[Insight-users] Computing the gradient in the Metric?

[motes motes]

Hm not sure I understand:

    dM/dx(T(p,x))  * dT/dp:



Don't you mean:

    dM/dx'  * dT(p,x)/dp:

where:

    x' = T(p,x)

corresponding to the transformed pixel in the moving image M?





On Sun, Sep 6, 2009 at 5:07 PM, Luke Bloy<luke.bloy at gmail.com> wrote:
> You use the chain rule.
>
> dt/dp = dM(T(p,x))/dp = dM/dx(T(p,x)) * dT/dp
>
> so dM/dx(T(p,x)) is the gradient of the moving image evaluated at T(p,x) and
> dT/dp is the jacobian of the transform.
>
> -luke
>
> motes motes wrote:
>>
>> Ok I am trying to do the metric E differentiation by hand:
>>
>> E = \sum[F(x) - M(T(p,x))]^2
>>   = t^2
>>
>> where:
>>
>> t = F(x) - M(T(p,x))
>>
>> dE/dp = 2t * dt/dp
>>
>> where:
>>
>> dt/dp = dM(T(p,x))/dp
>>
>> since the fixed image is treated as a constant because we are
>> differentiating with respect to the parameters p.
>>
>>
>> Now I think I understand why its the gradient of the moving image that
>> is used. But how are the Jacobian introduced in:
>>
>> dt/dp = dM(T(p,x))/dp
>>
>> ?
>>
>>
>>
>>
>> On Sun, Sep 6, 2009 at 3:30 PM, motes motes<mort.motes at gmail.com> wrote:
>>
>>>
>>> Thanks for the hint I have read those pages but I still have two
>>> questions to the below expression:
>>>
>>>         sum += 2.0 * diff * jacobian( dim, par ) * gradient[dim];
>>>
>>> 1) In the above expression gradient[dim] corresponds to the gradient
>>> for the current pixel in the moving image? On page 249 in the "Insight
>>> Into Images" this is the first term in (10.6). But why are the
>>> gradient of the moving image used to compute the derivative of the
>>> metric? I know this is a very basic question but I still cannot see
>>> the connection.
>>>
>>> 2) When I debug the code the jacobian actually just contains the
>>> weights (computed using the transform kernel) for each node in the
>>> support region. But are the jacobian not supposed to contain the first
>>> order partial derivatives? It seems a bit confusing that the jacobian
>>> is used as a container for the kernel transform weights.
>>>
>>>
>>>
>>>
>>>
>>> On Sun, Sep 6, 2009 at 1:01 PM, Neuner Markus<neuner.markus at gmx.net>
>>> wrote:
>>>
>>>>
>>>> Hi,
>>>>
>>>> To understand why the jacobian is used i would suggest to read pages
>>>> 249-251
>>>> in the Book "Insight Into Images" by Terry S. Yoo. Ther eis described
>>>> how
>>>> and why the jacobian is used.
>>>>
>>>> Greets
>>>>
>>>> motes motes wrote:
>>>>
>>>>>
>>>>> In itkMeanSquaresImageToImageMetric.txx the metric value and the
>>>>> gradient is computed. I pretty much understand how the metric value is
>>>>> computed but am a bif confused on how the gradient is computed:
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>     const RealType movingValue  = this->m_Interpolator->Evaluate(
>>>>> transformedPoint );
>>>>>     const TransformJacobianType & jacobian =
>>>>> this->m_Transform->GetJacobian( inputPoint );
>>>>>     const RealType fixedValue     = ti.Value();
>>>>>     this->m_NumberOfPixelsCounted++;
>>>>>     const RealType diff = movingValue - fixedValue;
>>>>>     measure += diff * diff;
>>>>>
>>>>>     for(unsigned int par=0; par<ParametersDimension; par++)
>>>>>       {
>>>>>       RealType sum = NumericTraits< RealType >::Zero;
>>>>>       for(unsigned int dim=0; dim<ImageDimension; dim++)
>>>>>         {
>>>>>         sum += 2.0 * diff * jacobian( dim, par ) * gradient[dim];
>>>>>         }
>>>>>       derivative[par] += sum;
>>>>>       }
>>>>>     }
>>>>>
>>>>> It basically comes down to this line:
>>>>>
>>>>>
>>>>>
>>>>>         sum += 2.0 * diff * jacobian( dim, par ) * gradient[dim];
>>>>>
>>>>> why is the jacobian multiplied with the current gradient?
>>>>> _____________________________________
>>>>> Powered by www.kitware.com
>>>>>
>>>>> Visit other Kitware open-source projects at
>>>>> http://www.kitware.com/opensource/opensource.html
>>>>>
>>>>> Please keep messages on-topic and check the ITK FAQ at:
>>>>> http://www.itk.org/Wiki/ITK_FAQ
>>>>>
>>>>> Follow this link to subscribe/unsubscribe:
>>>>> http://www.itk.org/mailman/listinfo/insight-users
>>>>>
>>>>>
>>>>>
>>
>> _____________________________________
>> Powered by www.kitware.com
>>
>> Visit other Kitware open-source projects at
>> http://www.kitware.com/opensource/opensource.html
>>
>> Please keep messages on-topic and check the ITK FAQ at:
>> http://www.itk.org/Wiki/ITK_FAQ
>>
>> Follow this link to subscribe/unsubscribe:
>> http://www.itk.org/mailman/listinfo/insight-users
>>
>