ABCF->ab-angle a

Percentage Accurate: 18.3% → 59.0%
Time: 24.4s
Alternatives: 11
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 11 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: 18.3% 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: 59.0% accurate, 0.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := -\sqrt{2}\\ t_1 := \left(4 \cdot A\right) \cdot C\\ t_2 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_1\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_1 - {B\_m}^{2}}\\ t_3 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;t\_2 \leq -2 \cdot 10^{-213}:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B\_m, A - C\right)\right)}{{B\_m}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot t\_0\\ \mathbf{elif}\;t\_2 \leq 10^{+111}:\\ \;\;\;\;{\left(\frac{t\_3}{-\sqrt{t\_3 \cdot \left(F \cdot \mathsf{fma}\left(C, 4, \frac{-{B\_m}^{2}}{A}\right)\right)}}\right)}^{-1}\\ \mathbf{elif}\;t\_2 \leq \infty:\\ \;\;\;\;\frac{C}{4 \cdot \left(A \cdot C\right) - {B\_m}^{2}} \cdot \sqrt{F \cdot \left(2 \cdot \frac{-4 \cdot \left(A \cdot \left(A - A\right)\right) + 2 \cdot {B\_m}^{2}}{C} + A \cdot -16\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot t\_0\\ \end{array} \end{array} \]
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 (- (sqrt 2.0)))
        (t_1 (* (* 4.0 A) C))
        (t_2
         (/
          (sqrt
           (*
            (* 2.0 (* (- (pow B_m 2.0) t_1) F))
            (+ (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
          (- t_1 (pow B_m 2.0))))
        (t_3 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= t_2 -2e-213)
     (*
      (sqrt
       (*
        F
        (/
         (+ A (+ C (hypot B_m (- A C))))
         (+ (pow B_m 2.0) (* -4.0 (* A C))))))
      t_0)
     (if (<= t_2 1e+111)
       (pow
        (/ t_3 (- (sqrt (* t_3 (* F (fma C 4.0 (/ (- (pow B_m 2.0)) A)))))))
        -1.0)
       (if (<= t_2 INFINITY)
         (*
          (/ C (- (* 4.0 (* A C)) (pow B_m 2.0)))
          (sqrt
           (*
            F
            (+
             (* 2.0 (/ (+ (* -4.0 (* A (- A A))) (* 2.0 (pow B_m 2.0))) C))
             (* A -16.0)))))
         (* (* (sqrt F) (sqrt (/ 1.0 B_m))) t_0))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = -sqrt(2.0);
	double t_1 = (4.0 * A) * C;
	double t_2 = sqrt(((2.0 * ((pow(B_m, 2.0) - t_1) * F)) * ((A + C) + sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / (t_1 - pow(B_m, 2.0));
	double t_3 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (t_2 <= -2e-213) {
		tmp = sqrt((F * ((A + (C + hypot(B_m, (A - C)))) / (pow(B_m, 2.0) + (-4.0 * (A * C)))))) * t_0;
	} else if (t_2 <= 1e+111) {
		tmp = pow((t_3 / -sqrt((t_3 * (F * fma(C, 4.0, (-pow(B_m, 2.0) / A)))))), -1.0);
	} else if (t_2 <= ((double) INFINITY)) {
		tmp = (C / ((4.0 * (A * C)) - pow(B_m, 2.0))) * sqrt((F * ((2.0 * (((-4.0 * (A * (A - A))) + (2.0 * pow(B_m, 2.0))) / C)) + (A * -16.0))));
	} else {
		tmp = (sqrt(F) * sqrt((1.0 / B_m))) * t_0;
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(-sqrt(2.0))
	t_1 = Float64(Float64(4.0 * A) * C)
	t_2 = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_1) * F)) * Float64(Float64(A + C) + sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / Float64(t_1 - (B_m ^ 2.0)))
	t_3 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if (t_2 <= -2e-213)
		tmp = Float64(sqrt(Float64(F * Float64(Float64(A + Float64(C + hypot(B_m, Float64(A - C)))) / Float64((B_m ^ 2.0) + Float64(-4.0 * Float64(A * C)))))) * t_0);
	elseif (t_2 <= 1e+111)
		tmp = Float64(t_3 / Float64(-sqrt(Float64(t_3 * Float64(F * fma(C, 4.0, Float64(Float64(-(B_m ^ 2.0)) / A))))))) ^ -1.0;
	elseif (t_2 <= Inf)
		tmp = Float64(Float64(C / Float64(Float64(4.0 * Float64(A * C)) - (B_m ^ 2.0))) * sqrt(Float64(F * Float64(Float64(2.0 * Float64(Float64(Float64(-4.0 * Float64(A * Float64(A - A))) + Float64(2.0 * (B_m ^ 2.0))) / C)) + Float64(A * -16.0)))));
	else
		tmp = Float64(Float64(sqrt(F) * sqrt(Float64(1.0 / B_m))) * t_0);
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = (-N[Sqrt[2.0], $MachinePrecision])}, Block[{t$95$1 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$2 = N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$1), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$1 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$2, -2e-213], N[(N[Sqrt[N[(F * N[(N[(A + N[(C + N[Sqrt[B$95$m ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision], If[LessEqual[t$95$2, 1e+111], N[Power[N[(t$95$3 / (-N[Sqrt[N[(t$95$3 * N[(F * N[(C * 4.0 + N[((-N[Power[B$95$m, 2.0], $MachinePrecision]) / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision], -1.0], $MachinePrecision], If[LessEqual[t$95$2, Infinity], N[(N[(C / N[(N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(F * N[(N[(2.0 * N[(N[(N[(-4.0 * N[(A * N[(A - A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(2.0 * N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision] + N[(A * -16.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[F], $MachinePrecision] * N[Sqrt[N[(1.0 / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$0), $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 := -\sqrt{2}\\
t_1 := \left(4 \cdot A\right) \cdot C\\
t_2 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_1\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_1 - {B\_m}^{2}}\\
t_3 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;t\_2 \leq -2 \cdot 10^{-213}:\\
\;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B\_m, A - C\right)\right)}{{B\_m}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot t\_0\\

\mathbf{elif}\;t\_2 \leq 10^{+111}:\\
\;\;\;\;{\left(\frac{t\_3}{-\sqrt{t\_3 \cdot \left(F \cdot \mathsf{fma}\left(C, 4, \frac{-{B\_m}^{2}}{A}\right)\right)}}\right)}^{-1}\\

\mathbf{elif}\;t\_2 \leq \infty:\\
\;\;\;\;\frac{C}{4 \cdot \left(A \cdot C\right) - {B\_m}^{2}} \cdot \sqrt{F \cdot \left(2 \cdot \frac{-4 \cdot \left(A \cdot \left(A - A\right)\right) + 2 \cdot {B\_m}^{2}}{C} + A \cdot -16\right)}\\

