Question

作为我项目的一部分，我需要在C ++中以封闭形式解决四次多项式。

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

我找到了几个链接到这个目的。其中一个是here。但它计算所有的根源，而我只想要真正的根源。算法主要使用法拉利的方法来减少顺序。

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solveCubic（），它提供了实数和虚数解决方案：

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

是否有人可以修改代码来解决实际根的四次多项式？

我想尽可能高效地实施它。顺便说一句，如果有人为此目的引入了一个有用的库，比如LAPACK，我会很感激（它似乎无法直接计算四次多项式的根）。

Answer 1

对于真正的根来说，以封闭形式求解这个方程的最有效方法可能是以封闭形式求解所有根，然后丢弃虚构的根。

您可能认为可以使用try / catch对来确定虚数是否正在出现，但这不是一个非常好的策略，因为您在计算实根时生成的某些中间值可能是虚构的。

因此，您可以尝试使用C ++复数库（请参阅here或here）进行计算。

之后，检查数字的虚部是否为非零，如果丢弃它。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。

C ++求解四次根（四阶多项式）

1 个答案: