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作为我项目的一部分，我需要在 C++中以封闭形式求解四次多项式。

A*x 4 + B*x 3 + C*x 2 + D*x + E = 0

我为此找到了几个链接。其中之一在这里。但它计算所有根，而我只想要真正的根。算法主要使用Ferrari的方法来降低阶数。

bool solveQuartic(double a, double b, double c, double d, double e, double &root)
{
// I switched to this method, and it seems to be more numerically stable.
// http://www.gamedev.n...topic_id=451048 

// When a or (a and b) are magnitudes of order smaller than C,D,E
// just ignore them entirely. This seems to happen because of numerical
// inaccuracies of the line-circle algorithm. I wanted a robust solver,
// so I put the fix here instead of there.
if(a == 0.0 || abs(a/b) < 1.0e-5 || abs(a/c) < 1.0e-5 || abs(a/d) < 1.0e-5)
    return solveCubic(b, c, d, e, root);

double B = b/a, C = c/a, D = d/a, E = e/a;
double BB = B*B;
double I = -3.0*BB*0.125 + C;
double J = BB*B*0.125 - B*C*0.5 + D;
double K = -3*BB*BB/256.0 + C*BB/16.0 - B*D*0.25 + E;

double z;
bool foundRoot2 = false, foundRoot3 = false, foundRoot4 = false, foundRoot5 = false;
if(solveCubic(1.0, I+I, I*I - 4*K, -(J*J), z))
{
    double value = z*z*z + z*z*(I+I) + z*(I*I - 4*K) - J*J;

    double p = sqrt(z);
    double r = -p;
    double q = (I + z - J/p)*0.5;
    double s = (I + z + J/p)*0.5;

    bool foundRoot = false, foundARoot;
    double aRoot;
    foundRoot = solveQuadratic(1.0, p, q, root);
    root -= B/4.0;

    foundARoot = solveQuadratic(1.0, r, s, aRoot);
    aRoot -= B/4.0;
    if((foundRoot && foundARoot && ((aRoot < root && aRoot >= 0.0) 
        || root < 0.0)) || (!foundRoot && foundARoot)) 
    {
        root = aRoot;
        foundRoot = true;
    }

    foundARoot = solveQuadraticOther(1.0, p, q, aRoot);
    aRoot -= B/4.0;
    if((foundRoot && foundARoot && ((aRoot < root && aRoot >= 0.0) 
        || root < 0.0)) || (!foundRoot && foundARoot)) 
    {
        root = aRoot;
        foundRoot = true;
    }

    foundARoot = solveQuadraticOther(1.0, r, s, aRoot);
    aRoot -= B/4.0;
    if((foundRoot && foundARoot && ((aRoot < root && aRoot >= 0.0) 
        || root < 0.0)) || (!foundRoot && foundARoot)) 
    {
        root = aRoot;
        foundRoot = true;
    }
    return foundRoot;
}
return false;
}

这使用了solveCubic()，它给出了实解和虚解：

bool solveCubic(double &a, double &b, double &c, double &d, double &root)
{
if(a == 0.0 || abs(a/b) < 1.0e-6)
    return solveQuadratic(b, c, d, root);

double B = b/a, C = c/a, D = d/a;

double Q = (B*B - C*3.0)/9.0, QQQ = Q*Q*Q;
double R = (2.0*B*B*B - 9.0*B*C + 27.0*D)/54.0, RR = R*R;

// 3 real roots
if(RR<QQQ)
{
    /* This sqrt and division is safe, since RR >= 0, so QQQ > RR,    */
    /* so QQQ > 0.  The acos is also safe, since RR/QQQ < 1, and      */
    /* thus R/sqrt(QQQ) < 1.                                     */
    double theta = acos(R/sqrt(QQQ));
    /* This sqrt is safe, since QQQ >= 0, and thus Q >= 0             */
    double r1, r2, r3;
    r1 = r2 = r3 = -2.0*sqrt(Q);
    r1 *= cos(theta/3.0);
    r2 *= cos((theta+2*PI)/3.0);
    r3 *= cos((theta-2*PI)/3.0);

    r1 -= B/3.0;
    r2 -= B/3.0;
    r3 -= B/3.0; 

    root = 1000000.0;

    if(r1 >= 0.0) root = r1;
    if(r2 >= 0.0 && r2 < root) root = r2;
    if(r3 >= 0.0 && r3 < root) root = r3;

    return true;
}
// 1 real root
else
{
    double A2 = -pow(fabs®+sqrt(RR-QQQ),1.0/3.0);
    if (A2!=0.0) {
        if (R<0.0) A2 = -A2; 
        root = A2 + Q/A2; 
    }
    root -= B/3.0;
    return true;
}
}

以下是一些解释代码的链接。solveCubic和solveQuartic

有没有人可以修改代码来求解实根的四次多项式？

我想尽可能有效地实现它。顺便说一句，如果有人为此目的引入了一个有用的库，例如 LAPACK，我将不胜感激（它似乎无法直接计算四次多项式的根）。

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1 回答 1

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对于实根以封闭形式求解该方程的最有效方法可能是对所有根以封闭形式求解它，然后丢弃虚构的根。

您可能认为您可以使用 try/catch 对来确定是否出现虚数，但这不是一个很好的策略，因为您在计算实根时生成的一些中间值可能是虚数。

因此，您可以尝试使用 C++ 复数库（参见此处或此处）进行计算。

然后，检查数字的虚部是否非零，如果是则丢弃它。但请记住，浮点数学是不精确的，因此“零”包括一系列非常接近于零的数字。

于 2012-10-20T19:15:56.037 回答