\mathbf{else}:\\
\;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -1.9999999999999999e-213

    1. Initial program 33.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in F around 0 45.4%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F \cdot \left(A + \left(C + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg45.4%

        \[\leadsto \color{blue}{-\sqrt{\frac{F \cdot \left(A + \left(C + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}} \]
    5. Simplified79.8%

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

    if -1.9999999999999999e-213 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < 9.99999999999999957e110

    1. Initial program 23.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified26.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C + \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in A around -inf 33.8%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \color{blue}{\left(-1 \cdot \frac{{B}^{2}}{A} + 4 \cdot C\right)}}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. clear-num33.9%

        \[\leadsto \color{blue}{\frac{1}{\frac{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(-1 \cdot \frac{{B}^{2}}{A} + 4 \cdot C\right)}}}} \]
      2. inv-pow33.9%

        \[\leadsto \color{blue}{{\left(\frac{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(-1 \cdot \frac{{B}^{2}}{A} + 4 \cdot C\right)}}\right)}^{-1}} \]
      3. associate-*l*33.9%

        \[\leadsto {\left(\frac{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(-1 \cdot \frac{{B}^{2}}{A} + 4 \cdot C\right)\right)}}}\right)}^{-1} \]
      4. +-commutative33.9%

        \[\leadsto {\left(\frac{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \color{blue}{\left(4 \cdot C + -1 \cdot \frac{{B}^{2}}{A}\right)}\right)}}\right)}^{-1} \]
      5. *-commutative33.9%

        \[\leadsto {\left(\frac{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(\color{blue}{C \cdot 4} + -1 \cdot \frac{{B}^{2}}{A}\right)\right)}}\right)}^{-1} \]
      6. fma-define33.9%

        \[\leadsto {\left(\frac{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \color{blue}{\mathsf{fma}\left(C, 4, -1 \cdot \frac{{B}^{2}}{A}\right)}\right)}}\right)}^{-1} \]
      7. mul-1-neg33.9%

        \[\leadsto {\left(\frac{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \mathsf{fma}\left(C, 4, \color{blue}{-\frac{{B}^{2}}{A}}\right)\right)}}\right)}^{-1} \]
    6. Applied egg-rr33.9%

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

    if 9.99999999999999957e110 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < +inf.0

    1. Initial program 9.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified34.5%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 4.0%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in F around 0 39.9%

      \[\leadsto \color{blue}{\frac{C}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \cdot \sqrt{F \cdot \left(-16 \cdot A + 2 \cdot \frac{-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}}{C}\right)}} \]

    if +inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)))

    1. Initial program 0.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 18.7%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg18.7%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified18.7%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. pow1/218.7%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      2. div-inv18.7%

        \[\leadsto -{\color{blue}{\left(F \cdot \frac{1}{B}\right)}}^{0.5} \cdot \sqrt{2} \]
      3. unpow-prod-down30.1%

        \[\leadsto -\color{blue}{\left({F}^{0.5} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
      4. pow1/230.1%

        \[\leadsto -\left(\color{blue}{\sqrt{F}} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right) \cdot \sqrt{2} \]
    7. Applied egg-rr30.1%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
    8. Step-by-step derivation
      1. unpow1/230.1%

        \[\leadsto -\left(\sqrt{F} \cdot \color{blue}{\sqrt{\frac{1}{B}}}\right) \cdot \sqrt{2} \]
    9. Simplified30.1%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right)} \cdot \sqrt{2} \]
  3. Recombined 4 regimes into one program.
  4. Final simplification48.3%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -2 \cdot 10^{-213}:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B, A - C\right)\right)}{{B}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq 10^{+111}:\\ \;\;\;\;{\left(\frac{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \mathsf{fma}\left(C, 4, \frac{-{B}^{2}}{A}\right)\right)}}\right)}^{-1}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{C}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \cdot \sqrt{F \cdot \left(2 \cdot \frac{-4 \cdot \left(A \cdot \left(A - A\right)\right) + 2 \cdot {B}^{2}}{C} + A \cdot -16\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 57.5% accurate, 1.2× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := -\sqrt{2}\\ t_1 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;B\_m \leq 9.2 \cdot 10^{-275}:\\ \;\;\;\;\sqrt{t\_1 \cdot \left(F \cdot \mathsf{fma}\left(C, 4, \frac{-{B\_m}^{2}}{A}\right)\right)} \cdot \frac{-1}{t\_1}\\ \mathbf{elif}\;B\_m \leq 3.9 \cdot 10^{-66}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{elif}\;B\_m \leq 2.1 \cdot 10^{+133}:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B\_m, A - C\right)\right)}{{B\_m}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot t\_0\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot t\_0\\ \end{array} \end{array} \]
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 (- (sqrt 2.0))) (t_1 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= B_m 9.2e-275)
     (* (sqrt (* t_1 (* F (fma C 4.0 (/ (- (pow B_m 2.0)) A))))) (/ -1.0 t_1))
     (if (<= B_m 3.9e-66)
       (* 0.25 (/ (sqrt (* F (* A -16.0))) A))
       (if (<= B_m 2.1e+133)
         (*
          (sqrt
           (*
            F
            (/
             (+ A (+ C (hypot B_m (- A C))))
             (+ (pow B_m 2.0) (* -4.0 (* A C))))))
          t_0)
         (* (* (sqrt F) (sqrt (/ 1.0 B_m))) t_0))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = -sqrt(2.0);
	double t_1 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (B_m <= 9.2e-275) {
		tmp = sqrt((t_1 * (F * fma(C, 4.0, (-pow(B_m, 2.0) / A))))) * (-1.0 / t_1);
	} else if (B_m <= 3.9e-66) {
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	} else if (B_m <= 2.1e+133) {
		tmp = sqrt((F * ((A + (C + hypot(B_m, (A - C)))) / (pow(B_m, 2.0) + (-4.0 * (A * C)))))) * t_0;
	} else {
		tmp = (sqrt(F) * sqrt((1.0 / B_m))) * t_0;
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(-sqrt(2.0))
	t_1 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if (B_m <= 9.2e-275)
		tmp = Float64(sqrt(Float64(t_1 * Float64(F * fma(C, 4.0, Float64(Float64(-(B_m ^ 2.0)) / A))))) * Float64(-1.0 / t_1));
	elseif (B_m <= 3.9e-66)
		tmp = Float64(0.25 * Float64(sqrt(Float64(F * Float64(A * -16.0))) / A));
	elseif (B_m <= 2.1e+133)
		tmp = Float64(sqrt(Float64(F * Float64(Float64(A + Float64(C + hypot(B_m, Float64(A - C)))) / Float64((B_m ^ 2.0) + Float64(-4.0 * Float64(A * C)))))) * t_0);
	else
		tmp = Float64(Float64(sqrt(F) * sqrt(Float64(1.0 / B_m))) * t_0);
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = (-N[Sqrt[2.0], $MachinePrecision])}, Block[{t$95$1 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B$95$m, 9.2e-275], N[(N[Sqrt[N[(t$95$1 * N[(F * N[(C * 4.0 + N[((-N[Power[B$95$m, 2.0], $MachinePrecision]) / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[B$95$m, 3.9e-66], N[(0.25 * N[(N[Sqrt[N[(F * N[(A * -16.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision], If[LessEqual[B$95$m, 2.1e+133], N[(N[Sqrt[N[(F * N[(N[(A + N[(C + N[Sqrt[B$95$m ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision], N[(N[(N[Sqrt[F], $MachinePrecision] * N[Sqrt[N[(1.0 / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$0), $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 := -\sqrt{2}\\
t_1 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;B\_m \leq 9.2 \cdot 10^{-275}:\\
\;\;\;\;\sqrt{t\_1 \cdot \left(F \cdot \mathsf{fma}\left(C, 4, \frac{-{B\_m}^{2}}{A}\right)\right)} \cdot \frac{-1}{t\_1}\\

