ABCF->ab-angle b

Percentage Accurate: 18.8% → 47.9%
Time: 24.8s
Alternatives: 7
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 7 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: 18.8% 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: 47.9% accurate, 1.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := A - \mathsf{hypot}\left(B_m, A\right)\\ t_1 := \mathsf{fma}\left(B_m, B_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B_m}^{2} \leq 2 \cdot 10^{-56}:\\ \;\;\;\;\frac{-\sqrt{t_1 \cdot \left(F \cdot \left(A \cdot 4\right)\right)}}{t_1}\\ \mathbf{elif}\;{B_m}^{2} \leq 2 \cdot 10^{+116}:\\ \;\;\;\;\frac{-\sqrt{\left(t_1 \cdot F\right) \cdot \left(2 \cdot t_0\right)}}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B_m} \cdot \left(-\sqrt{F \cdot t_0}\right)\\ \end{array} \end{array} \]
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 (- A (hypot B_m A))) (t_1 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 2e-56)
     (/ (- (sqrt (* t_1 (* F (* A 4.0))))) t_1)
     (if (<= (pow B_m 2.0) 2e+116)
       (/ (- (sqrt (* (* t_1 F) (* 2.0 t_0)))) t_1)
       (* (/ (sqrt 2.0) B_m) (- (sqrt (* F t_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) {
	double t_0 = A - hypot(B_m, A);
	double t_1 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 2e-56) {
		tmp = -sqrt((t_1 * (F * (A * 4.0)))) / t_1;
	} else if (pow(B_m, 2.0) <= 2e+116) {
		tmp = -sqrt(((t_1 * F) * (2.0 * t_0))) / t_1;
	} else {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * t_0));
	}
	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(A - hypot(B_m, A))
	t_1 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-56)
		tmp = Float64(Float64(-sqrt(Float64(t_1 * Float64(F * Float64(A * 4.0))))) / t_1);
	elseif ((B_m ^ 2.0) <= 2e+116)
		tmp = Float64(Float64(-sqrt(Float64(Float64(t_1 * F) * Float64(2.0 * t_0)))) / t_1);
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * t_0))));
	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[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $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-56], N[((-N[Sqrt[N[(t$95$1 * N[(F * N[(A * 4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+116], N[((-N[Sqrt[N[(N[(t$95$1 * F), $MachinePrecision] * N[(2.0 * t$95$0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * t$95$0), $MachinePrecision]], $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 := A - \mathsf{hypot}\left(B_m, A\right)\\
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^{-56}:\\
\;\;\;\;\frac{-\sqrt{t_1 \cdot \left(F \cdot \left(A \cdot 4\right)\right)}}{t_1}\\

\mathbf{elif}\;{B_m}^{2} \leq 2 \cdot 10^{+116}:\\
\;\;\;\;\frac{-\sqrt{\left(t_1 \cdot F\right) \cdot \left(2 \cdot t_0\right)}}{t_1}\\

\mathbf{else}:\\
\;\;\;\;\frac{\sqrt{2}}{B_m} \cdot \left(-\sqrt{F \cdot t_0}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (pow.f64 B 2) < 2.0000000000000001e-56

    1. Initial program 20.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified29.9%

      \[\leadsto \color{blue}{\frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Taylor expanded in C around inf 28.3%

      \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{A}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    4. Step-by-step derivation
      1. expm1-log1p-u27.5%

        \[\leadsto \frac{-\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + A\right)\right)}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. expm1-udef15.5%

        \[\leadsto \frac{-\color{blue}{\left(e^{\mathsf{log1p}\left(\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + A\right)\right)}\right)} - 1\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. associate-*l*15.5%

        \[\leadsto \frac{-\left(e^{\mathsf{log1p}\left(\sqrt{\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + A\right)\right)\right)}}\right)} - 1\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. count-215.5%

        \[\leadsto \frac{-\left(e^{\mathsf{log1p}\left(\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \color{blue}{\left(2 \cdot A\right)}\right)\right)}\right)} - 1\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Applied egg-rr15.5%

      \[\leadsto \frac{-\color{blue}{\left(e^{\mathsf{log1p}\left(\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(2 \cdot A\right)\right)\right)}\right)} - 1\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Step-by-step derivation
      1. expm1-def29.9%

        \[\leadsto \frac{-\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(2 \cdot A\right)\right)\right)}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. expm1-log1p30.7%

