ABCF->ab-angle b

Percentage Accurate: 18.6% → 52.6%
Time: 23.8s
Alternatives: 6
Speedup: 2.0×

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 6 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: 52.6% accurate, 0.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ t_1 := F \cdot t\_0\\ t_2 := A + \left(C - \mathsf{hypot}\left(B\_m, A - C\right)\right)\\ t_3 := -t\_0\\ t_4 := \left(4 \cdot A\right) \cdot C\\ t_5 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_4\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_4 - {B\_m}^{2}}\\ \mathbf{if}\;t\_5 \leq -\infty:\\ \;\;\;\;\sqrt{F \cdot \frac{t\_2}{\mathsf{fma}\left(-4, A \cdot C, {B\_m}^{2}\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{elif}\;t\_5 \leq -2 \cdot 10^{-175}:\\ \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot t\_2\right)}}{t\_3}\\ \mathbf{elif}\;t\_5 \leq \infty:\\ \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)}}{t\_3}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
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 (fma B_m B_m (* A (* C -4.0))))
        (t_1 (* F t_0))
        (t_2 (+ A (- C (hypot B_m (- A C)))))
        (t_3 (- t_0))
        (t_4 (* (* 4.0 A) C))
        (t_5
         (/
          (sqrt
           (*
            (* 2.0 (* (- (pow B_m 2.0) t_4) F))
            (- (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
          (- t_4 (pow B_m 2.0)))))
   (if (<= t_5 (- INFINITY))
     (* (sqrt (* F (/ t_2 (fma -4.0 (* A C) (pow B_m 2.0))))) (- (sqrt 2.0)))
     (if (<= t_5 -2e-175)
       (/ (sqrt (* t_1 (* 2.0 t_2))) t_3)
       (if (<= t_5 INFINITY)
         (/
          (sqrt (* t_1 (* 2.0 (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C)))))))
          t_3)
         (* (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 = fma(B_m, B_m, (A * (C * -4.0)));
	double t_1 = F * t_0;
	double t_2 = A + (C - hypot(B_m, (A - C)));
	double t_3 = -t_0;
	double t_4 = (4.0 * A) * C;
	double t_5 = sqrt(((2.0 * ((pow(B_m, 2.0) - t_4) * F)) * ((A + C) - sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / (t_4 - pow(B_m, 2.0));
	double tmp;
	if (t_5 <= -((double) INFINITY)) {
		tmp = sqrt((F * (t_2 / fma(-4.0, (A * C), pow(B_m, 2.0))))) * -sqrt(2.0);
	} else if (t_5 <= -2e-175) {
		tmp = sqrt((t_1 * (2.0 * t_2))) / t_3;
	} else if (t_5 <= ((double) INFINITY)) {
		tmp = sqrt((t_1 * (2.0 * (A + (A + (-0.5 * (pow(B_m, 2.0) / C))))))) / t_3;
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	t_1 = Float64(F * t_0)
	t_2 = Float64(A + Float64(C - hypot(B_m, Float64(A - C))))
	t_3 = Float64(-t_0)
	t_4 = Float64(Float64(4.0 * A) * C)
	t_5 = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_4) * F)) * Float64(Float64(A + C) - sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / Float64(t_4 - (B_m ^ 2.0)))
	tmp = 0.0
	if (t_5 <= Float64(-Inf))
		tmp = Float64(sqrt(Float64(F * Float64(t_2 / fma(-4.0, Float64(A * C), (B_m ^ 2.0))))) * Float64(-sqrt(2.0)));
	elseif (t_5 <= -2e-175)
		tmp = Float64(sqrt(Float64(t_1 * Float64(2.0 * t_2))) / t_3);
	elseif (t_5 <= Inf)
		tmp = Float64(sqrt(Float64(t_1 * Float64(2.0 * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C))))))) / t_3);
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(A + N[(C - N[Sqrt[B$95$m ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = (-t$95$0)}, Block[{t$95$4 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$5 = N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$4), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$4 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$5, (-Infinity)], N[(N[Sqrt[N[(F * N[(t$95$2 / N[(-4.0 * N[(A * C), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision], If[LessEqual[t$95$5, -2e-175], N[(N[Sqrt[N[(t$95$1 * N[(2.0 * t$95$2), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$3), $MachinePrecision], If[LessEqual[t$95$5, Infinity], N[(N[Sqrt[N[(t$95$1 * N[(2.0 * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$3), $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 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
t_1 := F \cdot t\_0\\
t_2 := A + \left(C - \mathsf{hypot}\left(B\_m, A - C\right)\right)\\
t_3 := -t\_0\\
t_4 := \left(4 \cdot A\right) \cdot C\\
t_5 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_4\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_4 - {B\_m}^{2}}\\
\mathbf{if}\;t\_5 \leq -\infty:\\
\;\;\;\;\sqrt{F \cdot \frac{t\_2}{\mathsf{fma}\left(-4, A \cdot C, {B\_m}^{2}\right)}} \cdot \left(-\sqrt{2}\right)\\

