Gehua,
Thanks for your reply!
As I use the second edition of H&Z book, the algorithm
on page 114 is 4.3.
As for the number of residuals, I still do not quite
understand why it's 4n. Since each point
correspondence gives 2 residuals, i.e the squared
distances in 2 images are d(x_i, xhat_i)^2 and
d(xpri_i, xhatpri_i)^2.
it ought to be 2n.
below is the code segment of function f in my class
derived from vnl_least_squares_function:
void XXX:f(const vnl_vector<double>& x,
vnl_vector<double>& fx)
{
......
for ( int i = 0; i < n; i++ ) {
...
fx[2*i] = squared_distance(x1[i], x1_h[i]);
fx[2*i+1] = squared_distance(x2[i], x2_h[i]);
}
}
I will be very grateful if you would be patient to
give me a further explaination.
Marc.
 Gehua Yang <yangg2@...> wrote:
> First of all, Page 114, or Algorithm 3.7 is the Gold
> Standard Algorithm for
> estimating an AFFINE transformation, not homography.
>
> If homography is to be estimated, refer to Algorithm
> 3.3 on Page 98. There
> are two choices: Sampson error or Gold Standard
> error. For Sampson error,
> there are only 9 parameters(or one can choose other
> parameterization). As
> for Gold Standard error, there are 2n+9 variables.
> But there are 4n
> residuals:
> sum d(x_i, xhat_i)^2 + d(xpri_i, xhatpri_i)^2
> Recall that LM takes a vector of residuals before
> squaring. for each d(.)
> function, there are two residuals. Another way to
> look at it is that each
> ideal point x hat brings in two parameters, but
> provides 4 constraints.
>
>
> Gehua
>
>  Original Message 
> From: Marc Anderson
> To: vxlusers@...
> Sent: Tuesday, November 16, 2004 9:58 PM
> Subject: [Vxlusers] Premise of
> vnl_levenberg_marquardt hold back Gold
> standard?
>
>
> > Hi, all vxl guys!
>
> > The class vnl_levenberg_marquardt checks if the
> number of parameters is
> > less than that of residuals before carrying out
> the minimization process,
> > otherwise it'll returns false. This requirement
> prevents the estimation of
> > homography between two images using the Gold
> Standard algorithm(H&Z book,
> > p114), in which case the number of parameters is
> 2n+9 while the number of
> > residuals is n. ( where n is the number of point
> correspondences ). Would
> > someone get the idea to figure this issue out?
> Thanks!
>
>
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