\mathbf{elif}\;B\_m \leq 3.9 \cdot 10^{-66}:\\
\;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\

\mathbf{elif}\;B\_m \leq 2.1 \cdot 10^{+133}:\\
\;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B\_m, A - C\right)\right)}{{B\_m}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot t\_0\\

\mathbf{else}:\\
\;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if B < 9.19999999999999959e-275

    1. Initial program 16.5%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified25.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C + \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in A around -inf 17.5%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \color{blue}{\left(-1 \cdot \frac{{B}^{2}}{A} + 4 \cdot C\right)}}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. div-inv17.5%

        \[\leadsto \color{blue}{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(-1 \cdot \frac{{B}^{2}}{A} + 4 \cdot C\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
      2. associate-*l*17.6%

        \[\leadsto \sqrt{\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(-1 \cdot \frac{{B}^{2}}{A} + 4 \cdot C\right)\right)}} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. +-commutative17.6%

        \[\leadsto \sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \color{blue}{\left(4 \cdot C + -1 \cdot \frac{{B}^{2}}{A}\right)}\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. *-commutative17.6%

        \[\leadsto \sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(\color{blue}{C \cdot 4} + -1 \cdot \frac{{B}^{2}}{A}\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      5. fma-define17.6%

        \[\leadsto \sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \color{blue}{\mathsf{fma}\left(C, 4, -1 \cdot \frac{{B}^{2}}{A}\right)}\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      6. mul-1-neg17.6%

        \[\leadsto \sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \mathsf{fma}\left(C, 4, \color{blue}{-\frac{{B}^{2}}{A}}\right)\right)} \cdot \frac{1}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Applied egg-rr17.6%

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

    if 9.19999999999999959e-275 < B < 3.89999999999999983e-66

    1. Initial program 27.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified36.3%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 21.2%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in B around 0 36.4%

      \[\leadsto \color{blue}{0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)} \]
    6. Step-by-step derivation
      1. pow136.4%

        \[\leadsto \color{blue}{{\left(0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)\right)}^{1}} \]
    7. Applied egg-rr36.4%

      \[\leadsto \color{blue}{{\left(0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}\right)}^{1}} \]
    8. Step-by-step derivation
      1. unpow136.4%

        \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}} \]
    9. Simplified36.4%

      \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, F \cdot \frac{0}{C}, F \cdot \left(A \cdot -16\right)\right)}}{A}} \]
    10. Taylor expanded in F around 0 36.4%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{-16 \cdot \left(A \cdot F\right)}}}{A} \]
    11. Step-by-step derivation
      1. associate-*r*36.4%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot F}}}{A} \]
      2. *-commutative36.4%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(A \cdot -16\right)} \cdot F}}{A} \]
      3. *-commutative36.4%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]
    12. Simplified36.4%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]

    if 3.89999999999999983e-66 < B < 2.1e133

    1. Initial program 20.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in F around 0 31.4%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F \cdot \left(A + \left(C + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg31.4%

        \[\leadsto \color{blue}{-\sqrt{\frac{F \cdot \left(A + \left(C + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}} \]
    5. Simplified56.4%

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

    if 2.1e133 < B

    1. Initial program 2.5%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 40.8%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg40.8%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified40.8%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. pow1/240.8%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      2. div-inv40.8%

        \[\leadsto -{\color{blue}{\left(F \cdot \frac{1}{B}\right)}}^{0.5} \cdot \sqrt{2} \]
      3. unpow-prod-down70.2%

        \[\leadsto -\color{blue}{\left({F}^{0.5} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
      4. pow1/270.2%

        \[\leadsto -\left(\color{blue}{\sqrt{F}} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right) \cdot \sqrt{2} \]
    7. Applied egg-rr70.2%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
    8. Step-by-step derivation
      1. unpow1/270.2%