\mathbf{elif}\;t\_5 \leq -2 \cdot 10^{-175}:\\
\;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot t\_2\right)}}{t\_3}\\

\mathbf{elif}\;t\_5 \leq \infty:\\
\;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)}}{t\_3}\\

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

    1. Initial program 3.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in F around 0 9.7%

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg9.7%

        \[\leadsto \color{blue}{-\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \sqrt{2}} \]
      2. *-commutative9.7%

        \[\leadsto -\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}}} \]
      3. associate-/l*22.0%

        \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{F \cdot \frac{\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}}} \]
      4. associate--l+22.0%

        \[\leadsto -\sqrt{2} \cdot \sqrt{F \cdot \frac{\color{blue}{A + \left(C - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \]
      5. unpow222.0%

        \[\leadsto -\sqrt{2} \cdot \sqrt{F \cdot \frac{A + \left(C - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \]
      6. unpow222.0%

        \[\leadsto -\sqrt{2} \cdot \sqrt{F \cdot \frac{A + \left(C - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \]
      7. hypot-undefine69.1%

        \[\leadsto -\sqrt{2} \cdot \sqrt{F \cdot \frac{A + \left(C - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \]
      8. cancel-sign-sub-inv69.1%

        \[\leadsto -\sqrt{2} \cdot \sqrt{F \cdot \frac{A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)}{\color{blue}{{B}^{2} + \left(-4\right) \cdot \left(A \cdot C\right)}}} \]
    5. Simplified69.1%

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

    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))) < -2e-175

    1. Initial program 98.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified98.1%

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

    if -2e-175 < (/.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 14.8%

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

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 32.6%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{C} - -1 \cdot A\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. mul-1-neg32.6%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-0.5 \cdot \frac{{B}^{2}}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified32.6%

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

    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 1.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg1.6%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative1.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow21.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow21.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define21.1%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified21.1%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 4 regimes into one program.
  4. Final simplification44.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -\infty:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)}{\mathsf{fma}\left(-4, A \cdot C, {B}^{2}\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -2 \cdot 10^{-175}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 46.8% accurate, 1.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ t_1 := A - \mathsf{hypot}\left(B\_m, A\right)\\ t_2 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-61}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+197}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_2\right) \cdot \left(2 \cdot t\_1\right)}}{-t\_2}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot t\_1} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C))
        (t_1 (- A (hypot B_m A)))
        (t_2 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 2e-61)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (if (<= (pow B_m 2.0) 2e+197)
       (/ (sqrt (* (* F t_2) (* 2.0 t_1))) (- t_2))
       (* (sqrt (* F t_1)) (/ (sqrt 2.0) (- B_m)))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double t_1 = A - hypot(B_m, A);
	double t_2 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 2e-61) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else if (pow(B_m, 2.0) <= 2e+197) {
		tmp = sqrt(((F * t_2) * (2.0 * t_1))) / -t_2;
	} else {
		tmp = sqrt((F * t_1)) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	t_1 = Float64(A - hypot(B_m, A))
	t_2 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-61)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	elseif ((B_m ^ 2.0) <= 2e+197)
		tmp = Float64(sqrt(Float64(Float64(F * t_2) * Float64(2.0 * t_1))) / Float64(-t_2));
	else
		tmp = Float64(sqrt(Float64(F * t_1)) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$1 = N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-61], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+197], N[(N[Sqrt[N[(N[(F * t$95$2), $MachinePrecision] * N[(2.0 * t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$2)), $MachinePrecision], N[(N[Sqrt[N[(F * t$95$1), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
t_1 := A - \mathsf{hypot}\left(B\_m, A\right)\\
t_2 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-61}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{elif}\;{B\_m}^{2} \leq 2 \cdot 10^{+197}:\\
\;\;\;\;\frac{\sqrt{\left(F \cdot t\_2\right) \cdot \left(2 \cdot t\_1\right)}}{-t\_2}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 2.0000000000000001e-61

    1. Initial program 15.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 23.5%

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

    if 2.0000000000000001e-61 < (pow.f64 B #s(literal 2 binary64)) < 1.9999999999999999e197

    1. Initial program 35.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified48.7%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around 0 34.8%

      \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{-1 \cdot \sqrt{{A}^{2} + {B}^{2}}}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    5. Step-by-step derivation
      1. mul-1-neg34.8%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \color{blue}{\left(-\sqrt{{A}^{2} + {B}^{2}}\right)}\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      2. +-commutative34.8%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      3. unpow234.8%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      4. unpow234.8%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
      5. hypot-define41.4%

        \[\leadsto \frac{\sqrt{\left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot F\right) \cdot \left(2 \cdot \left(A + \left(-\color{blue}{\mathsf{hypot}\left(B, A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
    6. Simplified41.4%

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

    if 1.9999999999999999e197 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 8.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 8.2%

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

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

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

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

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define30.7%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified30.7%

