ABCF->ab-angle b

Percentage Accurate: 18.6% → 48.4%
Time: 26.4s
Alternatives: 15
Speedup: 3.0×

Specification

?
\[\begin{array}{l} \\ \begin{array}{l} t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ \frac{-\sqrt{\left(2 \cdot \left(t\_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t\_0} \end{array} \end{array} \]
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (pow B 2.0) (* (* 4.0 A) C))))
   (/
    (-
     (sqrt
      (*
       (* 2.0 (* t_0 F))
       (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
    t_0)))
double code(double A, double B, double C, double F) {
	double t_0 = pow(B, 2.0) - ((4.0 * A) * C);
	return -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / t_0;
}
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b ** 2.0d0) - ((4.0d0 * a) * c)
    code = -sqrt(((2.0d0 * (t_0 * f)) * ((a + c) - sqrt((((a - c) ** 2.0d0) + (b ** 2.0d0)))))) / t_0
end function
public static double code(double A, double B, double C, double F) {
	double t_0 = Math.pow(B, 2.0) - ((4.0 * A) * C);
	return -Math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / t_0;
}
def code(A, B, C, F):
	t_0 = math.pow(B, 2.0) - ((4.0 * A) * C)
	return -math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / t_0
function code(A, B, C, F)
	t_0 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	return Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_0 * F)) * Float64(Float64(A + C) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0))))))) / t_0)
end
function tmp = code(A, B, C, F)
	t_0 = (B ^ 2.0) - ((4.0 * A) * C);
	tmp = -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / t_0;
end
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$0 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\
\frac{-\sqrt{\left(2 \cdot \left(t\_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t\_0}
\end{array}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 15 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 18.6% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ \frac{-\sqrt{\left(2 \cdot \left(t\_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t\_0} \end{array} \end{array} \]
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (pow B 2.0) (* (* 4.0 A) C))))
   (/
    (-
     (sqrt
      (*
       (* 2.0 (* t_0 F))
       (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
    t_0)))
double code(double A, double B, double C, double F) {
	double t_0 = pow(B, 2.0) - ((4.0 * A) * C);
	return -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / t_0;
}
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b ** 2.0d0) - ((4.0d0 * a) * c)
    code = -sqrt(((2.0d0 * (t_0 * f)) * ((a + c) - sqrt((((a - c) ** 2.0d0) + (b ** 2.0d0)))))) / t_0
end function
public static double code(double A, double B, double C, double F) {
	double t_0 = Math.pow(B, 2.0) - ((4.0 * A) * C);
	return -Math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / t_0;
}
def code(A, B, C, F):
	t_0 = math.pow(B, 2.0) - ((4.0 * A) * C)
	return -math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / t_0
function code(A, B, C, F)
	t_0 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	return Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_0 * F)) * Float64(Float64(A + C) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0))))))) / t_0)
end
function tmp = code(A, B, C, F)
	t_0 = (B ^ 2.0) - ((4.0 * A) * C);
	tmp = -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / t_0;
end
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$0 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\
\frac{-\sqrt{\left(2 \cdot \left(t\_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t\_0}
\end{array}
\end{array}

Alternative 1: 48.4% accurate, 0.7× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ t_1 := \mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)\\ t_2 := C + \left(A - \mathsf{hypot}\left(B\_m, A - C\right)\right)\\ t_3 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-59}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sqrt{F \cdot t\_2} \cdot \sqrt{2 \cdot t\_1}}{-t\_1}\\ \mathbf{elif}\;{B\_m}^{2} \leq 10^{+62}:\\ \;\;\;\;\sqrt{t\_3 \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{t\_3}\\ \mathbf{elif}\;{B\_m}^{2} \leq 4 \cdot 10^{+260}:\\ \;\;\;\;\sqrt{2} \cdot \left(-\sqrt{F \cdot \frac{t\_2}{\mathsf{fma}\left(-4, A \cdot C, {B\_m}^{2}\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)}}{B\_m}\right)\right)\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C))
        (t_1 (fma C (* A -4.0) (pow B_m 2.0)))
        (t_2 (+ C (- A (hypot B_m (- A C)))))
        (t_3 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 5e-59)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (if (<= (pow B_m 2.0) 5e-8)
       (/ (* (sqrt (* F t_2)) (sqrt (* 2.0 t_1))) (- t_1))
       (if (<= (pow B_m 2.0) 1e+62)
         (*
          (sqrt (* t_3 (* F (* 2.0 (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C))))))))
          (/ -1.0 t_3))
         (if (<= (pow B_m 2.0) 4e+260)
           (*
            (sqrt 2.0)
            (- (sqrt (* F (/ t_2 (fma -4.0 (* A C) (pow B_m 2.0)))))))
           (-
            (expm1
             (log1p (/ (sqrt (* (* 2.0 F) (- A (hypot B_m A)))) B_m))))))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double t_1 = fma(C, (A * -4.0), pow(B_m, 2.0));
	double t_2 = C + (A - hypot(B_m, (A - C)));
	double t_3 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 5e-59) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else if (pow(B_m, 2.0) <= 5e-8) {
		tmp = (sqrt((F * t_2)) * sqrt((2.0 * t_1))) / -t_1;
	} else if (pow(B_m, 2.0) <= 1e+62) {
		tmp = sqrt((t_3 * (F * (2.0 * (A + (A + (-0.5 * (pow(B_m, 2.0) / C)))))))) * (-1.0 / t_3);
	} else if (pow(B_m, 2.0) <= 4e+260) {
		tmp = sqrt(2.0) * -sqrt((F * (t_2 / fma(-4.0, (A * C), pow(B_m, 2.0)))));
	} else {
		tmp = -expm1(log1p((sqrt(((2.0 * F) * (A - hypot(B_m, A)))) / B_m)));
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	t_1 = fma(C, Float64(A * -4.0), (B_m ^ 2.0))
	t_2 = Float64(C + Float64(A - hypot(B_m, Float64(A - C))))
	t_3 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-59)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	elseif ((B_m ^ 2.0) <= 5e-8)
		tmp = Float64(Float64(sqrt(Float64(F * t_2)) * sqrt(Float64(2.0 * t_1))) / Float64(-t_1));
	elseif ((B_m ^ 2.0) <= 1e+62)
		tmp = Float64(sqrt(Float64(t_3 * Float64(F * Float64(2.0 * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C)))))))) * Float64(-1.0 / t_3));
	elseif ((B_m ^ 2.0) <= 4e+260)
		tmp = Float64(sqrt(2.0) * Float64(-sqrt(Float64(F * Float64(t_2 / fma(-4.0, Float64(A * C), (B_m ^ 2.0)))))));
	else
		tmp = Float64(-expm1(log1p(Float64(sqrt(Float64(Float64(2.0 * F) * Float64(A - hypot(B_m, A)))) / B_m))));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$1 = N[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(C + N[(A - N[Sqrt[B$95$m ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-59], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-8], N[(N[(N[Sqrt[N[(F * t$95$2), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(2.0 * t$95$1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / (-t$95$1)), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e+62], N[(N[Sqrt[N[(t$95$3 * N[(F * N[(2.0 * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / t$95$3), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 4e+260], N[(N[Sqrt[2.0], $MachinePrecision] * (-N[Sqrt[N[(F * N[(t$95$2 / N[(-4.0 * N[(A * C), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision], (-N[(Exp[N[Log[1 + N[(N[Sqrt[N[(N[(2.0 * F), $MachinePrecision] * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / B$95$m), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision])]]]]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
t_1 := \mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)\\
t_2 := C + \left(A - \mathsf{hypot}\left(B\_m, A - C\right)\right)\\
t_3 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-59}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{-8}:\\
\;\;\;\;\frac{\sqrt{F \cdot t\_2} \cdot \sqrt{2 \cdot t\_1}}{-t\_1}\\

