Alglib: solving A * x = b in a least squares sense - c++

I have a somewhat complicated algorithm that requires the fitting of a quadric to a set of points. This quadric is given by its parametrization (u, v, f(u,v)), where f(u,v) = au^2+bv^2+cuv+du+ev+f.
The coefficients of the f(u,v) function need to be found since I have a set of exactly 6 constraints this function should obey. The problem is that this set of constraints, although yielding a problem like A*x = b, is not completely well behaved to guarantee a unique solution.
Thus, to cut it short, I'd like to use alglib's facilities to somehow either determine A's pseudoinverse or directly find the best fit for the x vector.
Apart from computing the SVD, is there a more direct algorithm implemented in this library that can solve a system in a least squares sense (again, apart from the SVD or from using the naive inv(transpose(A)*A)*transpose(A)*b formula for general least squares problems where A is not a square matrix?

Found the answer through some careful documentation browsing:
rmatrixsolvels( A, noRows, noCols, b, singularValueThreshold, info, solverReport, x)
The documentation states the the singular value threshold is a clamping threshold that sets any singular value from the SVD decomposition S matrix to 0 if that value is below it. Thus it should be a scalar between 0 and 1.
Hopefully, it will help someone else too.

Related

How to optimize nonlinear funtion with some constraint in c++

I want to find a variable in C++ that allows a given nonlinear formula to have a maximum value in a constraint.
It is to calculate the maximum value of the formula below in the given constraint in C++.
You can also use the library.(e.g. Nlopt)
formula : ln(1+ax+by+c*z)
a, b, c are numbers input by the user
x, y, z are variables to be derived
variable constraint is that x, y, z are positive and x+y+z<=1
This can actually be transformed into a linear optimization problem.
max ln(1+ax+by+cz) <--> max (ax+by+cz) s.t. ax+by+cz > -1
This means that it is a linear optimization problem (with one more constraint) that you can easily handle with whatever C++ methods together with your Convex Optimization knowledge.
Reminders to write a good code:
Check the validity of input value.
Since the input value can be negative, you need to consider this circumstance which can yield different results.
P.S.
This problem seems to be Off Topic on SO.
If it is your homework, it is for your own good to write the code yourself. Besides, we do not have enough time to write that for you.
This should have been a comment if I had more reputation.

How to extract matrixL() and matrixU() when using Eigen::CholmodSupernodalLLT?

I'm trying to use Eigen::CholmodSupernodalLLT for Cholesky decomposition, however, it seems that I could not get matrixL() and matrixU(). How can I extract matrixL() and matrixU() from Eigen::CholmodSupernodalLLT for future use?
A partial answer to integrate what others have said.
Consider Y ~ MultivariateNormal(0, A). One may want to (1) evaluate the (log-)likelihood (a multivariate normal density), (2) sample from such density.
For (1), it is necessary to solve Ax = b where A is symmetric positive-definite, and compute its log-determinant. (2) requires L such that A = L * L.transpose() since Y ~ MultivariateNormal(0, A) can be found as Y = L u where u ~ MultivariateNormal(0, I).
A Cholesky LLT or LDLT decomposition is useful because chol(A) can be used for both purposes. Solving Ax=b is easy given the decomposition, andthe (log)determinant can be easily derived from the (sum)product of the (log-)components of D or the diagonal of L. By definition L can then be used for sampling.
So, in Eigen one can use:
Eigen::SimplicialLDLT solver(A) (or Eigen::SimplicialLLT), when solver.solve(b) and calculate the determinant using solver.vectorD().diag(). Useful because if A is a covariance matrix, then solver can be used for likelihood evaluations, and matrixL() for sampling.
Eigen::CholmodDecomposition does not give access to matrixL() or vectorD() but exposes .logDeterminant() to achieve the (1) goal but not (2).
Eigen::PardisoLDLT does not give access to matrixL() or vectorD() and does not expose a way to get the determinant.
In some applications, step (2) - sampling - can be done at a later stage so Eigen::CholmodDecomposition is enough. At least in my configuration, Eigen::CholmodDecomposition works 2 to 5 times faster than Eigen::SimplicialLDLT (I guess because of the permutations done under the hood to facilitate parallelization)
Example: in Bayesian spatial Gaussian process regression, the spatial random effects can be integrated out and do not need to be sampled. So MCMC can proceed swiftly with Eigen::CholmodDecomposition to achieve convergence for the uknown parameters. The spatial random effects can then be recovered in parallel using Eigen::SimplicialLDLT. Typically this is only a small part of the computations but having matrixL() directly from CholmodDecomposition would simplify them a bit.
You cannot do this using the given class. The class you are referencing is equotation solver (which indeed uses cholesky decomposition). To decompose your matrix you should rather use Eigen::LLT. Code example from their website:
MatrixXd A(3,3);
A << 4,-1,2, -1,6,0, 2,0,5;
LLT<MatrixXd> lltOfA(A);
MatrixXd L = lltOfA.matrixL();
MatrixXd U = lltOfA.matrixU();
As reported somewhere else, e.g., it cannot be done easily.
I am copying a possible recommendation (answered by Gael Guennebaud himself), even if somewhat old:
If you really need access to the factor to do your own cooking, then
better use the built-in SimplicialL{D}LT<> class. Extracting the
factors from the supernodal internal represations of Cholmod/Pardiso
is indeed not straightforward and very rarely needed. We have to
check, but if Cholmod/Pardiso provide routines to manipulate the
factors, like applying it to a vector, then we could let
matrix{L,U}() return a pseudo expression wrapping these routines.
Developing code for extracting this is likely beyond SO, and probably a topic for a feature request.
Of course, the solution with LLT is at hand (but not the topic of the OP).

