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Posted to user@commons.apache.org by andrea antonello <an...@gmail.com> on 2013/12/05 14:31:58 UTC
[math] solving bivariate quadratic equations
Hi, I need a hint about how to solve an equation of the type:
Z = aX^2 + bY^2 + cXY + dX + eY + f
Is an example that handles such a case available? A testcase code maybe?
Thanks for any hint,
Andrea
PS: is there a way to search the mailinglist for topics? I wasn't able
to check if this had already been asked.
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Re: [math] solving bivariate quadratic equations
Posted by andrea antonello <an...@gmail.com>.
Hi Gilles
[...]
> Independently of which method is more effective, what Luc outlined
> should certainly be doable with the official release, not just the
> development version.
thanks for pointing this out, I assume the development version has
just refactored to a different position the various pieces (the new
org.apache.commons.math3.fitting.leastsquares package).
Best regards,
Andrea
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Re: [math] solving bivariate quadratic equations
Posted by Gilles <gi...@harfang.homelinux.org>.
Hi.
> [...]
> I implemented it and it works fine.
>
> One thing that for sure is positive of this method is the fact that
> it
> can be performed with the current available 3.2 release, which is
> available in maven, whereas for the Optimization solution the
> development version is needed.
Independently of which method is more effective, what Luc outlined
should certainly be doable with the official release, not just the
development version.
Best regards,
Gilles
> [...]
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Re: [math] solving bivariate quadratic equations
Posted by andrea antonello <an...@gmail.com>.
Hi Ted,
thanks a lot for this comment.
I implemented it and it works fine.
One thing that for sure is positive of this method is the fact that it
can be performed with the current available 3.2 release, which is
available in maven, whereas for the Optimization solution the
development version is needed.
Anyways I really need to say that I am amazed by the degree of quality
help one gets from this mailinglist.
Thanks a ton,
Best regards,
Andrea
On Sun, Dec 8, 2013 at 12:48 AM, Ted Dunning <te...@gmail.com> wrote:
> I think that this can actually be solved using linear methods (for the
> quadratic error).
>
> The values of x, x^2, xy, y, y^2 and 1 can be considered coefficients of
> the terms a, b, c, d, e and f. Linear least squares systems of this form
> can be solved directly without use of an optimizer by using QR
> decomposition of the normal equation.
>
> To do this, form the matrix A where each row i is an observation and
> columns are values of x_i, x_i^2, x_i y_i and so on. The rightmost column
> should be the constant 1. We want to solve
>
> A u = z
>
> where u = [a, b, c, d, e, f]' and z = [z_1 ... z_n]' for the solution u
> that results in the smallest error in z. One simple way to do this is to
> note that setting the derivative of this squared error to zero is
> equivalent to solving this equation
>
> A' A u = A' z
>
> exactly. You can go ahead and do this, but if you use the QR
> decomposition, you get more stable results. The QR decomposition breaks A
> into two matrices, Q and R. Q is a orthornormal matrix which means that Q'
> Q = I and R is upper triangular. The normal equation can be re-written as
>
> (Q R)' (Q R) u = (Q R)' z
> R' R u = R' Q' z
>
> This allows back-substitution to be used to solve for u.
>
> The particularly nice thing about QR decomposition used in this way is that
> all of the QR implementations that I know of have a solve method.
>
> If you form the matrix A and vector z, then the code that you want in
> commons math is something kind of like this:
>
> DecompositionSolver<http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/linear/DecompositionSolver.html>solver
> = new
> RRQRDecomposition<http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/linear/RRQRDecomposition.html>
> (A).getSolver<http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/linear/RRQRDecomposition.html#getSolver()>
> ();
> RealVector u =
> solver.solve<http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/linear/DecompositionSolver.html#solve(org.apache.commons.math3.linear.RealMatrix)>
> (z);
>
>
>
>
> On Sat, Dec 7, 2013 at 7:44 AM, andrea antonello <andrea.antonello@gmail.com
>> wrote:
>
>> >> I didn't expect to have to use weights for this calculation, is there
>> >> a way to define the weights properly (for dummies like me)?
>> >
>> > The weights are mainly used as relatively to each other, if some points
>> > are considered to be more accurate than others. Typically, each point
>> > should have a weight related to the variance of the error for the point,
>> > or to the unit of the measurements (for example when mixing distances
>> > and angles in the residuals, which is not your case).
>> >
>> > If all your points are measured in the same way, they all have the same
>> > variance and hence they should share the same weight. Then, as the
>> > meaning of the weight is only relative, if they all have the same weight
>> > you may well use simply 1.0 as the global weight for everyone.
>>
>> That all makes perfectly sense, thank you very much!
>>
>> Best regards,
>> Andrea
>>
>>
>> >
>> > best regards,
>> > Luc
>> >
>> >>
>> >> Thanks,
>> >> Andrea
>> >>
>> >>
>> >>
>> >>
>> >> On Fri, Dec 6, 2013 at 10:58 AM, Luc Maisonobe <lu...@spaceroots.org>
>> wrote:
>> >>> Hi Andrea,
>> >>>
>> >>> Le 06/12/2013 10:02, andrea antonello a écrit :
>> >>>> Hi Ted,
>> >>>> thanks for the reply.
>> >>>>
>> >>>>> How would you like to handle the fact that you may have an infinite
>> number
>> >>>>> of solutions? Will you be happy with any of them? Or do you
>> somehow want
>> >>>>> to find all of them?
>> >>>>
>> >>>> I am afraid my math knowledge goes only to a certain point here (too
>> >>>> few it seems).
>> >>>> In my usecase I am trying to define a terrain model from sparse points
>> >>>> (lidar elevation points).
>> >>>> So my idea was to get a set of points (given X, Y, Z) to define the
>> >>>> parameters of the equation and then be able to calculate an elevation
>> >>>> (Z) for any position I need interpolated.
>> >>>>
>> >>>> I would then expect to have an exact value once the parameters are
>> defined.
