ABCF->ab-angle b

Percentage Accurate: 18.6% → 49.8%
Time: 26.2s
Alternatives: 16
Speedup: 3.0×

Specification

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

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

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

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

Accuracy vs Speed?

Herbie found 16 alternatives:

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

Initial Program: 18.6% accurate, 1.0× speedup?

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

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

Alternative 1: 49.8% accurate, 0.4× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := C \cdot \left(4 \cdot A\right)\\ t_1 := t\_0 - {B\_m}^{2}\\ t_2 := {B\_m}^{2} - t\_0\\ t_3 := 2 \cdot t\_2\\ t_4 := \frac{\sqrt{\left(2 \cdot \left(F \cdot t\_2\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_1}\\ \mathbf{if}\;t\_4 \leq -5 \cdot 10^{-200}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(B\_m, A - C\right)\right)} \cdot \sqrt{t\_3}}{t\_1}\\ \mathbf{elif}\;t\_4 \leq \infty:\\ \;\;\;\;\frac{\sqrt{t\_3 \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)}}{t\_1}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* C (* 4.0 A)))
        (t_1 (- t_0 (pow B_m 2.0)))
        (t_2 (- (pow B_m 2.0) t_0))
        (t_3 (* 2.0 t_2))
        (t_4
         (/
          (sqrt
           (*
            (* 2.0 (* F t_2))
            (- (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
          t_1)))
   (if (<= t_4 -5e-200)
     (/ (* (sqrt (* F (- (+ A C) (hypot B_m (- A C))))) (sqrt t_3)) t_1)
     (if (<= t_4 INFINITY)
       (/ (sqrt (* t_3 (* F (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C))))))) t_1)
       (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m)))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = C * (4.0 * A);
	double t_1 = t_0 - pow(B_m, 2.0);
	double t_2 = pow(B_m, 2.0) - t_0;
	double t_3 = 2.0 * t_2;
	double t_4 = sqrt(((2.0 * (F * t_2)) * ((A + C) - sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / t_1;
	double tmp;
	if (t_4 <= -5e-200) {
		tmp = (sqrt((F * ((A + C) - hypot(B_m, (A - C))))) * sqrt(t_3)) / t_1;
	} else if (t_4 <= ((double) INFINITY)) {
		tmp = sqrt((t_3 * (F * (A + (A + (-0.5 * (pow(B_m, 2.0) / C))))))) / t_1;
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = C * (4.0 * A);
	double t_1 = t_0 - Math.pow(B_m, 2.0);
	double t_2 = Math.pow(B_m, 2.0) - t_0;
	double t_3 = 2.0 * t_2;
	double t_4 = Math.sqrt(((2.0 * (F * t_2)) * ((A + C) - Math.sqrt((Math.pow(B_m, 2.0) + Math.pow((A - C), 2.0)))))) / t_1;
	double tmp;
	if (t_4 <= -5e-200) {
		tmp = (Math.sqrt((F * ((A + C) - Math.hypot(B_m, (A - C))))) * Math.sqrt(t_3)) / t_1;
	} else if (t_4 <= Double.POSITIVE_INFINITY) {
		tmp = Math.sqrt((t_3 * (F * (A + (A + (-0.5 * (Math.pow(B_m, 2.0) / C))))))) / t_1;
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = C * (4.0 * A)
	t_1 = t_0 - math.pow(B_m, 2.0)
	t_2 = math.pow(B_m, 2.0) - t_0
	t_3 = 2.0 * t_2
	t_4 = math.sqrt(((2.0 * (F * t_2)) * ((A + C) - math.sqrt((math.pow(B_m, 2.0) + math.pow((A - C), 2.0)))))) / t_1
	tmp = 0
	if t_4 <= -5e-200:
		tmp = (math.sqrt((F * ((A + C) - math.hypot(B_m, (A - C))))) * math.sqrt(t_3)) / t_1
	elif t_4 <= math.inf:
		tmp = math.sqrt((t_3 * (F * (A + (A + (-0.5 * (math.pow(B_m, 2.0) / C))))))) / t_1
	else:
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(C * Float64(4.0 * A))
	t_1 = Float64(t_0 - (B_m ^ 2.0))
	t_2 = Float64((B_m ^ 2.0) - t_0)
	t_3 = Float64(2.0 * t_2)
	t_4 = Float64(sqrt(Float64(Float64(2.0 * Float64(F * t_2)) * Float64(Float64(A + C) - sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / t_1)
	tmp = 0.0
	if (t_4 <= -5e-200)
		tmp = Float64(Float64(sqrt(Float64(F * Float64(Float64(A + C) - hypot(B_m, Float64(A - C))))) * sqrt(t_3)) / t_1);
	elseif (t_4 <= Inf)
		tmp = Float64(sqrt(Float64(t_3 * Float64(F * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C))))))) / t_1);
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = C * (4.0 * A);
	t_1 = t_0 - (B_m ^ 2.0);
	t_2 = (B_m ^ 2.0) - t_0;
	t_3 = 2.0 * t_2;
	t_4 = sqrt(((2.0 * (F * t_2)) * ((A + C) - sqrt(((B_m ^ 2.0) + ((A - C) ^ 2.0)))))) / t_1;
	tmp = 0.0;
	if (t_4 <= -5e-200)
		tmp = (sqrt((F * ((A + C) - hypot(B_m, (A - C))))) * sqrt(t_3)) / t_1;
	elseif (t_4 <= Inf)
		tmp = sqrt((t_3 * (F * (A + (A + (-0.5 * ((B_m ^ 2.0) / C))))))) / t_1;
	else
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	end
	tmp_2 = tmp;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision]}, Block[{t$95$3 = N[(2.0 * t$95$2), $MachinePrecision]}, Block[{t$95$4 = N[(N[Sqrt[N[(N[(2.0 * N[(F * t$95$2), $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] / t$95$1), $MachinePrecision]}, If[LessEqual[t$95$4, -5e-200], N[(N[(N[Sqrt[N[(F * N[(N[(A + C), $MachinePrecision] - N[Sqrt[B$95$m ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Sqrt[t$95$3], $MachinePrecision]), $MachinePrecision] / t$95$1), $MachinePrecision], If[LessEqual[t$95$4, Infinity], N[(N[Sqrt[N[(t$95$3 * N[(F * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$1), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := C \cdot \left(4 \cdot A\right)\\
t_1 := t\_0 - {B\_m}^{2}\\
t_2 := {B\_m}^{2} - t\_0\\
t_3 := 2 \cdot t\_2\\
t_4 := \frac{\sqrt{\left(2 \cdot \left(F \cdot t\_2\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_1}\\
\mathbf{if}\;t\_4 \leq -5 \cdot 10^{-200}:\\
\;\;\;\;\frac{\sqrt{F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(B\_m, A - C\right)\right)} \cdot \sqrt{t\_3}}{t\_1}\\

