求封闭式多项式根的算法
Algorithm for closed-form polynomial root finding
我正在寻找一种 稳健 算法（或描述算法的论文），它可以找到多项式的根（最好是到第 4 个 debree，但任何东西都可以）使用封闭形式的解决方案。我只对真正的根源感兴趣。
我第一次尝试求解二次方程涉及到这个（我也有类似风格的三次/四次方程代码，但现在让我们关注二次方程）：
/**
* @brief a simple quadratic equation solver
*
* With double-precision floating-point, this reaches 1e-12 worst-case and 1e-15 average
* precision of the roots (the value of the function in the roots). The roots can be however
* quite far from the true roots, up to 1e-10 worst-case and 1e-18 average absolute difference
* for cases when two roots exist. If only a single root exists, the worst-case precision is
* 1e-13 and average-case precision is 1e-18.
*
* With single-precision floating-point, this reaches 1e-3 worst-case and 1e-7 average
* precision of the roots (the value of the function in the roots). The roots can be however
* quite far from the true roots, up to 1e-1 worst-case and 1e-10 average absolute difference
* for cases when two roots exist. If only a single root exists, the worst-case precision is
* 1e+2 (!) and average-case precision is 1e-2. Do not use single-precision floating point,
* except if pressed by time.
*
* All the precision measurements are scaled by the maximum absolute coefficient value.
*
* @tparam T is data type of the arguments (default double)
* @tparam b_sort_roots is root sorting flag (if set, the roots are
* given in ascending (not absolute) value; default true)
* @tparam n_2nd_order_coeff_log10_thresh is base 10 logarithm of threshold
* on the first coefficient (if below threshold, the equation is a linear one; default -6)
* @tparam n_zero_discriminant_log10_thresh is base 10 logarithm of threshold
* on the discriminant (if below negative threshold, the equation does not
* have a real root, if below threshold, the equation has just a single solution; default -6)
*/
template <class T = double, const bool b_sort_roots = true,
const int n_2nd_order_coeff_log10_thresh = -6,
const int n_zero_discriminant_log10_thresh = -6>
class CQuadraticEq {
protected:
T a; /**< @brief the 2nd order coefficient */
T b; /**< @brief the 1st order coefficient */
T c; /**< @brief 0th order coefficient */
T p_real_root[2]; /**< @brief list of the roots (real parts) */
//T p_im_root[2]; // imaginary part of the roots
size_t n_real_root_num; /**< @brief number of real roots */
public:
/**
* @brief default constructor; solves for roots of \f$ax^2 + bx + c = 0\f$
*
* This finds roots of the given equation. It tends to find two identical roots instead of one, rather
* than missing one of two different roots - the number of roots found is therefore orientational,
* as the roots might have the same value.
*
* @param[in] _a is the 2nd order coefficient
* @param[in] _b is the 1st order coefficient
* @param[in] _c is 0th order coefficient
*/
CQuadraticEq(T _a, T _b, T _c) // ax2 + bx + c = 0
:a(_a), b(_b), c(_c)
{
T _aa = fabs(_a);
if(_aa < f_Power_Static(10, n_2nd_order_coeff_log10_thresh)) { // otherwise division by a yields large numbers, this is then more precise
p_real_root[0] = -_c / _b;
//p_im_root[0] = 0;
n_real_root_num = 1;
return;
}
// a simple linear equation
if(_aa < 1) { // do not divide always, that makes it worse
_b /= _a;
_c /= _a;
_a = 1;
// could copy the code here and optimize away division by _a (optimizing compiler might do it for us)
}
// improve numerical stability if the coeffs are very small
const double f_thresh = f_Power_Static(10, n_zero_discriminant_log10_thresh);
double f_disc = _b * _b - 4 * _a * _c;
if(f_disc < -f_thresh) // only really negative
n_real_root_num = 0; // only two complex roots
else if(/*fabs(f_disc) < f_thresh*/f_disc <= f_thresh) { // otherwise gives problems for double root situations
p_real_root[0] = T(-_b / (2 * _a));
n_real_root_num = 1;
} else {
f_disc = sqrt(f_disc);
int i = (b_sort_roots)? ((_a > 0)? 0 : 1) : 0; // produce sorted roots, if required
p_real_root[i] = T((-_b - f_disc) / (2 * _a));
p_real_root[1 - i] = T((-_b + f_disc) / (2 * _a));
//p_im_root[0] = 0;
//p_im_root[1] = 0;
n_real_root_num = 2;
}
}
/**
* @brief gets number of real roots
* @return Returns number of real roots (0 to 2).