>> >>>>
>> >>>> I am not sure how to find the parameters though...
>> >>>
>> >>> For this use case, I would suggest to use any kind of multivariate
>> >>> optimizer. Your Z variable is linear in 6 unknown parameters (a, b, c,
>> >>> d, e and f). What you want is find values for these 6 parameters such
>> >>> that the quadratic sum of residuals is minimum, i.e. you want to
>> minimize:
>> >>>
>> >>> sum (Z_i - Z(X_i, Y_i))^2
>> >>>
>> >>> Where X_i, Y_i, Z_i are the coordinates for point i and Z(X_i, Y_i) is
>> >>> the theoretical value from you model equation, which depends on the 6
>> >>> parameters.
>> >>>
>> >>> As Z is linear in the 6 parameters, the sum above is quadratic, so your
>> >>> problem is a least squares problem. Take a look at the
>> >>> fitting.leastsquares package. You have the choice among two solvers
>> >>> there: Gauss-Newton and Levenberg-Marquardt. As you problem is simple
>> >>> (linear Z), both should give almost immediate convergence, pick up one
>> >>> as you want. Start with something roughly along this line (of course,
>> >>> you should change the convergence threshold to something meaningful for
>> >>> your data):
>> >>>
>> >>> // your points
>> >>> final double[] xArray = ...;
>> >>> final double[] yArray = ...;
>> >>> final double[] zArray = ...;
>> >>>
>> >>> // start values for a, b, c, d, e, f, all set to 0.0
>> >>> double[] initialGuess = new double[6];
>> >>>
>> >>> // model: this computes the theoretical z values, given a current
>> >>> // estimate for the parameters a, b, c ...
>> >>> MultivariateVectorFunction model =
>> >>> new MultivariateVectorFunction() {
>> >>> double[] value(double[] parameters) {
>> >>> double[] computedZ = new double[zArray.length];
>> >>> for (int i = 0; i < computedZ.length; ++i) {
>> >>> double x = xArray[i];
>> >>> double y = yArray[i];
>> >>> computedZ[i] = parameter[0] * x * x +
>> >>> parameter[1] * y * y +
>> >>> parameter[2] * x * y +
>> >>> parameter[3] * x +
>> >>> parameter[4] * y +
>> >>> parameter[5];
>> >>> }
>> >>> return computedZ;
>> >>> }
>> >>> };
>> >>>
>> >>> // jacobian: this computes the Jacobian of the model
>> >>> MultivariateMatrixFunction jacobian =
>> >>> new MultivariateMatrixFunction() {
>> >>> double[][] value(double[] parameters) {
>> >>> double[][] jacobian =
>> >>> new double[zArray.length][parameters.length];
>> >>> for (int i = 0; i < computedZ.length; ++i) {
>> >>> double x = xArray[i];
>> >>> double y = yArray[i];
>> >>> jacobian[i][0] = x * x;
>> >>> jacobian[i][1] = y * y;
>> >>> jacobian[i][2] = x * y;
>> >>> jacobian[i][3] = x;
>> >>> jacobian[i][4] = y;
>> >>> jacobian[i][5] = 1;
>> >>> }
>> >>> return jacobian;
>> >>> }
>> >>> };
>> >>>
>> >>> // convergence checker
>> >>> ConvergenceChecker<PointVectorValuePair> checker =
>> >>> new SimpleVectorValueChecker(1.0e-6, 1.0e-10);
>> >>>
>> >>> // configure optimizer
>> >>> LevenbergMarquardtOptimizer optimizer =
>> >>> new LevenbergMarquardtOptimizer().
>> >>> withModelAndJacobian(model, jacobian).
>> >>> withTarget(zArray).
>> >>> withStartPoint(initialGuess).
>> >>> withConvergenceChecker(checker);
>> >>>
>> >>> // perform fitting
>> >>> PointVectorValuePair result = optimizer.optimize();
>> >>>
>> >>> // extract parameters
>> >>> double[] parameters = result.getPoint();
>> >>>
>> >>>
>> >>> BEware I just wrote the previous code in the mail, it certainly has
>> >>> syntax errors and missing parts, it should only be seen as a skeleton.
>> >>>
>> >>> best regards,
>> >>> Luc
>> >>>
>> >>>
>> >>>>
>> >>>> Cheers,
>> >>>> Andrea
>> >>>>
>> >>>>
>> >>>>
>> >>>>>
>> >>>>>
>> >>>>>
>> >>>>> On Thu, Dec 5, 2013 at 5:31 AM, andrea antonello <
>> andrea.antonello@gmail.com
>> >>>>>> wrote:
>> >>>>>
>> >>>>>> Hi, I need a hint about how to solve an equation of the type:
>> >>>>>> Z = aX^2 + bY^2 + cXY + dX + eY + f
>> >>>>>>
>> >>>>>> Is an example that handles such a case available? A testcase code
>> maybe?
>> >>>>>>
>> >>>>>> Thanks for any hint,
>> >>>>>> Andrea
>> >>>>>>
>> >>>>>> PS: is there a way to search the mailinglist for topics? I wasn't
>> able
>> >>>>>> to check if this had already been asked.
>> >>>>>>
>> >>>>>>
>> ---------------------------------------------------------------------
>> >>>>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>> >>>>>> For additional commands, e-mail: user-help@commons.apache.org
>> >>>>>>
>> >>>>>>
>> >>>>
>> >>>> ---------------------------------------------------------------------
>> >>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>> >>>> For additional commands, e-mail: user-help@commons.apache.org
>> >>>>
>> >>>>
>> >>>
>> >>>
>> >>> ---------------------------------------------------------------------
>> >>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>> >>> For additional commands, e-mail: user-help@commons.apache.org
>> >>>
>> >>
>> >> ---------------------------------------------------------------------
>> >> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>> >> For additional commands, e-mail: user-help@commons.apache.org
>> >>
>> >>
>> >>
>> >
>> >
>> > ---------------------------------------------------------------------
>> > To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>> > For additional commands, e-mail: user-help@commons.apache.org
>> >
>>
>> ---------------------------------------------------------------------
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>>
>>
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Re: [math] solving bivariate quadratic equations
Posted by Ted Dunning <te...@gmail.com>.