\mathbf{elif}\;t\_4 \leq \infty:\\
\;\;\;\;\frac{\sqrt{t\_3 \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)}}{t\_1}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (/.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))) < -4.99999999999999991e-200

    1. Initial program 44.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied egg-rr53.8%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot F\right) \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    4. Step-by-step derivation
      1. associate-*l*54.0%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. hypot-undefine44.0%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\sqrt{\left(A - C\right) \cdot \left(A - C\right) + B \cdot B}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      3. unpow244.0%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{\left(A - C\right)}^{2}} + B \cdot B}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      4. unpow244.0%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      5. +-commutative44.0%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      6. unpow244.0%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      7. unpow244.0%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      8. hypot-undefine54.0%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    5. Simplified54.0%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    6. Step-by-step derivation
      1. pow1/254.0%

        \[\leadsto \frac{\color{blue}{{\left(\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}^{0.5}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. *-commutative54.0%

        \[\leadsto \frac{{\color{blue}{\left(\left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right) \cdot \left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right)\right)}}^{0.5}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      3. unpow-prod-down74.9%

        \[\leadsto \frac{\color{blue}{{\left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}^{0.5} \cdot {\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right)}^{0.5}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      4. pow1/274.9%

        \[\leadsto \frac{\color{blue}{\sqrt{F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)}} \cdot {\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right)}^{0.5}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      5. pow1/274.9%

        \[\leadsto \frac{\sqrt{F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)} \cdot \color{blue}{\sqrt{2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    7. Applied egg-rr74.9%

      \[\leadsto \frac{\color{blue}{\sqrt{F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)} \cdot \sqrt{2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    8. Step-by-step derivation
      1. associate-+r-74.0%

        \[\leadsto \frac{\sqrt{F \cdot \color{blue}{\left(\left(C + A\right) - \mathsf{hypot}\left(B, A - C\right)\right)}} \cdot \sqrt{2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    9. Simplified74.0%

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

    if -4.99999999999999991e-200 < (/.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 20.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied egg-rr31.0%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot F\right) \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    4. Step-by-step derivation
      1. associate-*l*32.9%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. hypot-undefine24.8%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\sqrt{\left(A - C\right) \cdot \left(A - C\right) + B \cdot B}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      3. unpow224.8%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{\left(A - C\right)}^{2}} + B \cdot B}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      4. unpow224.8%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      5. +-commutative24.8%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      6. unpow224.8%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      7. unpow224.8%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      8. hypot-undefine32.9%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    5. Simplified32.9%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    6. Taylor expanded in C around inf 24.7%

      \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \color{blue}{\left(\left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right) - -1 \cdot A\right)}\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    7. Step-by-step derivation
      1. mul-1-neg24.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(\left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right) - \color{blue}{\left(-A\right)}\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    8. Simplified24.7%

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

    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 C around 0 2.1%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg2.1%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative2.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow22.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow22.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define19.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified19.1%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification38.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - C \cdot \left(4 \cdot A\right)\right)\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}} \leq -5 \cdot 10^{-200}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(B, A - C\right)\right)} \cdot \sqrt{2 \cdot \left({B}^{2} - C \cdot \left(4 \cdot A\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - C \cdot \left(4 \cdot A\right)\right)\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(4 \cdot A\right)\right)\right) \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 44.4% accurate, 1.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \frac{\sqrt{2}}{-B\_m}\\ \mathbf{if}\;{B\_m}^{2} \leq 10^{-176}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 10^{-80} \lor \neg \left({B\_m}^{2} \leq 10^{-9}\right):\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot t\_0\\ \mathbf{else}:\\ \;\;\;\;\sqrt{-0.5 \cdot \frac{{B\_m}^{2} \cdot F}{C}} \cdot t\_0\\ \end{array} \end{array} \]
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) (- B_m))))
   (if (<= (pow B_m 2.0) 1e-176)
     (/
      (sqrt (* (* A -8.0) (* C (* F (+ A A)))))
      (- (* C (* 4.0 A)) (pow B_m 2.0)))
     (if (or (<= (pow B_m 2.0) 1e-80) (not (<= (pow B_m 2.0) 1e-9)))
       (* (sqrt (* F (- A (hypot B_m A)))) t_0)
       (* (sqrt (* -0.5 (/ (* (pow B_m 2.0) F) C))) 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) / -B_m;
	double tmp;
	if (pow(B_m, 2.0) <= 1e-176) {
		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - pow(B_m, 2.0));
	} else if ((pow(B_m, 2.0) <= 1e-80) || !(pow(B_m, 2.0) <= 1e-9)) {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * t_0;
	} else {
		tmp = sqrt((-0.5 * ((pow(B_m, 2.0) * F) / C))) * t_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 t_0 = Math.sqrt(2.0) / -B_m;
	double tmp;
	if (Math.pow(B_m, 2.0) <= 1e-176) {
		tmp = Math.sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - Math.pow(B_m, 2.0));
	} else if ((Math.pow(B_m, 2.0) <= 1e-80) || !(Math.pow(B_m, 2.0) <= 1e-9)) {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * t_0;
	} else {
		tmp = Math.sqrt((-0.5 * ((Math.pow(B_m, 2.0) * F) / C))) * t_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):
	t_0 = math.sqrt(2.0) / -B_m
	tmp = 0
	if math.pow(B_m, 2.0) <= 1e-176:
		tmp = math.sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - math.pow(B_m, 2.0))
	elif (math.pow(B_m, 2.0) <= 1e-80) or not (math.pow(B_m, 2.0) <= 1e-9):
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * t_0
	else:
		tmp = math.sqrt((-0.5 * ((math.pow(B_m, 2.0) * F) / C))) * 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) / Float64(-B_m))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 1e-176)
		tmp = Float64(sqrt(Float64(Float64(A * -8.0) * Float64(C * Float64(F * Float64(A + A))))) / Float64(Float64(C * Float64(4.0 * A)) - (B_m ^ 2.0)));
	elseif (((B_m ^ 2.0) <= 1e-80) || !((B_m ^ 2.0) <= 1e-9))
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * t_0);
	else
		tmp = Float64(sqrt(Float64(-0.5 * Float64(Float64((B_m ^ 2.0) * F) / C))) * t_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)
	t_0 = sqrt(2.0) / -B_m;
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 1e-176)
		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - (B_m ^ 2.0));
	elseif (((B_m ^ 2.0) <= 1e-80) || ~(((B_m ^ 2.0) <= 1e-9)))
		tmp = sqrt((F * (A - hypot(B_m, A)))) * t_0;
	else
		tmp = sqrt((-0.5 * (((B_m ^ 2.0) * F) / C))) * t_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_] := Block[{t$95$0 = N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-176], N[(N[Sqrt[N[(N[(A * -8.0), $MachinePrecision] * N[(C * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[Or[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-80], N[Not[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-9]], $MachinePrecision]], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision], N[(N[Sqrt[N[(-0.5 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] * F), $MachinePrecision] / C), $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 := \frac{\sqrt{2}}{-B\_m}\\
\mathbf{if}\;{B\_m}^{2} \leq 10^{-176}:\\
\;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\