*/
size_t n_RealRoot_Num() const
{
_ASSERTE(n_real_root_num >= 0);
return n_real_root_num;
}
/**
* @brief gets value of a real root
* @param[in] n_index is zero-based index of the root
* @return Returns value of the specified root.
*/
T f_RealRoot(size_t n_index) const
{
_ASSERTE(n_index < 2 && n_index < n_real_root_num);
return p_real_root[n_index];
}
/**
* @brief evaluates the equation for a given argument
* @param[in] f_x is value of the argument \f$x\f$
* @return Returns value of \f$ax^2 + bx + c\f$.
*/
T operator ()(T f_x) const
{
T f_x2 = f_x * f_x;
return f_x2 * a + f_x * b + c;
}
};
代码很烂，我讨厌所有的门槛。但是对于根在 [-100, 100] 区间的随机方程，这还不错：
root response precision 1e-100: 6315 cases
root response precision 1e-19: 2 cases
root response precision 1e-17: 2 cases
root response precision 1e-16: 6 cases
root response precision 1e-15: 6333 cases
root response precision 1e-14: 3765 cases
root response precision 1e-13: 241 cases
root response precision 1e-12: 3 cases
2-root solution precision 1e-100: 5353 cases
2-root solution precision 1e-19: 656 cases
2-root solution precision 1e-18: 4481 cases
2-root solution precision 1e-17: 2312 cases
2-root solution precision 1e-16: 455 cases
2-root solution precision 1e-15: 68 cases
2-root solution precision 1e-14: 7 cases
2-root solution precision 1e-13: 2 cases
1-root solution precision 1e-100: 3022 cases
1-root solution precision 1e-19: 38 cases
1-root solution precision 1e-18: 197 cases
1-root solution precision 1e-17: 68 cases
1-root solution precision 1e-16: 7 cases
1-root solution precision 1e-15: 1 cases
请注意，此精度与系数的大小有关，系数的大小通常在 10^6 范围内（因此最终精度远非完美，但可能大部分可用）。然而，如果没有阈值，它几乎没有用。
我尝试过使用多精度算法，通常效果很好，但往往会拒绝许多根，这仅仅是因为多项式的系数不是多精度并且某些多项式无法精确表示（如果有双在二阶多项式中求根，它主要要么将它分成两个根（我不介意）或者说根本没有根）。如果我想恢复甚至可能稍微不精确的根，我的代码会变得复杂且充满阈值。
到目前为止，我尝试过使用 CCmath，但要么我无法正确使用它，要么精度非常差。此外，它在 plrt().
中使用迭代（非封闭形式）求解器
我尝试过使用 GNU 科学库 gsl_poly_solve_quadratic() 但这似乎是一种幼稚的方法，并且在数值上不是很稳定。
天真地使用 std::complex 数字也被证明是一个非常糟糕的主意，因为精度和速度都可能很糟糕（尤其是三次/四次方程，其中代码包含大量超越函数）。
将根恢复为复数是唯一的出路吗？这样就不会遗漏任何根，用户可以 select 根需要多精确（因此忽略不太精确的根中的小虚部）。
这并没有真正回答您的问题，但我认为您可以改进现有的内容，因为您目前在 b^2 >> ac 时遇到 'loss of significance' 问题。在这种情况下，您最终会得到一个类似于 (-b + (b + eps))/(2 * a) 的公式，其中 b 的取消可能会丢失 eps.
中的许多有效数字。
处理此问题的正确方法是使用 'normal' 方程求二次方程的一个根，使用鲜为人知的 'alternative' 或 'upside down' 方程求另一个根。您选择哪条路取决于 _b.