I think that this can actually be solved using linear methods (for the
quadratic error).
The values of x, x^2, xy, y, y^2 and 1 can be considered coefficients of
the terms a, b, c, d, e and f. Linear least squares systems of this form
can be solved directly without use of an optimizer by using QR
decomposition of the normal equation.
To do this, form the matrix A where each row i is an observation and
columns are values of x_i, x_i^2, x_i y_i and so on. The rightmost column
should be the constant 1. We want to solve
A u = z
where u = [a, b, c, d, e, f]' and z = [z_1 ... z_n]' for the solution u
that results in the smallest error in z. One simple way to do this is to
note that setting the derivative of this squared error to zero is
equivalent to solving this equation
A' A u = A' z
exactly. You can go ahead and do this, but if you use the QR
decomposition, you get more stable results. The QR decomposition breaks A
into two matrices, Q and R. Q is a orthornormal matrix which means that Q'
Q = I and R is upper triangular. The normal equation can be re-written as
(Q R)' (Q R) u = (Q R)' z
R' R u = R' Q' z
This allows back-substitution to be used to solve for u.
The particularly nice thing about QR decomposition used in this way is that
all of the QR implementations that I know of have a solve method.
If you form the matrix A and vector z, then the code that you want in
commons math is something kind of like this:
DecompositionSolver<http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/linear/DecompositionSolver.html>solver
= new
RRQRDecomposition<http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/linear/RRQRDecomposition.html>
(A).getSolver<http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/linear/RRQRDecomposition.html#getSolver()>
();
RealVector u =
solver.solve<http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/linear/DecompositionSolver.html#solve(org.apache.commons.math3.linear.RealMatrix)>
(z);
On Sat, Dec 7, 2013 at 7:44 AM, andrea antonello <andrea.antonello@gmail.com
> wrote:
> >> I didn't expect to have to use weights for this calculation, is there
> >> a way to define the weights properly (for dummies like me)?
> >
> > The weights are mainly used as relatively to each other, if some points
> > are considered to be more accurate than others. Typically, each point
> > should have a weight related to the variance of the error for the point,
> > or to the unit of the measurements (for example when mixing distances
> > and angles in the residuals, which is not your case).
> >
> > If all your points are measured in the same way, they all have the same
> > variance and hence they should share the same weight. Then, as the
> > meaning of the weight is only relative, if they all have the same weight
> > you may well use simply 1.0 as the global weight for everyone.
>
> That all makes perfectly sense, thank you very much!
>
> Best regards,
> Andrea
>
>
> >
> > best regards,
> > Luc
> >
> >>
> >> Thanks,
> >> Andrea
> >>
> >>
> >>
> >>
> >> On Fri, Dec 6, 2013 at 10:58 AM, Luc Maisonobe <lu...@spaceroots.org>
> wrote:
> >>> Hi Andrea,
> >>>
> >>> Le 06/12/2013 10:02, andrea antonello a écrit :
> >>>> Hi Ted,
> >>>> thanks for the reply.
> >>>>
> >>>>> How would you like to handle the fact that you may have an infinite
> number
> >>>>> of solutions? Will you be happy with any of them? Or do you
> somehow want
> >>>>> to find all of them?
> >>>>
> >>>> I am afraid my math knowledge goes only to a certain point here (too
> >>>> few it seems).
> >>>> In my usecase I am trying to define a terrain model from sparse points
> >>>> (lidar elevation points).
> >>>> So my idea was to get a set of points (given X, Y, Z) to define the
> >>>> parameters of the equation and then be able to calculate an elevation
> >>>> (Z) for any position I need interpolated.
> >>>>
> >>>> I would then expect to have an exact value once the parameters are
> defined.
> >>>>
> >>>> I am not sure how to find the parameters though...
> >>>
> >>> For this use case, I would suggest to use any kind of multivariate
> >>> optimizer. Your Z variable is linear in 6 unknown parameters (a, b, c,
> >>> d, e and f). What you want is find values for these 6 parameters such
> >>> that the quadratic sum of residuals is minimum, i.e. you want to
> minimize:
> >>>
> >>> sum (Z_i - Z(X_i, Y_i))^2
> >>>
> >>> Where X_i, Y_i, Z_i are the coordinates for point i and Z(X_i, Y_i) is
> >>> the theoretical value from you model equation, which depends on the 6
> >>> parameters.
> >>>
> >>> As Z is linear in the 6 parameters, the sum above is quadratic, so your
> >>> problem is a least squares problem. Take a look at the
> >>> fitting.leastsquares package. You have the choice among two solvers
> >>> there: Gauss-Newton and Levenberg-Marquardt. As you problem is simple
> >>> (linear Z), both should give almost immediate convergence, pick up one
> >>> as you want. Start with something roughly along this line (of course,
> >>> you should change the convergence threshold to something meaningful for
> >>> your data):
> >>>
> >>> // your points
> >>> final double[] xArray = ...;
> >>> final double[] yArray = ...;
> >>> final double[] zArray = ...;
> >>>
> >>> // start values for a, b, c, d, e, f, all set to 0.0
> >>> double[] initialGuess = new double[6];
> >>>
> >>> // model: this computes the theoretical z values, given a current
> >>> // estimate for the parameters a, b, c ...
