使用自定义张量数据结构 boost::odeint 的错误输出

Question

基于@Piotr Skotnicki 的回答我定义了以下数据类型：

#ifndef TENSOR_HPP
#define TENSOR_HPP

#include <vector>
#include <array>
#include <boost/numeric/odeint.hpp>

namespace nex {

template <size_t...> struct seq {};
template <size_t N, size_t... Is> struct gen_seq : gen_seq<N-1, N-1, Is...> {};
template <size_t... Is> struct gen_seq<0, Is...> { using type = seq<Is...>; };

template<typename T, typename S>
class tensor;

template<typename T, size_t... Is>
class tensor<T, seq<Is...>>
{

    typedef std::vector<T> vector;

public:

    // constructor
    tensor ();

    tensor (decltype(Is)...);

    // iterators
    using iterator       =  typename vector::iterator;
    using const_iterator =  typename vector::const_iterator;

    iterator begin()
    { return m_v.begin(); }

    const_iterator begin() const
    { return m_v.begin(); }

    iterator end() 
    { return m_v.end(); }

    const_iterator end() const
    { return m_v.end(); }

    // resizing
    size_t size() const; 

    void resize( const size_t  );

    // operators

    const T& operator() (decltype(Is)...) const;

    T& operator() (decltype(Is)...); 

private:

    // helper functions
    template <size_t I>
    auto index(const std::array<size_t, sizeof...(Is)>& a) const
    -> typename std::enable_if<I == 0, size_t>::type 
    {
        return a[I];
    }

    template <size_t I>
    auto index(const std::array<size_t, sizeof...(Is)>& a) const
    -> typename std::enable_if<I != 0, size_t>::type 
    {
        return index<I-1>(a) * s[I] + a[I];
    }

    size_t mult(size_t N) 
    {
        return N;
    }

    template <typename... S>
    size_t mult(size_t N, S... Ns) 
    {
        return N * mult(Ns...);
    }

    vector m_v;
    const std::array<size_t, sizeof...(Is)> s;

    }; // class tensor

 template<typename T, size_t... Is    
 tensor<T, seq<Is...>> :: tensor ()
   : m_v(), s{ { 0 } }
{}

template<typename T, size_t... Is>
tensor<T, seq<Is...>> :: tensor (decltype(Is)... size)
   : m_v( mult( size... ) ), s{ { size... } }
{}

template<typename T, size_t... Is>
size_t tensor<T, seq<Is...>> :: size () const 
{ return m_v.size(); }

template<typename T, size_t... Is>
void tensor<T, seq<Is...>> :: resize (const size_t n)
{ m_v.resize( n ); }

template<typename T, size_t... Is>
const T& tensor<T, seq<Is...>> :: operator() (decltype(Is)... n) const
{ return m_v.at(index<sizeof...(Is)-1>( { { n... } } ) ); }

template<typename T, size_t... Is>    
T& tensor<T, seq<Is...>> :: operator() (decltype(Is)... n) 
{ return m_v.at(index<sizeof...(Is)-1>( { { n... } } ) ); }

} // namespace nex

namespace boost { namespace numeric { namespace odeint {

template<typename T, size_t... Is>
struct is_resizeable< nex::tensor<T, nex::seq<Is...>> >
{
    typedef boost::true_type type;
    static const bool value = type::value;
};
} // namespace odeint
} // namespace numeric
} // namespace boost

template <typename T, std::size_t N>
using tensor = nex::tensor<T, typename nex::gen_seq<N>::type>;

#endif // TENSOR_HPP

然后我想在boost::odeint中使用这个数据类型来集成非常基本的问题

#include "tensor.hpp"

typedef tensor<double, 1> state_type;

void lorenz( const state_type &x , state_type &dxdt , const double t )
{

    dxdt(0) = x(0);
    dxdt(1) = x(1);
    dxdt(2) = x(2);
}

void write_lorenz( const state_type &x , const double t )
{
    std::cout << t << '\t' << x(0) << '\t' << x(1) << '\t' << x(2) << std::endl;
}


using namespace boost::numeric::odeint;

int main()
{
    state_type x(3);
    x(0) = 1.0 ; x(1) = 10.0 ; x(2) = 10.0;


    integrate_const( runge_kutta4< state_type >() , lorenz , x , 0.0 , 1.0 , 0.1, write_lorenz );
}

它工作得很好，即使对于不同的方程组（比如洛伦兹吸引子）也是如此。不幸的是，当我想使用张量的相同容量但定义不同时，一切都出错了

#include "tensor.hpp"

typedef tensor<double, 2> state_type;

void lorenz( const state_type &x , state_type &dxdt , const double t )
{

    dxdt(0,0) = x(0,0);
    dxdt(1,0) = x(1,0);
    dxdt(2,0) = x(2,0);
}

void write_lorenz( const state_type &x , const double t )
{
    std::cout << t << '\t' << x(0,0) << '\t' << x(1,0) << '\t' << x(2,0) << std::endl;
}


using namespace boost::numeric::odeint;

int main()
{
    state_type x(3,1);
    x(0,0) = 1.0 ; x(1,0) = 10.0 ; x(2,0) = 10.0;


    integrate_const( runge_kutta4< state_type >() , lorenz , x , 0.0 , 1.0 , 0.1, write_lorenz );
}

