c++ - 编译时 std::ratio 平方根的有理逼近

标签 c++ c++11 recursion template-meta-programming

我正在尝试找到 std::ratio 的平方根的有理近似值在编译时。导出 ellipsoid parameters 将非常有用来自为坐标转换定义的参数，这些参数本身定义为 std::ratio .

有一个question about finding powers/roots of std::ratio ，但作为该问题的条件，如果比率没有整数根，则可以失败，这与我想要的相反。相反，我想找到最接近合理的近似值。

我提出了以下基于 Newton-Raphson Method 计算根的元程序，已知仅需几次迭代即可产生(相对)准确的结果:

namespace detail
{
    // implementation of ratio_sqrt
    // N is an std::ratio, not an int
    template<class N , class K = std::ratio<4>, std::intmax_t RecursionDepth = 5>
    struct ratio_sqrt_impl
    {
        static_assert(/* N is std::ratio */);

        // Recursive Newton-Raphson
        // EQUATION: K_{n+1} = (K_{n} - N / K_{n}) / 2
        // WHERE:
        // K_{n+1} : square root approximation
        // K_{n}   : previous square root approximation
        // N       : ratio whose square root we are finding
        using type = typename ratio_sqrt_impl<N, std::ratio_subtract<K,
          std::ratio_divide<std::ratio_subtract<std::ratio_multiply<K, K>, N>, 
          std::ratio_multiply<std::ratio<2>, K>>>, RecursionDepth - 1>::type;
    };
    template<class N, class K>
    struct ratio_sqrt_impl<N, K, 1>
    {
        using type = K;
    };
}

template<class Ratio>
using ratio_sqrt = typename detail::ratio_sqrt_impl<Ratio>::type;

使用示例:

// Error calculations
using rt2 = ratio_sqrt<std::ratio<2>>;
std::cout << (sqrt(2) - ((double)rt2::num / rt2::den))/sqrt(2) << std::endl;

scalar_t result = pow<2>(scalar_t((double)rt2::num / rt2::den));
std::cout << (2 - result.toDouble()) / 2 << std::endl;

using rt4 = ratio_sqrt<std::ratio<4>>;
std::cout << (sqrt(4) - ((double)rt4::num / rt4::den)) / sqrt(4) << std::endl;

using rt10 = ratio_sqrt<std::ratio<10>>;
std::cout << (sqrt(10) - ((double)rt10::num / rt10::den)) / sqrt(10) << std::endl;

产生结果:

1.46538e-05 // sqrt(2)

4.64611e-08 // sqrt(4)

2.38737e-15 // sqrt(10)

这对于某些应用程序来说肯定是不错的。

问题

  1. 这里最大的问题是固定的递归深度。这些比率变得很大，非常快，所以对于 > 100 的根，这会疯狂地溢出。然而，递归深度太小，你会失去所有的准确性。

有没有一种好方法可以使递归适应溢出深度限制，然后将类型设置为在此之前的一两次迭代？ (我说了几次迭代，因为将余量保持在整数大小以便稍后进行进一步计算可能会很好)

  1. 4 的初始条件似乎非常神奇，因为它可以为 < 100 的根产生最低的误差，但是否有更系统的方法来设置它？

编辑:

我没有使用 constexpr 寻找任何解决方案，因为我必须支持的编译器并不统一拥有它。

增加递归深度的问题是 std::ratio 的 num/denom仅在几次递归后溢出。表示的平方根的准确性实际上还可以，但我需要找到一个通用的解决方案，将递归深度限制在比率不会溢出的点(因此，不会编译)。例如。 ratio_sqrt<std::ratio<2>>可以在溢出前到达深度 5，但是 ratio_sqrt<std::ratio<1000>>限制为 4 个。

