ABCF->ab-angle a

Percentage Accurate: 18.8% → 56.8%
Time: 30.7s
Alternatives: 13
Speedup: 6.0×

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 13 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: 56.8% accurate, 1.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B\_m}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\log 2 \cdot 0.5}}{B\_m} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B\_m, C\right)} \cdot \left(-\sqrt{F}\right)\right)\\ \end{array} \end{array} \]
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) 2e+23)
   (/
    (* 2.0 (sqrt (* C (* F (- (pow B_m 2.0) (* 4.0 (* C A)))))))
    (- (* C (* 4.0 A)) (pow B_m 2.0)))
   (*
    (/ (exp (* (log 2.0) 0.5)) B_m)
    (* (sqrt (+ C (hypot B_m C))) (- (sqrt F))))))
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+23) {
		tmp = (2.0 * sqrt((C * (F * (pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - pow(B_m, 2.0));
	} else {
		tmp = (exp((log(2.0) * 0.5)) / B_m) * (sqrt((C + hypot(B_m, C))) * -sqrt(F));
	}
	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 tmp;
	if (Math.pow(B_m, 2.0) <= 2e+23) {
		tmp = (2.0 * Math.sqrt((C * (F * (Math.pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - Math.pow(B_m, 2.0));
	} else {
		tmp = (Math.exp((Math.log(2.0) * 0.5)) / B_m) * (Math.sqrt((C + Math.hypot(B_m, C))) * -Math.sqrt(F));
	}
	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):
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e+23:
		tmp = (2.0 * math.sqrt((C * (F * (math.pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - math.pow(B_m, 2.0))
	else:
		tmp = (math.exp((math.log(2.0) * 0.5)) / B_m) * (math.sqrt((C + math.hypot(B_m, C))) * -math.sqrt(F))
	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+23)
		tmp = Float64(Float64(2.0 * sqrt(Float64(C * Float64(F * Float64((B_m ^ 2.0) - Float64(4.0 * Float64(C * A))))))) / Float64(Float64(C * Float64(4.0 * A)) - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(exp(Float64(log(2.0) * 0.5)) / B_m) * Float64(sqrt(Float64(C + hypot(B_m, C))) * Float64(-sqrt(F))));
	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)
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e+23)
		tmp = (2.0 * sqrt((C * (F * ((B_m ^ 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - (B_m ^ 2.0));
	else
		tmp = (exp((log(2.0) * 0.5)) / B_m) * (sqrt((C + hypot(B_m, C))) * -sqrt(F));
	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_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+23], N[(N[(2.0 * N[Sqrt[N[(C * N[(F * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Exp[N[(N[Log[2.0], $MachinePrecision] * 0.5), $MachinePrecision]], $MachinePrecision] / B$95$m), $MachinePrecision] * N[(N[Sqrt[N[(C + N[Sqrt[B$95$m ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[F], $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 2 \cdot 10^{+23}:\\
\;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B\_m}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\frac{e^{\log 2 \cdot 0.5}}{B\_m} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B\_m, C\right)} \cdot \left(-\sqrt{F}\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1.9999999999999998e23

    1. Initial program 16.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 27.9%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Taylor expanded in F around 0 28.6%

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

    if 1.9999999999999998e23 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 12.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg12.6%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    5. Simplified12.6%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    6. Step-by-step derivation
      1. pow1/212.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{{\left(F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)\right)}^{0.5}} \]
      2. *-commutative12.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot {\color{blue}{\left(\left(C + \sqrt{{B}^{2} + {C}^{2}}\right) \cdot F\right)}}^{0.5} \]
      3. unpow-prod-down15.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\left({\left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}^{0.5} \cdot {F}^{0.5}\right)} \]
      4. pow1/215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\color{blue}{\sqrt{C + \sqrt{{B}^{2} + {C}^{2}}}} \cdot {F}^{0.5}\right) \]
      5. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \sqrt{\color{blue}{B \cdot B} + {C}^{2}}} \cdot {F}^{0.5}\right) \]
      6. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \sqrt{B \cdot B + \color{blue}{C \cdot C}}} \cdot {F}^{0.5}\right) \]
      7. hypot-define40.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \color{blue}{\mathsf{hypot}\left(B, C\right)}} \cdot {F}^{0.5}\right) \]
      8. pow1/240.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \color{blue}{\sqrt{F}}\right) \]
    7. Applied egg-rr40.4%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \sqrt{F}\right)} \]
    8. Step-by-step derivation
      1. pow1/240.4%

        \[\leadsto -\frac{\color{blue}{{2}^{0.5}}}{B} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \sqrt{F}\right) \]
      2. pow-to-exp40.5%

        \[\leadsto -\frac{\color{blue}{e^{\log 2 \cdot 0.5}}}{B} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \sqrt{F}\right) \]
    9. Applied egg-rr40.5%