\mathbf{elif}\;{B\_m}^{2} \leq 10^{-80} \lor \neg \left({B\_m}^{2} \leq 10^{-9}\right):\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot t\_0\\

\mathbf{else}:\\
\;\;\;\;\sqrt{-0.5 \cdot \frac{{B\_m}^{2} \cdot F}{C}} \cdot t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1e-176

    1. Initial program 22.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied egg-rr32.7%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot F\right) \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    4. Step-by-step derivation
      1. associate-*l*34.1%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. hypot-undefine25.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\sqrt{\left(A - C\right) \cdot \left(A - C\right) + B \cdot B}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      3. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{\left(A - C\right)}^{2}} + B \cdot B}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      4. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      5. +-commutative25.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      6. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      7. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      8. hypot-undefine34.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    5. Simplified34.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    6. Taylor expanded in C around inf 15.8%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    7. Step-by-step derivation
      1. associate-*r*15.8%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. mul-1-neg15.8%

        \[\leadsto \frac{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    8. Simplified15.8%

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

    if 1e-176 < (pow.f64 B #s(literal 2 binary64)) < 9.99999999999999961e-81 or 1.00000000000000006e-9 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 17.5%

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

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

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative11.6%

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

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

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define24.5%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified24.5%

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

    if 9.99999999999999961e-81 < (pow.f64 B #s(literal 2 binary64)) < 1.00000000000000006e-9

    1. Initial program 15.5%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 9.2%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg9.2%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
      2. unpow29.2%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)} \]
      3. unpow29.2%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
      4. hypot-define9.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
    5. Simplified9.7%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
    6. Taylor expanded in C around inf 15.4%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{-0.5 \cdot \frac{{B}^{2} \cdot F}{C}}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification21.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 10^{-176}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 10^{-80} \lor \neg \left({B}^{2} \leq 10^{-9}\right):\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{-0.5 \cdot \frac{{B}^{2} \cdot F}{C}} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 45.2% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := C \cdot \left(4 \cdot A\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 10^{-176}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left({B\_m}^{2} - t\_0\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* C (* 4.0 A))))
   (if (<= (pow B_m 2.0) 1e-176)
     (/
      (sqrt (* (* 2.0 (- (pow B_m 2.0) t_0)) (* F (* 2.0 A))))
      (- t_0 (pow B_m 2.0)))
     (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = C * (4.0 * A);
	double tmp;
	if (pow(B_m, 2.0) <= 1e-176) {
		tmp = sqrt(((2.0 * (pow(B_m, 2.0) - t_0)) * (F * (2.0 * A)))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = C * (4.0 * A);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 1e-176) {
		tmp = Math.sqrt(((2.0 * (Math.pow(B_m, 2.0) - t_0)) * (F * (2.0 * A)))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = C * (4.0 * A)
	tmp = 0
	if math.pow(B_m, 2.0) <= 1e-176:
		tmp = math.sqrt(((2.0 * (math.pow(B_m, 2.0) - t_0)) * (F * (2.0 * A)))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(C * Float64(4.0 * A))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 1e-176)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64((B_m ^ 2.0) - t_0)) * Float64(F * Float64(2.0 * A)))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = C * (4.0 * A);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 1e-176)
		tmp = sqrt(((2.0 * ((B_m ^ 2.0) - t_0)) * (F * (2.0 * A)))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	end
	tmp_2 = tmp;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-176], N[(N[Sqrt[N[(N[(2.0 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision]), $MachinePrecision] * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := C \cdot \left(4 \cdot A\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 10^{-176}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left({B\_m}^{2} - t\_0\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)}}{t\_0 - {B\_m}^{2}}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1e-176

    1. Initial program 22.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied egg-rr32.7%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot F\right) \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    4. Step-by-step derivation
      1. associate-*l*34.1%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. hypot-undefine25.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\sqrt{\left(A - C\right) \cdot \left(A - C\right) + B \cdot B}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      3. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{\left(A - C\right)}^{2}} + B \cdot B}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      4. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      5. +-commutative25.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      6. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      7. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      8. hypot-undefine34.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    5. Simplified34.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    6. Taylor expanded in A around -inf 21.0%

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

    if 1e-176 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 17.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 11.4%

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

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative11.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow211.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow211.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define23.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified23.3%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification22.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 10^{-176}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(4 \cdot A\right)\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 4: 45.3% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := C \cdot \left(4 \cdot A\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 10^{-176}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B\_m}^{2} - t\_0\right)\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* C (* 4.0 A))))
   (if (<= (pow B_m 2.0) 1e-176)
     (/
      (sqrt (* (* 2.0 (* F (- (pow B_m 2.0) t_0))) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = C * (4.0 * A);
	double tmp;
	if (pow(B_m, 2.0) <= 1e-176) {
		tmp = sqrt(((2.0 * (F * (pow(B_m, 2.0) - t_0))) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = C * (4.0 * A);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 1e-176) {
		tmp = Math.sqrt(((2.0 * (F * (Math.pow(B_m, 2.0) - t_0))) * (2.0 * A))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = C * (4.0 * A)
	tmp = 0
	if math.pow(B_m, 2.0) <= 1e-176:
		tmp = math.sqrt(((2.0 * (F * (math.pow(B_m, 2.0) - t_0))) * (2.0 * A))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(C * Float64(4.0 * A))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 1e-176)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(F * Float64((B_m ^ 2.0) - t_0))) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = C * (4.0 * A);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 1e-176)
		tmp = sqrt(((2.0 * (F * ((B_m ^ 2.0) - t_0))) * (2.0 * A))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	end
	tmp_2 = tmp;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-176], N[(N[Sqrt[N[(N[(2.0 * N[(F * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := C \cdot \left(4 \cdot A\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 10^{-176}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B\_m}^{2} - t\_0\right)\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1e-176

    1. Initial program 22.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 18.9%

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

    if 1e-176 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 17.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 11.4%

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

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative11.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow211.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow211.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define23.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified23.3%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification21.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 10^{-176}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - C \cdot \left(4 \cdot A\right)\right)\right)\right) \cdot \left(2 \cdot A\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 44.3% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 10^{-176}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 1e-176)
   (/
    (sqrt (* (* A -8.0) (* C (* F (+ A A)))))
    (- (* C (* 4.0 A)) (pow B_m 2.0)))
   (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m)))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 1e-176) {
		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double tmp;
	if (Math.pow(B_m, 2.0) <= 1e-176) {
		tmp = Math.sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	tmp = 0
	if math.pow(B_m, 2.0) <= 1e-176:
		tmp = math.sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 1e-176)
		tmp = Float64(sqrt(Float64(Float64(A * -8.0) * Float64(C * Float64(F * Float64(A + A))))) / Float64(Float64(C * Float64(4.0 * A)) - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 1e-176)
		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - (B_m ^ 2.0));
	else
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	end
	tmp_2 = tmp;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-176], N[(N[Sqrt[N[(N[(A * -8.0), $MachinePrecision] * N[(C * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
\mathbf{if}\;{B\_m}^{2} \leq 10^{-176}:\\
\;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 1e-176

    1. Initial program 22.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied egg-rr32.7%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot F\right) \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    4. Step-by-step derivation
      1. associate-*l*34.1%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. hypot-undefine25.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\sqrt{\left(A - C\right) \cdot \left(A - C\right) + B \cdot B}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      3. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{\left(A - C\right)}^{2}} + B \cdot B}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      4. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      5. +-commutative25.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      6. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      7. unpow225.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      8. hypot-undefine34.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    5. Simplified34.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    6. Taylor expanded in C around inf 15.8%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    7. Step-by-step derivation
      1. associate-*r*15.8%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. mul-1-neg15.8%

        \[\leadsto \frac{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    8. Simplified15.8%

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

    if 1e-176 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 17.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 11.4%

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

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative11.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow211.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow211.4%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define23.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified23.3%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification20.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 10^{-176}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 6: 44.3% accurate, 1.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \frac{\sqrt{2}}{-B\_m}\\ \mathbf{if}\;B\_m \leq 3.1 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B\_m}^{2}\right)}\\ \mathbf{elif}\;B\_m \leq 1.95 \cdot 10^{-38} \lor \neg \left(B\_m \leq 0.000102\right):\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot t\_0\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(-0.5 \cdot \frac{{B\_m}^{2}}{C}\right)} \cdot t\_0\\ \end{array} \end{array} \]
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) (- B_m))))
   (if (<= B_m 3.1e-85)
     (/
      (sqrt (* 2.0 (* -4.0 (* A (* C (* F (+ A A)))))))
      (fma A (* 4.0 C) (- (pow B_m 2.0))))
     (if (or (<= B_m 1.95e-38) (not (<= B_m 0.000102)))
       (* (sqrt (* F (- A (hypot B_m A)))) t_0)
       (* (sqrt (* F (* -0.5 (/ (pow B_m 2.0) C)))) 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) / -B_m;
	double tmp;
	if (B_m <= 3.1e-85) {
		tmp = sqrt((2.0 * (-4.0 * (A * (C * (F * (A + A))))))) / fma(A, (4.0 * C), -pow(B_m, 2.0));
	} else if ((B_m <= 1.95e-38) || !(B_m <= 0.000102)) {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * t_0;
	} else {
		tmp = sqrt((F * (-0.5 * (pow(B_m, 2.0) / C)))) * 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) / Float64(-B_m))
	tmp = 0.0
	if (B_m <= 3.1e-85)
		tmp = Float64(sqrt(Float64(2.0 * Float64(-4.0 * Float64(A * Float64(C * Float64(F * Float64(A + A))))))) / fma(A, Float64(4.0 * C), Float64(-(B_m ^ 2.0))));
	elseif ((B_m <= 1.95e-38) || !(B_m <= 0.000102))
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * t_0);
	else
		tmp = Float64(sqrt(Float64(F * Float64(-0.5 * Float64((B_m ^ 2.0) / C)))) * 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[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]}, If[LessEqual[B$95$m, 3.1e-85], N[(N[Sqrt[N[(2.0 * N[(-4.0 * N[(A * N[(C * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(A * N[(4.0 * C), $MachinePrecision] + (-N[Power[B$95$m, 2.0], $MachinePrecision])), $MachinePrecision]), $MachinePrecision], If[Or[LessEqual[B$95$m, 1.95e-38], N[Not[LessEqual[B$95$m, 0.000102]], $MachinePrecision]], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $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 := \frac{\sqrt{2}}{-B\_m}\\
\mathbf{if}\;B\_m \leq 3.1 \cdot 10^{-85}:\\
\;\;\;\;\frac{\sqrt{2 \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B\_m}^{2}\right)}\\

\mathbf{elif}\;B\_m \leq 1.95 \cdot 10^{-38} \lor \neg \left(B\_m \leq 0.000102\right):\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot t\_0\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot \left(-0.5 \cdot \frac{{B\_m}^{2}}{C}\right)} \cdot t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if B < 3.1000000000000002e-85

    1. Initial program 22.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified30.1%

      \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 10.7%

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

    if 3.1000000000000002e-85 < B < 1.95e-38 or 1.01999999999999999e-4 < B

    1. Initial program 12.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 21.3%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg21.3%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative21.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow221.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow221.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define45.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified45.7%