        \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(2 \cdot A\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. associate-*r*30.7%

        \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \color{blue}{\left(\left(2 \cdot 2\right) \cdot A\right)}\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. metadata-eval30.7%

        \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(\color{blue}{4} \cdot A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    7. Simplified30.7%

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

    if 2.0000000000000001e-56 < (pow.f64 B 2) < 2.00000000000000003e116

    1. Initial program 53.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified58.9%

      \[\leadsto \color{blue}{\frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Taylor expanded in C around 0 43.4%

      \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{-1 \cdot \sqrt{{A}^{2} + {B}^{2}}}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg43.4%

        \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-\sqrt{{A}^{2} + {B}^{2}}\right)}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. +-commutative43.4%

        \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. unpow243.4%

        \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. unpow243.4%

        \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      5. hypot-def48.7%

        \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\color{blue}{\mathsf{hypot}\left(B, A\right)}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Simplified48.7%

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

    if 2.00000000000000003e116 < (pow.f64 B 2)

    1. Initial program 12.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified7.2%

      \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)}} \]
    3. Taylor expanded in C around 0 14.2%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. associate-*r*14.2%

        \[\leadsto \color{blue}{\left(-1 \cdot \frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. mul-1-neg14.2%

        \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right)} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)} \]
      3. +-commutative14.2%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      4. unpow214.2%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      5. unpow214.2%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      6. hypot-def28.8%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified28.8%

      \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification32.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-56}:\\ \;\;\;\;\frac{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A \cdot 4\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;{B}^{2} \leq 2 \cdot 10^{+116}:\\ \;\;\;\;\frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)\\ \end{array} \]

Alternative 2: 47.1% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(B_m, B_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B_m}^{2} \leq 5 \cdot 10^{-61}:\\ \;\;\;\;\frac{-\sqrt{t_0 \cdot \left(F \cdot \left(A \cdot 4\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B_m} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B_m, A\right)\right)}\right)\\ \end{array} \end{array} \]
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 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 5e-61)
     (/ (- (sqrt (* t_0 (* F (* A 4.0))))) t_0)
     (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (- A (hypot B_m A)))))))))
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 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 5e-61) {
		tmp = -sqrt((t_0 * (F * (A * 4.0)))) / t_0;
	} else {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - hypot(B_m, A))));
	}
	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 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-61)
		tmp = Float64(Float64(-sqrt(Float64(t_0 * Float64(F * Float64(A * 4.0))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(A - hypot(B_m, A))))));
	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[(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-61], N[((-N[Sqrt[N[(t$95$0 * N[(F * N[(A * 4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $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 := \mathsf{fma}\left(B_m, B_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;{B_m}^{2} \leq 5 \cdot 10^{-61}:\\
\;\;\;\;\frac{-\sqrt{t_0 \cdot \left(F \cdot \left(A \cdot 4\right)\right)}}{t_0}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B 2) < 4.9999999999999999e-61

    1. Initial program 20.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified30.4%

      \[\leadsto \color{blue}{\frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Taylor expanded in C around inf 28.8%

      \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{A}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    4. Step-by-step derivation
      1. expm1-log1p-u27.9%

        \[\leadsto \frac{-\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + A\right)\right)}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. expm1-udef15.7%

        \[\leadsto \frac{-\color{blue}{\left(e^{\mathsf{log1p}\left(\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + A\right)\right)}\right)} - 1\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. associate-*l*15.7%

        \[\leadsto \frac{-\left(e^{\mathsf{log1p}\left(\sqrt{\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(A + A\right)\right)\right)}}\right)} - 1\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. count-215.7%

        \[\leadsto \frac{-\left(e^{\mathsf{log1p}\left(\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \color{blue}{\left(2 \cdot A\right)}\right)\right)}\right)} - 1\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Applied egg-rr15.7%

      \[\leadsto \frac{-\color{blue}{\left(e^{\mathsf{log1p}\left(\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(2 \cdot A\right)\right)\right)}\right)} - 1\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Step-by-step derivation
      1. expm1-def30.4%

        \[\leadsto \frac{-\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(2 \cdot A\right)\right)\right)}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. expm1-log1p31.2%