的符号
按照以下这几行对代码进行更改应该会减少由此产生的错误。
if( _b > 0 ) {
p_real_root[i] = T((-_b - f_disc) / (2 * _a));
p_real_root[1 - i] = T((2 * _c) / (-_b - f_disc));
}
else{
p_real_root[i] = T((2 * _c) / (-_b + f_disc));
p_real_root[1 - i] = T((-_b + f_disc) / (2 * _a));
}
我正在寻找一种 稳健 算法（或描述算法的论文），它可以找到多项式的根（最好是到第 4 个 debree，但任何东西都可以）使用封闭形式的解决方案。我只对真正的根源感兴趣。
我第一次尝试求解二次方程涉及到这个（我也有类似风格的三次/四次方程代码，但现在让我们关注二次方程）：
/**
* @brief a simple quadratic equation solver
*
* With double-precision floating-point, this reaches 1e-12 worst-case and 1e-15 average
* precision of the roots (the value of the function in the roots). The roots can be however
* quite far from the true roots, up to 1e-10 worst-case and 1e-18 average absolute difference
* for cases when two roots exist. If only a single root exists, the worst-case precision is
* 1e-13 and average-case precision is 1e-18.
*
* With single-precision floating-point, this reaches 1e-3 worst-case and 1e-7 average
* precision of the roots (the value of the function in the roots). The roots can be however
* quite far from the true roots, up to 1e-1 worst-case and 1e-10 average absolute difference
* for cases when two roots exist. If only a single root exists, the worst-case precision is
* 1e+2 (!) and average-case precision is 1e-2. Do not use single-precision floating point,
* except if pressed by time.
*
* All the precision measurements are scaled by the maximum absolute coefficient value.
*
* @tparam T is data type of the arguments (default double)
* @tparam b_sort_roots is root sorting flag (if set, the roots are
* given in ascending (not absolute) value; default true)
* @tparam n_2nd_order_coeff_log10_thresh is base 10 logarithm of threshold
* on the first coefficient (if below threshold, the equation is a linear one; default -6)
* @tparam n_zero_discriminant_log10_thresh is base 10 logarithm of threshold
* on the discriminant (if below negative threshold, the equation does not
* have a real root, if below threshold, the equation has just a single solution; default -6)
*/
template <class T = double, const bool b_sort_roots = true,
const int n_2nd_order_coeff_log10_thresh = -6,
const int n_zero_discriminant_log10_thresh = -6>
class CQuadraticEq {
protected:
T a; /**< @brief the 2nd order coefficient */
T b; /**< @brief the 1st order coefficient */
T c; /**< @brief 0th order coefficient */
T p_real_root[2]; /**< @brief list of the roots (real parts) */
//T p_im_root[2]; // imaginary part of the roots
size_t n_real_root_num; /**< @brief number of real roots */
public:
/**
* @brief default constructor; solves for roots of \f$ax^2 + bx + c = 0\f$
*
* This finds roots of the given equation. It tends to find two identical roots instead of one, rather
* than missing one of two different roots - the number of roots found is therefore orientational,
* as the roots might have the same value.
*
* @param[in] _a is the 2nd order coefficient
* @param[in] _b is the 1st order coefficient
* @param[in] _c is 0th order coefficient
*/
CQuadraticEq(T _a, T _b, T _c) // ax2 + bx + c = 0
:a(_a), b(_b), c(_c)
{
T _aa = fabs(_a);
if(_aa < f_Power_Static(10, n_2nd_order_coeff_log10_thresh)) { // otherwise division by a yields large numbers, this is then more precise
p_real_root[0] = -_c / _b;
//p_im_root[0] = 0;
n_real_root_num = 1;
return;
}
// a simple linear equation
if(_aa < 1) { // do not divide always, that makes it worse
_b /= _a;
_c /= _a;
_a = 1;
// could copy the code here and optimize away division by _a (optimizing compiler might do it for us)
}
// improve numerical stability if the coeffs are very small
const double f_thresh = f_Power_Static(10, n_zero_discriminant_log10_thresh);
double f_disc = _b * _b - 4 * _a * _c;
if(f_disc < -f_thresh) // only really negative
n_real_root_num = 0; // only two complex roots
else if(/*fabs(f_disc) < f_thresh*/f_disc <= f_thresh) { // otherwise gives problems for double root situations
p_real_root[0] = T(-_b / (2 * _a));
n_real_root_num = 1;
} else {
f_disc = sqrt(f_disc);
int i = (b_sort_roots)? ((_a > 0)? 0 : 1) : 0; // produce sorted roots, if required
p_real_root[i] = T((-_b - f_disc) / (2 * _a));
p_real_root[1 - i] = T((-_b + f_disc) / (2 * _a));
//p_im_root[0] = 0;
//p_im_root[1] = 0;
n_real_root_num = 2;
}
}
/**
* @brief gets number of real roots
* @return Returns number of real roots (0 to 2).
