[Insight-users] Level Set / Fast Marching explanations

Chris McIntosh cmcintos.sfu at gmail.com
Fri Jun 8 18:11:48 EDT 2007



gatts gatsu wrote:
>> The velocity at which the wave front (surface or contour) is moving 
>> at each grid location.
>
> Let's say that you are using a series of DICOM (representing for 
> instance a vessel).
> If you take as the level set the contour of this vessel in the first 
> picture. To calculate this velocity : is it totally independant of the 
> next "slides" (only using the result of the gradient?) ?

That depends on what algorithm you use to calculate the velocity. It 
could be based on the image, or only on the shape of the surface, or a 
set of images, etc.

>>>
>>> I tried to look at this following url to understand better : 
>>> http://math.berkeley.edu/~sethian/2006/Explanations/level_set_explain.html 
>>>
>>>
>>> If you look at the xample with the cone-shaped surface and the 
>>> circle. The starting front is a circle. The fronts at the next steps 
>>> are the intersection between the surface of the cone-shaped surface 
>>> and the starting front plane.  It seems that in the next steps, the 
>>> front will still be a circle. Why is it keeping the same form if the 
>>> algorithm allows us to get random forms ?
>>
>> Not random forms, forms driven by the image and internal smoothness 
>> forces.
>>
>
> What I meant is for instance : the circle can change form transforming 
> into half a circle and the other half some kind of rectangle (so that 
> it's still related with the previous form). However, how will it be 
> detected with a cone-shaped surface?

Ah, that example is just one of a surface evolving without any kind of 
image forces. In that example, the velocity image would be all negative 
one. The rectangle you see
is just a plane being used to visualize where zero is. The red curve, is 
where then where the signed distance transform (SDF) has a value equal 
to zero.

Think of it this way. You can visualize a 2D image as either a planar 
image where brighter values on the screen mean higher values in the 
image. Or you can
plot it in 3D where higher values (in the Z-dimension) correspond to 
higher values from the image.

What you see in his video is the latter option. So logically, 
subtracting one from everything will "lower" it in the Z-axis which in 
the 2D view would correspond to
expanding the contour (red) outwards.

>
>
> Concerning the Fast Marching Segmentation (9.3.1 in itksoftwareguide)
> "The output of the FastMarchingImageFilter
> is a time-crossing map that indicates, for each pixel, how much time 
> it would take for
> the front to arrive at the pixel location."
> I understand for the stopping value since it's a time crossing map. 
> But once again, how does it calculate the time of propagation between 
> the front to the top left of the image for instance ?

To my understanding, perhaps somebody will correct me, the output of 
that function is an unsigned distance transform. It calculates the 
values by simulating the propagation
of the front.

>
> "In the current example we decided to use a Sigmoid function since
> it offers a good deal of control parameters that can be customized to 
> shape a nice speed image.
> The mapping should be done in such a way that the propagation speed of 
> the front will be very
> low close to high image gradients while it will move rather fast in 
> low gradient areas."
> We use the sigmoid on the result of the gradient filter. We put the 
> output between 0 and 1.
> It is said to give a nice speed image. Is the value "1" high enough to 
> provide a high speed image ?

Depends on the function being used, and the other parameters involved it 
should be. To be honest, I'm not overly familiar with how
this particular example. Best thing to do is try with various images, 
and examples.

There's also ITK-SNAP (an application) which uses levelsets and is easy 
to setup and test.

Best thing though, is to read about the math behind the whole thing, 
then the programs will seem straightforward.

>
> Thanks again for taking time to answer to this post.

Your welcome.

-Chris


> Windows Live Spaces : créez votre blog à votre image ! 
> http://www.windowslive.fr/spaces


More information about the Insight-users mailing list