Least Squares Solution of Overdetermined Linear Algebraic Equation Ax = By

I have a linear algebraic equation of the form Ax=By. Where A is a matrix of 6x5, x is vector of size 5, B a matrix of 6x6 and y vector of size 6. A, B and y are known variables and their values are accessed in real time coming from the sensors. x is unknown and has to find. One solution is to find Least Square Estimation that is x = [(A^T*A)^-1]*(A^T)B*y. This is conventional solution of linear algebraic equations. I used Eigen QR Decomposition to solve this as below
matrixA = getMatrixA();
matrixB = getMatrixB();
vectorY = getVectorY();
//LSE Solution
Eigen::ColPivHouseholderQR<Eigen::MatrixXd> dec1(matrixA);
vectorX = dec1.solve(matrixB*vectorY);//
Everything is fine until now. But when I check the errore = Ax-By, its not zero always. Error is not very big but even not ignorable. Is there any other type of decomposition which is more reliable? I have gone through one of the page but could not understand the meaning or how to implement this. Below are lines from the reference how to solve the problem. Could anybody suggest me how to implement this?
The solution of such equations Ax = Byis obtained by forming the error vector e = Ax-By and the finding the unknown vector x that minimizes the weighted error (e^T*W*e), where W is a weighting matrix. For simplicity, this weighting matrix is chosen to be of the form W = K*S, where S is a constant diagonal scaling matrix, and K is scalar weight. Hence the solution to the equation becomes
x = [(A^T*W*A)^-1]*(A^T)*W*B*y
I did not understand how to form the matrix W.
Your statement " But when I check the error e = Ax-By, its not zero always. " almost always will be true, regardless of your technique, or what weighting you choose. When you have an over-described system, you are basically trying to fit a straight line to a slew of points. Unless, by chance, all the points can be placed exactly on a single perfectly straight line, there will be some error. So no matter what technique you use to choose the line, (weights and so on) you will always have some error if the points are not colinear. The alternative would be to use some kind of spline, or in higher dimensions to allow for warping. In those cases, you can choose to fit all the points exactly to a more complicated shape, and hence result with 0 error.
So the choice of a weight matrix simply changes which straight line you will use by giving each point a slightly different weight. So it will not ever completely remove the error. But if you had a few particular points that you care more about than the others, you can give the error on those points higher weight when choosing the least square error fit.
For spline fitting see:
http://en.wikipedia.org/wiki/Spline_interpolation
For the really nicest spline curve interpolation you can use Centripital Catmull-Rom, which in addition to finding a curve to fit all the points, will prevent unnecessary loops and self intersections that can sometimes come up during abrupt changes in the data direction.
Catmull-rom curve with no cusps and no self-intersections