        \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot \left(2 \cdot A\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. associate-*r*31.2%

        \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \color{blue}{\left(\left(2 \cdot 2\right) \cdot A\right)}\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. metadata-eval31.2%

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

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

    if 4.9999999999999999e-61 < (pow.f64 B 2)

    1. Initial program 23.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified19.1%

      \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)}} \]
    3. Taylor expanded in C around 0 16.4%

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

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

        \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right)} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)} \]
      3. +-commutative16.4%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      4. unpow216.4%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      5. unpow216.4%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      6. hypot-def27.1%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified27.1%

      \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification28.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-61}:\\ \;\;\;\;\frac{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A \cdot 4\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)\\ \end{array} \]

Alternative 3: 45.5% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B_m}^{2} \leq 5 \cdot 10^{-156}:\\ \;\;\;\;\frac{-\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B_m}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B_m} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B_m, A\right)\right)}\right)\\ \end{array} \end{array} \]
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
 (if (<= (pow B_m 2.0) 5e-156)
   (/
    (- (sqrt (* -8.0 (* (* A C) (* F (+ A A))))))
    (fma A (* C -4.0) (pow B_m 2.0)))
   (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (- A (hypot B_m A))))))))
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) <= 5e-156) {
		tmp = -sqrt((-8.0 * ((A * C) * (F * (A + A))))) / fma(A, (C * -4.0), pow(B_m, 2.0));
	} else {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - hypot(B_m, A))));
	}
	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) <= 5e-156)
		tmp = Float64(Float64(-sqrt(Float64(-8.0 * Float64(Float64(A * C) * Float64(F * Float64(A + A)))))) / fma(A, Float64(C * -4.0), (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(A - hypot(B_m, A))))));
	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], 5e-156], N[((-N[Sqrt[N[(-8.0 * N[(N[(A * C), $MachinePrecision] * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(A * N[(C * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $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 5 \cdot 10^{-156}:\\
\;\;\;\;\frac{-\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B_m}^{2}\right)}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B 2) < 5.00000000000000007e-156

    1. Initial program 19.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified28.1%

      \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)}} \]
    3. Taylor expanded in C around inf 26.5%

      \[\leadsto \frac{-\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)} \]
    4. Step-by-step derivation
      1. associate-*r*28.3%

        \[\leadsto \frac{-\sqrt{-8 \cdot \color{blue}{\left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)} \]
      2. sub-neg28.3%

        \[\leadsto \frac{-\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \color{blue}{\left(A + \left(--1 \cdot A\right)\right)}\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)} \]
      3. mul-1-neg28.3%

        \[\leadsto \frac{-\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + \left(-\color{blue}{\left(-A\right)}\right)\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)} \]
      4. remove-double-neg28.3%

        \[\leadsto \frac{-\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + \color{blue}{A}\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)} \]
    5. Simplified28.3%

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

    if 5.00000000000000007e-156 < (pow.f64 B 2)

    1. Initial program 23.5%

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

      \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)}} \]
    3. Taylor expanded in C around 0 15.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. associate-*r*15.6%

        \[\leadsto \color{blue}{\left(-1 \cdot \frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. mul-1-neg15.6%

        \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right)} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)} \]
      3. +-commutative15.6%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      4. unpow215.6%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      5. unpow215.6%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      6. hypot-def25.3%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified25.3%

      \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification26.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-156}:\\ \;\;\;\;\frac{-\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)\\ \end{array} \]