        \[\leadsto -\left(\sqrt{F} \cdot \color{blue}{\sqrt{\frac{1}{B}}}\right) \cdot \sqrt{2} \]
    9. Simplified70.2%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right)} \cdot \sqrt{2} \]
  3. Recombined 4 regimes into one program.
  4. Final simplification36.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 9.2 \cdot 10^{-275}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \mathsf{fma}\left(C, 4, \frac{-{B}^{2}}{A}\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;B \leq 3.9 \cdot 10^{-66}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{elif}\;B \leq 2.1 \cdot 10^{+133}:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B, A - C\right)\right)}{{B}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 57.7% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := -\sqrt{2}\\ t_1 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;B\_m \leq 1.4 \cdot 10^{-280}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_1\right) \cdot \left(4 \cdot C\right)}}{-t\_1}\\ \mathbf{elif}\;B\_m \leq 3.9 \cdot 10^{-66}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{elif}\;B\_m \leq 3.2 \cdot 10^{+134}:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B\_m, A - C\right)\right)}{{B\_m}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot t\_0\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot t\_0\\ \end{array} \end{array} \]
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 (- (sqrt 2.0))) (t_1 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= B_m 1.4e-280)
     (/ (sqrt (* (* F t_1) (* 4.0 C))) (- t_1))
     (if (<= B_m 3.9e-66)
       (* 0.25 (/ (sqrt (* F (* A -16.0))) A))
       (if (<= B_m 3.2e+134)
         (*
          (sqrt
           (*
            F
            (/
             (+ A (+ C (hypot B_m (- A C))))
             (+ (pow B_m 2.0) (* -4.0 (* A C))))))
          t_0)
         (* (* (sqrt F) (sqrt (/ 1.0 B_m))) t_0))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = -sqrt(2.0);
	double t_1 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (B_m <= 1.4e-280) {
		tmp = sqrt(((F * t_1) * (4.0 * C))) / -t_1;
	} else if (B_m <= 3.9e-66) {
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	} else if (B_m <= 3.2e+134) {
		tmp = sqrt((F * ((A + (C + hypot(B_m, (A - C)))) / (pow(B_m, 2.0) + (-4.0 * (A * C)))))) * t_0;
	} else {
		tmp = (sqrt(F) * sqrt((1.0 / B_m))) * t_0;
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(-sqrt(2.0))
	t_1 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if (B_m <= 1.4e-280)
		tmp = Float64(sqrt(Float64(Float64(F * t_1) * Float64(4.0 * C))) / Float64(-t_1));
	elseif (B_m <= 3.9e-66)
		tmp = Float64(0.25 * Float64(sqrt(Float64(F * Float64(A * -16.0))) / A));
	elseif (B_m <= 3.2e+134)
		tmp = Float64(sqrt(Float64(F * Float64(Float64(A + Float64(C + hypot(B_m, Float64(A - C)))) / Float64((B_m ^ 2.0) + Float64(-4.0 * Float64(A * C)))))) * t_0);
	else
		tmp = Float64(Float64(sqrt(F) * sqrt(Float64(1.0 / B_m))) * t_0);
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = (-N[Sqrt[2.0], $MachinePrecision])}, Block[{t$95$1 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B$95$m, 1.4e-280], N[(N[Sqrt[N[(N[(F * t$95$1), $MachinePrecision] * N[(4.0 * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$1)), $MachinePrecision], If[LessEqual[B$95$m, 3.9e-66], N[(0.25 * N[(N[Sqrt[N[(F * N[(A * -16.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision], If[LessEqual[B$95$m, 3.2e+134], N[(N[Sqrt[N[(F * N[(N[(A + N[(C + N[Sqrt[B$95$m ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision], N[(N[(N[Sqrt[F], $MachinePrecision] * N[Sqrt[N[(1.0 / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * t$95$0), $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 := -\sqrt{2}\\
t_1 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;B\_m \leq 1.4 \cdot 10^{-280}:\\
\;\;\;\;\frac{\sqrt{\left(F \cdot t\_1\right) \cdot \left(4 \cdot C\right)}}{-t\_1}\\

\mathbf{elif}\;B\_m \leq 3.9 \cdot 10^{-66}:\\
\;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\

\mathbf{elif}\;B\_m \leq 3.2 \cdot 10^{+134}:\\
\;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B\_m, A - C\right)\right)}{{B\_m}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot t\_0\\

\mathbf{else}:\\
\;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if B < 1.40000000000000009e-280

    1. Initial program 16.8%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified25.6%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C + \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in A around -inf 15.3%

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

    if 1.40000000000000009e-280 < B < 3.89999999999999983e-66

    1. Initial program 26.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified34.5%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 22.0%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in B around 0 36.3%

      \[\leadsto \color{blue}{0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)} \]
    6. Step-by-step derivation
      1. pow136.3%

        \[\leadsto \color{blue}{{\left(0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)\right)}^{1}} \]
    7. Applied egg-rr36.4%

      \[\leadsto \color{blue}{{\left(0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}\right)}^{1}} \]
    8. Step-by-step derivation
      1. unpow136.4%

        \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}} \]
    9. Simplified36.4%

      \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, F \cdot \frac{0}{C}, F \cdot \left(A \cdot -16\right)\right)}}{A}} \]
    10. Taylor expanded in F around 0 36.4%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{-16 \cdot \left(A \cdot F\right)}}}{A} \]
    11. Step-by-step derivation
      1. associate-*r*36.4%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot F}}}{A} \]
      2. *-commutative36.4%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(A \cdot -16\right)} \cdot F}}{A} \]
      3. *-commutative36.4%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]
    12. Simplified36.4%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]

    if 3.89999999999999983e-66 < B < 3.2000000000000001e134

    1. Initial program 20.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in F around 0 31.4%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F \cdot \left(A + \left(C + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg31.4%

        \[\leadsto \color{blue}{-\sqrt{\frac{F \cdot \left(A + \left(C + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}} \]
    5. Simplified56.4%

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

    if 3.2000000000000001e134 < B

    1. Initial program 2.5%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 40.8%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg40.8%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified40.8%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. pow1/240.8%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      2. div-inv40.8%

        \[\leadsto -{\color{blue}{\left(F \cdot \frac{1}{B}\right)}}^{0.5} \cdot \sqrt{2} \]
      3. unpow-prod-down70.2%

        \[\leadsto -\color{blue}{\left({F}^{0.5} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
      4. pow1/270.2%

        \[\leadsto -\left(\color{blue}{\sqrt{F}} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right) \cdot \sqrt{2} \]
    7. Applied egg-rr70.2%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
    8. Step-by-step derivation
      1. unpow1/270.2%

        \[\leadsto -\left(\sqrt{F} \cdot \color{blue}{\sqrt{\frac{1}{B}}}\right) \cdot \sqrt{2} \]
    9. Simplified70.2%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right)} \cdot \sqrt{2} \]
  3. Recombined 4 regimes into one program.
  4. Final simplification35.3%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.4 \cdot 10^{-280}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(4 \cdot C\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;B \leq 3.9 \cdot 10^{-66}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{elif}\;B \leq 3.2 \cdot 10^{+134}:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B, A - C\right)\right)}{{B}^{2} + -4 \cdot \left(A \cdot C\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 4: 55.6% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{+150}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 5e+150)
   (* 0.25 (/ (sqrt (* F (* A -16.0))) A))
   (* (* (sqrt F) (sqrt (/ 1.0 B_m))) (- (sqrt 2.0)))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 5e+150) {
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	} else {
		tmp = (sqrt(F) * sqrt((1.0 / B_m))) * -sqrt(2.0);
	}
	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) <= 5d+150) then
        tmp = 0.25d0 * (sqrt((f * (a * (-16.0d0)))) / a)
    else
        tmp = (sqrt(f) * sqrt((1.0d0 / b_m))) * -sqrt(2.0d0)
    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) <= 5e+150) {
		tmp = 0.25 * (Math.sqrt((F * (A * -16.0))) / A);
	} else {
		tmp = (Math.sqrt(F) * Math.sqrt((1.0 / B_m))) * -Math.sqrt(2.0);
	}
	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) <= 5e+150:
		tmp = 0.25 * (math.sqrt((F * (A * -16.0))) / A)
	else:
		tmp = (math.sqrt(F) * math.sqrt((1.0 / B_m))) * -math.sqrt(2.0)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e+150)
		tmp = Float64(0.25 * Float64(sqrt(Float64(F * Float64(A * -16.0))) / A));
	else
		tmp = Float64(Float64(sqrt(F) * sqrt(Float64(1.0 / B_m))) * Float64(-sqrt(2.0)));
	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) <= 5e+150)
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	else
		tmp = (sqrt(F) * sqrt((1.0 / B_m))) * -sqrt(2.0);
	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], 5e+150], N[(0.25 * N[(N[Sqrt[N[(F * N[(A * -16.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[F], $MachinePrecision] * N[Sqrt[N[(1.0 / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
\mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{+150}:\\
\;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\

