[CppAD] LuSolve nan

[Kasper Kristensen]

Hello all,

I have attached a small code example which tapes the mapping
A->log(abs(det(A))) for a 1x1 matrix A=1 using LuSolve.
I was expecting the tape to be valid for all non-zero matrices A.
However if I evaluate the tape with a matrix for which det(A)<0 the
evaluation gives nan (but the Jacobian is correct!).

Best regards
Kasper

PROGRAM OUTPUT:

A=2
{ 0.693147 }{ 0.5 }
A=-2
{ nan }{ -0.5 }

CODE:

# include <CppAD/CppAD.h>
using namespace CppAD;
int main()
{
   // dimension of the matrix
   size_t n = 1;

   // Dummy arguments to LuSolve
   vector<AD<double> > B(0);
   vector<AD<double> > X(0);

   // construct ADFun object F:A->log(abs(det(A)))
   vector<AD<double> > A(n);
   vector<AD<double> > y(n);
   A[0]=1;
   Independent(A);
   AD<double> logdet;
   LuSolve(n, 0, A, B, X, logdet);
   y[0]=logdet;
   ADFun<double> F(A,y);

   // evaluate the tape
   vector<double> A0(n);
   A0[0]=2;
   std::cout << "A=2\n";
   std::cout << F.Forward(0,A0) ;
   std::cout << F.Jacobian(A0) ;
   A0[0]=-2;
   std::cout << "\nA=-2\n";
   std::cout << F.Forward(0,A0) ;
   std::cout << F.Jacobian(A0) << "\n";

}