我正在尝试修改示例here:
# include <cppad/cppad.hpp>
namespace { // ---------------------------------------------------------
// define the template function JacobianCases<Vector> in empty namespace
template <typename Vector>
bool JacobianCases()
{ bool ok = true;
using CppAD::AD;
using CppAD::NearEqual;
double eps99 = 99.0 * std::numeric_limits<double>::epsilon();
using CppAD::exp;
using CppAD::sin;
using CppAD::cos;
// domain space vector
size_t n = 2;
CPPAD_TESTVECTOR(AD<double>) X(n);
X[0] = 1.;
X[1] = 2.;
// declare independent variables and starting recording
CppAD::Independent(X);
// a calculation between the domain and range values
AD<double> Square = X[0] * X[0];
// range space vector
size_t m = 3;
CPPAD_TESTVECTOR(AD<double>) Y(m);
Y[0] = Square * exp( X[1] );
Y[1] = Square * sin( X[1] );
Y[2] = Square * cos( X[1] );
// create f: X -> Y and stop tape recording
CppAD::ADFun<double> f(X, Y);
// new value for the independent variable vector
Vector x(n);
x[0] = 2.;
x[1] = 1.;
// compute the derivative at this x
Vector jac( m * n );
jac = f.Jacobian(x);
/*
F'(x) = [ 2 * x[0] * exp(x[1]) , x[0] * x[0] * exp(x[1]) ]
[ 2 * x[0] * sin(x[1]) , x[0] * x[0] * cos(x[1]) ]
[ 2 * x[0] * cos(x[1]) , -x[0] * x[0] * sin(x[i]) ]
*/
ok &= NearEqual( 2.*x[0]*exp(x[1]), jac[0*n+0], eps99, eps99);
ok &= NearEqual( 2.*x[0]*sin(x[1]), jac[1*n+0], eps99, eps99);
ok &= NearEqual( 2.*x[0]*cos(x[1]), jac[2*n+0], eps99, eps99);
ok &= NearEqual( x[0] * x[0] *exp(x[1]), jac[0*n+1], eps99, eps99);
ok &= NearEqual( x[0] * x[0] *cos(x[1]), jac[1*n+1], eps99, eps99);
ok &= NearEqual(-x[0] * x[0] *sin(x[1]), jac[2*n+1], eps99, eps99);
return ok;
}
} // End empty namespace
# include <vector>
# include <valarray>
bool Jacobian(void)
{ bool ok = true;
// Run with Vector equal to three different cases
// all of which are Simple Vectors with elements of type double.
ok &= JacobianCases< CppAD::vector <double> >();
ok &= JacobianCases< std::vector <double> >();
ok &= JacobianCases< std::valarray <double> >();
return ok;
}
我试图通过以下方式修改它:
设G是本例中计算的雅可比jac
,在行中:
jac = f.Jacobian(x);
并且,如示例中所示,让X
成为自变量。我想构造一个新函数H
,它是jac
的函数,即H(jacobian(X))
=某事,这样H是自动不同的。一个例子可能是H(X) = jacobian( jacobian(X)[0])
,即jacobian(X)
w.r.t X
的第一个元素的jacobian(种类的二阶导数)。
问题在于此处所写的jac
类型为Vector
,这是原始double
上的参数化类型,而不是AD<double>
。据我所知,这意味着输出不是自动不同的。
我正在寻找一些关于是否可以在更大的操作中使用雅可比行列的建议,并采用更大操作的雅可比行列式(与任何算术运算符不同)或者如果这是不可能的话。
编辑:这已经被提了一次赏金,但是我再次提出来看看是否有更好的解决方案,因为我觉得这很重要。更清楚一点,“正确”答案所需的要素是:a)计算任意阶导数的方法。
b)一种不必先验地指定衍生物顺序的智能方法。如果必须在编译时知道最大阶导数,则无法通过算法确定导数的阶数。此外,在当前给出的答案中指定一个非常大的订单将导致内存分配问题,并且我认为会出现性能问题。
c)从最终用户中抽象出衍生品订单的模板。这很重要,因为很难跟踪所需衍生物的顺序。如果b)得到解决,这可能是“免费”的。
如果有人能够解决这个问题,那将是一项非常有用的贡献和非常有用的操作。
答案 0 :(得分:5)
如果要嵌套函数,则应嵌套AD&lt;&gt;同样。您可以将Jacobians作为其他函数嵌套,例如,请参阅下面的代码片段,它通过嵌套Jacobian计算双重导数
#include <cstring>
#include <iostream> // standard input/output
#include <vector> // standard vector