\mathbf{else}:\\
\;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot \left(-\sqrt{2}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 5.00000000000000009e150

    1. Initial program 24.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified35.6%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 14.8%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in B around 0 29.9%

      \[\leadsto \color{blue}{0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)} \]
    6. Step-by-step derivation
      1. pow129.9%

        \[\leadsto \color{blue}{{\left(0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)\right)}^{1}} \]
    7. Applied egg-rr30.0%

      \[\leadsto \color{blue}{{\left(0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}\right)}^{1}} \]
    8. Step-by-step derivation
      1. unpow130.0%

        \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}} \]
    9. Simplified30.0%

      \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, F \cdot \frac{0}{C}, F \cdot \left(A \cdot -16\right)\right)}}{A}} \]
    10. Taylor expanded in F around 0 30.0%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{-16 \cdot \left(A \cdot F\right)}}}{A} \]
    11. Step-by-step derivation
      1. associate-*r*30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot F}}}{A} \]
      2. *-commutative30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(A \cdot -16\right)} \cdot F}}{A} \]
      3. *-commutative30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]
    12. Simplified30.0%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]

    if 5.00000000000000009e150 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 2.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 26.7%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg26.7%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified26.7%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. pow1/226.7%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      2. div-inv26.7%

        \[\leadsto -{\color{blue}{\left(F \cdot \frac{1}{B}\right)}}^{0.5} \cdot \sqrt{2} \]
      3. unpow-prod-down39.7%

        \[\leadsto -\color{blue}{\left({F}^{0.5} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
      4. pow1/239.7%

        \[\leadsto -\left(\color{blue}{\sqrt{F}} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right) \cdot \sqrt{2} \]
    7. Applied egg-rr39.7%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
    8. Step-by-step derivation
      1. unpow1/239.7%

        \[\leadsto -\left(\sqrt{F} \cdot \color{blue}{\sqrt{\frac{1}{B}}}\right) \cdot \sqrt{2} \]
    9. Simplified39.7%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right)} \cdot \sqrt{2} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification33.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{+150}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 55.6% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{+150}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F}}{\sqrt{B\_m}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 5e+150)
   (* 0.25 (/ (sqrt (* F (* A -16.0))) A))
   (* (/ (sqrt F) (sqrt B_m)) (- (sqrt 2.0)))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 5e+150) {
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	} else {
		tmp = (sqrt(F) / sqrt(B_m)) * -sqrt(2.0);
	}
	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) <= 5d+150) then
        tmp = 0.25d0 * (sqrt((f * (a * (-16.0d0)))) / a)
    else
        tmp = (sqrt(f) / sqrt(b_m)) * -sqrt(2.0d0)
    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) <= 5e+150) {
		tmp = 0.25 * (Math.sqrt((F * (A * -16.0))) / A);
	} else {
		tmp = (Math.sqrt(F) / Math.sqrt(B_m)) * -Math.sqrt(2.0);
	}
	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) <= 5e+150:
		tmp = 0.25 * (math.sqrt((F * (A * -16.0))) / A)
	else:
		tmp = (math.sqrt(F) / math.sqrt(B_m)) * -math.sqrt(2.0)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e+150)
		tmp = Float64(0.25 * Float64(sqrt(Float64(F * Float64(A * -16.0))) / A));
	else
		tmp = Float64(Float64(sqrt(F) / sqrt(B_m)) * Float64(-sqrt(2.0)));
	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) <= 5e+150)
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	else
		tmp = (sqrt(F) / sqrt(B_m)) * -sqrt(2.0);
	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], 5e+150], N[(0.25 * N[(N[Sqrt[N[(F * N[(A * -16.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[F], $MachinePrecision] / N[Sqrt[B$95$m], $MachinePrecision]), $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
\mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{+150}:\\
\;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 5.00000000000000009e150

    1. Initial program 24.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified35.6%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 14.8%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in B around 0 29.9%

      \[\leadsto \color{blue}{0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)} \]
    6. Step-by-step derivation
      1. pow129.9%

        \[\leadsto \color{blue}{{\left(0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)\right)}^{1}} \]
    7. Applied egg-rr30.0%

      \[\leadsto \color{blue}{{\left(0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}\right)}^{1}} \]
    8. Step-by-step derivation
      1. unpow130.0%

        \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}} \]
    9. Simplified30.0%

      \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, F \cdot \frac{0}{C}, F \cdot \left(A \cdot -16\right)\right)}}{A}} \]
    10. Taylor expanded in F around 0 30.0%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{-16 \cdot \left(A \cdot F\right)}}}{A} \]
    11. Step-by-step derivation
      1. associate-*r*30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot F}}}{A} \]
      2. *-commutative30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(A \cdot -16\right)} \cdot F}}{A} \]
      3. *-commutative30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]
    12. Simplified30.0%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]

    if 5.00000000000000009e150 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 2.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 26.7%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg26.7%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified26.7%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. sqrt-div39.7%

        \[\leadsto -\color{blue}{\frac{\sqrt{F}}{\sqrt{B}}} \cdot \sqrt{2} \]
    7. Applied egg-rr39.7%