> >>> MultivariateVectorFunction model =
> >>> new MultivariateVectorFunction() {
> >>> double[] value(double[] parameters) {
> >>> double[] computedZ = new double[zArray.length];
> >>> for (int i = 0; i < computedZ.length; ++i) {
> >>> double x = xArray[i];
> >>> double y = yArray[i];
> >>> computedZ[i] = parameter[0] * x * x +
> >>> parameter[1] * y * y +
> >>> parameter[2] * x * y +
> >>> parameter[3] * x +
> >>> parameter[4] * y +
> >>> parameter[5];
> >>> }
> >>> return computedZ;
> >>> }
> >>> };
> >>>
> >>> // jacobian: this computes the Jacobian of the model
> >>> MultivariateMatrixFunction jacobian =
> >>> new MultivariateMatrixFunction() {
> >>> double[][] value(double[] parameters) {
> >>> double[][] jacobian =
> >>> new double[zArray.length][parameters.length];
> >>> for (int i = 0; i < computedZ.length; ++i) {
> >>> double x = xArray[i];
> >>> double y = yArray[i];
> >>> jacobian[i][0] = x * x;
> >>> jacobian[i][1] = y * y;
> >>> jacobian[i][2] = x * y;
> >>> jacobian[i][3] = x;
> >>> jacobian[i][4] = y;
> >>> jacobian[i][5] = 1;
> >>> }
> >>> return jacobian;
> >>> }
> >>> };
> >>>
> >>> // convergence checker
> >>> ConvergenceChecker<PointVectorValuePair> checker =
> >>> new SimpleVectorValueChecker(1.0e-6, 1.0e-10);
> >>>
> >>> // configure optimizer
> >>> LevenbergMarquardtOptimizer optimizer =
> >>> new LevenbergMarquardtOptimizer().
> >>> withModelAndJacobian(model, jacobian).
> >>> withTarget(zArray).
> >>> withStartPoint(initialGuess).
> >>> withConvergenceChecker(checker);
> >>>
> >>> // perform fitting
> >>> PointVectorValuePair result = optimizer.optimize();
> >>>
> >>> // extract parameters
> >>> double[] parameters = result.getPoint();
> >>>
> >>>
> >>> BEware I just wrote the previous code in the mail, it certainly has
> >>> syntax errors and missing parts, it should only be seen as a skeleton.
> >>>
> >>> best regards,
> >>> Luc
> >>>
> >>>
> >>>>
> >>>> Cheers,
> >>>> Andrea
> >>>>
> >>>>
> >>>>
> >>>>>
> >>>>>
> >>>>>
> >>>>> On Thu, Dec 5, 2013 at 5:31 AM, andrea antonello <
> andrea.antonello@gmail.com
> >>>>>> wrote:
> >>>>>
> >>>>>> Hi, I need a hint about how to solve an equation of the type:
> >>>>>> Z = aX^2 + bY^2 + cXY + dX + eY + f
> >>>>>>
> >>>>>> Is an example that handles such a case available? A testcase code
> maybe?
> >>>>>>
> >>>>>> Thanks for any hint,
> >>>>>> Andrea
> >>>>>>
> >>>>>> PS: is there a way to search the mailinglist for topics? I wasn't
> able
> >>>>>> to check if this had already been asked.
> >>>>>>
> >>>>>>
> ---------------------------------------------------------------------
> >>>>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
> >>>>>> For additional commands, e-mail: user-help@commons.apache.org
> >>>>>>
> >>>>>>
> >>>>
> >>>> ---------------------------------------------------------------------
> >>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
> >>>> For additional commands, e-mail: user-help@commons.apache.org
> >>>>
> >>>>
> >>>
> >>>
> >>> ---------------------------------------------------------------------
> >>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
> >>> For additional commands, e-mail: user-help@commons.apache.org
> >>>
> >>
> >> ---------------------------------------------------------------------
> >> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
> >> For additional commands, e-mail: user-help@commons.apache.org
> >>
> >>
> >>
> >
> >
> > ---------------------------------------------------------------------
> > To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
> > For additional commands, e-mail: user-help@commons.apache.org
> >
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
> For additional commands, e-mail: user-help@commons.apache.org
>
>
Re: [math] solving bivariate quadratic equations
Posted by andrea antonello <an...@gmail.com>.
>> I didn't expect to have to use weights for this calculation, is there
>> a way to define the weights properly (for dummies like me)?
>
> The weights are mainly used as relatively to each other, if some points
> are considered to be more accurate than others. Typically, each point
> should have a weight related to the variance of the error for the point,
> or to the unit of the measurements (for example when mixing distances
> and angles in the residuals, which is not your case).
>
> If all your points are measured in the same way, they all have the same
> variance and hence they should share the same weight. Then, as the
> meaning of the weight is only relative, if they all have the same weight
> you may well use simply 1.0 as the global weight for everyone.
That all makes perfectly sense, thank you very much!
Best regards,
Andrea
>
> best regards,
> Luc
>
>>
>> Thanks,
>> Andrea
>>
>>
>>
>>
>> On Fri, Dec 6, 2013 at 10:58 AM, Luc Maisonobe <lu...@spaceroots.org> wrote:
>>> Hi Andrea,
>>>
>>> Le 06/12/2013 10:02, andrea antonello a écrit :
>>>> Hi Ted,
>>>> thanks for the reply.
>>>>
>>>>> How would you like to handle the fact that you may have an infinite number
>>>>> of solutions? Will you be happy with any of them? Or do you somehow want
>>>>> to find all of them?
>>>>
>>>> I am afraid my math knowledge goes only to a certain point here (too
>>>> few it seems).
>>>> In my usecase I am trying to define a terrain model from sparse points
>>>> (lidar elevation points).
>>>> So my idea was to get a set of points (given X, Y, Z) to define the
>>>> parameters of the equation and then be able to calculate an elevation
>>>> (Z) for any position I need interpolated.
>>>>
>>>> I would then expect to have an exact value once the parameters are defined.
>>>>
>>>> I am not sure how to find the parameters though...
>>>
>>> For this use case, I would suggest to use any kind of multivariate
>>> optimizer. Your Z variable is linear in 6 unknown parameters (a, b, c,
>>> d, e and f). What you want is find values for these 6 parameters such
>>> that the quadratic sum of residuals is minimum, i.e. you want to minimize:
>>>
>>> sum (Z_i - Z(X_i, Y_i))^2
>>>
>>> Where X_i, Y_i, Z_i are the coordinates for point i and Z(X_i, Y_i) is
>>> the theoretical value from you model equation, which depends on the 6
>>> parameters.