Alternative 4: 42.8% accurate, 2.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;B_m \leq 6 \cdot 10^{-77}:\\ \;\;\;\;\frac{-\sqrt{\left(\left(A \cdot -4\right) \cdot \left(C \cdot F\right)\right) \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B_m, B_m, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B_m} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B_m, A\right)\right)}\right)\\ \end{array} \end{array} \]
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
 (if (<= B_m 6e-77)
   (/
    (- (sqrt (* (* (* A -4.0) (* C F)) (* 2.0 (+ A A)))))
    (fma B_m B_m (* A (* C -4.0))))
   (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (- A (hypot B_m A))))))))
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 (B_m <= 6e-77) {
		tmp = -sqrt((((A * -4.0) * (C * F)) * (2.0 * (A + A)))) / fma(B_m, B_m, (A * (C * -4.0)));
	} else {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - hypot(B_m, A))));
	}
	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 <= 6e-77)
		tmp = Float64(Float64(-sqrt(Float64(Float64(Float64(A * -4.0) * Float64(C * F)) * Float64(2.0 * Float64(A + A))))) / fma(B_m, B_m, Float64(A * Float64(C * -4.0))));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(A - hypot(B_m, A))))));
	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[B$95$m, 6e-77], N[((-N[Sqrt[N[(N[(N[(A * -4.0), $MachinePrecision] * N[(C * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $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 \leq 6 \cdot 10^{-77}:\\
\;\;\;\;\frac{-\sqrt{\left(\left(A \cdot -4\right) \cdot \left(C \cdot F\right)\right) \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B_m, B_m, A \cdot \left(C \cdot -4\right)\right)}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if B < 6.00000000000000033e-77

    1. Initial program 20.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified28.5%

      \[\leadsto \color{blue}{\frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Taylor expanded in C around inf 19.3%

      \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{A}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    4. Taylor expanded in B around 0 16.2%

      \[\leadsto \frac{-\sqrt{\color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)} \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. associate-*r*16.2%

        \[\leadsto \frac{-\sqrt{\color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot F\right)\right)} \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. *-commutative16.2%

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

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

    if 6.00000000000000033e-77 < B

    1. Initial program 23.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified20.5%

      \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)}} \]
    3. Taylor expanded in C around 0 28.3%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. associate-*r*28.3%

        \[\leadsto \color{blue}{\left(-1 \cdot \frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. mul-1-neg28.3%

        \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right)} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)} \]
      3. +-commutative28.3%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      4. unpow228.3%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      5. unpow228.3%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      6. hypot-def46.0%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified46.0%

      \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification26.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 6 \cdot 10^{-77}:\\ \;\;\;\;\frac{-\sqrt{\left(\left(A \cdot -4\right) \cdot \left(C \cdot F\right)\right) \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)\\ \end{array} \]

Alternative 5: 38.5% accurate, 2.8× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;B_m \leq 7 \cdot 10^{-37}:\\ \;\;\;\;\frac{-\sqrt{\left(\left(A \cdot -4\right) \cdot \left(C \cdot F\right)\right) \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B_m, B_m, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2}}{B_m} \cdot \sqrt{F \cdot \left(-B_m\right)}\\ \end{array} \end{array} \]
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
 (if (<= B_m 7e-37)
   (/
    (- (sqrt (* (* (* A -4.0) (* C F)) (* 2.0 (+ A A)))))
    (fma B_m B_m (* A (* C -4.0))))
   (* (/ (- (sqrt 2.0)) B_m) (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) {
	double tmp;
	if (B_m <= 7e-37) {
		tmp = -sqrt((((A * -4.0) * (C * F)) * (2.0 * (A + A)))) / fma(B_m, B_m, (A * (C * -4.0)));
	} else {
		tmp = (-sqrt(2.0) / B_m) * sqrt((F * -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 <= 7e-37)
		tmp = Float64(Float64(-sqrt(Float64(Float64(Float64(A * -4.0) * Float64(C * F)) * Float64(2.0 * Float64(A + A))))) / fma(B_m, B_m, Float64(A * Float64(C * -4.0))));
	else
		tmp = Float64(Float64(Float64(-sqrt(2.0)) / B_m) * sqrt(Float64(F * 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[B$95$m, 7e-37], N[((-N[Sqrt[N[(N[(N[(A * -4.0), $MachinePrecision] * N[(C * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[((-N[Sqrt[2.0], $MachinePrecision]) / B$95$m), $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])\\
\\
\begin{array}{l}
\mathbf{if}\;B_m \leq 7 \cdot 10^{-37}:\\
\;\;\;\;\frac{-\sqrt{\left(\left(A \cdot -4\right) \cdot \left(C \cdot F\right)\right) \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B_m, B_m, A \cdot \left(C \cdot -4\right)\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{-\sqrt{2}}{B_m} \cdot \sqrt{F \cdot \left(-B_m\right)}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if B < 7.0000000000000003e-37