      \[\leadsto -\frac{\color{blue}{e^{\log 2 \cdot 0.5}}}{B} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \sqrt{F}\right) \]
  3. Recombined 2 regimes into one program.
  4. Final simplification34.3%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\log 2 \cdot 0.5}}{B} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \left(-\sqrt{F}\right)\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 56.8% accurate, 1.2× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B\_m}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{C + \mathsf{hypot}\left(B\_m, C\right)} \cdot \sqrt{F}\right) \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
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+23)
   (/
    (* 2.0 (sqrt (* C (* F (- (pow B_m 2.0) (* 4.0 (* C A)))))))
    (- (* C (* 4.0 A)) (pow B_m 2.0)))
   (* (* (sqrt (+ C (hypot B_m C))) (sqrt F)) (/ (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+23) {
		tmp = (2.0 * sqrt((C * (F * (pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - pow(B_m, 2.0));
	} else {
		tmp = (sqrt((C + hypot(B_m, C))) * sqrt(F)) * (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 tmp;
	if (Math.pow(B_m, 2.0) <= 2e+23) {
		tmp = (2.0 * Math.sqrt((C * (F * (Math.pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - Math.pow(B_m, 2.0));
	} else {
		tmp = (Math.sqrt((C + Math.hypot(B_m, C))) * Math.sqrt(F)) * (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):
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e+23:
		tmp = (2.0 * math.sqrt((C * (F * (math.pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - math.pow(B_m, 2.0))
	else:
		tmp = (math.sqrt((C + math.hypot(B_m, C))) * math.sqrt(F)) * (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)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e+23)
		tmp = Float64(Float64(2.0 * sqrt(Float64(C * Float64(F * Float64((B_m ^ 2.0) - Float64(4.0 * Float64(C * A))))))) / Float64(Float64(C * Float64(4.0 * A)) - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(sqrt(Float64(C + hypot(B_m, C))) * sqrt(F)) * 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)
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e+23)
		tmp = (2.0 * sqrt((C * (F * ((B_m ^ 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - (B_m ^ 2.0));
	else
		tmp = (sqrt((C + hypot(B_m, C))) * sqrt(F)) * (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_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+23], N[(N[(2.0 * N[Sqrt[N[(C * N[(F * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[N[(C + N[Sqrt[B$95$m ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Sqrt[F], $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^{+23}:\\
\;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B\_m}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\left(\sqrt{C + \mathsf{hypot}\left(B\_m, C\right)} \cdot \sqrt{F}\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)) < 1.9999999999999998e23

    1. Initial program 16.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 27.9%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Taylor expanded in F around 0 28.6%

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

    if 1.9999999999999998e23 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 12.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg12.6%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    5. Simplified12.6%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    6. Step-by-step derivation
      1. pow1/212.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{{\left(F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)\right)}^{0.5}} \]
      2. *-commutative12.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot {\color{blue}{\left(\left(C + \sqrt{{B}^{2} + {C}^{2}}\right) \cdot F\right)}}^{0.5} \]
      3. unpow-prod-down15.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\left({\left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}^{0.5} \cdot {F}^{0.5}\right)} \]
      4. pow1/215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\color{blue}{\sqrt{C + \sqrt{{B}^{2} + {C}^{2}}}} \cdot {F}^{0.5}\right) \]
      5. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \sqrt{\color{blue}{B \cdot B} + {C}^{2}}} \cdot {F}^{0.5}\right) \]
      6. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \sqrt{B \cdot B + \color{blue}{C \cdot C}}} \cdot {F}^{0.5}\right) \]
      7. hypot-define40.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \color{blue}{\mathsf{hypot}\left(B, C\right)}} \cdot {F}^{0.5}\right) \]
      8. pow1/240.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \color{blue}{\sqrt{F}}\right) \]
    7. Applied egg-rr40.4%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \sqrt{F}\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification34.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \sqrt{F}\right) \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 52.0% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B\_m}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B\_m + C}\right)\right)\\ \end{array} \end{array} \]
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) 2e+23)
   (/
    (* 2.0 (sqrt (* C (* F (- (pow B_m 2.0) (* 4.0 (* C A)))))))
    (- (* C (* 4.0 A)) (pow B_m 2.0)))
   (* (/ (sqrt 2.0) B_m) (* (sqrt F) (- (sqrt (+ B_m C)))))))
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+23) {
		tmp = (2.0 * sqrt((C * (F * (pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - pow(B_m, 2.0));
	} else {
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	}
	return tmp;
}
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
    real(8) :: tmp
    if ((b_m ** 2.0d0) <= 2d+23) then
        tmp = (2.0d0 * sqrt((c * (f * ((b_m ** 2.0d0) - (4.0d0 * (c * a))))))) / ((c * (4.0d0 * a)) - (b_m ** 2.0d0))
    else
        tmp = (sqrt(2.0d0) / b_m) * (sqrt(f) * -sqrt((b_m + c)))
    end if
    code = tmp
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) {
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e+23) {
		tmp = (2.0 * Math.sqrt((C * (F * (Math.pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - Math.pow(B_m, 2.0));
	} else {
		tmp = (Math.sqrt(2.0) / B_m) * (Math.sqrt(F) * -Math.sqrt((B_m + C)));
	}
	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):
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e+23:
		tmp = (2.0 * math.sqrt((C * (F * (math.pow(B_m, 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - math.pow(B_m, 2.0))
	else:
		tmp = (math.sqrt(2.0) / B_m) * (math.sqrt(F) * -math.sqrt((B_m + C)))
	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+23)
		tmp = Float64(Float64(2.0 * sqrt(Float64(C * Float64(F * Float64((B_m ^ 2.0) - Float64(4.0 * Float64(C * A))))))) / Float64(Float64(C * Float64(4.0 * A)) - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(sqrt(F) * Float64(-sqrt(Float64(B_m + C)))));
	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)
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e+23)
		tmp = (2.0 * sqrt((C * (F * ((B_m ^ 2.0) - (4.0 * (C * A))))))) / ((C * (4.0 * A)) - (B_m ^ 2.0));
	else
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	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_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+23], N[(N[(2.0 * N[Sqrt[N[(C * N[(F * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[(N[Sqrt[F], $MachinePrecision] * (-N[Sqrt[N[(B$95$m + C), $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 2 \cdot 10^{+23}:\\
\;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B\_m}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B\_m + C}\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1.9999999999999998e23

    1. Initial program 16.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 27.9%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Taylor expanded in F around 0 28.6%

