[Insight-users] Basic question concerning metric

[motes motes]

Ok so instead of using eg. gradient descent to estimate the transform
parameters for the displacement vector, "hardcoded" parameters are used
corresponding to the dx, dy values, right?

Another thing, why 50*50? Does these values correspond to milimiters or some
other unit? Could not find any documentation for this.



On Wed, Jul 1, 2009 at 10:32 PM, Luis Ibanez <luis.ibanez at kitware.com>wrote:

>
> Hi Motes,
>
>
>           Yes, you got it all right !
>
>
> This example is evaluating the Metric between
> one image and different translations of itself.
>
> You should find that the lowest metric value
> correspond to the displacement (0,0).
>
> The purpose of the exercise is to visualize
> the cost function (the Image Metric) for
> values in a grid of 50 x 50 positions in the
> parametric space.
>
> That's the kind of space that the Optimizers
> will have to navigate.
>
>
>     Regards,
>
>
>          Luis
>
>
> --------------------------------------------------------
> On Wed, Jul 1, 2009 at 4:27 PM, motes motes <mort.motes at gmail.com> wrote:
>
>> I am trying the metric example in the itkSoftwareGuide page 416. This is
>> what I understand so far:
>>
>> The following components are connected to the SSD metric:
>>
>> - Transform (translation)
>> - Interpolator (nearest neighbor)
>> - The fixed and moving images, in the example they are identical.
>>
>> Next the moving image is translated and for each translation the metric is
>> computed:
>>
>>     const int rangex = 50;
>>     const int rangey = 50;
>>     for(int dx = -rangex; dx <= rangex; dx++){
>>       for(int dy = -rangey; dy <= rangey; dy++){
>>         displacement[0] = dx;
>>         displacement[1] = dy;
>>         const double value = metric->GetValue(displacement);
>>         out << dx << "   "  << dy << "   " << value << std::endl;
>>       }
>>     }
>>
>> As I understand the call:
>>
>>   metric->GetValue(displacement);
>>
>> evaluates the metric for transform parameters contained in the
>> displacement vector using the underlying connected transform.
>>
>> Eg when dx=dy=50 we are actually testing how good the transform that
>> translates all pixels 50 units along the x and y axis aligns with  the fixed
>> image - which is not very good. The optimal solution is a transform that
>> translates all pixels 0 units along the x and y axis which makes sense since
>> the two images are identical.
>>
>> To put it another way: The fixed image is compared to 50*50 translated
>> versions of it self.
>>
>> Am I on the right track?
>>
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