    1. Initial program 20.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified28.9%

      \[\leadsto \color{blue}{\frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Taylor expanded in C around inf 20.4%

      \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{A}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    4. Taylor expanded in B around 0 16.3%

      \[\leadsto \frac{-\sqrt{\color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)} \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. associate-*r*16.3%

        \[\leadsto \frac{-\sqrt{\color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot F\right)\right)} \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. *-commutative16.3%

        \[\leadsto \frac{-\sqrt{\left(\left(-4 \cdot A\right) \cdot \color{blue}{\left(F \cdot C\right)}\right) \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified16.3%

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

    if 7.0000000000000003e-37 < B

    1. Initial program 24.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified19.3%

      \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)}} \]
    3. Taylor expanded in A around 0 28.3%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. associate-*r*28.3%

        \[\leadsto \color{blue}{\left(-1 \cdot \frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
      2. mul-1-neg28.3%

        \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right)} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)} \]
      3. unpow228.3%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)} \]
      4. unpow228.3%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
      5. hypot-def50.8%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
    5. Simplified50.8%

      \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
    6. Taylor expanded in C around 0 45.1%

      \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \color{blue}{\left(-1 \cdot B\right)}} \]
    7. Step-by-step derivation
      1. mul-1-neg45.1%

        \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}} \]
    8. Simplified45.1%

      \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification25.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 7 \cdot 10^{-37}:\\ \;\;\;\;\frac{-\sqrt{\left(\left(A \cdot -4\right) \cdot \left(C \cdot F\right)\right) \cdot \left(2 \cdot \left(A + A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(-B\right)}\\ \end{array} \]

Alternative 6: 26.5% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{-\sqrt{2}}{B_m} \cdot \sqrt{F \cdot \left(-B_m\right)} \end{array} \]
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 (- 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 * -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 * -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 * -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 * -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(Float64(-sqrt(2.0)) / B_m) * sqrt(Float64(F * 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) / B_m) * 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[((-N[Sqrt[2.0], $MachinePrecision]) / B$95$m), $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])\\
\\
\frac{-\sqrt{2}}{B_m} \cdot \sqrt{F \cdot \left(-B_m\right)}
\end{array}
Derivation
  1. Initial program 21.8%

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

    \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)}} \]
  3. Taylor expanded in A around 0 10.0%

    \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
  4. Step-by-step derivation
    1. associate-*r*10.0%

      \[\leadsto \color{blue}{\left(-1 \cdot \frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    2. mul-1-neg10.0%

      \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right)} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)} \]
    3. unpow210.0%

      \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)} \]
    4. unpow210.0%

      \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
    5. hypot-def17.2%

      \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
  5. Simplified17.2%

    \[\leadsto \color{blue}{\left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
  6. Taylor expanded in C around 0 15.6%

    \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \color{blue}{\left(-1 \cdot B\right)}} \]
  7. Step-by-step derivation
    1. mul-1-neg15.6%

      \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}} \]
  8. Simplified15.6%

    \[\leadsto \left(-\frac{\sqrt{2}}{B}\right) \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}} \]
  9. Final simplification15.6%

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

Alternative 7: 8.9% accurate, 5.8× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B_m}\right) \end{array} \]
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
 (* -2.0 (* (sqrt (* A F)) (/ 1.0 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 -2.0 * (sqrt((A * F)) * (1.0 / 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 = (-2.0d0) * (sqrt((a * f)) * (1.0d0 / 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 -2.0 * (Math.sqrt((A * F)) * (1.0 / 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 -2.0 * (math.sqrt((A * F)) * (1.0 / 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(-2.0 * Float64(sqrt(Float64(A * F)) * Float64(1.0 / 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 = -2.0 * (sqrt((A * F)) * (1.0 / 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[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] * N[(1.0 / 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])\\
\\
-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B_m}\right)
\end{array}
Derivation
  1. Initial program 21.8%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Simplified27.8%

    \[\leadsto \color{blue}{\frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
  3. Taylor expanded in C around inf 16.2%

    \[\leadsto \frac{-\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{A}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
  4. Taylor expanded in B around inf 2.8%

    \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)} \]
  5. Final simplification2.8%

    \[\leadsto -2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right) \]

Reproduce

?
herbie shell --seed 2023318 
(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))))