\mathbf{elif}\;{B\_m}^{2} \leq 10^{+62}:\\
\;\;\;\;\sqrt{t\_3 \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{t\_3}\\

\mathbf{elif}\;{B\_m}^{2} \leq 4 \cdot 10^{+260}:\\
\;\;\;\;\sqrt{2} \cdot \left(-\sqrt{F \cdot \frac{t\_2}{\mathsf{fma}\left(-4, A \cdot C, {B\_m}^{2}\right)}}\right)\\

\mathbf{else}:\\
\;\;\;\;-\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)}}{B\_m}\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 5 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000001e-59

    1. Initial program 27.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 33.7%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

    if 5.0000000000000001e-59 < (pow.f64 B #s(literal 2 binary64)) < 4.9999999999999998e-8

    1. Initial program 55.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified62.1%

      \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
    3. Add Preprocessing
    4. Step-by-step derivation
      1. pow1/262.1%

        \[\leadsto \frac{\color{blue}{{\left(F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)\right)}^{0.5}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      2. associate-*r*62.2%

        \[\leadsto \frac{{\color{blue}{\left(\left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      3. unpow-prod-down68.9%

        \[\leadsto \frac{\color{blue}{{\left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      4. associate-+r-69.1%

        \[\leadsto \frac{{\left(F \cdot \color{blue}{\left(\left(A + C\right) - \mathsf{hypot}\left(B, A - C\right)\right)}\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      5. hypot-undefine54.6%

        \[\leadsto \frac{{\left(F \cdot \left(\left(A + C\right) - \color{blue}{\sqrt{B \cdot B + \left(A - C\right) \cdot \left(A - C\right)}}\right)\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      6. unpow254.6%

        \[\leadsto \frac{{\left(F \cdot \left(\left(A + C\right) - \sqrt{\color{blue}{{B}^{2}} + \left(A - C\right) \cdot \left(A - C\right)}\right)\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      7. unpow254.6%

        \[\leadsto \frac{{\left(F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + \color{blue}{{\left(A - C\right)}^{2}}}\right)\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      8. +-commutative54.6%

        \[\leadsto \frac{{\left(F \cdot \left(\left(A + C\right) - \sqrt{\color{blue}{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      9. unpow254.6%

        \[\leadsto \frac{{\left(F \cdot \left(\left(A + C\right) - \sqrt{\color{blue}{\left(A - C\right) \cdot \left(A - C\right)} + {B}^{2}}\right)\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      10. unpow254.6%

        \[\leadsto \frac{{\left(F \cdot \left(\left(A + C\right) - \sqrt{\left(A - C\right) \cdot \left(A - C\right) + \color{blue}{B \cdot B}}\right)\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      11. hypot-define69.1%

        \[\leadsto \frac{{\left(F \cdot \left(\left(A + C\right) - \color{blue}{\mathsf{hypot}\left(A - C, B\right)}\right)\right)}^{0.5} \cdot {\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}^{0.5}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      12. pow1/269.1%

        \[\leadsto \frac{{\left(F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)\right)}^{0.5} \cdot \color{blue}{\sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    5. Applied egg-rr69.1%

      \[\leadsto \frac{\color{blue}{{\left(F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)\right)}^{0.5} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    6. Step-by-step derivation
      1. unpow1/269.1%

        \[\leadsto \frac{\color{blue}{\sqrt{F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)}} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      2. associate-+r-68.9%

        \[\leadsto \frac{\sqrt{F \cdot \color{blue}{\left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)}} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      3. +-commutative68.9%

        \[\leadsto \frac{\sqrt{F \cdot \color{blue}{\left(\left(C - \mathsf{hypot}\left(A - C, B\right)\right) + A\right)}} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      4. associate-+l-68.9%

        \[\leadsto \frac{\sqrt{F \cdot \color{blue}{\left(C - \left(\mathsf{hypot}\left(A - C, B\right) - A\right)\right)}} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      5. hypot-undefine54.5%

        \[\leadsto \frac{\sqrt{F \cdot \left(C - \left(\color{blue}{\sqrt{\left(A - C\right) \cdot \left(A - C\right) + B \cdot B}} - A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      6. unpow254.5%

        \[\leadsto \frac{\sqrt{F \cdot \left(C - \left(\sqrt{\color{blue}{{\left(A - C\right)}^{2}} + B \cdot B} - A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      7. unpow254.5%

        \[\leadsto \frac{\sqrt{F \cdot \left(C - \left(\sqrt{{\left(A - C\right)}^{2} + \color{blue}{{B}^{2}}} - A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      8. +-commutative54.5%

        \[\leadsto \frac{\sqrt{F \cdot \left(C - \left(\sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}} - A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      9. unpow254.5%

        \[\leadsto \frac{\sqrt{F \cdot \left(C - \left(\sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}} - A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      10. unpow254.5%

        \[\leadsto \frac{\sqrt{F \cdot \left(C - \left(\sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}} - A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      11. hypot-undefine68.9%

        \[\leadsto \frac{\sqrt{F \cdot \left(C - \left(\color{blue}{\mathsf{hypot}\left(B, A - C\right)} - A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    7. Simplified68.9%

      \[\leadsto \frac{\color{blue}{\sqrt{F \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]

    if 4.9999999999999998e-8 < (pow.f64 B #s(literal 2 binary64)) < 1.00000000000000004e62