>>>
>>> As Z is linear in the 6 parameters, the sum above is quadratic, so your
>>> problem is a least squares problem. Take a look at the
>>> fitting.leastsquares package. You have the choice among two solvers
>>> there: Gauss-Newton and Levenberg-Marquardt. As you problem is simple
>>> (linear Z), both should give almost immediate convergence, pick up one
>>> as you want. Start with something roughly along this line (of course,
>>> you should change the convergence threshold to something meaningful for
>>> your data):
>>>
>>> // your points
>>> final double[] xArray = ...;
>>> final double[] yArray = ...;
>>> final double[] zArray = ...;
>>>
>>> // start values for a, b, c, d, e, f, all set to 0.0
>>> double[] initialGuess = new double[6];
>>>
>>> // model: this computes the theoretical z values, given a current
>>> // estimate for the parameters a, b, c ...
>>> MultivariateVectorFunction model =
>>> new MultivariateVectorFunction() {
>>> double[] value(double[] parameters) {
>>> double[] computedZ = new double[zArray.length];
>>> for (int i = 0; i < computedZ.length; ++i) {
>>> double x = xArray[i];
>>> double y = yArray[i];
>>> computedZ[i] = parameter[0] * x * x +
>>> parameter[1] * y * y +
>>> parameter[2] * x * y +
>>> parameter[3] * x +
>>> parameter[4] * y +
>>> parameter[5];
>>> }
>>> return computedZ;
>>> }
>>> };
>>>
>>> // jacobian: this computes the Jacobian of the model
>>> MultivariateMatrixFunction jacobian =
>>> new MultivariateMatrixFunction() {
>>> double[][] value(double[] parameters) {
>>> double[][] jacobian =
>>> new double[zArray.length][parameters.length];
>>> for (int i = 0; i < computedZ.length; ++i) {
>>> double x = xArray[i];
>>> double y = yArray[i];
>>> jacobian[i][0] = x * x;
>>> jacobian[i][1] = y * y;
>>> jacobian[i][2] = x * y;
>>> jacobian[i][3] = x;
>>> jacobian[i][4] = y;
>>> jacobian[i][5] = 1;
>>> }
>>> return jacobian;
>>> }
>>> };
>>>
>>> // convergence checker
>>> ConvergenceChecker<PointVectorValuePair> checker =
>>> new SimpleVectorValueChecker(1.0e-6, 1.0e-10);
>>>
>>> // configure optimizer
>>> LevenbergMarquardtOptimizer optimizer =
>>> new LevenbergMarquardtOptimizer().
>>> withModelAndJacobian(model, jacobian).
>>> withTarget(zArray).
>>> withStartPoint(initialGuess).
>>> withConvergenceChecker(checker);
>>>
>>> // perform fitting
>>> PointVectorValuePair result = optimizer.optimize();
>>>
>>> // extract parameters
>>> double[] parameters = result.getPoint();
>>>
>>>
>>> BEware I just wrote the previous code in the mail, it certainly has
>>> syntax errors and missing parts, it should only be seen as a skeleton.
>>>
>>> best regards,
>>> Luc
>>>
>>>
>>>>
>>>> Cheers,
>>>> Andrea
>>>>
>>>>
>>>>
>>>>>
>>>>>
>>>>>
>>>>> On Thu, Dec 5, 2013 at 5:31 AM, andrea antonello <andrea.antonello@gmail.com
>>>>>> wrote:
>>>>>
>>>>>> Hi, I need a hint about how to solve an equation of the type:
>>>>>> Z = aX^2 + bY^2 + cXY + dX + eY + f
>>>>>>
>>>>>> Is an example that handles such a case available? A testcase code maybe?
>>>>>>
>>>>>> Thanks for any hint,
>>>>>> Andrea
>>>>>>
>>>>>> PS: is there a way to search the mailinglist for topics? I wasn't able
>>>>>> to check if this had already been asked.
>>>>>>
>>>>>> ---------------------------------------------------------------------
>>>>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>>>>>> For additional commands, e-mail: user-help@commons.apache.org
>>>>>>
>>>>>>
>>>>
>>>> ---------------------------------------------------------------------
>>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>>>> For additional commands, e-mail: user-help@commons.apache.org
>>>>
>>>>
>>>
>>>
>>> ---------------------------------------------------------------------
>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>>> For additional commands, e-mail: user-help@commons.apache.org
>>>
>>
>> ---------------------------------------------------------------------
>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>> For additional commands, e-mail: user-help@commons.apache.org
>>
>>
>>
>
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Re: [math] solving bivariate quadratic equations
Posted by Luc Maisonobe <lu...@spaceroots.org>.
Le 06/12/2013 15:26, andrea antonello a écrit :
> Hi Luc,
> thanks a ton for your example code, much appreciated.
>
> It took me a little while to understand that this works only with the
> development version :)
>
> I was able to get it working on a small sample of data with results in
> the expected range.
> I didn't check yet really on the result validity because I got a NPE
> because of missing weights.
> In the testcaes I then saw:
> .withWeight(new DiagonalMatrix(weights))
> where the weights are calculated for each point as 1/Z_i.
>
> I didn't expect to have to use weights for this calculation, is there
> a way to define the weights properly (for dummies like me)?
The weights are mainly used as relatively to each other, if some points
are considered to be more accurate than others. Typically, each point
should have a weight related to the variance of the error for the point,
or to the unit of the measurements (for example when mixing distances
and angles in the residuals, which is not your case).
If all your points are measured in the same way, they all have the same
variance and hence they should share the same weight. Then, as the
meaning of the weight is only relative, if they all have the same weight
you may well use simply 1.0 as the global weight for everyone.
best regards,
Luc
>
> Thanks,
> Andrea
>
>
>
>
> On Fri, Dec 6, 2013 at 10:58 AM, Luc Maisonobe <lu...@spaceroots.org> wrote:
>> Hi Andrea,
>>
>> Le 06/12/2013 10:02, andrea antonello a écrit :
>>> Hi Ted,
>>> thanks for the reply.