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

    if 1.9999999999999998e23 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 12.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg12.6%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    5. Simplified12.6%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    6. Step-by-step derivation
      1. pow1/212.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{{\left(F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)\right)}^{0.5}} \]
      2. *-commutative12.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot {\color{blue}{\left(\left(C + \sqrt{{B}^{2} + {C}^{2}}\right) \cdot F\right)}}^{0.5} \]
      3. unpow-prod-down15.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\left({\left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}^{0.5} \cdot {F}^{0.5}\right)} \]
      4. pow1/215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\color{blue}{\sqrt{C + \sqrt{{B}^{2} + {C}^{2}}}} \cdot {F}^{0.5}\right) \]
      5. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \sqrt{\color{blue}{B \cdot B} + {C}^{2}}} \cdot {F}^{0.5}\right) \]
      6. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \sqrt{B \cdot B + \color{blue}{C \cdot C}}} \cdot {F}^{0.5}\right) \]
      7. hypot-define40.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \color{blue}{\mathsf{hypot}\left(B, C\right)}} \cdot {F}^{0.5}\right) \]
      8. pow1/240.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \color{blue}{\sqrt{F}}\right) \]
    7. Applied egg-rr40.4%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \sqrt{F}\right)} \]
    8. Taylor expanded in C around 0 33.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{\color{blue}{B + C}} \cdot \sqrt{F}\right) \]
  3. Recombined 2 regimes into one program.
  4. Final simplification31.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{2 \cdot \sqrt{C \cdot \left(F \cdot \left({B}^{2} - 4 \cdot \left(C \cdot A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B + C}\right)\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 4: 52.0% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := C \cdot \left(4 \cdot A\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B\_m}^{2} - t\_0\right)\right)\right) \cdot \left(2 \cdot C\right)}}{t\_0 - B\_m \cdot B\_m}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B\_m + C}\right)\right)\\ \end{array} \end{array} \]
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 (* C (* 4.0 A))))
   (if (<= (pow B_m 2.0) 2e+23)
     (/
      (sqrt (* (* 2.0 (* F (- (pow B_m 2.0) t_0))) (* 2.0 C)))
      (- t_0 (* B_m B_m)))
     (* (/ (sqrt 2.0) B_m) (* (sqrt F) (- (sqrt (+ B_m C))))))))
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 = C * (4.0 * A);
	double tmp;
	if (pow(B_m, 2.0) <= 2e+23) {
		tmp = sqrt(((2.0 * (F * (pow(B_m, 2.0) - t_0))) * (2.0 * C))) / (t_0 - (B_m * B_m));
	} else {
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	}
	return tmp;
}
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
    real(8) :: t_0
    real(8) :: tmp
    t_0 = c * (4.0d0 * a)
    if ((b_m ** 2.0d0) <= 2d+23) then
        tmp = sqrt(((2.0d0 * (f * ((b_m ** 2.0d0) - t_0))) * (2.0d0 * c))) / (t_0 - (b_m * b_m))
    else
        tmp = (sqrt(2.0d0) / b_m) * (sqrt(f) * -sqrt((b_m + c)))
    end if
    code = tmp
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) {
	double t_0 = C * (4.0 * A);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e+23) {
		tmp = Math.sqrt(((2.0 * (F * (Math.pow(B_m, 2.0) - t_0))) * (2.0 * C))) / (t_0 - (B_m * B_m));
	} else {
		tmp = (Math.sqrt(2.0) / B_m) * (Math.sqrt(F) * -Math.sqrt((B_m + C)));
	}
	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 = C * (4.0 * A)
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e+23:
		tmp = math.sqrt(((2.0 * (F * (math.pow(B_m, 2.0) - t_0))) * (2.0 * C))) / (t_0 - (B_m * B_m))
	else:
		tmp = (math.sqrt(2.0) / B_m) * (math.sqrt(F) * -math.sqrt((B_m + C)))
	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(C * Float64(4.0 * A))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e+23)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(F * Float64((B_m ^ 2.0) - t_0))) * Float64(2.0 * C))) / Float64(t_0 - Float64(B_m * B_m)));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(sqrt(F) * Float64(-sqrt(Float64(B_m + C)))));
	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 = C * (4.0 * A);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e+23)
		tmp = sqrt(((2.0 * (F * ((B_m ^ 2.0) - t_0))) * (2.0 * C))) / (t_0 - (B_m * B_m));
	else
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	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[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+23], N[(N[Sqrt[N[(N[(2.0 * N[(F * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[(N[Sqrt[F], $MachinePrecision] * (-N[Sqrt[N[(B$95$m + C), $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 := C \cdot \left(4 \cdot A\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+23}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B\_m}^{2} - t\_0\right)\right)\right) \cdot \left(2 \cdot C\right)}}{t\_0 - B\_m \cdot B\_m}\\

\mathbf{else}:\\
\;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B\_m + C}\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1.9999999999999998e23

    1. Initial program 16.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 27.9%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Step-by-step derivation
      1. unpow227.9%

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot C\right)}}{\color{blue}{B \cdot B} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied egg-rr27.9%