    1. Initial program 9.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified18.2%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 16.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C} - -1 \cdot A\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. associate-*r/16.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\color{blue}{\frac{-0.5 \cdot {B}^{2}}{C}} - -1 \cdot A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. mul-1-neg16.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified16.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    7. Step-by-step derivation
      1. div-inv16.3%

        \[\leadsto \color{blue}{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
      2. associate-*l*16.3%

        \[\leadsto \sqrt{\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)}} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. associate-/l*16.3%

        \[\leadsto \sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(\color{blue}{-0.5 \cdot \frac{{B}^{2}}{C}} - \left(-A\right)\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    8. Applied egg-rr16.3%

      \[\leadsto \color{blue}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(-0.5 \cdot \frac{{B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]

    if 1.00000000000000004e62 < (pow.f64 B #s(literal 2 binary64)) < 4.00000000000000026e260

    1. Initial program 30.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in F around 0 43.7%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}\right)} \]
    4. Simplified64.0%

      \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{F \cdot \frac{C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)}{\mathsf{fma}\left(-4, C \cdot A, {B}^{2}\right)}}} \]

    if 4.00000000000000026e260 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 4.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 3.9%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg3.9%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative3.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow23.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow23.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define22.2%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified22.2%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
    6. Step-by-step derivation
      1. expm1-log1p-u22.0%

        \[\leadsto -\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)\right)} \]
      2. expm1-undefine3.5%

        \[\leadsto -\color{blue}{\left(e^{\mathsf{log1p}\left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)} - 1\right)} \]
      3. associate-*l/3.5%

        \[\leadsto -\left(e^{\mathsf{log1p}\left(\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}}\right)} - 1\right) \]
      4. pow1/23.5%

        \[\leadsto -\left(e^{\mathsf{log1p}\left(\frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\right)} - 1\right) \]
      5. pow1/23.5%

        \[\leadsto -\left(e^{\mathsf{log1p}\left(\frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}^{0.5}}}{B}\right)} - 1\right) \]
      6. pow-prod-down3.5%

        \[\leadsto -\left(e^{\mathsf{log1p}\left(\frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}^{0.5}}}{B}\right)} - 1\right) \]
    7. Applied egg-rr3.5%

      \[\leadsto -\color{blue}{\left(e^{\mathsf{log1p}\left(\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}^{0.5}}{B}\right)} - 1\right)} \]
    8. Step-by-step derivation
      1. expm1-define22.0%

        \[\leadsto -\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}^{0.5}}{B}\right)\right)} \]
      2. unpow1/222.0%

        \[\leadsto -\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}}{B}\right)\right) \]
      3. associate-*r*22.0%

        \[\leadsto -\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\color{blue}{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}}{B}\right)\right) \]
    9. Simplified22.0%

      \[\leadsto -\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\right)\right)} \]
  3. Recombined 5 regimes into one program.
  4. Final simplification36.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-59}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}\\ \mathbf{elif}\;{B}^{2} \leq 10^{+62}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;{B}^{2} \leq 4 \cdot 10^{+260}:\\ \;\;\;\;\sqrt{2} \cdot \left(-\sqrt{F \cdot \frac{C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)}{\mathsf{fma}\left(-4, A \cdot C, {B}^{2}\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\right)\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 48.6% accurate, 0.7× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ t_1 := A - \mathsf{hypot}\left(B\_m, A\right)\\ t_2 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-59}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_2\right) \cdot \left(2 \cdot t\_1\right)}}{-t\_2}\\ \mathbf{elif}\;{B\_m}^{2} \leq 10^{+62}:\\ \;\;\;\;\sqrt{t\_2 \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{t\_2}\\ \mathbf{elif}\;{B\_m}^{2} \leq 4 \cdot 10^{+260}:\\ \;\;\;\;\sqrt{2} \cdot \left(-\sqrt{F \cdot \frac{C + \left(A - \mathsf{hypot}\left(B\_m, A - C\right)\right)}{\mathsf{fma}\left(-4, A \cdot C, {B\_m}^{2}\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\left(2 \cdot F\right) \cdot t\_1}}{B\_m}\right)\right)\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C))
        (t_1 (- A (hypot B_m A)))
        (t_2 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 5e-59)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (if (<= (pow B_m 2.0) 5e-8)
       (/ (sqrt (* (* F t_2) (* 2.0 t_1))) (- t_2))
       (if (<= (pow B_m 2.0) 1e+62)
         (*
          (sqrt (* t_2 (* F (* 2.0 (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C))))))))
          (/ -1.0 t_2))
         (if (<= (pow B_m 2.0) 4e+260)
           (*
            (sqrt 2.0)
            (-
             (sqrt
              (*
               F
               (/
                (+ C (- A (hypot B_m (- A C))))
                (fma -4.0 (* A C) (pow B_m 2.0)))))))
           (- (expm1 (log1p (/ (sqrt (* (* 2.0 F) t_1)) B_m))))))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double t_1 = A - hypot(B_m, A);
	double t_2 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 5e-59) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else if (pow(B_m, 2.0) <= 5e-8) {
		tmp = sqrt(((F * t_2) * (2.0 * t_1))) / -t_2;
	} else if (pow(B_m, 2.0) <= 1e+62) {
		tmp = sqrt((t_2 * (F * (2.0 * (A + (A + (-0.5 * (pow(B_m, 2.0) / C)))))))) * (-1.0 / t_2);
	} else if (pow(B_m, 2.0) <= 4e+260) {
		tmp = sqrt(2.0) * -sqrt((F * ((C + (A - hypot(B_m, (A - C)))) / fma(-4.0, (A * C), pow(B_m, 2.0)))));
	} else {
		tmp = -expm1(log1p((sqrt(((2.0 * F) * t_1)) / B_m)));
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	t_1 = Float64(A - hypot(B_m, A))
	t_2 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-59)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	elseif ((B_m ^ 2.0) <= 5e-8)
		tmp = Float64(sqrt(Float64(Float64(F * t_2) * Float64(2.0 * t_1))) / Float64(-t_2));
	elseif ((B_m ^ 2.0) <= 1e+62)
		tmp = Float64(sqrt(Float64(t_2 * Float64(F * Float64(2.0 * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C)))))))) * Float64(-1.0 / t_2));
	elseif ((B_m ^ 2.0) <= 4e+260)
		tmp = Float64(sqrt(2.0) * Float64(-sqrt(Float64(F * Float64(Float64(C + Float64(A - hypot(B_m, Float64(A - C)))) / fma(-4.0, Float64(A * C), (B_m ^ 2.0)))))));
	else
		tmp = Float64(-expm1(log1p(Float64(sqrt(Float64(Float64(2.0 * F) * t_1)) / B_m))));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$1 = N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-59], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-8], N[(N[Sqrt[N[(N[(F * t$95$2), $MachinePrecision] * N[(2.0 * t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$2)), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e+62], N[(N[Sqrt[N[(t$95$2 * N[(F * N[(2.0 * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / t$95$2), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 4e+260], N[(N[Sqrt[2.0], $MachinePrecision] * (-N[Sqrt[N[(F * N[(N[(C + N[(A - N[Sqrt[B$95$m ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(-4.0 * N[(A * C), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision], (-N[(Exp[N[Log[1 + N[(N[Sqrt[N[(N[(2.0 * F), $MachinePrecision] * t$95$1), $MachinePrecision]], $MachinePrecision] / B$95$m), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision])]]]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
t_1 := A - \mathsf{hypot}\left(B\_m, A\right)\\
t_2 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-59}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{-8}:\\
\;\;\;\;\frac{\sqrt{\left(F \cdot t\_2\right) \cdot \left(2 \cdot t\_1\right)}}{-t\_2}\\