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

    if 1.95e-38 < B < 1.01999999999999999e-4

    1. Initial program 16.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 17.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg17.0%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
      2. unpow217.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)} \]
      3. unpow217.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
      4. hypot-define17.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
    5. Simplified17.3%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
    6. Taylor expanded in C around inf 14.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C}\right)}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification21.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.1 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)}\\ \mathbf{elif}\;B \leq 1.95 \cdot 10^{-38} \lor \neg \left(B \leq 0.000102\right):\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(-0.5 \cdot \frac{{B}^{2}}{C}\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 7: 44.3% accurate, 1.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \frac{\sqrt{2}}{-B\_m}\\ \mathbf{if}\;B\_m \leq 3.1 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\ \mathbf{elif}\;B\_m \leq 3.6 \cdot 10^{-38} \lor \neg \left(B\_m \leq 9.5 \cdot 10^{-5}\right):\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot t\_0\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(-0.5 \cdot \frac{{B\_m}^{2}}{C}\right)} \cdot t\_0\\ \end{array} \end{array} \]
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) (- B_m))))
   (if (<= B_m 3.1e-85)
     (/
      (sqrt (* (* A -8.0) (* C (* F (+ A A)))))
      (- (* C (* 4.0 A)) (pow B_m 2.0)))
     (if (or (<= B_m 3.6e-38) (not (<= B_m 9.5e-5)))
       (* (sqrt (* F (- A (hypot B_m A)))) t_0)
       (* (sqrt (* F (* -0.5 (/ (pow B_m 2.0) C)))) 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) / -B_m;
	double tmp;
	if (B_m <= 3.1e-85) {
		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - pow(B_m, 2.0));
	} else if ((B_m <= 3.6e-38) || !(B_m <= 9.5e-5)) {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * t_0;
	} else {
		tmp = sqrt((F * (-0.5 * (pow(B_m, 2.0) / C)))) * t_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 t_0 = Math.sqrt(2.0) / -B_m;
	double tmp;
	if (B_m <= 3.1e-85) {
		tmp = Math.sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - Math.pow(B_m, 2.0));
	} else if ((B_m <= 3.6e-38) || !(B_m <= 9.5e-5)) {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * t_0;
	} else {
		tmp = Math.sqrt((F * (-0.5 * (Math.pow(B_m, 2.0) / C)))) * t_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):
	t_0 = math.sqrt(2.0) / -B_m
	tmp = 0
	if B_m <= 3.1e-85:
		tmp = math.sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - math.pow(B_m, 2.0))
	elif (B_m <= 3.6e-38) or not (B_m <= 9.5e-5):
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * t_0
	else:
		tmp = math.sqrt((F * (-0.5 * (math.pow(B_m, 2.0) / C)))) * 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) / Float64(-B_m))
	tmp = 0.0
	if (B_m <= 3.1e-85)
		tmp = Float64(sqrt(Float64(Float64(A * -8.0) * Float64(C * Float64(F * Float64(A + A))))) / Float64(Float64(C * Float64(4.0 * A)) - (B_m ^ 2.0)));
	elseif ((B_m <= 3.6e-38) || !(B_m <= 9.5e-5))
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * t_0);
	else
		tmp = Float64(sqrt(Float64(F * Float64(-0.5 * Float64((B_m ^ 2.0) / C)))) * t_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)
	t_0 = sqrt(2.0) / -B_m;
	tmp = 0.0;
	if (B_m <= 3.1e-85)
		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - (B_m ^ 2.0));
	elseif ((B_m <= 3.6e-38) || ~((B_m <= 9.5e-5)))
		tmp = sqrt((F * (A - hypot(B_m, A)))) * t_0;
	else
		tmp = sqrt((F * (-0.5 * ((B_m ^ 2.0) / C)))) * t_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_] := Block[{t$95$0 = N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]}, If[LessEqual[B$95$m, 3.1e-85], N[(N[Sqrt[N[(N[(A * -8.0), $MachinePrecision] * N[(C * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[Or[LessEqual[B$95$m, 3.6e-38], N[Not[LessEqual[B$95$m, 9.5e-5]], $MachinePrecision]], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $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 := \frac{\sqrt{2}}{-B\_m}\\
\mathbf{if}\;B\_m \leq 3.1 \cdot 10^{-85}:\\
\;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\

\mathbf{elif}\;B\_m \leq 3.6 \cdot 10^{-38} \lor \neg \left(B\_m \leq 9.5 \cdot 10^{-5}\right):\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot t\_0\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot \left(-0.5 \cdot \frac{{B\_m}^{2}}{C}\right)} \cdot t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if B < 3.1000000000000002e-85

    1. Initial program 22.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied egg-rr29.3%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot F\right) \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    4. Step-by-step derivation
      1. associate-*l*30.1%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. hypot-undefine23.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\sqrt{\left(A - C\right) \cdot \left(A - C\right) + B \cdot B}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      3. unpow223.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{\left(A - C\right)}^{2}} + B \cdot B}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      4. unpow223.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      5. +-commutative23.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      6. unpow223.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      7. unpow223.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      8. hypot-undefine30.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    5. Simplified30.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    6. Taylor expanded in C around inf 10.7%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    7. Step-by-step derivation
      1. associate-*r*10.7%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. mul-1-neg10.7%

        \[\leadsto \frac{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    8. Simplified10.7%

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

    if 3.1000000000000002e-85 < B < 3.6000000000000001e-38 or 9.5000000000000005e-5 < B

    1. Initial program 12.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 21.3%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg21.3%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative21.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow221.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow221.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define45.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified45.7%

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

    if 3.6000000000000001e-38 < B < 9.5000000000000005e-5

    1. Initial program 16.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 17.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg17.0%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
      2. unpow217.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)} \]
      3. unpow217.0%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
      4. hypot-define17.3%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
    5. Simplified17.3%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
    6. Taylor expanded in C around inf 14.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C}\right)}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification21.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.1 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{elif}\;B \leq 3.6 \cdot 10^{-38} \lor \neg \left(B \leq 9.5 \cdot 10^{-5}\right):\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(-0.5 \cdot \frac{{B}^{2}}{C}\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 8: 39.8% accurate, 2.8× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;B\_m \leq 1.1 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B\_m, C\right)\right)\right)}}{-B\_m}\\ \end{array} \end{array} \]
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 (<= B_m 1.1e-85)
   (/
    (sqrt (* (* A -8.0) (* C (* F (+ A A)))))
    (- (* C (* 4.0 A)) (pow B_m 2.0)))
   (/ (sqrt (* 2.0 (* F (- C (hypot B_m C))))) (- B_m))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (B_m <= 1.1e-85) {
		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - pow(B_m, 2.0));
	} else {
		tmp = sqrt((2.0 * (F * (C - hypot(B_m, C))))) / -B_m;
	}
	return tmp;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double tmp;
	if (B_m <= 1.1e-85) {
		tmp = Math.sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((2.0 * (F * (C - Math.hypot(B_m, C))))) / -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 B_m <= 1.1e-85:
		tmp = math.sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((2.0 * (F * (C - math.hypot(B_m, C))))) / -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 <= 1.1e-85)
		tmp = Float64(sqrt(Float64(Float64(A * -8.0) * Float64(C * Float64(F * Float64(A + A))))) / Float64(Float64(C * Float64(4.0 * A)) - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(2.0 * Float64(F * Float64(C - hypot(B_m, C))))) / Float64(-B_m));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	tmp = 0.0;
	if (B_m <= 1.1e-85)
		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / ((C * (4.0 * A)) - (B_m ^ 2.0));
	else
		tmp = sqrt((2.0 * (F * (C - hypot(B_m, C))))) / -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[B$95$m, 1.1e-85], N[(N[Sqrt[N[(N[(A * -8.0), $MachinePrecision] * N[(C * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(C * N[(4.0 * A), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(2.0 * N[(F * N[(C - N[Sqrt[B$95$m ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-B$95$m)), $MachinePrecision]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
\mathbf{if}\;B\_m \leq 1.1 \cdot 10^{-85}:\\
\;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B\_m}^{2}}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if B < 1.1e-85

    1. Initial program 22.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied egg-rr29.3%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot F\right) \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    4. Step-by-step derivation
      1. associate-*l*30.1%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. hypot-undefine23.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\sqrt{\left(A - C\right) \cdot \left(A - C\right) + B \cdot B}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      3. unpow223.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{\left(A - C\right)}^{2}} + B \cdot B}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      4. unpow223.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      5. +-commutative23.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      6. unpow223.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      7. unpow223.7%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      8. hypot-undefine30.1%

        \[\leadsto \frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    5. Simplified30.1%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)}} \]
    6. Taylor expanded in C around inf 10.7%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    7. Step-by-step derivation
      1. associate-*r*10.7%

        \[\leadsto \frac{\sqrt{\color{blue}{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
      2. mul-1-neg10.7%

        \[\leadsto \frac{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)} \]
    8. Simplified10.7%