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

    if 1.9999999999999998e23 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 12.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg12.6%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    5. Simplified12.6%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    6. Step-by-step derivation
      1. pow1/212.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{{\left(F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)\right)}^{0.5}} \]
      2. *-commutative12.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot {\color{blue}{\left(\left(C + \sqrt{{B}^{2} + {C}^{2}}\right) \cdot F\right)}}^{0.5} \]
      3. unpow-prod-down15.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\left({\left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}^{0.5} \cdot {F}^{0.5}\right)} \]
      4. pow1/215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\color{blue}{\sqrt{C + \sqrt{{B}^{2} + {C}^{2}}}} \cdot {F}^{0.5}\right) \]
      5. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \sqrt{\color{blue}{B \cdot B} + {C}^{2}}} \cdot {F}^{0.5}\right) \]
      6. unpow215.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \sqrt{B \cdot B + \color{blue}{C \cdot C}}} \cdot {F}^{0.5}\right) \]
      7. hypot-define40.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \color{blue}{\mathsf{hypot}\left(B, C\right)}} \cdot {F}^{0.5}\right) \]
      8. pow1/240.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \color{blue}{\sqrt{F}}\right) \]
    7. Applied egg-rr40.4%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\left(\sqrt{C + \mathsf{hypot}\left(B, C\right)} \cdot \sqrt{F}\right)} \]
    8. Taylor expanded in C around 0 33.8%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \left(\sqrt{\color{blue}{B + C}} \cdot \sqrt{F}\right) \]
  3. Recombined 2 regimes into one program.
  4. Final simplification30.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - C \cdot \left(4 \cdot A\right)\right)\right)\right) \cdot \left(2 \cdot C\right)}}{C \cdot \left(4 \cdot A\right) - B \cdot B}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B + C}\right)\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 51.3% accurate, 1.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := C \cdot \left(4 \cdot A\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B\_m}^{2} - t\_0\right)\right)\right) \cdot \left(2 \cdot C\right)}}{t\_0 - B\_m \cdot B\_m}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{2}{B\_m}} \cdot \left(-\sqrt{F}\right)\\ \end{array} \end{array} \]
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 (* C (* 4.0 A))))
   (if (<= (pow B_m 2.0) 2e+23)
     (/
      (sqrt (* (* 2.0 (* F (- (pow B_m 2.0) t_0))) (* 2.0 C)))
      (- t_0 (* B_m B_m)))
     (* (sqrt (/ 2.0 B_m)) (- (sqrt F))))))
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 = C * (4.0 * A);
	double tmp;
	if (pow(B_m, 2.0) <= 2e+23) {
		tmp = sqrt(((2.0 * (F * (pow(B_m, 2.0) - t_0))) * (2.0 * C))) / (t_0 - (B_m * B_m));
	} else {
		tmp = sqrt((2.0 / B_m)) * -sqrt(F);
	}
	return tmp;
}
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
    real(8) :: t_0
    real(8) :: tmp
    t_0 = c * (4.0d0 * a)
    if ((b_m ** 2.0d0) <= 2d+23) then
        tmp = sqrt(((2.0d0 * (f * ((b_m ** 2.0d0) - t_0))) * (2.0d0 * c))) / (t_0 - (b_m * b_m))
    else
        tmp = sqrt((2.0d0 / b_m)) * -sqrt(f)
    end if
    code = tmp
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) {
	double t_0 = C * (4.0 * A);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e+23) {
		tmp = Math.sqrt(((2.0 * (F * (Math.pow(B_m, 2.0) - t_0))) * (2.0 * C))) / (t_0 - (B_m * B_m));
	} else {
		tmp = Math.sqrt((2.0 / B_m)) * -Math.sqrt(F);
	}
	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 = C * (4.0 * A)
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e+23:
		tmp = math.sqrt(((2.0 * (F * (math.pow(B_m, 2.0) - t_0))) * (2.0 * C))) / (t_0 - (B_m * B_m))
	else:
		tmp = math.sqrt((2.0 / B_m)) * -math.sqrt(F)
	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(C * Float64(4.0 * A))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e+23)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(F * Float64((B_m ^ 2.0) - t_0))) * Float64(2.0 * C))) / Float64(t_0 - Float64(B_m * B_m)));
	else
		tmp = Float64(sqrt(Float64(2.0 / B_m)) * Float64(-sqrt(F)));
	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 = C * (4.0 * A);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e+23)
		tmp = sqrt(((2.0 * (F * ((B_m ^ 2.0) - t_0))) * (2.0 * C))) / (t_0 - (B_m * B_m));
	else
		tmp = sqrt((2.0 / B_m)) * -sqrt(F);
	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[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+23], N[(N[Sqrt[N[(N[(2.0 * N[(F * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(2.0 / B$95$m), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[F], $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 := C \cdot \left(4 \cdot A\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+23}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B\_m}^{2} - t\_0\right)\right)\right) \cdot \left(2 \cdot C\right)}}{t\_0 - B\_m \cdot B\_m}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1.9999999999999998e23

    1. Initial program 16.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 27.9%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Step-by-step derivation
      1. unpow227.9%

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot C\right)}}{\color{blue}{B \cdot B} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied egg-rr27.9%

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

    if 1.9999999999999998e23 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 14.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 21.1%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg21.1%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. *-commutative21.1%

        \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    5. Simplified21.1%

      \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    6. Step-by-step derivation
      1. *-commutative21.1%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. pow1/221.1%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      3. pow1/221.1%

        \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
      4. pow-prod-down21.2%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    7. Applied egg-rr21.2%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    8. Step-by-step derivation
      1. unpow1/221.2%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    9. Simplified21.2%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    10. Step-by-step derivation
      1. associate-*l/21.2%

        \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
    11. Applied egg-rr21.2%

      \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
    12. Step-by-step derivation
      1. associate-/l*21.1%

        \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
    13. Simplified21.1%

      \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
    14. Step-by-step derivation
      1. pow1/221.1%

        \[\leadsto -\color{blue}{{\left(F \cdot \frac{2}{B}\right)}^{0.5}} \]
      2. *-commutative21.1%

        \[\leadsto -{\color{blue}{\left(\frac{2}{B} \cdot F\right)}}^{0.5} \]
      3. unpow-prod-down33.6%

        \[\leadsto -\color{blue}{{\left(\frac{2}{B}\right)}^{0.5} \cdot {F}^{0.5}} \]
      4. pow1/233.6%

        \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}}} \cdot {F}^{0.5} \]
      5. pow1/233.6%