\mathbf{elif}\;{B\_m}^{2} \leq 10^{+62}:\\
\;\;\;\;\sqrt{t\_2 \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{t\_2}\\

\mathbf{elif}\;{B\_m}^{2} \leq 4 \cdot 10^{+260}:\\
\;\;\;\;\sqrt{2} \cdot \left(-\sqrt{F \cdot \frac{C + \left(A - \mathsf{hypot}\left(B\_m, A - C\right)\right)}{\mathsf{fma}\left(-4, A \cdot C, {B\_m}^{2}\right)}}\right)\\

\mathbf{else}:\\
\;\;\;\;-\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\left(2 \cdot F\right) \cdot t\_1}}{B\_m}\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 5 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000001e-59

    1. Initial program 27.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 33.7%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

    if 5.0000000000000001e-59 < (pow.f64 B #s(literal 2 binary64)) < 4.9999999999999998e-8

    1. Initial program 55.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified62.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around 0 49.2%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{-1 \cdot \sqrt{{A}^{2} + {B}^{2}}}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. mul-1-neg49.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-\sqrt{{A}^{2} + {B}^{2}}\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. +-commutative49.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. unpow249.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. unpow249.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      5. hypot-define56.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\color{blue}{\mathsf{hypot}\left(B, A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified56.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-\mathsf{hypot}\left(B, A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]

    if 4.9999999999999998e-8 < (pow.f64 B #s(literal 2 binary64)) < 1.00000000000000004e62

    1. Initial program 9.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified18.2%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 16.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C} - -1 \cdot A\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. associate-*r/16.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\color{blue}{\frac{-0.5 \cdot {B}^{2}}{C}} - -1 \cdot A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. mul-1-neg16.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified16.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    7. Step-by-step derivation
      1. div-inv16.3%

        \[\leadsto \color{blue}{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
      2. associate-*l*16.3%

        \[\leadsto \sqrt{\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)}} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. associate-/l*16.3%

        \[\leadsto \sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(\color{blue}{-0.5 \cdot \frac{{B}^{2}}{C}} - \left(-A\right)\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    8. Applied egg-rr16.3%

      \[\leadsto \color{blue}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(-0.5 \cdot \frac{{B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]

    if 1.00000000000000004e62 < (pow.f64 B #s(literal 2 binary64)) < 4.00000000000000026e260

    1. Initial program 30.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in F around 0 43.7%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}\right)} \]
    4. Simplified64.0%

      \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{F \cdot \frac{C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)}{\mathsf{fma}\left(-4, C \cdot A, {B}^{2}\right)}}} \]

    if 4.00000000000000026e260 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 4.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 3.9%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg3.9%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative3.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow23.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow23.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define22.2%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified22.2%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
    6. Step-by-step derivation
      1. expm1-log1p-u22.0%

        \[\leadsto -\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)\right)} \]
      2. expm1-undefine3.5%

        \[\leadsto -\color{blue}{\left(e^{\mathsf{log1p}\left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)} - 1\right)} \]
      3. associate-*l/3.5%

        \[\leadsto -\left(e^{\mathsf{log1p}\left(\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}}\right)} - 1\right) \]
      4. pow1/23.5%

        \[\leadsto -\left(e^{\mathsf{log1p}\left(\frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\right)} - 1\right) \]
      5. pow1/23.5%

        \[\leadsto -\left(e^{\mathsf{log1p}\left(\frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}^{0.5}}}{B}\right)} - 1\right) \]
      6. pow-prod-down3.5%

        \[\leadsto -\left(e^{\mathsf{log1p}\left(\frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}^{0.5}}}{B}\right)} - 1\right) \]
    7. Applied egg-rr3.5%

      \[\leadsto -\color{blue}{\left(e^{\mathsf{log1p}\left(\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}^{0.5}}{B}\right)} - 1\right)} \]
    8. Step-by-step derivation
      1. expm1-define22.0%

        \[\leadsto -\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}^{0.5}}{B}\right)\right)} \]
      2. unpow1/222.0%

        \[\leadsto -\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}}{B}\right)\right) \]
      3. associate-*r*22.0%

        \[\leadsto -\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\color{blue}{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}}{B}\right)\right) \]
    9. Simplified22.0%