>>>
>>>> How would you like to handle the fact that you may have an infinite number
>>>> of solutions? Will you be happy with any of them? Or do you somehow want
>>>> to find all of them?
>>>
>>> I am afraid my math knowledge goes only to a certain point here (too
>>> few it seems).
>>> In my usecase I am trying to define a terrain model from sparse points
>>> (lidar elevation points).
>>> So my idea was to get a set of points (given X, Y, Z) to define the
>>> parameters of the equation and then be able to calculate an elevation
>>> (Z) for any position I need interpolated.
>>>
>>> I would then expect to have an exact value once the parameters are defined.
>>>
>>> I am not sure how to find the parameters though...
>>
>> For this use case, I would suggest to use any kind of multivariate
>> optimizer. Your Z variable is linear in 6 unknown parameters (a, b, c,
>> d, e and f). What you want is find values for these 6 parameters such
>> that the quadratic sum of residuals is minimum, i.e. you want to minimize:
>>
>> sum (Z_i - Z(X_i, Y_i))^2
>>
>> Where X_i, Y_i, Z_i are the coordinates for point i and Z(X_i, Y_i) is
>> the theoretical value from you model equation, which depends on the 6
>> parameters.
>>
>> As Z is linear in the 6 parameters, the sum above is quadratic, so your
>> problem is a least squares problem. Take a look at the
>> fitting.leastsquares package. You have the choice among two solvers
>> there: Gauss-Newton and Levenberg-Marquardt. As you problem is simple
>> (linear Z), both should give almost immediate convergence, pick up one
>> as you want. Start with something roughly along this line (of course,
>> you should change the convergence threshold to something meaningful for
>> your data):
>>
>> // your points
>> final double[] xArray = ...;
>> final double[] yArray = ...;
>> final double[] zArray = ...;
>>
>> // start values for a, b, c, d, e, f, all set to 0.0
>> double[] initialGuess = new double[6];
>>
>> // model: this computes the theoretical z values, given a current
>> // estimate for the parameters a, b, c ...
>> MultivariateVectorFunction model =
>> new MultivariateVectorFunction() {
>> double[] value(double[] parameters) {
>> double[] computedZ = new double[zArray.length];
>> for (int i = 0; i < computedZ.length; ++i) {
>> double x = xArray[i];
>> double y = yArray[i];
>> computedZ[i] = parameter[0] * x * x +
>> parameter[1] * y * y +
>> parameter[2] * x * y +
>> parameter[3] * x +
>> parameter[4] * y +
>> parameter[5];
>> }
>> return computedZ;
>> }
>> };
>>
>> // jacobian: this computes the Jacobian of the model
>> MultivariateMatrixFunction jacobian =
>> new MultivariateMatrixFunction() {
>> double[][] value(double[] parameters) {
>> double[][] jacobian =
>> new double[zArray.length][parameters.length];
>> for (int i = 0; i < computedZ.length; ++i) {
>> double x = xArray[i];
>> double y = yArray[i];
>> jacobian[i][0] = x * x;
>> jacobian[i][1] = y * y;
>> jacobian[i][2] = x * y;
>> jacobian[i][3] = x;
>> jacobian[i][4] = y;
>> jacobian[i][5] = 1;
>> }
>> return jacobian;
>> }
>> };
>>
>> // convergence checker
>> ConvergenceChecker<PointVectorValuePair> checker =
>> new SimpleVectorValueChecker(1.0e-6, 1.0e-10);
>>
>> // configure optimizer
>> LevenbergMarquardtOptimizer optimizer =
>> new LevenbergMarquardtOptimizer().
>> withModelAndJacobian(model, jacobian).
>> withTarget(zArray).
>> withStartPoint(initialGuess).
>> withConvergenceChecker(checker);
>>
>> // perform fitting
>> PointVectorValuePair result = optimizer.optimize();
>>
>> // extract parameters
>> double[] parameters = result.getPoint();
>>
>>
>> BEware I just wrote the previous code in the mail, it certainly has
>> syntax errors and missing parts, it should only be seen as a skeleton.
>>
>> best regards,
>> Luc
>>
>>
>>>
>>> Cheers,
>>> Andrea
>>>
>>>
>>>
>>>>
>>>>
>>>>
>>>> On Thu, Dec 5, 2013 at 5:31 AM, andrea antonello <andrea.antonello@gmail.com
>>>>> wrote:
>>>>
>>>>> Hi, I need a hint about how to solve an equation of the type:
>>>>> Z = aX^2 + bY^2 + cXY + dX + eY + f
>>>>>
>>>>> Is an example that handles such a case available? A testcase code maybe?
>>>>>
>>>>> Thanks for any hint,
>>>>> Andrea
>>>>>
>>>>> PS: is there a way to search the mailinglist for topics? I wasn't able
>>>>> to check if this had already been asked.
>>>>>
>>>>> ---------------------------------------------------------------------
>>>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>>>>> For additional commands, e-mail: user-help@commons.apache.org
>>>>>
>>>>>
>>>
>>> ---------------------------------------------------------------------
>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>>> For additional commands, e-mail: user-help@commons.apache.org
>>>
>>>
>>
>>
>> ---------------------------------------------------------------------
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>>
>
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Re: [math] solving bivariate quadratic equations
Posted by andrea antonello <an...@gmail.com>.
Hi Luc,
thanks a ton for your example code, much appreciated.
It took me a little while to understand that this works only with the
development version :)
I was able to get it working on a small sample of data with results in
the expected range.
I didn't check yet really on the result validity because I got a NPE
because of missing weights.
In the testcaes I then saw:
.withWeight(new DiagonalMatrix(weights))
where the weights are calculated for each point as 1/Z_i.