        \[\leadsto -\sqrt{\frac{2}{B}} \cdot \color{blue}{\sqrt{F}} \]
    15. Applied egg-rr33.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}} \cdot \sqrt{F}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification30.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+23}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - C \cdot \left(4 \cdot A\right)\right)\right)\right) \cdot \left(2 \cdot C\right)}}{C \cdot \left(4 \cdot A\right) - B \cdot B}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{2}{B}} \cdot \left(-\sqrt{F}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 6: 41.8% accurate, 2.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+120}:\\ \;\;\;\;-2 \cdot \sqrt{\frac{C \cdot F}{{B\_m}^{2} - 4 \cdot \left(C \cdot A\right)}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F} \cdot \left(-\sqrt{\frac{2}{B\_m}}\right)\\ \end{array} \end{array} \]
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 (<= (pow B_m 2.0) 2e+120)
   (* -2.0 (sqrt (/ (* C F) (- (pow B_m 2.0) (* 4.0 (* C A))))))
   (* (sqrt F) (- (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+120) {
		tmp = -2.0 * sqrt(((C * F) / (pow(B_m, 2.0) - (4.0 * (C * A)))));
	} else {
		tmp = sqrt(F) * -sqrt((2.0 / B_m));
	}
	return tmp;
}
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
    real(8) :: tmp
    if ((b_m ** 2.0d0) <= 2d+120) then
        tmp = (-2.0d0) * sqrt(((c * f) / ((b_m ** 2.0d0) - (4.0d0 * (c * a)))))
    else
        tmp = sqrt(f) * -sqrt((2.0d0 / b_m))
    end if
    code = tmp
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) {
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e+120) {
		tmp = -2.0 * Math.sqrt(((C * F) / (Math.pow(B_m, 2.0) - (4.0 * (C * A)))));
	} else {
		tmp = Math.sqrt(F) * -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):
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e+120:
		tmp = -2.0 * math.sqrt(((C * F) / (math.pow(B_m, 2.0) - (4.0 * (C * A)))))
	else:
		tmp = math.sqrt(F) * -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)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e+120)
		tmp = Float64(-2.0 * sqrt(Float64(Float64(C * F) / Float64((B_m ^ 2.0) - Float64(4.0 * Float64(C * A))))));
	else
		tmp = Float64(sqrt(F) * Float64(-sqrt(Float64(2.0 / 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)
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e+120)
		tmp = -2.0 * sqrt(((C * F) / ((B_m ^ 2.0) - (4.0 * (C * A)))));
	else
		tmp = sqrt(F) * -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_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+120], N[(-2.0 * N[Sqrt[N[(N[(C * F), $MachinePrecision] / N[(N[Power[B$95$m, 2.0], $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[F], $MachinePrecision] * (-N[Sqrt[N[(2.0 / 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])\\
\\
\begin{array}{l}
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+120}:\\
\;\;\;\;-2 \cdot \sqrt{\frac{C \cdot F}{{B\_m}^{2} - 4 \cdot \left(C \cdot A\right)}}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 2e120

    1. Initial program 17.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 26.3%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Taylor expanded in F around 0 16.8%

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

    if 2e120 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 12.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 21.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg21.0%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. *-commutative21.0%

        \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    5. Simplified21.0%

      \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    6. Step-by-step derivation
      1. *-commutative21.0%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. pow1/221.0%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      3. pow1/221.0%

        \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
      4. pow-prod-down21.1%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    7. Applied egg-rr21.1%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    8. Step-by-step derivation
      1. unpow1/221.1%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    9. Simplified21.1%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    10. Step-by-step derivation
      1. associate-*l/21.1%

        \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
    11. Applied egg-rr21.1%

      \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
    12. Step-by-step derivation
      1. associate-/l*21.0%

        \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
    13. Simplified21.0%

      \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
    14. Step-by-step derivation
      1. pow1/221.0%

        \[\leadsto -\color{blue}{{\left(F \cdot \frac{2}{B}\right)}^{0.5}} \]
      2. *-commutative21.0%

        \[\leadsto -{\color{blue}{\left(\frac{2}{B} \cdot F\right)}}^{0.5} \]
      3. unpow-prod-down34.7%

        \[\leadsto -\color{blue}{{\left(\frac{2}{B}\right)}^{0.5} \cdot {F}^{0.5}} \]
      4. pow1/234.7%

        \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}}} \cdot {F}^{0.5} \]
      5. pow1/234.7%

        \[\leadsto -\sqrt{\frac{2}{B}} \cdot \color{blue}{\sqrt{F}} \]
    15. Applied egg-rr34.7%

      \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}} \cdot \sqrt{F}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification24.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+120}:\\ \;\;\;\;-2 \cdot \sqrt{\frac{C \cdot F}{{B}^{2} - 4 \cdot \left(C \cdot A\right)}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F} \cdot \left(-\sqrt{\frac{2}{B}}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 7: 49.0% accurate, 2.8× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;B\_m \leq 45000000000000:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot C\right) \cdot \left(2 \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{2}{B\_m}} \cdot \left(-\sqrt{F}\right)\\ \end{array} \end{array} \]
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 45000000000000.0)
   (/
    (sqrt (* (* 2.0 C) (* 2.0 (* -4.0 (* A (* C F))))))
    (- (* C (* 4.0 A)) (pow B_m 2.0)))
   (* (sqrt (/ 2.0 B_m)) (- (sqrt F)))))
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 <= 45000000000000.0) {
		tmp = sqrt(((2.0 * C) * (2.0 * (-4.0 * (A * (C * F)))))) / ((C * (4.0 * A)) - pow(B_m, 2.0));
	} else {
		tmp = sqrt((2.0 / B_m)) * -sqrt(F);
	}
	return tmp;
}
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
    real(8) :: tmp
    if (b_m <= 45000000000000.0d0) then
        tmp = sqrt(((2.0d0 * c) * (2.0d0 * ((-4.0d0) * (a * (c * f)))))) / ((c * (4.0d0 * a)) - (b_m ** 2.0d0))
    else
        tmp = sqrt((2.0d0 / b_m)) * -sqrt(f)
    end if
    code = tmp
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) {
	double tmp;
	if (B_m <= 45000000000000.0) {
		tmp = Math.sqrt(((2.0 * C) * (2.0 * (-4.0 * (A * (C * F)))))) / ((C * (4.0 * A)) - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((2.0 / B_m)) * -Math.sqrt(F);
	}
	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):
	tmp = 0
	if B_m <= 45000000000000.0:
		tmp = math.sqrt(((2.0 * C) * (2.0 * (-4.0 * (A * (C * F)))))) / ((C * (4.0 * A)) - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((2.0 / B_m)) * -math.sqrt(F)
	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 <= 45000000000000.0)
		tmp = Float64(sqrt(Float64(Float64(2.0 * C) * Float64(2.0 * Float64(-4.0 * Float64(A * Float64(C * F)))))) / Float64(Float64(C * Float64(4.0 * A)) - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(2.0 / B_m)) * Float64(-sqrt(F)));
	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)
	tmp = 0.0;
	if (B_m <= 45000000000000.0)
		tmp = sqrt(((2.0 * C) * (2.0 * (-4.0 * (A * (C * F)))))) / ((C * (4.0 * A)) - (B_m ^ 2.0));
	else
		tmp = sqrt((2.0 / B_m)) * -sqrt(F);
	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_] := If[LessEqual[B$95$m, 45000000000000.0], N[(N[Sqrt[N[(N[(2.0 * C), $MachinePrecision] * N[(2.0 * N[(-4.0 * N[(A * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(2.0 / B$95$m), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[F], $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 45000000000000:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot C\right) \cdot \left(2 \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if B < 4.5e13