最佳答案

问题是您使用牛顿法计算平方根，如果您想获得数值近似值，这是正确的，但是，如果您想要找到最佳有理近似值，则必须使用 continued fraction .这是我的程序的结果:

hidden $ g++ -std=c++11 sqrt.cpp && ./a.out
sqrt(2/1) ~ 239/169, error=1.23789e-05, eps=0.0001
sqrt(2/1) ~ 114243/80782, error=5.41782e-11, eps=1e-10
sqrt(2/1) ~ 3880899/2744210, error=4.68514e-14, eps=1e-13
sqrt(2/1) ~ 131836323/93222358, error=0, eps=1e-16
sqrt(2/10001) ~ 1/71, error=5.69215e-05, eps=0.0001
sqrt(2/10001) ~ 1977/139802, error=2.18873e-11, eps=1e-10
sqrt(2/10001) ~ 13860/980099, error=7.36043e-15, eps=1e-13
sqrt(2/10001) ~ 1950299/137913860, error=3.64292e-17, eps=1e-16
sqrt(10001/2) ~ 495/7, error=7.21501e-05, eps=0.0001
sqrt(10001/2) ~ 980099/13860, error=3.68061e-11, eps=1e-10
sqrt(10001/2) ~ 415701778/5878617, error=1.42109e-14, eps=1e-13
sqrt(10001/2) ~ 970297515/13721393, error=0, eps=1e-16
sqrt(1060/83) ~ 461/129, error=2.19816e-05, eps=0.0001
sqrt(1060/83) ~ 2139943/598809, error=9.07718e-13, eps=1e-10
sqrt(1060/83) ~ 6448815/1804538, error=1.77636e-14, eps=1e-13
sqrt(1060/83) ~ 545951360/152770699, error=4.44089e-16, eps=1e-16
sqrt(1/12494234) ~ 1/3534, error=5.75083e-08, eps=0.0001
sqrt(1/12494234) ~ 32/113111, error=2.9907e-11, eps=1e-10
sqrt(1/12494234) ~ 419/1481047, error=6.02961e-14, eps=1e-13
sqrt(1/12494234) ~ 129879/459085688, error=4.49944e-18, eps=1e-16
sqrt(82378/1) ~ 18369/64, error=5.40142e-05, eps=0.0001
sqrt(82378/1) ~ 37361979/130174, error=1.16529e-11, eps=1e-10
sqrt(82378/1) ~ 1710431766/5959367, error=5.68434e-14, eps=1e-13
sqrt(82378/1) ~ 15563177213/54224136, error=0, eps=1e-16
sqrt(68389/3346222) ~ 197/1378, error=4.13769e-07, eps=0.0001
sqrt(68389/3346222) ~ 17801/124517, error=2.17069e-11, eps=1e-10
sqrt(68389/3346222) ~ 581697/4068938, error=4.30211e-15, eps=1e-13
sqrt(68389/3346222) ~ 16237871/113583000, error=2.77556e-17, eps=1e-16
sqrt(2/72) ~ 1/6, error=0, eps=0.0001
sqrt(10000/1) ~ 100/1, error=0, eps=0.0001
sqrt(0/20) ~ 0/1, error=0, eps=0.0001

我的程序使用(几乎)最小的分子和分母找到满足误差范围的近似值。如果您在我的代码中将 int 更改为更长的整数类型，它可以使用更小的 eps。

我的代码(它在 g++-4.8.4 下编译，恕我直言，如果允许使用 constexpr，代码会简单得多):

#include <cstdint>
#include <cmath>
#include <iostream>
#include <ratio>
#include <type_traits>
#include <utility>

using namespace std;

using Zero = ratio<0>;
using One = ratio<1>;
template <typename R> using Square = ratio_multiply<R, R>;

// Find the largest integer N such that Predicate<N>::value is true.
template <template <intmax_t N> class Predicate, typename Enabled = void>
struct BinarySearch {
  template <intmax_t N>
  struct SafeDouble_ {
    const intmax_t static value = 2 * N;
    static_assert(value > 0, "Overflows when computing 2 * N");
  };

  template <intmax_t Lower, intmax_t Upper, typename Enabled1 = void>
  struct DoubleSidedSearch_ : DoubleSidedSearch_<Lower, Lower+(Upper-Lower)/2> {};

  template <intmax_t Lower, intmax_t Upper>
  struct DoubleSidedSearch_<Lower, Upper, typename enable_if<Upper-Lower==1>::type> : integral_constant<intmax_t, Lower> {};

  template <intmax_t Lower, intmax_t Upper>
  struct DoubleSidedSearch_<Lower, Upper, typename enable_if<(Upper-Lower>1 && Predicate<Lower+(Upper-Lower)/2>::value)>::type>
      : DoubleSidedSearch_<Lower+(Upper-Lower)/2, Upper> {};