      \[\leadsto -\color{blue}{\frac{\sqrt{F}}{\sqrt{B}}} \cdot \sqrt{2} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification33.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{+150}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F}}{\sqrt{B}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 6: 55.0% accurate, 1.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;B\_m \leq 1.46 \cdot 10^{-280}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(4 \cdot C\right)}}{-t\_0}\\ \mathbf{elif}\;B\_m \leq 1.45 \cdot 10^{+83}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{elif}\;B\_m \leq 5.6 \cdot 10^{+169}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(C + \mathsf{hypot}\left(B\_m, C\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= B_m 1.46e-280)
     (/ (sqrt (* (* F t_0) (* 4.0 C))) (- t_0))
     (if (<= B_m 1.45e+83)
       (* 0.25 (/ (sqrt (* F (* A -16.0))) A))
       (if (<= B_m 5.6e+169)
         (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (+ C (hypot B_m C))))))
         (* (* (sqrt F) (sqrt (/ 1.0 B_m))) (- (sqrt 2.0))))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (B_m <= 1.46e-280) {
		tmp = sqrt(((F * t_0) * (4.0 * C))) / -t_0;
	} else if (B_m <= 1.45e+83) {
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	} else if (B_m <= 5.6e+169) {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (C + hypot(B_m, C))));
	} else {
		tmp = (sqrt(F) * sqrt((1.0 / B_m))) * -sqrt(2.0);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if (B_m <= 1.46e-280)
		tmp = Float64(sqrt(Float64(Float64(F * t_0) * Float64(4.0 * C))) / Float64(-t_0));
	elseif (B_m <= 1.45e+83)
		tmp = Float64(0.25 * Float64(sqrt(Float64(F * Float64(A * -16.0))) / A));
	elseif (B_m <= 5.6e+169)
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(C + hypot(B_m, C))))));
	else
		tmp = Float64(Float64(sqrt(F) * sqrt(Float64(1.0 / B_m))) * Float64(-sqrt(2.0)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B$95$m, 1.46e-280], N[(N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(4.0 * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], If[LessEqual[B$95$m, 1.45e+83], N[(0.25 * N[(N[Sqrt[N[(F * N[(A * -16.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision], If[LessEqual[B$95$m, 5.6e+169], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(C + N[Sqrt[B$95$m ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision], N[(N[(N[Sqrt[F], $MachinePrecision] * N[Sqrt[N[(1.0 / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;B\_m \leq 1.46 \cdot 10^{-280}:\\
\;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(4 \cdot C\right)}}{-t\_0}\\

\mathbf{elif}\;B\_m \leq 1.45 \cdot 10^{+83}:\\
\;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\

\mathbf{elif}\;B\_m \leq 5.6 \cdot 10^{+169}:\\
\;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(C + \mathsf{hypot}\left(B\_m, C\right)\right)}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot \left(-\sqrt{2}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if B < 1.46e-280

    1. Initial program 16.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified25.5%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C + \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in A around -inf 15.3%

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

    if 1.46e-280 < B < 1.45e83

    1. Initial program 27.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified38.7%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 16.6%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in B around 0 29.2%

      \[\leadsto \color{blue}{0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)} \]
    6. Step-by-step derivation
      1. pow129.2%

        \[\leadsto \color{blue}{{\left(0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)\right)}^{1}} \]
    7. Applied egg-rr29.3%

      \[\leadsto \color{blue}{{\left(0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}\right)}^{1}} \]
    8. Step-by-step derivation
      1. unpow129.3%

        \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}} \]
    9. Simplified29.3%

      \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, F \cdot \frac{0}{C}, F \cdot \left(A \cdot -16\right)\right)}}{A}} \]
    10. Taylor expanded in F around 0 29.3%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{-16 \cdot \left(A \cdot F\right)}}}{A} \]
    11. Step-by-step derivation
      1. associate-*r*29.3%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot F}}}{A} \]
      2. *-commutative29.3%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(A \cdot -16\right)} \cdot F}}{A} \]
      3. *-commutative29.3%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]
    12. Simplified29.3%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]

    if 1.45e83 < B < 5.6000000000000003e169

    1. Initial program 5.5%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 23.0%

      \[\leadsto \frac{\color{blue}{-1 \cdot \left(\left(B \cdot \sqrt{2}\right) \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Step-by-step derivation
      1. mul-1-neg23.0%

        \[\leadsto \frac{\color{blue}{-\left(B \cdot \sqrt{2}\right) \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Simplified23.0%

      \[\leadsto \frac{\color{blue}{-\left(B \cdot \sqrt{2}\right) \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Taylor expanded in A around 0 23.4%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    7. Step-by-step derivation
      1. mul-1-neg23.4%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
      2. unpow223.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)} \]
      3. unpow223.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
      4. hypot-undefine43.5%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
    8. Simplified43.5%

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

    if 5.6000000000000003e169 < B

    1. Initial program 0.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 45.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg45.0%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified45.0%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. pow1/245.0%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      2. div-inv45.0%

        \[\leadsto -{\color{blue}{\left(F \cdot \frac{1}{B}\right)}}^{0.5} \cdot \sqrt{2} \]
      3. unpow-prod-down83.1%

        \[\leadsto -\color{blue}{\left({F}^{0.5} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
      4. pow1/283.1%

        \[\leadsto -\left(\color{blue}{\sqrt{F}} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right) \cdot \sqrt{2} \]
    7. Applied egg-rr83.1%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
    8. Step-by-step derivation
      1. unpow1/283.1%

        \[\leadsto -\left(\sqrt{F} \cdot \color{blue}{\sqrt{\frac{1}{B}}}\right) \cdot \sqrt{2} \]
    9. Simplified83.1%

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

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.46 \cdot 10^{-280}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(4 \cdot C\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;B \leq 1.45 \cdot 10^{+83}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{elif}\;B \leq 5.6 \cdot 10^{+169}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(C + \mathsf{hypot}\left(B, C\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 7: 55.5% accurate, 2.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;B\_m \leq 1.45 \cdot 10^{+83}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{elif}\;B\_m \leq 9.5 \cdot 10^{+169}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(C + \mathsf{hypot}\left(B\_m, C\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
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 1.45e+83)
   (* 0.25 (/ (sqrt (* F (* A -16.0))) A))
   (if (<= B_m 9.5e+169)
     (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (+ C (hypot B_m C))))))
     (* (* (sqrt F) (sqrt (/ 1.0 B_m))) (- (sqrt 2.0))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (B_m <= 1.45e+83) {
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	} else if (B_m <= 9.5e+169) {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (C + hypot(B_m, C))));
	} else {
		tmp = (sqrt(F) * sqrt((1.0 / B_m))) * -sqrt(2.0);
	}
	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 (B_m <= 1.45e+83) {
		tmp = 0.25 * (Math.sqrt((F * (A * -16.0))) / A);
	} else if (B_m <= 9.5e+169) {
		tmp = (Math.sqrt(2.0) / B_m) * -Math.sqrt((F * (C + Math.hypot(B_m, C))));
	} else {
		tmp = (Math.sqrt(F) * Math.sqrt((1.0 / B_m))) * -Math.sqrt(2.0);
	}
	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 <= 1.45e+83:
		tmp = 0.25 * (math.sqrt((F * (A * -16.0))) / A)
	elif B_m <= 9.5e+169:
		tmp = (math.sqrt(2.0) / B_m) * -math.sqrt((F * (C + math.hypot(B_m, C))))
	else:
		tmp = (math.sqrt(F) * math.sqrt((1.0 / B_m))) * -math.sqrt(2.0)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if (B_m <= 1.45e+83)
		tmp = Float64(0.25 * Float64(sqrt(Float64(F * Float64(A * -16.0))) / A));
	elseif (B_m <= 9.5e+169)
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(C + hypot(B_m, C))))));
	else
		tmp = Float64(Float64(sqrt(F) * sqrt(Float64(1.0 / B_m))) * Float64(-sqrt(2.0)));
	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 <= 1.45e+83)
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	elseif (B_m <= 9.5e+169)
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (C + hypot(B_m, C))));
	else
		tmp = (sqrt(F) * sqrt((1.0 / B_m))) * -sqrt(2.0);
	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, 1.45e+83], N[(0.25 * N[(N[Sqrt[N[(F * N[(A * -16.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision], If[LessEqual[B$95$m, 9.5e+169], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(C + N[Sqrt[B$95$m ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision], N[(N[(N[Sqrt[F], $MachinePrecision] * N[Sqrt[N[(1.0 / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * (-N[Sqrt[2.0], $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 1.45 \cdot 10^{+83}:\\
\;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\