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

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-61}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 2 \cdot 10^{+197}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 46.9% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ \mathbf{if}\;{B\_m}^{2} \leq 4 \cdot 10^{-45}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C)))
   (if (<= (pow B_m 2.0) 4e-45)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double tmp;
	if (pow(B_m, 2.0) <= 4e-45) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double tmp;
	if (Math.pow(B_m, 2.0) <= 4e-45) {
		tmp = Math.sqrt(((2.0 * ((Math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = (4.0 * A) * C
	tmp = 0
	if math.pow(B_m, 2.0) <= 4e-45:
		tmp = math.sqrt(((2.0 * ((math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 4e-45)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = (4.0 * A) * C;
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 4e-45)
		tmp = sqrt(((2.0 * (((B_m ^ 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	end
	tmp_2 = tmp;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 4e-45], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \left(4 \cdot A\right) \cdot C\\
\mathbf{if}\;{B\_m}^{2} \leq 4 \cdot 10^{-45}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 3.99999999999999994e-45

    1. Initial program 15.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 24.9%

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

    if 3.99999999999999994e-45 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 19.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 10.9%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg10.9%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative10.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow210.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow210.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define25.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified25.6%

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

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 4 \cdot 10^{-45}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 4: 46.0% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 4 \cdot 10^{-45}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 4e-45)
   (/
    (sqrt (* -8.0 (* (* A C) (* F (+ A A)))))
    (- (fma C (* A -4.0) (pow B_m 2.0))))
   (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m)))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 4e-45) {
		tmp = sqrt((-8.0 * ((A * C) * (F * (A + A))))) / -fma(C, (A * -4.0), pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 4e-45)
		tmp = Float64(sqrt(Float64(-8.0 * Float64(Float64(A * C) * Float64(F * Float64(A + A))))) / Float64(-fma(C, Float64(A * -4.0), (B_m ^ 2.0))));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 4e-45], N[(N[Sqrt[N[(-8.0 * N[(N[(A * C), $MachinePrecision] * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-N[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision])), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
\mathbf{if}\;{B\_m}^{2} \leq 4 \cdot 10^{-45}:\\
\;\;\;\;\frac{\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 3.99999999999999994e-45

    1. Initial program 15.7%

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

      \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 20.9%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    5. Step-by-step derivation
      1. associate-*r*21.0%

        \[\leadsto \frac{\sqrt{-8 \cdot \color{blue}{\left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      2. *-commutative21.0%

        \[\leadsto \frac{\sqrt{-8 \cdot \left(\color{blue}{\left(C \cdot A\right)} \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
      3. mul-1-neg21.0%

        \[\leadsto \frac{\sqrt{-8 \cdot \left(\left(C \cdot A\right) \cdot \left(F \cdot \left(A - \color{blue}{\left(-A\right)}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    6. Simplified21.0%

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

    if 3.99999999999999994e-45 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 19.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 10.9%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg10.9%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
      2. +-commutative10.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
      3. unpow210.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
      4. unpow210.9%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
      5. hypot-define25.6%

        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
    5. Simplified25.6%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification23.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 4 \cdot 10^{-45}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 31.9% accurate, 2.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
}
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}
\end{array}
Derivation
  1. Initial program 17.9%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in C around 0 7.6%

    \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
  4. Step-by-step derivation
    1. mul-1-neg7.6%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    2. +-commutative7.6%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)} \]
    3. unpow27.6%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
    4. unpow27.6%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
    5. hypot-define16.3%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)} \]
  5. Simplified16.3%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}} \]
  6. Final simplification16.3%

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

Alternative 6: 1.6% accurate, 3.1× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{2} \cdot \sqrt{\frac{F}{B\_m}} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (* (sqrt 2.0) (sqrt (/ F B_m))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt(2.0) * sqrt((F / B_m));
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = sqrt(2.0d0) * sqrt((f / b_m))
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt(2.0) * Math.sqrt((F / B_m));
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt(2.0) * math.sqrt((F / B_m))
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(2.0) * sqrt(Float64(F / B_m)))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt(2.0) * sqrt((F / B_m));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[2.0], $MachinePrecision] * N[Sqrt[N[(F / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\sqrt{2} \cdot \sqrt{\frac{F}{B\_m}}
\end{array}
Derivation
  1. Initial program 17.9%

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

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

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

      \[\leadsto -\sqrt{\frac{F}{B}} \cdot \left(\color{blue}{\left(\sqrt{-1} \cdot \sqrt{-1}\right)} \cdot \sqrt{2}\right) \]
    3. rem-square-sqrt2.0%

      \[\leadsto -\sqrt{\frac{F}{B}} \cdot \left(\color{blue}{-1} \cdot \sqrt{2}\right) \]
  5. Simplified2.0%

    \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \left(-1 \cdot \sqrt{2}\right)} \]
  6. Final simplification2.0%

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

Reproduce

?
herbie shell --seed 2024089 
(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))))