    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 A around -inf 20.4%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Taylor expanded in B around 0 19.0%

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

    if 4.5e13 < B

    1. Initial program 16.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 37.1%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg37.1%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. *-commutative37.1%

        \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    5. Simplified37.1%

      \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    6. Step-by-step derivation
      1. *-commutative37.1%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. pow1/237.1%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      3. pow1/237.1%

        \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
      4. pow-prod-down37.1%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    7. Applied egg-rr37.1%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    8. Step-by-step derivation
      1. unpow1/237.1%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    9. Simplified37.1%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    10. Step-by-step derivation
      1. associate-*l/37.1%

        \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
    11. Applied egg-rr37.1%

      \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
    12. Step-by-step derivation
      1. associate-/l*37.1%

        \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
    13. Simplified37.1%

      \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
    14. Step-by-step derivation
      1. pow1/237.1%

        \[\leadsto -\color{blue}{{\left(F \cdot \frac{2}{B}\right)}^{0.5}} \]
      2. *-commutative37.1%

        \[\leadsto -{\color{blue}{\left(\frac{2}{B} \cdot F\right)}}^{0.5} \]
      3. unpow-prod-down62.1%

        \[\leadsto -\color{blue}{{\left(\frac{2}{B}\right)}^{0.5} \cdot {F}^{0.5}} \]
      4. pow1/262.1%

        \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}}} \cdot {F}^{0.5} \]
      5. pow1/262.1%

        \[\leadsto -\sqrt{\frac{2}{B}} \cdot \color{blue}{\sqrt{F}} \]
    15. Applied egg-rr62.1%

      \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}} \cdot \sqrt{F}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification30.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 45000000000000:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot C\right) \cdot \left(2 \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{2}{B}} \cdot \left(-\sqrt{F}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 8: 38.3% accurate, 2.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;F \leq 7.5 \cdot 10^{+26}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(C + \mathsf{hypot}\left(B\_m, C\right)\right)\right)}}{-B\_m}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F} \cdot \left(-\sqrt{\frac{2}{B\_m}}\right)\\ \end{array} \end{array} \]
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 (<= F 7.5e+26)
   (/ (sqrt (* 2.0 (* F (+ C (hypot B_m C))))) (- B_m))
   (* (sqrt F) (- (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 (F <= 7.5e+26) {
		tmp = sqrt((2.0 * (F * (C + hypot(B_m, C))))) / -B_m;
	} else {
		tmp = sqrt(F) * -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 tmp;
	if (F <= 7.5e+26) {
		tmp = Math.sqrt((2.0 * (F * (C + Math.hypot(B_m, C))))) / -B_m;
	} else {
		tmp = Math.sqrt(F) * -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):
	tmp = 0
	if F <= 7.5e+26:
		tmp = math.sqrt((2.0 * (F * (C + math.hypot(B_m, C))))) / -B_m
	else:
		tmp = math.sqrt(F) * -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)
	tmp = 0.0
	if (F <= 7.5e+26)
		tmp = Float64(sqrt(Float64(2.0 * Float64(F * Float64(C + hypot(B_m, C))))) / Float64(-B_m));
	else
		tmp = Float64(sqrt(F) * Float64(-sqrt(Float64(2.0 / 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)
	tmp = 0.0;
	if (F <= 7.5e+26)
		tmp = sqrt((2.0 * (F * (C + hypot(B_m, C))))) / -B_m;
	else
		tmp = sqrt(F) * -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_] := If[LessEqual[F, 7.5e+26], N[(N[Sqrt[N[(2.0 * N[(F * N[(C + N[Sqrt[B$95$m ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-B$95$m)), $MachinePrecision], N[(N[Sqrt[F], $MachinePrecision] * (-N[Sqrt[N[(2.0 / 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])\\
\\
\begin{array}{l}
\mathbf{if}\;F \leq 7.5 \cdot 10^{+26}:\\
\;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(C + \mathsf{hypot}\left(B\_m, C\right)\right)\right)}}{-B\_m}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if F < 7.49999999999999941e26

    1. Initial program 17.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 11.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg11.6%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    5. Simplified11.6%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
    6. Step-by-step derivation
      1. neg-sub011.6%

        \[\leadsto \color{blue}{0 - \frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
      2. associate-*l/11.6%

        \[\leadsto 0 - \color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}{B}} \]
      3. pow1/211.6%

        \[\leadsto 0 - \frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}{B} \]
      4. pow1/211.6%

        \[\leadsto 0 - \frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)\right)}^{0.5}}}{B} \]
      5. pow-prod-down11.7%

        \[\leadsto 0 - \frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)\right)\right)}^{0.5}}}{B} \]
      6. unpow211.7%

        \[\leadsto 0 - \frac{{\left(2 \cdot \left(F \cdot \left(C + \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)\right)}^{0.5}}{B} \]
      7. unpow211.7%

        \[\leadsto 0 - \frac{{\left(2 \cdot \left(F \cdot \left(C + \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)\right)}^{0.5}}{B} \]
      8. hypot-define23.9%

        \[\leadsto 0 - \frac{{\left(2 \cdot \left(F \cdot \left(C + \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)\right)}^{0.5}}{B} \]
    7. Applied egg-rr23.9%

      \[\leadsto \color{blue}{0 - \frac{{\left(2 \cdot \left(F \cdot \left(C + \mathsf{hypot}\left(B, C\right)\right)\right)\right)}^{0.5}}{B}} \]
    8. Step-by-step derivation
      1. neg-sub023.9%

        \[\leadsto \color{blue}{-\frac{{\left(2 \cdot \left(F \cdot \left(C + \mathsf{hypot}\left(B, C\right)\right)\right)\right)}^{0.5}}{B}} \]
      2. distribute-neg-frac223.9%