  template <intmax_t Lower, typename Enabled1 = void>
  struct SingleSidedSearch_ : DoubleSidedSearch_<Lower, SafeDouble_<Lower>::value> {};

  template <intmax_t Lower>
  struct SingleSidedSearch_<Lower, typename enable_if<Predicate<SafeDouble_<Lower>::value>::value>::type>
      : SingleSidedSearch_<SafeDouble_<Lower>::value> {};

  const static intmax_t value = SingleSidedSearch_<1>::value;
};

template <template <intmax_t N> class Predicate>
struct BinarySearch<Predicate, typename enable_if<!Predicate<1>::value>::type> : integral_constant<intmax_t, 0> {};

// Find largest integer N such that N<=sqrt(R)
template <typename R>
struct Integer {
  template <intmax_t N> using Predicate_ = ratio_less_equal<ratio<N>, ratio_divide<R, ratio<N>>>;
  const static intmax_t value = BinarySearch<Predicate_>::value;
};

template <typename R>
struct IsPerfectSquare {
  const static intmax_t DenSqrt_ = Integer<ratio<R::den>>::value;
  const static intmax_t NumSqrt_ = Integer<ratio<R::num>>::value;
  const static bool value = DenSqrt_ * DenSqrt_ == R::den && NumSqrt_ * NumSqrt_ == R::num;
  using Sqrt = ratio<NumSqrt_, DenSqrt_>;
};

// Represents sqrt(P)-Q.
template <typename Tp, typename Tq>
struct Remainder {
  using P = Tp;
  using Q = Tq;
};

// Represents 1/R = I + Rem where R is a Remainder.
template <typename R>
struct Reciprocal {
  using P_ = typename R::P;
  using Q_ = typename R::Q;
  using Den_ = ratio_subtract<P_, Square<Q_>>;
  using A_ = ratio_divide<Q_, Den_>;
  using B_ = ratio_divide<P_, Square<Den_>>;
  const static intmax_t I_ = (A_::num + Integer<ratio_multiply<B_, Square<ratio<A_::den>>>>::value) / A_::den;
  using I = ratio<I_>;
  using Rem = Remainder<B_, ratio_subtract<I, A_>>;
};

// Expands sqrt(R) to continued fraction:
// f(x)=C1+1/(C2+1/(C3+1/(...+1/(Cn+x)))) = (U*x+V)/(W*x+1) and sqrt(R)=f(Rem).
// The error |f(Rem)-V| = |(U-W*V)x/(W*x+1)| <= |U-W*V|*Rem <= |U-W*V|/I' where
// I' is the integer part of reciprocal of Rem.
template <typename R, intmax_t N>
struct ContinuedFraction {
  template <typename T>
  using Abs_ = typename conditional<ratio_less<T, Zero>::value, ratio_subtract<Zero, T>, T>::type;

  using Last_ = ContinuedFraction<R, N-1>;
  using Reciprocal_ = Reciprocal<typename Last_::Rem>;
  using Rem = typename Reciprocal_::Rem;
  using I_ = typename Reciprocal_::I;
  using Den_ = ratio_add<typename Last_::W, I_>;
  using U = ratio_divide<typename Last_::V, Den_>;
  using V = ratio_divide<ratio_add<typename Last_::U, ratio_multiply<typename Last_::V, I_>>, Den_>;
  using W = ratio_divide<One, Den_>;
  using Error = Abs_<ratio_divide<ratio_subtract<U, ratio_multiply<V, W>>, typename Reciprocal<Rem>::I>>;
};

template <typename R>
struct ContinuedFraction<R, 1> {
  using U = One;
  using V = ratio<Integer<R>::value>;
  using W = Zero;
  using Rem = Remainder<R, V>;
  using Error = ratio_divide<One, typename Reciprocal<Rem>::I>;
};

template <typename R, typename Eps, intmax_t N=1, typename Enabled = void>
struct Sqrt_ : Sqrt_<R, Eps, N+1> {};

template <typename R, typename Eps, intmax_t N>
struct Sqrt_<R, Eps, N, typename enable_if<ratio_less_equal<typename ContinuedFraction<R, N>::Error, Eps>::value>::type> {
  using type = typename ContinuedFraction<R, N>::V;
};