您可以运行自己编写代码并检查 x(1,0) x(2,0) 是否完全没有变化。我认为 operator() 的适当定义足以让 odeint 正常工作。我对 boost::odeint 知之甚少，所以这可能只是一个基本错误。我指望你的帮助。

更新

当我使用 std::array 将张量定义为固定大小的数组时，它工作正常。但是我不需要添加 boost::is_resizeable。我认为调整大小需要不同的定义（我真的不知道）。

Answer 1

的确，问题出在调整大小上。目前，调整张量的大小只会改变基础向量的大小 m_v。虽然这确保了正确的内存分配，但这还不够。您还需要正确设置维度，即成员 s。你也可以这样想：只给定一个整数是不可能调整多维张量的大小的，你需要设置每个维度的大小。唯一的例外是一维张量，对于它来说一切正常，对于任何更高的维度，这都必须失败。

以下 tensor.hpp 解决了这个问题并且在我的快速测试中似乎运行良好：

#ifndef TENSOR_HPP
#define TENSOR_HPP

#include <vector>
#include <array>
#include <boost/numeric/odeint.hpp>

namespace nex {

template <size_t...> struct seq {};
template <size_t N, size_t... Is> struct gen_seq : gen_seq<N-1, N-1, Is...> {};
template <size_t... Is> struct gen_seq<0, Is...> { using type = seq<Is...>; };

template<typename T, typename S>
class tensor;

template<typename T, size_t... Is>
class tensor<T, seq<Is...>>
{

    typedef std::vector<T> vector;

public:

    // constructor
    tensor ();

    tensor (decltype(Is)...);

    // iterators
    using iterator       =  typename vector::iterator;
    using const_iterator =  typename vector::const_iterator;

    iterator begin()
    { return m_v.begin(); }

    const_iterator begin() const
    { return m_v.begin(); }

    iterator end() 
    { return m_v.end(); }

    const_iterator end() const
    { return m_v.end(); }

    // resizing
    size_t size() const; 

    void resize( const tensor<T, seq<Is...>>& );

    // operators

    const T& operator() (decltype(Is)...) const;

    T& operator() (decltype(Is)...); 

private:

    // helper functions
    template <size_t I>
    auto index(const std::array<size_t, sizeof...(Is)>& a) const
    -> typename std::enable_if<I == 0, size_t>::type 
    {
        return a[I];
    }

    template <size_t I>
    auto index(const std::array<size_t, sizeof...(Is)>& a) const
    -> typename std::enable_if<I != 0, size_t>::type 
    {
        return index<I-1>(a) * s[I] + a[I];
    }

    size_t mult(size_t N) 
    {
        return N;
    }

    template <typename... S>
    size_t mult(size_t N, S... Ns) 
    {
        return N * mult(Ns...);
    }

    vector m_v;
    std::array<size_t, sizeof...(Is)> s;

    }; // class tensor

    template<typename T, size_t... Is>
 tensor<T, seq<Is...>> :: tensor ()
        : m_v(), s{ { 0 } }
{}

template<typename T, size_t... Is>
tensor<T, seq<Is...>> :: tensor (decltype(Is)... size)
   : m_v( mult( size... ) ), s{ { size... } }
{}

template<typename T, size_t... Is>
size_t tensor<T, seq<Is...>> :: size () const 
{ return m_v.size(); }

template<typename T, size_t... Is>
void tensor<T, seq<Is...>> :: resize (const tensor<T, seq<Is...>> &x)
{ 
    m_v.resize( x.size() );
    s = x.s;
}

template<typename T, size_t... Is>
const T& tensor<T, seq<Is...>> :: operator() (decltype(Is)... n) const
{ return m_v.at(index<sizeof...(Is)-1>( { { n... } } ) ); }

template<typename T, size_t... Is>    
T& tensor<T, seq<Is...>> :: operator() (decltype(Is)... n) 
{ return m_v.at(index<sizeof...(Is)-1>( { { n... } } ) ); }

} // namespace nex

namespace boost { namespace numeric { namespace odeint {

template<typename T, size_t... Is>
struct is_resizeable< nex::tensor<T, nex::seq<Is...>> >
{
    typedef boost::true_type type;
    static const bool value = type::value;
};

template<typename T, size_t... Is>
struct resize_impl< nex::tensor<T, nex::seq<Is...>>, nex::tensor<T, nex::seq<Is...>> >
{
    typedef nex::tensor<T, nex::seq<Is...>> state_type;
    static void resize( state_type &x1 , const state_type &x2 )
    {
        x1.resize(x2);
    }
};

} // namespace odeint
} // namespace numeric
} // namespace boost

template <typename T, std::size_t N>
using tensor = nex::tensor<T, typename nex::gen_seq<N>::type>;

#endif // TENSOR_HPP

使用自定义张量数据结构 boost::odeint 的错误输出

wrong output of boost::odeint with custom tensor data structure

boost

c++11

odeint

更新