I didn't expect to have to use weights for this calculation, is there
a way to define the weights properly (for dummies like me)?
Thanks,
Andrea
On Fri, Dec 6, 2013 at 10:58 AM, Luc Maisonobe <lu...@spaceroots.org> wrote:
> Hi Andrea,
>
> Le 06/12/2013 10:02, andrea antonello a écrit :
>> Hi Ted,
>> thanks for the reply.
>>
>>> How would you like to handle the fact that you may have an infinite number
>>> of solutions? Will you be happy with any of them? Or do you somehow want
>>> to find all of them?
>>
>> I am afraid my math knowledge goes only to a certain point here (too
>> few it seems).
>> In my usecase I am trying to define a terrain model from sparse points
>> (lidar elevation points).
>> So my idea was to get a set of points (given X, Y, Z) to define the
>> parameters of the equation and then be able to calculate an elevation
>> (Z) for any position I need interpolated.
>>
>> I would then expect to have an exact value once the parameters are defined.
>>
>> I am not sure how to find the parameters though...
>
> For this use case, I would suggest to use any kind of multivariate
> optimizer. Your Z variable is linear in 6 unknown parameters (a, b, c,
> d, e and f). What you want is find values for these 6 parameters such
> that the quadratic sum of residuals is minimum, i.e. you want to minimize:
>
> sum (Z_i - Z(X_i, Y_i))^2
>
> Where X_i, Y_i, Z_i are the coordinates for point i and Z(X_i, Y_i) is
> the theoretical value from you model equation, which depends on the 6
> parameters.
>
> As Z is linear in the 6 parameters, the sum above is quadratic, so your
> problem is a least squares problem. Take a look at the
> fitting.leastsquares package. You have the choice among two solvers
> there: Gauss-Newton and Levenberg-Marquardt. As you problem is simple
> (linear Z), both should give almost immediate convergence, pick up one
> as you want. Start with something roughly along this line (of course,
> you should change the convergence threshold to something meaningful for
> your data):
>
> // your points
> final double[] xArray = ...;
> final double[] yArray = ...;
> final double[] zArray = ...;
>
> // start values for a, b, c, d, e, f, all set to 0.0
> double[] initialGuess = new double[6];
>
> // model: this computes the theoretical z values, given a current
> // estimate for the parameters a, b, c ...
> MultivariateVectorFunction model =
> new MultivariateVectorFunction() {
> double[] value(double[] parameters) {
> double[] computedZ = new double[zArray.length];
> for (int i = 0; i < computedZ.length; ++i) {
> double x = xArray[i];
> double y = yArray[i];
> computedZ[i] = parameter[0] * x * x +
> parameter[1] * y * y +
> parameter[2] * x * y +
> parameter[3] * x +
> parameter[4] * y +
> parameter[5];
> }
> return computedZ;
> }
> };
>
> // jacobian: this computes the Jacobian of the model
> MultivariateMatrixFunction jacobian =
> new MultivariateMatrixFunction() {
> double[][] value(double[] parameters) {
> double[][] jacobian =
> new double[zArray.length][parameters.length];
> for (int i = 0; i < computedZ.length; ++i) {
> double x = xArray[i];
> double y = yArray[i];
> jacobian[i][0] = x * x;
> jacobian[i][1] = y * y;
> jacobian[i][2] = x * y;
> jacobian[i][3] = x;
> jacobian[i][4] = y;
> jacobian[i][5] = 1;
> }
> return jacobian;
> }
> };
>
> // convergence checker
> ConvergenceChecker<PointVectorValuePair> checker =
> new SimpleVectorValueChecker(1.0e-6, 1.0e-10);
>
> // configure optimizer
> LevenbergMarquardtOptimizer optimizer =
> new LevenbergMarquardtOptimizer().
> withModelAndJacobian(model, jacobian).
> withTarget(zArray).
> withStartPoint(initialGuess).
> withConvergenceChecker(checker);
>
> // perform fitting
> PointVectorValuePair result = optimizer.optimize();
>
> // extract parameters
> double[] parameters = result.getPoint();
>
>
> BEware I just wrote the previous code in the mail, it certainly has
> syntax errors and missing parts, it should only be seen as a skeleton.
>
> best regards,
> Luc
>
>
>>
>> Cheers,
>> Andrea
>>
>>
>>
>>>
>>>
>>>
>>> On Thu, Dec 5, 2013 at 5:31 AM, andrea antonello <andrea.antonello@gmail.com
>>>> wrote:
>>>
>>>> Hi, I need a hint about how to solve an equation of the type:
>>>> Z = aX^2 + bY^2 + cXY + dX + eY + f
>>>>
>>>> Is an example that handles such a case available? A testcase code maybe?
>>>>
>>>> Thanks for any hint,
>>>> Andrea
>>>>
>>>> PS: is there a way to search the mailinglist for topics? I wasn't able
>>>> to check if this had already been asked.
>>>>
>>>> ---------------------------------------------------------------------
>>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>>>> For additional commands, e-mail: user-help@commons.apache.org
>>>>
>>>>
>>
>> ---------------------------------------------------------------------
>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>> For additional commands, e-mail: user-help@commons.apache.org
>>
>>
>
>
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>
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Re: [math] solving bivariate quadratic equations
Posted by Luc Maisonobe <lu...@spaceroots.org>.
Hi Andrea,
Le 06/12/2013 10:02, andrea antonello a écrit :
> Hi Ted,
> thanks for the reply.
>
>> How would you like to handle the fact that you may have an infinite number
>> of solutions? Will you be happy with any of them? Or do you somehow want
>> to find all of them?
>
> I am afraid my math knowledge goes only to a certain point here (too
> few it seems).
> In my usecase I am trying to define a terrain model from sparse points
> (lidar elevation points).
> So my idea was to get a set of points (given X, Y, Z) to define the
> parameters of the equation and then be able to calculate an elevation
> (Z) for any position I need interpolated.