      \[\leadsto -\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\right)\right)} \]
  3. Recombined 5 regimes into one program.
  4. Final simplification35.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-59}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;{B}^{2} \leq 10^{+62}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;{B}^{2} \leq 4 \cdot 10^{+260}:\\ \;\;\;\;\sqrt{2} \cdot \left(-\sqrt{F \cdot \frac{C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)}{\mathsf{fma}\left(-4, A \cdot C, {B}^{2}\right)}}\right)\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\right)\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 47.4% accurate, 0.8× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ t_1 := A - \mathsf{hypot}\left(B\_m, A\right)\\ t_2 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ t_3 := -t\_2\\ t_4 := F \cdot t\_2\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-59}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sqrt{t\_4 \cdot \left(2 \cdot t\_1\right)}}{t\_3}\\ \mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+79}:\\ \;\;\;\;\frac{\sqrt{t\_4 \cdot \left(2 \cdot \left(A + \left(A + \frac{{B\_m}^{2} \cdot -0.5}{C}\right)\right)\right)}}{t\_3}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot t\_1} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C))
        (t_1 (- A (hypot B_m A)))
        (t_2 (fma B_m B_m (* A (* C -4.0))))
        (t_3 (- t_2))
        (t_4 (* F t_2)))
   (if (<= (pow B_m 2.0) 5e-59)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (if (<= (pow B_m 2.0) 5e-8)
       (/ (sqrt (* t_4 (* 2.0 t_1))) t_3)
       (if (<= (pow B_m 2.0) 2e+79)
         (/
          (sqrt (* t_4 (* 2.0 (+ A (+ A (/ (* (pow B_m 2.0) -0.5) C))))))
          t_3)
         (* (sqrt (* F t_1)) (/ (sqrt 2.0) (- B_m))))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double t_1 = A - hypot(B_m, A);
	double t_2 = fma(B_m, B_m, (A * (C * -4.0)));
	double t_3 = -t_2;
	double t_4 = F * t_2;
	double tmp;
	if (pow(B_m, 2.0) <= 5e-59) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else if (pow(B_m, 2.0) <= 5e-8) {
		tmp = sqrt((t_4 * (2.0 * t_1))) / t_3;
	} else if (pow(B_m, 2.0) <= 2e+79) {
		tmp = sqrt((t_4 * (2.0 * (A + (A + ((pow(B_m, 2.0) * -0.5) / C)))))) / t_3;
	} else {
		tmp = sqrt((F * t_1)) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	t_1 = Float64(A - hypot(B_m, A))
	t_2 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	t_3 = Float64(-t_2)
	t_4 = Float64(F * t_2)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-59)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	elseif ((B_m ^ 2.0) <= 5e-8)
		tmp = Float64(sqrt(Float64(t_4 * Float64(2.0 * t_1))) / t_3);
	elseif ((B_m ^ 2.0) <= 2e+79)
		tmp = Float64(sqrt(Float64(t_4 * Float64(2.0 * Float64(A + Float64(A + Float64(Float64((B_m ^ 2.0) * -0.5) / C)))))) / t_3);
	else
		tmp = Float64(sqrt(Float64(F * t_1)) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$1 = N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = (-t$95$2)}, Block[{t$95$4 = N[(F * t$95$2), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-59], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-8], N[(N[Sqrt[N[(t$95$4 * N[(2.0 * t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$3), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+79], N[(N[Sqrt[N[(t$95$4 * N[(2.0 * N[(A + N[(A + N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] * -0.5), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$3), $MachinePrecision], N[(N[Sqrt[N[(F * t$95$1), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]]]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
t_1 := A - \mathsf{hypot}\left(B\_m, A\right)\\
t_2 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
t_3 := -t\_2\\
t_4 := F \cdot t\_2\\
\mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-59}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{-8}:\\
\;\;\;\;\frac{\sqrt{t\_4 \cdot \left(2 \cdot t\_1\right)}}{t\_3}\\

\mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+79}:\\
\;\;\;\;\frac{\sqrt{t\_4 \cdot \left(2 \cdot \left(A + \left(A + \frac{{B\_m}^{2} \cdot -0.5}{C}\right)\right)\right)}}{t\_3}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot t\_1} \cdot \frac{\sqrt{2}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000001e-59

    1. Initial program 27.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 33.7%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

    if 5.0000000000000001e-59 < (pow.f64 B #s(literal 2 binary64)) < 4.9999999999999998e-8

    1. Initial program 55.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified62.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around 0 49.2%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{-1 \cdot \sqrt{{A}^{2} + {B}^{2}}}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. mul-1-neg49.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-\sqrt{{A}^{2} + {B}^{2}}\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. +-commutative49.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. unpow249.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. unpow249.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      5. hypot-define56.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\color{blue}{\mathsf{hypot}\left(B, A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified56.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-\mathsf{hypot}\left(B, A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]

    if 4.9999999999999998e-8 < (pow.f64 B #s(literal 2 binary64)) < 1.99999999999999993e79

    1. Initial program 13.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified20.4%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 13.0%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C} - -1 \cdot A\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. associate-*r/13.0%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\color{blue}{\frac{-0.5 \cdot {B}^{2}}{C}} - -1 \cdot A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. mul-1-neg13.0%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified13.0%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]

    if 1.99999999999999993e79 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 15.1%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg15.1%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative15.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define26.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified26.9%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 4 regimes into one program.
  4. Final simplification30.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-59}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;{B}^{2} \leq 2 \cdot 10^{+79}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A + \left(A + \frac{{B}^{2} \cdot -0.5}{C}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 4: 46.8% accurate, 0.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ t_1 := A - \mathsf{hypot}\left(B\_m, A\right)\\ t_2 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-59}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_2\right) \cdot \left(2 \cdot t\_1\right)}}{-t\_2}\\ \mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+67}:\\ \;\;\;\;\sqrt{\frac{F \cdot \left(2 \cdot A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)}{{B\_m}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot t\_1} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C))
        (t_1 (- A (hypot B_m A)))
        (t_2 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 5e-59)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (if (<= (pow B_m 2.0) 5e-8)
       (/ (sqrt (* (* F t_2) (* 2.0 t_1))) (- t_2))
       (if (<= (pow B_m 2.0) 2e+67)
         (*
          (sqrt
           (/
            (* F (+ (* 2.0 A) (* -0.5 (/ (pow B_m 2.0) C))))
            (+ (pow B_m 2.0) (* -4.0 (* A C)))))
          (- (sqrt 2.0)))
         (* (sqrt (* F t_1)) (/ (sqrt 2.0) (- B_m))))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double t_1 = A - hypot(B_m, A);
	double t_2 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 5e-59) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else if (pow(B_m, 2.0) <= 5e-8) {
		tmp = sqrt(((F * t_2) * (2.0 * t_1))) / -t_2;
	} else if (pow(B_m, 2.0) <= 2e+67) {
		tmp = sqrt(((F * ((2.0 * A) + (-0.5 * (pow(B_m, 2.0) / C)))) / (pow(B_m, 2.0) + (-4.0 * (A * C))))) * -sqrt(2.0);
	} else {
		tmp = sqrt((F * t_1)) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	t_1 = Float64(A - hypot(B_m, A))
	t_2 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-59)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	elseif ((B_m ^ 2.0) <= 5e-8)
		tmp = Float64(sqrt(Float64(Float64(F * t_2) * Float64(2.0 * t_1))) / Float64(-t_2));
	elseif ((B_m ^ 2.0) <= 2e+67)
		tmp = Float64(sqrt(Float64(Float64(F * Float64(Float64(2.0 * A) + Float64(-0.5 * Float64((B_m ^ 2.0) / C)))) / Float64((B_m ^ 2.0) + Float64(-4.0 * Float64(A * C))))) * Float64(-sqrt(2.0)));
	else
		tmp = Float64(sqrt(Float64(F * t_1)) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$1 = N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-59], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-8], N[(N[Sqrt[N[(N[(F * t$95$2), $MachinePrecision] * N[(2.0 * t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$2)), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+67], N[(N[Sqrt[N[(N[(F * N[(N[(2.0 * A), $MachinePrecision] + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision], N[(N[Sqrt[N[(F * t$95$1), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
t_1 := A - \mathsf{hypot}\left(B\_m, A\right)\\
t_2 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-59}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{-8}:\\
\;\;\;\;\frac{\sqrt{\left(F \cdot t\_2\right) \cdot \left(2 \cdot t\_1\right)}}{-t\_2}\\

\mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+67}:\\
\;\;\;\;\sqrt{\frac{F \cdot \left(2 \cdot A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)}{{B\_m}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot \left(-\sqrt{2}\right)\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot t\_1} \cdot \frac{\sqrt{2}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000001e-59

    1. Initial program 27.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 33.7%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

    if 5.0000000000000001e-59 < (pow.f64 B #s(literal 2 binary64)) < 4.9999999999999998e-8

    1. Initial program 55.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified62.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around 0 49.2%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{-1 \cdot \sqrt{{A}^{2} + {B}^{2}}}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. mul-1-neg49.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-\sqrt{{A}^{2} + {B}^{2}}\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. +-commutative49.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. unpow249.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. unpow249.2%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      5. hypot-define56.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\color{blue}{\mathsf{hypot}\left(B, A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified56.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-\mathsf{hypot}\left(B, A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]

    if 4.9999999999999998e-8 < (pow.f64 B #s(literal 2 binary64)) < 1.99999999999999997e67

    1. Initial program 8.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified15.7%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 13.8%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C} - -1 \cdot A\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. associate-*r/13.8%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\color{blue}{\frac{-0.5 \cdot {B}^{2}}{C}} - -1 \cdot A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. mul-1-neg13.8%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified13.8%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    7. Taylor expanded in F around 0 13.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F \cdot \left(-0.5 \cdot \frac{{B}^{2}}{C} + 2 \cdot A\right)}{-4 \cdot \left(A \cdot C\right) + {B}^{2}}} \cdot \sqrt{2}\right)} \]

    if 1.99999999999999997e67 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 15.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg15.0%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative15.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow215.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow215.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define26.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified26.7%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 4 regimes into one program.
  4. Final simplification30.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-59}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;{B}^{2} \leq 2 \cdot 10^{+67}:\\ \;\;\;\;\sqrt{\frac{F \cdot \left(2 \cdot A + -0.5 \cdot \frac{{B}^{2}}{C}\right)}{{B}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 47.1% accurate, 1.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ t_1 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-67}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+67}:\\ \;\;\;\;\sqrt{t\_1 \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{t\_1}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C)) (t_1 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 2e-67)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (if (<= (pow B_m 2.0) 2e+67)
       (*
        (sqrt (* t_1 (* F (* 2.0 (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C))))))))
        (/ -1.0 t_1))
       (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m)))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double t_1 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 2e-67) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else if (pow(B_m, 2.0) <= 2e+67) {
		tmp = sqrt((t_1 * (F * (2.0 * (A + (A + (-0.5 * (pow(B_m, 2.0) / C)))))))) * (-1.0 / t_1);
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	t_1 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-67)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	elseif ((B_m ^ 2.0) <= 2e+67)
		tmp = Float64(sqrt(Float64(t_1 * Float64(F * Float64(2.0 * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C)))))))) * Float64(-1.0 / t_1));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$1 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-67], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+67], N[(N[Sqrt[N[(t$95$1 * N[(F * N[(2.0 * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / t$95$1), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
t_1 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-67}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+67}:\\
\;\;\;\;\sqrt{t\_1 \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{t\_1}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1.99999999999999989e-67

    1. Initial program 26.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 33.4%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

    if 1.99999999999999989e-67 < (pow.f64 B #s(literal 2 binary64)) < 1.99999999999999997e67

    1. Initial program 29.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified36.5%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 21.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C} - -1 \cdot A\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. associate-*r/21.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\color{blue}{\frac{-0.5 \cdot {B}^{2}}{C}} - -1 \cdot A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. mul-1-neg21.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified21.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    7. Step-by-step derivation
      1. div-inv21.4%

        \[\leadsto \color{blue}{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
      2. associate-*l*20.9%

        \[\leadsto \sqrt{\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)}} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. associate-/l*20.9%

        \[\leadsto \sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(\color{blue}{-0.5 \cdot \frac{{B}^{2}}{C}} - \left(-A\right)\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    8. Applied egg-rr20.9%

      \[\leadsto \color{blue}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(-0.5 \cdot \frac{{B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]

    if 1.99999999999999997e67 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 15.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg15.0%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative15.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow215.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow215.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define26.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified26.7%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification29.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-67}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 2 \cdot 10^{+67}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 6: 47.1% accurate, 1.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ t_1 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-67}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+67}:\\ \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)\right)}}{-t\_1}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C)) (t_1 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 2e-67)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (if (<= (pow B_m 2.0) 2e+67)
       (/
        (sqrt (* t_1 (* F (* 2.0 (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C))))))))
        (- t_1))
       (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m)))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double t_1 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 2e-67) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else if (pow(B_m, 2.0) <= 2e+67) {
		tmp = sqrt((t_1 * (F * (2.0 * (A + (A + (-0.5 * (pow(B_m, 2.0) / C)))))))) / -t_1;
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	t_1 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-67)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	elseif ((B_m ^ 2.0) <= 2e+67)
		tmp = Float64(sqrt(Float64(t_1 * Float64(F * Float64(2.0 * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C)))))))) / Float64(-t_1));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$1 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-67], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+67], N[(N[Sqrt[N[(t$95$1 * N[(F * N[(2.0 * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$1)), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
t_1 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-67}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+67}:\\
\;\;\;\;\frac{\sqrt{t\_1 \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)\right)}}{-t\_1}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1.99999999999999989e-67

    1. Initial program 26.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 33.4%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

    if 1.99999999999999989e-67 < (pow.f64 B #s(literal 2 binary64)) < 1.99999999999999997e67