        \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(F \cdot \left(C + \mathsf{hypot}\left(B, C\right)\right)\right)\right)}^{0.5}}{-B}} \]
      3. unpow1/223.9%

        \[\leadsto \frac{\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(C + \mathsf{hypot}\left(B, C\right)\right)\right)}}}{-B} \]
    9. Simplified23.9%

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

    if 7.49999999999999941e26 < F

    1. Initial program 11.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 16.1%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg16.1%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. *-commutative16.1%

        \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    5. Simplified16.1%

      \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    6. Step-by-step derivation
      1. *-commutative16.1%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. pow1/216.4%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      3. pow1/216.4%

        \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
      4. pow-prod-down16.5%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    7. Applied egg-rr16.5%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    8. Step-by-step derivation
      1. unpow1/216.1%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    9. Simplified16.1%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    10. Step-by-step derivation
      1. associate-*l/16.1%

        \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
    11. Applied egg-rr16.1%

      \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
    12. Step-by-step derivation
      1. associate-/l*16.1%

        \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
    13. Simplified16.1%

      \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
    14. Step-by-step derivation
      1. pow1/216.4%

        \[\leadsto -\color{blue}{{\left(F \cdot \frac{2}{B}\right)}^{0.5}} \]
      2. *-commutative16.4%

        \[\leadsto -{\color{blue}{\left(\frac{2}{B} \cdot F\right)}}^{0.5} \]
      3. unpow-prod-down16.3%

        \[\leadsto -\color{blue}{{\left(\frac{2}{B}\right)}^{0.5} \cdot {F}^{0.5}} \]
      4. pow1/216.3%

        \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}}} \cdot {F}^{0.5} \]
      5. pow1/216.3%

        \[\leadsto -\sqrt{\frac{2}{B}} \cdot \color{blue}{\sqrt{F}} \]
    15. Applied egg-rr16.3%

      \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}} \cdot \sqrt{F}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification20.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;F \leq 7.5 \cdot 10^{+26}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(C + \mathsf{hypot}\left(B, C\right)\right)\right)}}{-B}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F} \cdot \left(-\sqrt{\frac{2}{B}}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 9: 34.7% accurate, 3.1× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{F} \cdot \left(-\sqrt{\frac{2}{B\_m}}\right) \end{array} \]
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 F) (- (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) * -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) * -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) * -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) * -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(F) * Float64(-sqrt(Float64(2.0 / 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 = sqrt(F) * -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[F], $MachinePrecision] * (-N[Sqrt[N[(2.0 / 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])\\
\\
\sqrt{F} \cdot \left(-\sqrt{\frac{2}{B\_m}}\right)
\end{array}
Derivation
  1. Initial program 15.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in B around inf 12.6%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg12.6%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  5. Simplified12.6%

    \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  6. Step-by-step derivation
    1. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. pow1/212.7%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
    3. pow1/212.7%

      \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
    4. pow-prod-down12.8%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  7. Applied egg-rr12.8%

    \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  8. Step-by-step derivation
    1. unpow1/212.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  9. Simplified12.6%

    \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  10. Step-by-step derivation
    1. associate-*l/12.6%

      \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
  11. Applied egg-rr12.6%

    \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
  12. Step-by-step derivation
    1. associate-/l*12.6%

      \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
  13. Simplified12.6%

    \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
  14. Step-by-step derivation
    1. pow1/212.8%

      \[\leadsto -\color{blue}{{\left(F \cdot \frac{2}{B}\right)}^{0.5}} \]
    2. *-commutative12.8%

      \[\leadsto -{\color{blue}{\left(\frac{2}{B} \cdot F\right)}}^{0.5} \]
    3. unpow-prod-down18.6%

      \[\leadsto -\color{blue}{{\left(\frac{2}{B}\right)}^{0.5} \cdot {F}^{0.5}} \]
    4. pow1/218.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}}} \cdot {F}^{0.5} \]
    5. pow1/218.6%

      \[\leadsto -\sqrt{\frac{2}{B}} \cdot \color{blue}{\sqrt{F}} \]
  15. Applied egg-rr18.6%

    \[\leadsto -\color{blue}{\sqrt{\frac{2}{B}} \cdot \sqrt{F}} \]
  16. Final simplification18.6%

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

Alternative 10: 27.2% accurate, 3.1× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{\left|F \cdot \frac{2}{B\_m}\right|} \end{array} \]
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 (fabs (* F (/ 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(fabs((F * (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(abs((f * (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(Math.abs((F * (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(math.fabs((F * (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(abs(Float64(F * Float64(2.0 / 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 = -sqrt(abs((F * (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[Sqrt[N[Abs[N[(F * N[(2.0 / 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])\\
\\
-\sqrt{\left|F \cdot \frac{2}{B\_m}\right|}
\end{array}
Derivation
  1. Initial program 15.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in B around inf 12.6%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg12.6%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  5. Simplified12.6%

    \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  6. Step-by-step derivation
    1. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. pow1/212.7%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
    3. pow1/212.7%

      \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
    4. pow-prod-down12.8%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  7. Applied egg-rr12.8%

    \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  8. Step-by-step derivation
    1. unpow1/212.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  9. Simplified12.6%

    \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  10. Step-by-step derivation
    1. associate-*l/12.6%

      \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
  11. Applied egg-rr12.6%

    \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
  12. Step-by-step derivation
    1. associate-/l*12.6%