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

    if 1.1e-85 < B

    1. Initial program 12.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 24.2%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg24.2%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
      2. unpow224.2%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)} \]
      3. unpow224.2%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
      4. hypot-define48.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
    5. Simplified48.9%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
    6. Step-by-step derivation
      1. associate-*l/48.9%

        \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
    7. Applied egg-rr48.9%

      \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
    8. Step-by-step derivation
      1. *-un-lft-identity48.9%

        \[\leadsto -\color{blue}{1 \cdot \frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
      2. sqrt-unprod49.0%

        \[\leadsto -1 \cdot \frac{\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}}{B} \]
    9. Applied egg-rr49.0%

      \[\leadsto -\color{blue}{1 \cdot \frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}{B}} \]
    10. Step-by-step derivation
      1. *-lft-identity49.0%

        \[\leadsto -\color{blue}{\frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}{B}} \]
    11. Simplified49.0%

      \[\leadsto -\color{blue}{\frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}{B}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification23.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.1 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{C \cdot \left(4 \cdot A\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 9: 26.7% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{-1}{\frac{B\_m}{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B\_m, C\right)\right)\right)}}} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (/ -1.0 (/ B_m (sqrt (* 2.0 (* F (- C (hypot B_m C))))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return -1.0 / (B_m / sqrt((2.0 * (F * (C - hypot(B_m, C))))));
}
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 -1.0 / (B_m / Math.sqrt((2.0 * (F * (C - Math.hypot(B_m, C))))));
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return -1.0 / (B_m / math.sqrt((2.0 * (F * (C - math.hypot(B_m, C))))))
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(-1.0 / Float64(B_m / sqrt(Float64(2.0 * Float64(F * Float64(C - hypot(B_m, C)))))))
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 = -1.0 / (B_m / sqrt((2.0 * (F * (C - hypot(B_m, C))))));
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[(-1.0 / N[(B$95$m / N[Sqrt[N[(2.0 * N[(F * N[(C - N[Sqrt[B$95$m ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{-1}{\frac{B\_m}{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B\_m, C\right)\right)\right)}}}
\end{array}
Derivation
  1. Initial program 19.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in A around 0 10.0%

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

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

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

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
    4. hypot-define18.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
  5. Simplified18.7%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
  6. Step-by-step derivation
    1. associate-*l/18.7%

      \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
  7. Applied egg-rr18.7%

    \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
  8. Step-by-step derivation
    1. clear-num18.7%

      \[\leadsto -\color{blue}{\frac{1}{\frac{B}{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}}} \]
    2. sqrt-unprod18.8%

      \[\leadsto -\frac{1}{\frac{B}{\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}}} \]
  9. Applied egg-rr18.8%

    \[\leadsto -\color{blue}{\frac{1}{\frac{B}{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}}} \]
  10. Final simplification18.8%

    \[\leadsto \frac{-1}{\frac{B}{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}} \]
  11. Add Preprocessing

Alternative 10: 26.7% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B\_m, C\right)\right)\right)}}{-B\_m} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (/ (sqrt (* 2.0 (* F (- C (hypot B_m C))))) (- 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 * (C - hypot(B_m, C))))) / -B_m;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt((2.0 * (F * (C - Math.hypot(B_m, C))))) / -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 * (C - math.hypot(B_m, C))))) / -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 * Float64(C - hypot(B_m, C))))) / Float64(-B_m))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt((2.0 * (F * (C - hypot(B_m, C))))) / -B_m;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(2.0 * N[(F * N[(C - N[Sqrt[B$95$m ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-B$95$m)), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B\_m, C\right)\right)\right)}}{-B\_m}
\end{array}
Derivation
  1. Initial program 19.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in A around 0 10.0%

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

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

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

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
    4. hypot-define18.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
  5. Simplified18.7%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
  6. Step-by-step derivation
    1. associate-*l/18.7%

      \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
  7. Applied egg-rr18.7%

    \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
  8. Step-by-step derivation
    1. *-un-lft-identity18.7%

      \[\leadsto -\color{blue}{1 \cdot \frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
    2. sqrt-unprod18.8%

      \[\leadsto -1 \cdot \frac{\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}}{B} \]
  9. Applied egg-rr18.8%

    \[\leadsto -\color{blue}{1 \cdot \frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}{B}} \]
  10. Step-by-step derivation
    1. *-lft-identity18.8%

      \[\leadsto -\color{blue}{\frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}{B}} \]
  11. Simplified18.8%

    \[\leadsto -\color{blue}{\frac{\sqrt{2 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}}{B}} \]
  12. Final simplification18.8%

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

Alternative 11: 26.4% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{B\_m \cdot \left(-F\right)}\right) \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (* (/ (sqrt 2.0) B_m) (- (sqrt (* B_m (- 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) {
	return (sqrt(2.0) / B_m) * -sqrt((B_m * -F));
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = (sqrt(2.0d0) / b_m) * -sqrt((b_m * -f))
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return (Math.sqrt(2.0) / B_m) * -Math.sqrt((B_m * -F));
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return (math.sqrt(2.0) / B_m) * -math.sqrt((B_m * -F))
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(B_m * Float64(-F)))))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = (sqrt(2.0) / B_m) * -sqrt((B_m * -F));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(B$95$m * (-F)), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{B\_m \cdot \left(-F\right)}\right)
\end{array}
Derivation
  1. Initial program 19.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in A around 0 10.0%

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

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

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

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
    4. hypot-define18.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
  5. Simplified18.7%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
  6. Taylor expanded in C around 0 15.7%

    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{-1 \cdot \left(B \cdot F\right)}} \]
  7. Step-by-step derivation
    1. associate-*r*15.7%

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

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{\left(-B\right)} \cdot F} \]
  8. Simplified15.7%

    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{\left(-B\right) \cdot F}} \]
  9. Final simplification15.7%

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

Alternative 12: 2.0% accurate, 5.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{{\left(2 \cdot \left(-4 \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C + C\right)\right)\right)\right)\right)}^{0.5}}{4 \cdot \left(A \cdot C\right)} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (/ (pow (* 2.0 (* -4.0 (* A (* (* C F) (+ C C))))) 0.5) (* 4.0 (* A C))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return pow((2.0 * (-4.0 * (A * ((C * F) * (C + C))))), 0.5) / (4.0 * (A * C));
}
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 * ((-4.0d0) * (a * ((c * f) * (c + c))))) ** 0.5d0) / (4.0d0 * (a * c))
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 * (-4.0 * (A * ((C * F) * (C + C))))), 0.5) / (4.0 * (A * C));
}
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 * (-4.0 * (A * ((C * F) * (C + C))))), 0.5) / (4.0 * (A * C))
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(-4.0 * Float64(A * Float64(Float64(C * F) * Float64(C + C))))) ^ 0.5) / Float64(4.0 * Float64(A * C)))
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 * (-4.0 * (A * ((C * F) * (C + C))))) ^ 0.5) / (4.0 * (A * C));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Power[N[(2.0 * N[(-4.0 * N[(A * N[(N[(C * F), $MachinePrecision] * N[(C + C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 0.5], $MachinePrecision] / N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{{\left(2 \cdot \left(-4 \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C + C\right)\right)\right)\right)\right)}^{0.5}}{4 \cdot \left(A \cdot C\right)}
\end{array}
Derivation
  1. Initial program 19.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Simplified25.7%