    1. Initial program 29.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified36.5%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 21.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C} - -1 \cdot A\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. associate-*r/21.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\color{blue}{\frac{-0.5 \cdot {B}^{2}}{C}} - -1 \cdot A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. mul-1-neg21.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified21.4%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    7. Step-by-step derivation
      1. *-un-lft-identity21.4%

        \[\leadsto \color{blue}{1 \cdot \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
      2. associate-*l*21.0%

        \[\leadsto 1 \cdot \frac{\sqrt{\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(\frac{-0.5 \cdot {B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)}}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. associate-/l*21.0%

        \[\leadsto 1 \cdot \frac{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(\color{blue}{-0.5 \cdot \frac{{B}^{2}}{C}} - \left(-A\right)\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    8. Applied egg-rr21.0%

      \[\leadsto \color{blue}{1 \cdot \frac{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(-0.5 \cdot \frac{{B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    9. Step-by-step derivation
      1. *-lft-identity21.0%

        \[\leadsto \color{blue}{\frac{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(-0.5 \cdot \frac{{B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    10. Simplified21.0%

      \[\leadsto \color{blue}{\frac{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(-0.5 \cdot \frac{{B}^{2}}{C} - \left(-A\right)\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]

    if 1.99999999999999997e67 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 15.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg15.0%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative15.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow215.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow215.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define26.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified26.7%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification29.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-67}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 2 \cdot 10^{+67}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 7: 46.7% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+79}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C)))
   (if (<= (pow B_m 2.0) 2e+79)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double tmp;
	if (pow(B_m, 2.0) <= 2e+79) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e+79) {
		tmp = Math.sqrt(((2.0 * ((Math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = (4.0 * A) * C
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e+79:
		tmp = math.sqrt(((2.0 * ((math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e+79)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = (4.0 * A) * C;
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e+79)
		tmp = sqrt(((2.0 * (((B_m ^ 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	end
	tmp_2 = tmp;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+79], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+79}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1.99999999999999993e79

    1. Initial program 27.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 30.0%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

    if 1.99999999999999993e79 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 15.1%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg15.1%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative15.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define26.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified26.9%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification28.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+79}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 8: 45.0% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-181}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 2e-181)
   (/
    (sqrt (* -8.0 (* (* A C) (* F (+ A A)))))
    (- (fma C (* A -4.0) (pow B_m 2.0))))
   (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m)))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 2e-181) {
		tmp = sqrt((-8.0 * ((A * C) * (F * (A + A))))) / -fma(C, (A * -4.0), pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-181)
		tmp = Float64(sqrt(Float64(-8.0 * Float64(Float64(A * C) * Float64(F * Float64(A + A))))) / Float64(-fma(C, Float64(A * -4.0), (B_m ^ 2.0))));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-181], N[(N[Sqrt[N[(-8.0 * N[(N[(A * C), $MachinePrecision] * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-N[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision])), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-181}:\\
\;\;\;\;\frac{\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 2.00000000000000009e-181

    1. Initial program 22.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified24.0%

      \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 32.0%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    5. Step-by-step derivation
      1. associate-*r*31.9%

        \[\leadsto \frac{\sqrt{-8 \cdot \color{blue}{\left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      2. *-commutative31.9%

        \[\leadsto \frac{\sqrt{-8 \cdot \left(\color{blue}{\left(C \cdot A\right)} \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      3. mul-1-neg31.9%

        \[\leadsto \frac{\sqrt{-8 \cdot \left(\left(C \cdot A\right) \cdot \left(F \cdot \left(A - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    6. Simplified31.9%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(\left(C \cdot A\right) \cdot \left(F \cdot \left(A - \left(-A\right)\right)\right)\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]

    if 2.00000000000000009e-181 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 22.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 12.8%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg12.8%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative12.8%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow212.8%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow212.8%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define20.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified20.7%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification24.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-181}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 9: 44.3% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-181}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 2e-181)
   (/
    (sqrt (* -8.0 (* A (* C (* F (+ A A))))))
    (- (fma C (* A -4.0) (pow B_m 2.0))))
   (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m)))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 2e-181) {
		tmp = sqrt((-8.0 * (A * (C * (F * (A + A)))))) / -fma(C, (A * -4.0), pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-181)
		tmp = Float64(sqrt(Float64(-8.0 * Float64(A * Float64(C * Float64(F * Float64(A + A)))))) / Float64(-fma(C, Float64(A * -4.0), (B_m ^ 2.0))));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-181], N[(N[Sqrt[N[(-8.0 * N[(A * N[(C * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-N[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision])), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-181}:\\
\;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 2.00000000000000009e-181

    1. Initial program 22.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified24.0%

      \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 32.0%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]

    if 2.00000000000000009e-181 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 22.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 12.8%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg12.8%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative12.8%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow212.8%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow212.8%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define20.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified20.7%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification25.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-181}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 10: 31.0% accurate, 2.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}
\end{array}
Derivation
  1. Initial program 22.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in C around 0 8.8%

    \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg8.8%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    2. +-commutative8.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
    3. unpow28.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
    4. unpow28.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
    5. hypot-define14.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
  5. Simplified14.7%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  6. Final simplification14.7%

    \[\leadsto \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B} \]
  7. Add Preprocessing

Alternative 11: 31.1% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)}}{-B\_m} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (/ (sqrt (* (* 2.0 F) (- A (hypot B_m A)))) (- B_m)))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt(((2.0 * F) * (A - hypot(B_m, A)))) / -B_m;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt(((2.0 * F) * (A - Math.hypot(B_m, A)))) / -B_m;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt(((2.0 * F) * (A - math.hypot(B_m, A)))) / -B_m
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(Float64(Float64(2.0 * F) * Float64(A - hypot(B_m, A)))) / Float64(-B_m))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt(((2.0 * F) * (A - hypot(B_m, A)))) / -B_m;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(N[(2.0 * F), $MachinePrecision] * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-B$95$m)), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)}}{-B\_m}
\end{array}
Derivation
  1. Initial program 22.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in C around 0 8.8%

    \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg8.8%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    2. +-commutative8.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
    3. unpow28.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
    4. unpow28.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
    5. hypot-define14.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
  5. Simplified14.7%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  6. Step-by-step derivation
    1. associate-*l/14.7%

      \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}} \]
    2. pow1/214.7%

      \[\leadsto -\frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B} \]
    3. pow1/214.8%

      \[\leadsto -\frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}^{0.5}}}{B} \]
    4. pow-prod-down14.4%

      \[\leadsto -\frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}^{0.5}}}{B} \]
  7. Applied egg-rr14.4%

    \[\leadsto -\color{blue}{\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}^{0.5}}{B}} \]
  8. Step-by-step derivation
    1. unpow1/214.4%

      \[\leadsto -\frac{\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}}{B} \]
    2. associate-*r*14.4%

      \[\leadsto -\frac{\sqrt{\color{blue}{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}}{B} \]
  9. Simplified14.4%

    \[\leadsto -\color{blue}{\frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}} \]
  10. Final simplification14.4%

    \[\leadsto \frac{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{-B} \]
  11. Add Preprocessing

