\left(\left(x \cdot y + \frac{z \cdot t}{16}\right) - \frac{a \cdot b}{4}\right) + c\mathsf{fma}\left(z, \frac{t}{16}, \mathsf{fma}\left(y, x, \mathsf{fma}\left(-\frac{a}{4}, b, c\right)\right)\right)double f(double x, double y, double z, double t, double a, double b, double c) {
double r180875 = x;
double r180876 = y;
double r180877 = r180875 * r180876;
double r180878 = z;
double r180879 = t;
double r180880 = r180878 * r180879;
double r180881 = 16.0;
double r180882 = r180880 / r180881;
double r180883 = r180877 + r180882;
double r180884 = a;
double r180885 = b;
double r180886 = r180884 * r180885;
double r180887 = 4.0;
double r180888 = r180886 / r180887;
double r180889 = r180883 - r180888;
double r180890 = c;
double r180891 = r180889 + r180890;
return r180891;
}
double f(double x, double y, double z, double t, double a, double b, double c) {
double r180892 = z;
double r180893 = t;
double r180894 = 16.0;
double r180895 = r180893 / r180894;
double r180896 = y;
double r180897 = x;
double r180898 = a;
double r180899 = 4.0;
double r180900 = r180898 / r180899;
double r180901 = -r180900;
double r180902 = b;
double r180903 = c;
double r180904 = fma(r180901, r180902, r180903);
double r180905 = fma(r180896, r180897, r180904);
double r180906 = fma(r180892, r180895, r180905);
return r180906;
}



Bits error versus x



Bits error versus y



Bits error versus z



Bits error versus t



Bits error versus a



Bits error versus b



Bits error versus c
Initial program 0.1
Simplified0.0
Final simplification0.0
herbie shell --seed 2020060 +o rules:numerics
(FPCore (x y z t a b c)
:name "Diagrams.Solve.Polynomial:quartForm from diagrams-solve-0.1, C"
:precision binary64
(+ (- (+ (* x y) (/ (* z t) 16)) (/ (* a b) 4)) c))