\mathbf{elif}\;B\_m \leq 9.5 \cdot 10^{+169}:\\
\;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(C + \mathsf{hypot}\left(B\_m, C\right)\right)}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B\_m}}\right) \cdot \left(-\sqrt{2}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if B < 1.45e83

    1. Initial program 21.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified30.8%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 12.3%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in B around 0 25.6%

      \[\leadsto \color{blue}{0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)} \]
    6. Step-by-step derivation
      1. pow125.6%

        \[\leadsto \color{blue}{{\left(0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)\right)}^{1}} \]
    7. Applied egg-rr25.6%

      \[\leadsto \color{blue}{{\left(0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}\right)}^{1}} \]
    8. Step-by-step derivation
      1. unpow125.6%

        \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}} \]
    9. Simplified25.6%

      \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, F \cdot \frac{0}{C}, F \cdot \left(A \cdot -16\right)\right)}}{A}} \]
    10. Taylor expanded in F around 0 25.6%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{-16 \cdot \left(A \cdot F\right)}}}{A} \]
    11. Step-by-step derivation
      1. associate-*r*25.6%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot F}}}{A} \]
      2. *-commutative25.6%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(A \cdot -16\right)} \cdot F}}{A} \]
      3. *-commutative25.6%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]
    12. Simplified25.6%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]

    if 1.45e83 < B < 9.4999999999999995e169

    1. Initial program 5.5%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 23.0%

      \[\leadsto \frac{\color{blue}{-1 \cdot \left(\left(B \cdot \sqrt{2}\right) \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Step-by-step derivation
      1. mul-1-neg23.0%

        \[\leadsto \frac{\color{blue}{-\left(B \cdot \sqrt{2}\right) \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Simplified23.0%

      \[\leadsto \frac{\color{blue}{-\left(B \cdot \sqrt{2}\right) \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Taylor expanded in A around 0 23.4%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    7. Step-by-step derivation
      1. mul-1-neg23.4%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
      2. unpow223.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)} \]
      3. unpow223.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
      4. hypot-undefine43.5%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
    8. Simplified43.5%

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

    if 9.4999999999999995e169 < B

    1. Initial program 0.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 45.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg45.0%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified45.0%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. pow1/245.0%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \sqrt{2} \]
      2. div-inv45.0%

        \[\leadsto -{\color{blue}{\left(F \cdot \frac{1}{B}\right)}}^{0.5} \cdot \sqrt{2} \]
      3. unpow-prod-down83.1%

        \[\leadsto -\color{blue}{\left({F}^{0.5} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
      4. pow1/283.1%

        \[\leadsto -\left(\color{blue}{\sqrt{F}} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right) \cdot \sqrt{2} \]
    7. Applied egg-rr83.1%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot {\left(\frac{1}{B}\right)}^{0.5}\right)} \cdot \sqrt{2} \]
    8. Step-by-step derivation
      1. unpow1/283.1%

        \[\leadsto -\left(\sqrt{F} \cdot \color{blue}{\sqrt{\frac{1}{B}}}\right) \cdot \sqrt{2} \]
    9. Simplified83.1%

      \[\leadsto -\color{blue}{\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right)} \cdot \sqrt{2} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification34.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.45 \cdot 10^{+83}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{elif}\;B \leq 9.5 \cdot 10^{+169}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(C + \mathsf{hypot}\left(B, C\right)\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\sqrt{F} \cdot \sqrt{\frac{1}{B}}\right) \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 8: 48.5% accurate, 2.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{+150}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{2 \cdot \frac{F}{B\_m}}\\ \end{array} \end{array} \]
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) 5e+150)
   (* 0.25 (/ (sqrt (* F (* A -16.0))) A))
   (- (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) {
	double tmp;
	if (pow(B_m, 2.0) <= 5e+150) {
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	} else {
		tmp = -sqrt((2.0 * (F / 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) <= 5d+150) then
        tmp = 0.25d0 * (sqrt((f * (a * (-16.0d0)))) / a)
    else
        tmp = -sqrt((2.0d0 * (f / 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) <= 5e+150) {
		tmp = 0.25 * (Math.sqrt((F * (A * -16.0))) / A);
	} else {
		tmp = -Math.sqrt((2.0 * (F / 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) <= 5e+150:
		tmp = 0.25 * (math.sqrt((F * (A * -16.0))) / A)
	else:
		tmp = -math.sqrt((2.0 * (F / B_m)))
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e+150)
		tmp = Float64(0.25 * Float64(sqrt(Float64(F * Float64(A * -16.0))) / A));
	else
		tmp = Float64(-sqrt(Float64(2.0 * Float64(F / 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) <= 5e+150)
		tmp = 0.25 * (sqrt((F * (A * -16.0))) / A);
	else
		tmp = -sqrt((2.0 * (F / 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], 5e+150], N[(0.25 * N[(N[Sqrt[N[(F * N[(A * -16.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision], (-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])\\
\\
\begin{array}{l}
\mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{+150}:\\
\;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 5.00000000000000009e150

    1. Initial program 24.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified35.6%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 14.8%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in B around 0 29.9%

      \[\leadsto \color{blue}{0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)} \]
    6. Step-by-step derivation
      1. pow129.9%

        \[\leadsto \color{blue}{{\left(0.25 \cdot \left(\frac{1}{A} \cdot \sqrt{-16 \cdot \left(A \cdot F\right) + -8 \cdot \frac{A \cdot \left(F \cdot \left(A + -1 \cdot A\right)\right)}{C}}\right)\right)}^{1}} \]
    7. Applied egg-rr30.0%

      \[\leadsto \color{blue}{{\left(0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}\right)}^{1}} \]
    8. Step-by-step derivation
      1. unpow130.0%