Two point boundary with odeint

I am trying to solve two point boundary problem with odeint. My equation has the form of
y'' + a*y' + b*y + c = 0
It is pretty trivial when I have boundary conditions of y(x_1) = y_1 , y'(x_2) = y_2, but when boundary conditions are y(x_1) = y_1 , y(x_2) = y_2 I am lost. Does anybody know the way to deal with problems like this with odeint or other scientific library?
In this case you need a shooting method. odeint does not have such a method, it solved the initial value problem (IVP) which is your first case. I think in the Numerical Recipies this method is explained and you can use Boost.Odeint to do the time stepping.
An alternative and more efficient method to solve this type of problem is finite differences or finite elements method. For finite differences you can check Numerical Recipes. For finite elements I recommend dealii library.
Another approach is to use b-splines: Assuming you do know the initial x0 and final xfinal points of integration, then you can expand the solution y(x) in a b-spline basis, defined over (x0,xfinal), i.e.
y(x)= \sum_{i=1}^n A_i*B_i(x),
where A_i are constant coefficients to be determined, and B_i(x) are b-spline basis (well defined polynomial functions, that can be differentiated numerically). For scientific applications you can find an implementation of b-splines in GSL.
With this substitution the boundary value problem is reduced to a linear problem, since (am using Einstein summation for repeated indices):
A_i*[ B_i''(x) + a*B_i'(x) + b*B_i(x)] + c =0
You can choose a set of points x and create a linear system from the above equation. You can find information for this type of method in the following review paper "Applications of B-splines in Atomic and Molecular Physics" - H Bachau, E Cormier, P Decleva, J E Hansen and F Martín
http://iopscience.iop.org/0034-4885/64/12/205/
I do not know of any library solving directly this problem, but there are several libraries for B-splines (I recommend GSL for your needs), that will allow you to form the linear system. See this stackoverflow question:
Spline, B-Spline and NURBS C++ library

Removing unsolvable equations from an underdetermined system

My program tries to solve a system of linear equations. In order to do that, it assembles matrix coeff_matrix and vector value_vector, and uses Eigen to solve them like:
Eigen::VectorXd sol_vector = coeff_matrix
.colPivHouseholderQr().solve(value_vector);
The problem is that the system can be both over- and under-determined. In the former case, Eigen either gives a correct or uncorrect solution, and I check the solution using coeff_matrix * sol_vector - value_vector.
However, please consider the following system of equations:
a + b - c = 0
c - d = 0
c = 11
- c + d = 0
In this particular case, Eigen solves the three latter equations correctly but also gives solutions for a and b.
What I would like to achieve is that only the equations which have only one solution would be solved, and the remaining ones (the first equation here) would be retained in the system.
In other words, I'm looking for a method to find out which equations can be solved in a given system of equations at the time, and which cannot because there will be more than one solution.
Could you suggest any good way of achieving that?
Edit: please note that in most cases the matrix won't be square. I've added one more row here just to note that over-determination can happen too.
I think what you want to is the singular value decomposition (SVD), which will give you exact what you want. After SVD, "the equations which have only one solution will be solved", and the solution is pseudoinverse. It will also give you the null space (where infinite solutions come from) and left null space (where inconsistency comes from, i.e. no solution).
Based on the SVD comment, I was able to do something like this:
Eigen::FullPivLU<Eigen::MatrixXd> lu = coeff_matrix.fullPivLu();
Eigen::VectorXd sol_vector = lu.solve(value_vector);
Eigen::VectorXd null_vector = lu.kernel().rowwise().sum();
AFAICS, the null_vector rows corresponding to single solutions are 0s while the ones corresponding to non-determinate solutions are 1s. I can reproduce this throughout all my examples with the default treshold Eigen has.
However, I'm not sure if I'm doing something correct or just noticed a random pattern.
What you need is to calculate the determinant of your system. If the determinant is 0, then you have an infinite number of solutions. If the determinant is very small, the solution exists, but I wouldn't trust the solution found by a computer (it will lead to numerical instabilities).
Here is a link to what is the determinant and how to calculate it: http://en.wikipedia.org/wiki/Determinant
Note that Gaussian elimination should also work: http://en.wikipedia.org/wiki/Gaussian_elimination
With this method, you end up with lines of 0s if there are an infinite number of solutions.
Edit
In case the matrix is not square, you first need to extract a square matrix. There are two cases:
You have more variables than equations: then you have either no solution, or an infinite number of them.
You have more equations than variables: in this case, find a square sub-matrix of non-null determinant. Solve for this matrix and check the solution. If the solution doesn't fit, it means you have no solution. If the solution fits, it means the extra equations were linearly-dependant on the extract ones.
In both case, before checking the dimension of the matrix, remove rows and columns with only 0s.
As for the gaussian elimination, it should work directly with non-square matrices. However, this time, you should check that the number of non-empty row (i.e. a row with some non-0 values) is equal to the number of variable. If it's less you have an infinite number of solution, and if it's more, you don't have any solutions.