template <typename R, typename Eps, typename Enabled = void>
struct Sqrt {
  static_assert(ratio_greater_equal<R, Zero>::value, "R can't be negative");
};

template <typename R, typename Eps>
struct Sqrt<R, Eps, typename enable_if<ratio_greater_equal<R, Zero>::value && IsPerfectSquare<R>::value>::type> {
  using type = typename IsPerfectSquare<R>::Sqrt;
};

template <typename R, typename Eps>
struct Sqrt<R, Eps, typename enable_if<(ratio_greater_equal<R, Zero>::value && !IsPerfectSquare<R>::value)>::type> : Sqrt_<R, Eps> {};

// Test finding sqrt(N/D) with error 1/Eps
template <intmax_t N, intmax_t D, intmax_t Eps>
void test() {
  using T = typename Sqrt<ratio<N, D>, ratio<1, Eps>>::type;
  cout << "sqrt(" << N << "/" << D << ") ~ " << T::num << "/" << T::den << ", "
       << "error=" << abs(sqrt(N/(double)D) - T::num/(double)T::den) << ", "
       << "eps=" << 1/(double)Eps << endl;
}

template <intmax_t N, intmax_t D>
void testAll() {
  test<N, D, 10000>();
  test<N, D, 10000000000>();
  test<N, D, 10000000000000>();
  test<N, D, 10000000000000000>();
}

int main() {
  testAll<2, 1>();
  testAll<2, 10001>();
  testAll<10001, 2>();

  testAll<1060, 83>();
  testAll<1, 12494234>();
  testAll<82378, 1>();
  testAll<68389, 3346222>();

  test<2, 72, 10000>();
  test<10000, 1, 10000>();
  test<0, 20, 10000>();
  // static assertion failure.
  // test<-10001, 2, 100000>();
}

为 MSVC 2013 修改了 BinarySearch

由于 Visual Studio 2013 编译器的模板推导实现中的错误，在该平台上构建时必须修改 BinarySearch:

template <template <std::intmax_t N> class Predicate, typename enabled = void>
struct BinarySearch {
    template <std::intmax_t N>
    struct SafeDouble_ {
        static const std::intmax_t value = 2 * N;
        static_assert(value > 0, "Overflows when computing 2 * N");
    };

    template <intmax_t Lower, intmax_t Upper, typename Condition1 = void, typename Condition2 = void>
    struct DoubleSidedSearch_ : DoubleSidedSearch_<Lower, Upper,
        typename std::conditional<(Upper - Lower == 1), std::true_type, std::false_type>::type,
        typename std::conditional<((Upper - Lower>1 && Predicate<Lower + (Upper - Lower) / 2>::value)), std::true_type, std::false_type>::type> {};

    template <intmax_t Lower, intmax_t Upper>
    struct DoubleSidedSearch_<Lower, Upper, std::false_type, std::false_type> : DoubleSidedSearch_<Lower, Lower + (Upper - Lower) / 2> {};

    template <intmax_t Lower, intmax_t Upper, typename Condition2>
    struct DoubleSidedSearch_<Lower, Upper, std::true_type, Condition2> : std::integral_constant<intmax_t, Lower>{};

    template <intmax_t Lower, intmax_t Upper, typename Condition1>
    struct DoubleSidedSearch_<Lower, Upper, Condition1, std::true_type> : DoubleSidedSearch_<Lower + (Upper - Lower) / 2, Upper>{};

    template <std::intmax_t Lower, class enabled1 = void>
    struct SingleSidedSearch_ : SingleSidedSearch_<Lower, typename std::conditional<Predicate<SafeDouble_<Lower>::value>::value, std::true_type, std::false_type>::type>{};

    template <std::intmax_t Lower>
    struct SingleSidedSearch_<Lower, std::false_type> : DoubleSidedSearch_<Lower, SafeDouble_<Lower>::value> {};

    template <std::intmax_t Lower>
    struct SingleSidedSearch_<Lower, std::true_type> : SingleSidedSearch_<SafeDouble_<Lower>::value>{};

    const static std::intmax_t value = SingleSidedSearch_<1>::value;
};

关于c++ - 编译时 std::ratio 平方根的有理逼近，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36321295/

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