    \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)}} \]
  3. Add Preprocessing
  4. Taylor expanded in A around inf 11.3%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  5. Step-by-step derivation
    1. associate-*r*11.3%

      \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
    2. mul-1-neg11.3%

      \[\leadsto \frac{\sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \color{blue}{\left(-C\right)}\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  6. Simplified11.3%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  7. Taylor expanded in A around inf 12.5%

    \[\leadsto \frac{\sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}{\color{blue}{4 \cdot \left(A \cdot C\right)}} \]
  8. Step-by-step derivation
    1. pow1/212.7%

      \[\leadsto \frac{\color{blue}{{\left(2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)\right)}^{0.5}}}{4 \cdot \left(A \cdot C\right)} \]
    2. associate-*l*12.7%

      \[\leadsto \frac{{\left(2 \cdot \color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)\right)}\right)}^{0.5}}{4 \cdot \left(A \cdot C\right)} \]
    3. associate-*r*12.7%

      \[\leadsto \frac{{\left(2 \cdot \left(-4 \cdot \left(A \cdot \color{blue}{\left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)}\right)\right)\right)}^{0.5}}{4 \cdot \left(A \cdot C\right)} \]
  9. Applied egg-rr12.7%

    \[\leadsto \frac{\color{blue}{{\left(2 \cdot \left(-4 \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)\right)\right)\right)}^{0.5}}}{4 \cdot \left(A \cdot C\right)} \]
  10. Final simplification12.7%

    \[\leadsto \frac{{\left(2 \cdot \left(-4 \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C + C\right)\right)\right)\right)\right)}^{0.5}}{4 \cdot \left(A \cdot C\right)} \]
  11. Add Preprocessing

Alternative 13: 1.6% accurate, 5.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{2 \cdot \left(\left(C + C\right) \cdot \left(\left(C \cdot F\right) \cdot \left(A \cdot -4\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (/ (sqrt (* 2.0 (* (+ C C) (* (* C F) (* A -4.0))))) (* 4.0 (* A C))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt((2.0 * ((C + C) * ((C * F) * (A * -4.0))))) / (4.0 * (A * C));
}
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 * ((c + c) * ((c * f) * (a * (-4.0d0)))))) / (4.0d0 * (a * c))
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 * ((C + C) * ((C * F) * (A * -4.0))))) / (4.0 * (A * C));
}
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 * ((C + C) * ((C * F) * (A * -4.0))))) / (4.0 * (A * C))
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(Float64(C + C) * Float64(Float64(C * F) * Float64(A * -4.0))))) / Float64(4.0 * Float64(A * C)))
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 * ((C + C) * ((C * F) * (A * -4.0))))) / (4.0 * (A * C));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(2.0 * N[(N[(C + C), $MachinePrecision] * N[(N[(C * F), $MachinePrecision] * N[(A * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{\sqrt{2 \cdot \left(\left(C + C\right) \cdot \left(\left(C \cdot F\right) \cdot \left(A \cdot -4\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)}
\end{array}
Derivation
  1. Initial program 19.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Simplified25.7%

    \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)}} \]
  3. Add Preprocessing
  4. Taylor expanded in A around inf 11.3%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  5. Step-by-step derivation
    1. associate-*r*11.3%

      \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
    2. mul-1-neg11.3%

      \[\leadsto \frac{\sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \color{blue}{\left(-C\right)}\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  6. Simplified11.3%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  7. Taylor expanded in A around inf 12.5%

    \[\leadsto \frac{\sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}{\color{blue}{4 \cdot \left(A \cdot C\right)}} \]
  8. Step-by-step derivation
    1. associate-*l*12.5%

      \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
    2. associate-*r*12.5%

      \[\leadsto \frac{\sqrt{2 \cdot \left(-4 \cdot \left(A \cdot \color{blue}{\left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)}\right)\right)}}{4 \cdot \left(A \cdot C\right)} \]
  9. Applied egg-rr12.5%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(-4 \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
  10. Step-by-step derivation
    1. associate-*r*12.5%

      \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
    2. associate-*r*13.7%

      \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(\left(-4 \cdot A\right) \cdot \left(C \cdot F\right)\right) \cdot \left(C - \left(-C\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
  11. Simplified13.7%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(\left(-4 \cdot A\right) \cdot \left(C \cdot F\right)\right) \cdot \left(C - \left(-C\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
  12. Final simplification13.7%

    \[\leadsto \frac{\sqrt{2 \cdot \left(\left(C + C\right) \cdot \left(\left(C \cdot F\right) \cdot \left(A \cdot -4\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \]
  13. Add Preprocessing

Alternative 14: 1.7% accurate, 5.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{2 \cdot \left(\left(A \cdot -4\right) \cdot \left(C \cdot \left(F \cdot \left(C + C\right)\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (/ (sqrt (* 2.0 (* (* A -4.0) (* C (* F (+ C C)))))) (* 4.0 (* A C))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt((2.0 * ((A * -4.0) * (C * (F * (C + C)))))) / (4.0 * (A * C));
}
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 * ((a * (-4.0d0)) * (c * (f * (c + c)))))) / (4.0d0 * (a * c))
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 * ((A * -4.0) * (C * (F * (C + C)))))) / (4.0 * (A * C));
}
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 * ((A * -4.0) * (C * (F * (C + C)))))) / (4.0 * (A * C))
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(Float64(A * -4.0) * Float64(C * Float64(F * Float64(C + C)))))) / Float64(4.0 * Float64(A * C)))
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 * ((A * -4.0) * (C * (F * (C + C)))))) / (4.0 * (A * C));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(2.0 * N[(N[(A * -4.0), $MachinePrecision] * N[(C * N[(F * N[(C + C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{\sqrt{2 \cdot \left(\left(A \cdot -4\right) \cdot \left(C \cdot \left(F \cdot \left(C + C\right)\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)}
\end{array}
Derivation
  1. Initial program 19.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Simplified25.7%

    \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)}} \]
  3. Add Preprocessing
  4. Taylor expanded in A around inf 11.3%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  5. Step-by-step derivation
    1. associate-*r*11.3%

