The VXL Project

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: vxl-users@...
Sent: Tuesday, November 16, 2004 9:58 PM
Subject: [Vxl-users] 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!

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: vxl-users@...
> Sent: Tuesday, November 16, 2004 9:58 PM
> Subject: [Vxl-users] 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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Hi, Marc

	I think your code would be modified as the follow which yields 4n residuals:
	
void XXX:f(const vnl_vector<double>& x,
vnl_vector<double>& fx)
{
    ......
    for ( int i = 0; i < n; i++ ) {
        ...
        fx[4*i] = (x1[i].first, x1_h[i].first)^2; 	
        fx[4*i+1] = (x1[i].second, x1_h[i].second)^2;
        fx[4*i+2] = (x2[i].first, x2_h[i].first)^2; 	
        fx[4*i+3] = (x2[i].second, x2_h[i].second)^2;
    }
}

in which,x1[i].first refers to the "x" element of a 2D point,and x1[i].second the "y" element.

Shawrie Lee
regards.

Recall that LM takes a vector of residuals BEFORE squaring. 

In this case, the function shall look like:

fx[4*i]   = x1[i][0]-x1_h[i][0];
fx[4*i+1] = x1[i][1]-x1_h[i][1];
fx[4*i+2] = x2[i][0]-x2_h[i][0];
fx[4*i+3] = x2[i][1]-x2_h[i][1];


> 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.
>