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

[Luke Bloy]

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?
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