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@(@\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }@)@
Source: LuInvert
# ifndef CPPAD_LU_INVERT_HPP
# define CPPAD_LU_INVERT_HPP
# include <cppad/core/cppad_assert.hpp>
# include <cppad/utility/check_simple_vector.hpp>
# include <cppad/utility/check_numeric_type.hpp>

namespace CppAD { // BEGIN CppAD namespace

// LuInvert
template <typename SizeVector, typename FloatVector>
void LuInvert(
     const SizeVector  &ip,
     const SizeVector  &jp,
     const FloatVector &LU,
     FloatVector       &B )
{     size_t k; // column index in X
     size_t p; // index along diagonal in LU
     size_t i; // row index in LU and X

     typedef typename FloatVector::value_type Float;

     // check numeric type specifications
     CheckNumericType<Float>();

     // check simple vector class specifications
     CheckSimpleVector<Float, FloatVector>();
     CheckSimpleVector<size_t, SizeVector>();

     Float etmp;

     size_t n = ip.size();
     CPPAD_ASSERT_KNOWN(
          size_t(jp.size()) == n,
          "Error in LuInvert: jp must have size equal to n * n"
     );
     CPPAD_ASSERT_KNOWN(
          size_t(LU.size()) == n * n,
          "Error in LuInvert: Lu must have size equal to n * m"
     );
     size_t m = size_t(B.size()) / n;
     CPPAD_ASSERT_KNOWN(
          size_t(B.size()) == n * m,
          "Error in LuSolve: B must have size equal to a multiple of n"
     );

     // temporary storage for reordered solution
     FloatVector x(n);

     // loop over equations
     for(k = 0; k < m; k++)
     {     // invert the equation c = L * b
          for(p = 0; p < n; p++)
          {     // solve for c[p]
               etmp = B[ ip[p] * m + k ] / LU[ ip[p] * n + jp[p] ];
               B[ ip[p] * m + k ] = etmp;
               // subtract off effect on other variables
               for(i = p+1; i < n; i++)
                    B[ ip[i] * m + k ] -=
                         etmp * LU[ ip[i] * n + jp[p] ];
          }

          // invert the equation x = U * c
          p = n;
          while( p > 0 )
          {     --p;
               etmp       = B[ ip[p] * m + k ];
               x[ jp[p] ] = etmp;
               for(i = 0; i < p; i++ )
                    B[ ip[i] * m + k ] -=
                         etmp * LU[ ip[i] * n + jp[p] ];
          }

          // copy reordered solution into B
          for(i = 0; i < n; i++)
               B[i * m + k] = x[i];
     }
     return;
}
} // END CppAD namespace
# endif

Input File: omh/lu_invert_hpp.omh