      \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
    2. mul-1-neg11.3%

      \[\leadsto \frac{\sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \color{blue}{\left(-C\right)}\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  6. Simplified11.3%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  7. Taylor expanded in A around inf 12.5%

    \[\leadsto \frac{\sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}{\color{blue}{4 \cdot \left(A \cdot C\right)}} \]
  8. Final simplification12.5%

    \[\leadsto \frac{\sqrt{2 \cdot \left(\left(A \cdot -4\right) \cdot \left(C \cdot \left(F \cdot \left(C + C\right)\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \]
  9. Add Preprocessing

Alternative 15: 1.7% accurate, 5.4× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(C + C\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (/ (sqrt (* (* A -8.0) (* C (* F (+ C C))))) (* 4.0 (* A C))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt(((A * -8.0) * (C * (F * (C + C))))) / (4.0 * (A * C));
}
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(((a * (-8.0d0)) * (c * (f * (c + c))))) / (4.0d0 * (a * c))
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(((A * -8.0) * (C * (F * (C + C))))) / (4.0 * (A * C));
}
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(((A * -8.0) * (C * (F * (C + C))))) / (4.0 * (A * C))
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(Float64(Float64(A * -8.0) * Float64(C * Float64(F * Float64(C + C))))) / Float64(4.0 * Float64(A * C)))
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(((A * -8.0) * (C * (F * (C + C))))) / (4.0 * (A * C));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(N[(A * -8.0), $MachinePrecision] * N[(C * N[(F * N[(C + C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(C + C\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)}
\end{array}
Derivation
  1. Initial program 19.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Simplified25.7%

    \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right) \cdot \left(F \cdot \left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)}} \]
  3. Add Preprocessing
  4. Taylor expanded in A around inf 11.3%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  5. Step-by-step derivation
    1. associate-*r*11.3%

      \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
    2. mul-1-neg11.3%

      \[\leadsto \frac{\sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \color{blue}{\left(-C\right)}\right)\right)\right)\right)}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  6. Simplified11.3%

    \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}}{\mathsf{fma}\left(A, 4 \cdot C, -{B}^{2}\right)} \]
  7. Taylor expanded in A around inf 12.5%

    \[\leadsto \frac{\sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}{\color{blue}{4 \cdot \left(A \cdot C\right)}} \]
  8. Step-by-step derivation
    1. *-un-lft-identity12.5%

      \[\leadsto \frac{\color{blue}{1 \cdot \sqrt{2 \cdot \left(\left(-4 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
    2. associate-*l*12.5%

      \[\leadsto \frac{1 \cdot \sqrt{2 \cdot \color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
    3. associate-*r*12.5%

      \[\leadsto \frac{1 \cdot \sqrt{2 \cdot \left(-4 \cdot \left(A \cdot \color{blue}{\left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)}\right)\right)}}{4 \cdot \left(A \cdot C\right)} \]
  9. Applied egg-rr12.5%

    \[\leadsto \frac{\color{blue}{1 \cdot \sqrt{2 \cdot \left(-4 \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
  10. Step-by-step derivation
    1. *-lft-identity12.5%

      \[\leadsto \frac{\color{blue}{\sqrt{2 \cdot \left(-4 \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
    2. associate-*r*12.5%

      \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot -4\right) \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
    3. metadata-eval12.5%

      \[\leadsto \frac{\sqrt{\color{blue}{-8} \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C - \left(-C\right)\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \]
    4. neg-mul-112.5%

      \[\leadsto \frac{\sqrt{-8 \cdot \left(A \cdot \left(\left(C \cdot F\right) \cdot \left(C - \color{blue}{-1 \cdot C}\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \]
    5. associate-*r*12.5%

      \[\leadsto \frac{\sqrt{-8 \cdot \left(A \cdot \color{blue}{\left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)}\right)}}{4 \cdot \left(A \cdot C\right)} \]
    6. associate-*r*12.5%

      \[\leadsto \frac{\sqrt{\color{blue}{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - -1 \cdot C\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
    7. neg-mul-112.5%

      \[\leadsto \frac{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \color{blue}{\left(-C\right)}\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \]
  11. Simplified12.5%

    \[\leadsto \frac{\color{blue}{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(C - \left(-C\right)\right)\right)\right)}}}{4 \cdot \left(A \cdot C\right)} \]
  12. Final simplification12.5%

    \[\leadsto \frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(C + C\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)} \]
  13. Add Preprocessing

Alternative 16: 1.0% accurate, 5.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{-2}{B\_m} \cdot \sqrt{C \cdot F} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (* (/ -2.0 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) {
	return (-2.0 / B_m) * sqrt((C * F));
}
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) / b_m) * sqrt((c * f))
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return (-2.0 / B_m) * Math.sqrt((C * F));
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return (-2.0 / B_m) * math.sqrt((C * F))
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 / B_m) * sqrt(Float64(C * F)))
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 / B_m) * sqrt((C * F));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[(-2.0 / B$95$m), $MachinePrecision] * N[Sqrt[N[(C * F), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\frac{-2}{B\_m} \cdot \sqrt{C \cdot F}
\end{array}
Derivation
  1. Initial program 19.2%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in A around 0 10.0%

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

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

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

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)} \]
    4. hypot-define18.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)} \]
  5. Simplified18.7%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}} \]
  6. Step-by-step derivation
    1. associate-*l/18.7%

      \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
  7. Applied egg-rr18.7%

    \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}}{B}} \]
  8. Taylor expanded in C around -inf 0.0%

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

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

      \[\leadsto -\left(-\frac{\color{blue}{\left(\sqrt{-1} \cdot \sqrt{-1}\right)} \cdot {\left(\sqrt{2}\right)}^{2}}{B} \cdot \sqrt{C \cdot F}\right) \]
    3. rem-square-sqrt4.3%

      \[\leadsto -\left(-\frac{\color{blue}{-1} \cdot {\left(\sqrt{2}\right)}^{2}}{B} \cdot \sqrt{C \cdot F}\right) \]
    4. unpow24.3%

      \[\leadsto -\left(-\frac{-1 \cdot \color{blue}{\left(\sqrt{2} \cdot \sqrt{2}\right)}}{B} \cdot \sqrt{C \cdot F}\right) \]
    5. rem-square-sqrt4.3%

      \[\leadsto -\left(-\frac{-1 \cdot \color{blue}{2}}{B} \cdot \sqrt{C \cdot F}\right) \]
    6. metadata-eval4.3%

      \[\leadsto -\left(-\frac{\color{blue}{-2}}{B} \cdot \sqrt{C \cdot F}\right) \]
  10. Simplified4.3%

    \[\leadsto -\color{blue}{\left(-\frac{-2}{B} \cdot \sqrt{C \cdot F}\right)} \]
  11. Final simplification4.3%

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

Reproduce

?
herbie shell --seed 2024096 
(FPCore (A B C F)
  :name "ABCF->ab-angle b"
  :precision binary64
  (/ (- (sqrt (* (* 2.0 (* (- (pow B 2.0) (* (* 4.0 A) C)) F)) (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))) (- (pow B 2.0) (* (* 4.0 A) C))))