Alternative 12: 26.4% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(A - B\_m\right)}\right) \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (- A B_m))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return (sqrt(2.0) / B_m) * -sqrt((F * (A - B_m)));
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = (sqrt(2.0d0) / b_m) * -sqrt((f * (a - b_m)))
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return (Math.sqrt(2.0) / B_m) * -Math.sqrt((F * (A - B_m)));
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return (math.sqrt(2.0) / B_m) * -math.sqrt((F * (A - B_m)))
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(A - B_m)))))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - B_m)));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(A - B\_m\right)}\right)
\end{array}
Derivation
  1. Initial program 22.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in C around 0 8.8%

    \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg8.8%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    2. +-commutative8.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
    3. unpow28.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
    4. unpow28.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
    5. hypot-define14.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
  5. Simplified14.7%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  6. Taylor expanded in A around 0 11.9%

    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}} \]
  7. Final simplification11.9%

    \[\leadsto \frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right) \]
  8. Add Preprocessing

Alternative 13: 26.2% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{F \cdot \left(-B\_m\right)} \cdot \frac{\sqrt{2}}{-B\_m} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (* (sqrt (* F (- B_m))) (/ (sqrt 2.0) (- B_m))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt((F * -B_m)) * (sqrt(2.0) / -B_m);
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = sqrt((f * -b_m)) * (sqrt(2.0d0) / -b_m)
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt((F * -B_m)) * (Math.sqrt(2.0) / -B_m);
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt((F * -B_m)) * (math.sqrt(2.0) / -B_m)
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(Float64(F * Float64(-B_m))) * Float64(sqrt(2.0) / Float64(-B_m)))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt((F * -B_m)) * (sqrt(2.0) / -B_m);
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(F * (-B$95$m)), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\sqrt{F \cdot \left(-B\_m\right)} \cdot \frac{\sqrt{2}}{-B\_m}
\end{array}
Derivation
  1. Initial program 22.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in C around 0 8.8%

    \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg8.8%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    2. +-commutative8.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
    3. unpow28.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
    4. unpow28.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
    5. hypot-define14.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
  5. Simplified14.7%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  6. Taylor expanded in A around 0 12.7%

    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(-1 \cdot B\right)}} \]
  7. Step-by-step derivation
    1. mul-1-neg12.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}} \]
  8. Simplified12.7%

    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}} \]
  9. Final simplification12.7%

    \[\leadsto \sqrt{F \cdot \left(-B\right)} \cdot \frac{\sqrt{2}}{-B} \]
  10. Add Preprocessing

Alternative 14: 0.0% accurate, 3.1× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{\frac{F}{A}} \cdot \left(-\sqrt{-1}\right) \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (* (sqrt (/ F A)) (- (sqrt -1.0))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt((F / A)) * -sqrt(-1.0);
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = sqrt((f / a)) * -sqrt((-1.0d0))
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt((F / A)) * -Math.sqrt(-1.0);
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt((F / A)) * -math.sqrt(-1.0)
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(Float64(F / A)) * Float64(-sqrt(-1.0)))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt((F / A)) * -sqrt(-1.0);
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(F / A), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[-1.0], $MachinePrecision])), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\sqrt{\frac{F}{A}} \cdot \left(-\sqrt{-1}\right)
\end{array}
Derivation
  1. Initial program 22.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Simplified21.8%

    \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
  3. Add Preprocessing
  4. Taylor expanded in C around -inf 11.2%

    \[\leadsto \frac{\sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
  5. Step-by-step derivation
    1. associate-*r*11.2%

      \[\leadsto \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot \left({C}^{2} \cdot F\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
  6. Simplified11.2%

    \[\leadsto \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot \left({C}^{2} \cdot F\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
  7. Taylor expanded in A around -inf 0.0%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{A}} \cdot \sqrt{-1}\right)} \]
  8. Step-by-step derivation
    1. mul-1-neg0.0%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{A}} \cdot \sqrt{-1}} \]
  9. Simplified0.0%

    \[\leadsto \color{blue}{-\sqrt{\frac{F}{A}} \cdot \sqrt{-1}} \]
  10. Final simplification0.0%

    \[\leadsto \sqrt{\frac{F}{A}} \cdot \left(-\sqrt{-1}\right) \]
  11. Add Preprocessing

Alternative 15: 1.6% accurate, 3.1× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{2} \cdot \sqrt{\frac{F}{B\_m}} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (* (sqrt 2.0) (sqrt (/ F B_m))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt(2.0) * sqrt((F / B_m));
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = sqrt(2.0d0) * sqrt((f / b_m))
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt(2.0) * Math.sqrt((F / B_m));
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt(2.0) * math.sqrt((F / B_m))
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(2.0) * sqrt(Float64(F / B_m)))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt(2.0) * sqrt((F / B_m));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[2.0], $MachinePrecision] * N[Sqrt[N[(F / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\sqrt{2} \cdot \sqrt{\frac{F}{B\_m}}
\end{array}
Derivation
  1. Initial program 22.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in B around -inf 0.0%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \left({\left(\sqrt{-1}\right)}^{2} \cdot \sqrt{2}\right)\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg0.0%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \left({\left(\sqrt{-1}\right)}^{2} \cdot \sqrt{2}\right)} \]
    2. unpow20.0%

      \[\leadsto -\sqrt{\frac{F}{B}} \cdot \left(\color{blue}{\left(\sqrt{-1} \cdot \sqrt{-1}\right)} \cdot \sqrt{2}\right) \]
    3. rem-square-sqrt1.9%

      \[\leadsto -\sqrt{\frac{F}{B}} \cdot \left(\color{blue}{-1} \cdot \sqrt{2}\right) \]
  5. Simplified1.9%

    \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \left(-1 \cdot \sqrt{2}\right)} \]
  6. Final simplification1.9%

    \[\leadsto \sqrt{2} \cdot \sqrt{\frac{F}{B}} \]
  7. Add Preprocessing

Reproduce

?
herbie shell --seed 2024080 
(FPCore (A B C F)
  :name "ABCF->ab-angle b"
  :precision binary64
  (/ (- (sqrt (* (* 2.0 (* (- (pow B 2.0) (* (* 4.0 A) C)) F)) (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))) (- (pow B 2.0) (* (* 4.0 A) C))))