>
> I would then expect to have an exact value once the parameters are defined.
>
> I am not sure how to find the parameters though...
For this use case, I would suggest to use any kind of multivariate
optimizer. Your Z variable is linear in 6 unknown parameters (a, b, c,
d, e and f). What you want is find values for these 6 parameters such
that the quadratic sum of residuals is minimum, i.e. you want to minimize:
sum (Z_i - Z(X_i, Y_i))^2
Where X_i, Y_i, Z_i are the coordinates for point i and Z(X_i, Y_i) is
the theoretical value from you model equation, which depends on the 6
parameters.
As Z is linear in the 6 parameters, the sum above is quadratic, so your
problem is a least squares problem. Take a look at the
fitting.leastsquares package. You have the choice among two solvers
there: Gauss-Newton and Levenberg-Marquardt. As you problem is simple
(linear Z), both should give almost immediate convergence, pick up one
as you want. Start with something roughly along this line (of course,
you should change the convergence threshold to something meaningful for
your data):
// your points
final double[] xArray = ...;
final double[] yArray = ...;
final double[] zArray = ...;
// start values for a, b, c, d, e, f, all set to 0.0
double[] initialGuess = new double[6];
// model: this computes the theoretical z values, given a current
// estimate for the parameters a, b, c ...
MultivariateVectorFunction model =
new MultivariateVectorFunction() {
double[] value(double[] parameters) {
double[] computedZ = new double[zArray.length];
for (int i = 0; i < computedZ.length; ++i) {
double x = xArray[i];
double y = yArray[i];
computedZ[i] = parameter[0] * x * x +
parameter[1] * y * y +
parameter[2] * x * y +
parameter[3] * x +
parameter[4] * y +
parameter[5];
}
return computedZ;
}
};
// jacobian: this computes the Jacobian of the model
MultivariateMatrixFunction jacobian =
new MultivariateMatrixFunction() {
double[][] value(double[] parameters) {
double[][] jacobian =
new double[zArray.length][parameters.length];
for (int i = 0; i < computedZ.length; ++i) {
double x = xArray[i];
double y = yArray[i];
jacobian[i][0] = x * x;
jacobian[i][1] = y * y;
jacobian[i][2] = x * y;
jacobian[i][3] = x;
jacobian[i][4] = y;
jacobian[i][5] = 1;
}
return jacobian;
}
};
// convergence checker
ConvergenceChecker<PointVectorValuePair> checker =
new SimpleVectorValueChecker(1.0e-6, 1.0e-10);
// configure optimizer
LevenbergMarquardtOptimizer optimizer =
new LevenbergMarquardtOptimizer().
withModelAndJacobian(model, jacobian).
withTarget(zArray).
withStartPoint(initialGuess).
withConvergenceChecker(checker);
// perform fitting
PointVectorValuePair result = optimizer.optimize();
// extract parameters
double[] parameters = result.getPoint();
BEware I just wrote the previous code in the mail, it certainly has
syntax errors and missing parts, it should only be seen as a skeleton.
best regards,
Luc
>
> Cheers,
> Andrea
>
>
>
>>
>>
>>
>> On Thu, Dec 5, 2013 at 5:31 AM, andrea antonello <andrea.antonello@gmail.com
>>> wrote:
>>
>>> Hi, I need a hint about how to solve an equation of the type:
>>> Z = aX^2 + bY^2 + cXY + dX + eY + f
>>>
>>> Is an example that handles such a case available? A testcase code maybe?
>>>
>>> Thanks for any hint,
>>> Andrea
>>>
>>> PS: is there a way to search the mailinglist for topics? I wasn't able
>>> to check if this had already been asked.
>>>
>>> ---------------------------------------------------------------------
>>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>>> For additional commands, e-mail: user-help@commons.apache.org
>>>
>>>
>
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>
>
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Re: [math] solving bivariate quadratic equations
Posted by andrea antonello <an...@gmail.com>.
Hi Ted,
thanks for the reply.
> How would you like to handle the fact that you may have an infinite number
> of solutions? Will you be happy with any of them? Or do you somehow want
> to find all of them?
I am afraid my math knowledge goes only to a certain point here (too
few it seems).
In my usecase I am trying to define a terrain model from sparse points
(lidar elevation points).
So my idea was to get a set of points (given X, Y, Z) to define the
parameters of the equation and then be able to calculate an elevation
(Z) for any position I need interpolated.
I would then expect to have an exact value once the parameters are defined.
I am not sure how to find the parameters though...
Cheers,
Andrea
>
>
>
> On Thu, Dec 5, 2013 at 5:31 AM, andrea antonello <andrea.antonello@gmail.com
>> wrote:
>
>> Hi, I need a hint about how to solve an equation of the type:
>> Z = aX^2 + bY^2 + cXY + dX + eY + f
>>
>> Is an example that handles such a case available? A testcase code maybe?
>>
>> Thanks for any hint,
>> Andrea
>>
>> PS: is there a way to search the mailinglist for topics? I wasn't able
>> to check if this had already been asked.
>>
>> ---------------------------------------------------------------------
>> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
>> For additional commands, e-mail: user-help@commons.apache.org
>>
>>
---------------------------------------------------------------------
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Re: [math] solving bivariate quadratic equations
Posted by Ted Dunning <te...@gmail.com>.
Andrea,
How would you like to handle the fact that you may have an infinite number
of solutions? Will you be happy with any of them? Or do you somehow want
to find all of them?
On Thu, Dec 5, 2013 at 5:31 AM, andrea antonello <andrea.antonello@gmail.com
> wrote:
> Hi, I need a hint about how to solve an equation of the type:
> Z = aX^2 + bY^2 + cXY + dX + eY + f
>
> Is an example that handles such a case available? A testcase code maybe?
>
> Thanks for any hint,
> Andrea
>
> PS: is there a way to search the mailinglist for topics? I wasn't able
> to check if this had already been asked.
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
> For additional commands, e-mail: user-help@commons.apache.org
>
>