        \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, A \cdot \frac{F \cdot \left(0 \cdot A\right)}{C}, \left(-16 \cdot A\right) \cdot F\right)}}{A}} \]
    9. Simplified30.0%

      \[\leadsto \color{blue}{0.25 \cdot \frac{\sqrt{\mathsf{fma}\left(-8, F \cdot \frac{0}{C}, F \cdot \left(A \cdot -16\right)\right)}}{A}} \]
    10. Taylor expanded in F around 0 30.0%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{-16 \cdot \left(A \cdot F\right)}}}{A} \]
    11. Step-by-step derivation
      1. associate-*r*30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(-16 \cdot A\right) \cdot F}}}{A} \]
      2. *-commutative30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{\left(A \cdot -16\right)} \cdot F}}{A} \]
      3. *-commutative30.0%

        \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]
    12. Simplified30.0%

      \[\leadsto 0.25 \cdot \frac{\sqrt{\color{blue}{F \cdot \left(A \cdot -16\right)}}}{A} \]

    if 5.00000000000000009e150 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 2.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 26.7%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg26.7%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified26.7%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. neg-sub026.7%

        \[\leadsto \color{blue}{0 - \sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
      2. sqrt-unprod26.8%

        \[\leadsto 0 - \color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    7. Applied egg-rr26.8%

      \[\leadsto \color{blue}{0 - \sqrt{\frac{F}{B} \cdot 2}} \]
    8. Step-by-step derivation
      1. neg-sub026.8%

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B} \cdot 2}} \]
    9. Simplified26.8%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B} \cdot 2}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification28.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{+150}:\\ \;\;\;\;0.25 \cdot \frac{\sqrt{F \cdot \left(A \cdot -16\right)}}{A}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{2 \cdot \frac{F}{B}}\\ \end{array} \]
  5. Add Preprocessing

Alternative 9: 28.2% accurate, 5.6× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;C \leq 5.5 \cdot 10^{+159}:\\ \;\;\;\;-{\left(2 \cdot \frac{F}{B\_m}\right)}^{0.5}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\frac{1}{B\_m} \cdot \sqrt{C \cdot F}\right)\\ \end{array} \end{array} \]
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 (<= C 5.5e+159)
   (- (pow (* 2.0 (/ F B_m)) 0.5))
   (* -2.0 (* (/ 1.0 B_m) (sqrt (* C 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 (C <= 5.5e+159) {
		tmp = -pow((2.0 * (F / B_m)), 0.5);
	} else {
		tmp = -2.0 * ((1.0 / B_m) * sqrt((C * 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 (c <= 5.5d+159) then
        tmp = -((2.0d0 * (f / b_m)) ** 0.5d0)
    else
        tmp = (-2.0d0) * ((1.0d0 / b_m) * sqrt((c * 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 (C <= 5.5e+159) {
		tmp = -Math.pow((2.0 * (F / B_m)), 0.5);
	} else {
		tmp = -2.0 * ((1.0 / B_m) * Math.sqrt((C * 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 C <= 5.5e+159:
		tmp = -math.pow((2.0 * (F / B_m)), 0.5)
	else:
		tmp = -2.0 * ((1.0 / B_m) * math.sqrt((C * 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 (C <= 5.5e+159)
		tmp = Float64(-(Float64(2.0 * Float64(F / B_m)) ^ 0.5));
	else
		tmp = Float64(-2.0 * Float64(Float64(1.0 / B_m) * sqrt(Float64(C * 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 (C <= 5.5e+159)
		tmp = -((2.0 * (F / B_m)) ^ 0.5);
	else
		tmp = -2.0 * ((1.0 / B_m) * sqrt((C * 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[C, 5.5e+159], (-N[Power[N[(2.0 * N[(F / B$95$m), $MachinePrecision]), $MachinePrecision], 0.5], $MachinePrecision]), N[(-2.0 * N[(N[(1.0 / B$95$m), $MachinePrecision] * N[Sqrt[N[(C * F), $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}\;C \leq 5.5 \cdot 10^{+159}:\\
\;\;\;\;-{\left(2 \cdot \frac{F}{B\_m}\right)}^{0.5}\\

\mathbf{else}:\\
\;\;\;\;-2 \cdot \left(\frac{1}{B\_m} \cdot \sqrt{C \cdot F}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if C < 5.4999999999999998e159

    1. Initial program 19.5%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 15.2%

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

        \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    5. Simplified15.2%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. sqrt-unprod15.2%

        \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
      2. pow1/215.3%

        \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
    7. Applied egg-rr15.3%

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

    if 5.4999999999999998e159 < C

    1. Initial program 1.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified26.6%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(A + \left(C + \mathsf{hypot}\left(B, C - A\right)\right)\right)}}{4 \cdot \left(A \cdot C\right) - {B}^{2}}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 1.2%

      \[\leadsto \frac{\sqrt{\color{blue}{{C}^{2} \cdot \left(-16 \cdot \left(A \cdot F\right) + 2 \cdot \frac{F \cdot \left(-4 \cdot \left(A \cdot \left(A + -1 \cdot A\right)\right) + 2 \cdot {B}^{2}\right)}{C}\right)}}}{4 \cdot \left(A \cdot C\right) - {B}^{2}} \]
    5. Taylor expanded in A around 0 12.9%

      \[\leadsto \color{blue}{-2 \cdot \left(\frac{1}{B} \cdot \sqrt{C \cdot F}\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification15.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 5.5 \cdot 10^{+159}:\\ \;\;\;\;-{\left(2 \cdot \frac{F}{B}\right)}^{0.5}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\frac{1}{B} \cdot \sqrt{C \cdot F}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 10: 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 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 B around inf 13.5%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg13.5%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
  5. Simplified13.5%

    \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
  6. Step-by-step derivation
    1. sqrt-unprod13.5%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
    2. pow1/213.6%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  7. Applied egg-rr13.6%

    \[\leadsto -\color{blue}{{\left(\frac{F}{B} \cdot 2\right)}^{0.5}} \]
  8. Final simplification13.6%

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

Alternative 11: 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 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 B around inf 13.5%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg13.5%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
  5. Simplified13.5%

    \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
  6. Step-by-step derivation
    1. neg-sub013.5%

      \[\leadsto \color{blue}{0 - \sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]
    2. sqrt-unprod13.5%

      \[\leadsto 0 - \color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]
  7. Applied egg-rr13.5%

    \[\leadsto \color{blue}{0 - \sqrt{\frac{F}{B} \cdot 2}} \]
  8. Step-by-step derivation
    1. neg-sub013.5%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B} \cdot 2}} \]
  9. Simplified13.5%

    \[\leadsto \color{blue}{-\sqrt{\frac{F}{B} \cdot 2}} \]
  10. Final simplification13.5%

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

Reproduce

?
herbie shell --seed 2024137 
(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))))