      \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
  13. Simplified12.6%

    \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
  14. Step-by-step derivation
    1. add-sqr-sqrt12.6%

      \[\leadsto -\sqrt{\color{blue}{\sqrt{F \cdot \frac{2}{B}} \cdot \sqrt{F \cdot \frac{2}{B}}}} \]
    2. pow1/212.6%

      \[\leadsto -\sqrt{\color{blue}{{\left(F \cdot \frac{2}{B}\right)}^{0.5}} \cdot \sqrt{F \cdot \frac{2}{B}}} \]
    3. pow1/212.8%

      \[\leadsto -\sqrt{{\left(F \cdot \frac{2}{B}\right)}^{0.5} \cdot \color{blue}{{\left(F \cdot \frac{2}{B}\right)}^{0.5}}} \]
    4. pow-prod-down11.8%

      \[\leadsto -\sqrt{\color{blue}{{\left(\left(F \cdot \frac{2}{B}\right) \cdot \left(F \cdot \frac{2}{B}\right)\right)}^{0.5}}} \]
    5. pow211.8%

      \[\leadsto -\sqrt{{\color{blue}{\left({\left(F \cdot \frac{2}{B}\right)}^{2}\right)}}^{0.5}} \]
    6. associate-*r/11.8%

      \[\leadsto -\sqrt{{\left({\color{blue}{\left(\frac{F \cdot 2}{B}\right)}}^{2}\right)}^{0.5}} \]
  15. Applied egg-rr11.8%

    \[\leadsto -\sqrt{\color{blue}{{\left({\left(\frac{F \cdot 2}{B}\right)}^{2}\right)}^{0.5}}} \]
  16. Step-by-step derivation
    1. unpow1/211.8%

      \[\leadsto -\sqrt{\color{blue}{\sqrt{{\left(\frac{F \cdot 2}{B}\right)}^{2}}}} \]
    2. unpow211.8%

      \[\leadsto -\sqrt{\sqrt{\color{blue}{\frac{F \cdot 2}{B} \cdot \frac{F \cdot 2}{B}}}} \]
    3. rem-sqrt-square21.5%

      \[\leadsto -\sqrt{\color{blue}{\left|\frac{F \cdot 2}{B}\right|}} \]
    4. associate-/l*21.5%

      \[\leadsto -\sqrt{\left|\color{blue}{F \cdot \frac{2}{B}}\right|} \]
  17. Simplified21.5%

    \[\leadsto -\sqrt{\color{blue}{\left|F \cdot \frac{2}{B}\right|}} \]
  18. Add Preprocessing

Alternative 11: 27.0% accurate, 5.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -{\left(2 \cdot \frac{F}{B\_m}\right)}^{0.5} \end{array} \]
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 (- (pow (* 2.0 (/ F B_m)) 0.5)))
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 -pow((2.0 * (F / B_m)), 0.5);
}
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 * (f / b_m)) ** 0.5d0)
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.pow((2.0 * (F / B_m)), 0.5);
}
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.pow((2.0 * (F / B_m)), 0.5)
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(2.0 * Float64(F / B_m)) ^ 0.5))
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 * (F / B_m)) ^ 0.5);
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[Power[N[(2.0 * N[(F / B$95$m), $MachinePrecision]), $MachinePrecision], 0.5], $MachinePrecision])
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
-{\left(2 \cdot \frac{F}{B\_m}\right)}^{0.5}
\end{array}
Derivation
  1. Initial program 15.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in B around inf 12.6%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg12.6%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  5. Simplified12.6%

    \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  6. Step-by-step derivation
    1. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. pow1/212.7%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
    3. pow1/212.7%

      \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
    4. pow-prod-down12.8%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  7. Applied egg-rr12.8%

    \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  8. Final simplification12.8%

    \[\leadsto -{\left(2 \cdot \frac{F}{B}\right)}^{0.5} \]
  9. Add Preprocessing

Alternative 12: 27.0% accurate, 6.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{2 \cdot \frac{F}{B\_m}} \end{array} \]
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 (/ 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 * (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 * (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 * (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 * (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(Float64(2.0 * 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 * (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[Sqrt[N[(2.0 * 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 \frac{F}{B\_m}}
\end{array}
Derivation
  1. Initial program 15.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in B around inf 12.6%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg12.6%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  5. Simplified12.6%

    \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  6. Step-by-step derivation
    1. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. pow1/212.7%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
    3. pow1/212.7%

      \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
    4. pow-prod-down12.8%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  7. Applied egg-rr12.8%

    \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  8. Step-by-step derivation
    1. unpow1/212.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  9. Simplified12.6%

    \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  10. Final simplification12.6%

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

Alternative 13: 27.0% accurate, 6.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{F \cdot \frac{2}{B\_m}} \end{array} \]
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 (/ 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 * (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 * (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 * (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 * (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(2.0 / 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 = -sqrt((F * (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[Sqrt[N[(F * N[(2.0 / 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])\\
\\
-\sqrt{F \cdot \frac{2}{B\_m}}
\end{array}
Derivation
  1. Initial program 15.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in B around inf 12.6%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg12.6%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  5. Simplified12.6%

    \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
  6. Step-by-step derivation
    1. *-commutative12.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. pow1/212.7%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
    3. pow1/212.7%

      \[\leadsto -{\left(\frac{F}{B}\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}} \]
    4. pow-prod-down12.8%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  7. Applied egg-rr12.8%

    \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  8. Step-by-step derivation
    1. unpow1/212.6%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  9. Simplified12.6%

    \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  10. Step-by-step derivation
    1. associate-*l/12.6%

      \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
  11. Applied egg-rr12.6%

    \[\leadsto -\sqrt{\color{blue}{\frac{F \cdot 2}{B}}} \]
  12. Step-by-step derivation
    1. associate-/l*12.6%

      \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
  13. Simplified12.6%

    \[\leadsto -\sqrt{\color{blue}{F \cdot \frac{2}{B}}} \]
  14. Add Preprocessing

Reproduce

?
herbie shell --seed 2024114 
(FPCore (A B C F)
  :name "ABCF->ab-angle a"
  :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))))