when we provide a
>         solution to a registration, we provide the
> solution that will be applied to the "moving" image to
> put it into the space of the fixed image.  In my own
> code using ITK, when I apply a translation of [-3,5]
> to an image and then register it, the solution is
> [-3,5].  If I were to apply that to the "moving"
> image, the result would be a translation of [-6,10],
> so at least the way that I am used to doing it, I
> would have to take the negative of the solution to put
> the moving into the fixed space.  I checked the
> software guide and found that this is consistent with
> the examples.  Once again, am I all screwed up with
> this??
> 
>   Thanks for the help
> 
>    
> 
>   david
> 
>          
> 
>     
> 
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