*/
size_t n_RealRoot_Num() const
{
_ASSERTE(n_real_root_num >= 0);
return n_real_root_num;
}
/**
* @brief gets value of a real root
* @param[in] n_index is zero-based index of the root
* @return Returns value of the specified root.
*/
T f_RealRoot(size_t n_index) const
{
_ASSERTE(n_index < 2 && n_index < n_real_root_num);
return p_real_root[n_index];
}
/**
* @brief evaluates the equation for a given argument
* @param[in] f_x is value of the argument \f$x\f$
* @return Returns value of \f$ax^2 + bx + c\f$.
*/
T operator ()(T f_x) const
{
T f_x2 = f_x * f_x;
return f_x2 * a + f_x * b + c;
}
};
代码很烂，我讨厌所有的门槛。但是对于根在 [-100, 100] 区间的随机方程，这还不错：
root response precision 1e-100: 6315 cases
root response precision 1e-19: 2 cases
root response precision 1e-17: 2 cases
root response precision 1e-16: 6 cases
root response precision 1e-15: 6333 cases
root response precision 1e-14: 3765 cases
root response precision 1e-13: 241 cases
root response precision 1e-12: 3 cases
2-root solution precision 1e-100: 5353 cases
2-root solution precision 1e-19: 656 cases
2-root solution precision 1e-18: 4481 cases
2-root solution precision 1e-17: 2312 cases
2-root solution precision 1e-16: 455 cases
2-root solution precision 1e-15: 68 cases
2-root solution precision 1e-14: 7 cases
2-root solution precision 1e-13: 2 cases
1-root solution precision 1e-100: 3022 cases
1-root solution precision 1e-19: 38 cases
1-root solution precision 1e-18: 197 cases
1-root solution precision 1e-17: 68 cases
1-root solution precision 1e-16: 7 cases
1-root solution precision 1e-15: 1 cases
请注意，此精度与系数的大小有关，系数的大小通常在 10^6 范围内（因此最终精度远非完美，但可能大部分可用）。然而，如果没有阈值，它几乎没有用。
我尝试过使用多精度算法，通常效果很好，但往往会拒绝许多根，这仅仅是因为多项式的系数不是多精度并且某些多项式无法精确表示（如果有双在二阶多项式中求根，它主要要么将它分成两个根（我不介意）或者说根本没有根）。如果我想恢复甚至可能稍微不精确的根，我的代码会变得复杂且充满阈值。
到目前为止，我尝试过使用 CCmath，但要么我无法正确使用它，要么精度非常差。此外，它在 plrt().
我尝试过使用 GNU 科学库 gsl_poly_solve_quadratic() 但这似乎是一种幼稚的方法，并且在数值上不是很稳定。
天真地使用 std::complex 数字也被证明是一个非常糟糕的主意，因为精度和速度都可能很糟糕（尤其是三次/四次方程，其中代码包含大量超越函数）。
将根恢复为复数是唯一的出路吗？这样就不会遗漏任何根，用户可以 select 根需要多精确（因此忽略不太精确的根中的小虚部）。
这并没有真正回答您的问题，但我认为您可以改进现有的内容，因为您目前在 b^2 >> ac 时遇到 'loss of significance' 问题。在这种情况下，您最终会得到一个类似于 (-b + (b + eps))/(2 * a) 的公式，其中 b 的取消可能会丢失 eps.
处理此问题的正确方法是使用 'normal' 方程求二次方程的一个根，使用鲜为人知的 'alternative' 或 'upside down' 方程求另一个根。您选择哪条路取决于 _b.
按照以下这几行对代码进行更改应该会减少由此产生的错误。
if( _b > 0 ) {
p_real_root[i] = T((-_b - f_disc) / (2 * _a));
p_real_root[1 - i] = T((2 * _c) / (-_b - f_disc));
}
else{
p_real_root[i] = T((2 * _c) / (-_b + f_disc));
p_real_root[1 - i] = T((-_b + f_disc) / (2 * _a));
}