#include <cppad/cppad.hpp> // the CppAD package http://www.coin-or.org/CppAD/
// main program
int main(void)
{ using CppAD::AD; // use AD as abbreviation for CppAD::AD
using std::vector; // use vector as abbreviation for std::vector
size_t i; // a temporary index
// domain space vector
auto Square = [](auto t){return t*t;};
vector< AD<AD<double>> > X(1); // vector of domain space variables
// declare independent variables and start recording operation sequence
CppAD::Independent(X);
// range space vector
vector< AD<AD<double>> > Y(1); // vector of ranges space variables
Y[0] = Square(X[0]); // value during recording of operations
// store operation sequence in f: X -> Y and stop recording
CppAD::ADFun<AD<double>> f(X, Y);
// compute derivative using operation sequence stored in f
vector<AD<double>> jac(1); // Jacobian of f (m by n matrix)
vector<AD<double>> x(1); // domain space vector
CppAD::Independent(x);
jac = f.Jacobian(x); // Jacobian for operation sequence
CppAD::ADFun<double> f2(x, jac);
vector<double> result(1);
vector<double> x_res(1);
x_res[0]=15.;
result=f2.Jacobian(x_res);
// print the results
std::cout << "f'' computed by CppAD = " << result[0] << std::endl;
}
作为旁注,由于实现表达模板和自动区分的C ++ 14或11变得更容易并且可以用更少的努力完成,如图所示。在这个视频结束时https://www.youtube.com/watch?v=cC9MtflQ_nI(抱歉质量很差)。如果我必须实现相当简单的符号操作,我将从头开始使用现代C ++:您可以编写更简单的代码,并且您可以轻松地理解错误。
修改强> 推广用于构建任意顺序导数的示例可以是模板元编程练习。下面的代码段显示可以使用模板递归
#include <cstring>
#include <iostream>
#include <vector>
#include <cppad/cppad.hpp>
using CppAD::AD;
using std::vector;
template<typename T>
struct remove_ad{
using type=T;
};
template<typename T>
struct remove_ad<AD<T>>{
using type=T;
};
template<int N>
struct derivative{
using type = AD<typename derivative<N-1>::type >;
static constexpr int order = N;
};
template<>
struct derivative<0>{
using type = double;
static constexpr int order = 0;
};
template<typename T>
struct Jac{
using value_type = typename remove_ad<typename T::type>::type;
template<typename P, typename Q>
auto operator()(P & X, Q & Y){
CppAD::ADFun<value_type> f(X, Y);
vector<value_type> jac(1);
vector<value_type> x(1);
CppAD::Independent(x);
jac = f.Jacobian(x);
return Jac<derivative<T::order-1>>{}(x, jac);
}
};
template<>
struct Jac<derivative<1>>{
using value_type = derivative<0>::type;
template<typename P, typename Q>
auto operator()(P & x, Q & jac){
CppAD::ADFun<value_type> f2(x, jac);
vector<value_type> res(1);
vector<value_type> x_res(1);
x_res[0]=15.;
return f2.Jacobian(x_res);
}
};
int main(void)
{
constexpr int order=4;
auto Square = [](auto t){return t*t;};
vector< typename derivative<order>::type > X(1);
vector< typename derivative<order>::type > Y(1);
CppAD::Independent(X);
Y[0] = Square(X[0]);
auto result = Jac<derivative<order>>{}(X, Y);
std::cout << "f'' computed by CppAD = " << result[0] << std::endl;
}
答案 1 :(得分:1)
CppAD中有一项新功能,可以消除对AD