ABCF->ab-angle b

Percentage Accurate: 18.6% → 46.1%
Time: 31.3s
Alternatives: 9
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 9 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: 46.1% accurate, 0.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 0.0001:\\ \;\;\;\;\frac{-1}{\frac{t\_0}{\sqrt{\left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right) \cdot \left(2 \cdot t\_0\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}\\ \end{array} \end{array} \]
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 C (* A -4.0) (pow B_m 2.0))))
   (if (<= (pow B_m 2.0) 0.0001)
     (/
      -1.0
      (/
       t_0
       (sqrt (* (* F (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C))))) (* 2.0 t_0)))))
     (/ (pow (* 2.0 (* F (- A (hypot A B_m)))) 0.5) (- 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(C, (A * -4.0), pow(B_m, 2.0));
	double tmp;
	if (pow(B_m, 2.0) <= 0.0001) {
		tmp = -1.0 / (t_0 / sqrt(((F * (A + (A + (-0.5 * (pow(B_m, 2.0) / C))))) * (2.0 * t_0))));
	} else {
		tmp = pow((2.0 * (F * (A - hypot(A, B_m)))), 0.5) / -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(C, Float64(A * -4.0), (B_m ^ 2.0))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 0.0001)
		tmp = Float64(-1.0 / Float64(t_0 / sqrt(Float64(Float64(F * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C))))) * Float64(2.0 * t_0)))));
	else
		tmp = Float64((Float64(2.0 * Float64(F * Float64(A - hypot(A, B_m)))) ^ 0.5) / 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[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 0.0001], N[(-1.0 / N[(t$95$0 / N[Sqrt[N[(N[(F * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * t$95$0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Power[N[(2.0 * N[(F * N[(A - N[Sqrt[A ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 0.5], $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}
t_0 := \mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 0.0001:\\
\;\;\;\;\frac{-1}{\frac{t\_0}{\sqrt{\left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right) \cdot \left(2 \cdot t\_0\right)}}}\\

\mathbf{else}:\\
\;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B 2) < 1.00000000000000005e-4

    1. Initial program 23.2%

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

      \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
    3. Add Preprocessing
    4. Step-by-step derivation
      1. clear-num28.9%

        \[\leadsto \color{blue}{\frac{1}{\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}}} \]
      2. inv-pow28.9%

        \[\leadsto \color{blue}{{\left(\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}\right)}^{-1}} \]
    5. Applied egg-rr28.0%

      \[\leadsto \color{blue}{{\left(\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}}\right)}^{-1}} \]
    6. Step-by-step derivation
      1. unpow-128.0%

        \[\leadsto \color{blue}{\frac{1}{\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}}}} \]
    7. Simplified28.0%

      \[\leadsto \color{blue}{\frac{1}{\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}}}} \]
    8. Taylor expanded in C around inf 30.3%

      \[\leadsto \frac{1}{\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(F \cdot \color{blue}{\left(\left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right) - -1 \cdot A\right)}\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}}} \]

    if 1.00000000000000005e-4 < (pow.f64 B 2)

    1. Initial program 14.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 14.2%

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

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    5. Simplified14.2%

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

        \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]
      2. pow1/214.2%

        \[\leadsto -\frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B} \]
      3. pow1/214.2%

        \[\leadsto -\frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)}^{0.5}}}{B} \]
      4. pow-prod-down14.2%

        \[\leadsto -\frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)\right)}^{0.5}}}{B} \]
      5. unpow214.2%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)\right)}^{0.5}}{B} \]
      6. unpow214.2%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)\right)}^{0.5}}{B} \]
      7. hypot-define28.8%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)\right)}^{0.5}}{B} \]
    7. Applied egg-rr28.8%

      \[\leadsto -\color{blue}{\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{B}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification29.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 0.0001:\\ \;\;\;\;\frac{-1}{\frac{\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 46.3% accurate, 1.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 0.0001:\\ \;\;\;\;\frac{-1}{\frac{t\_0}{\sqrt{\left(2 \cdot t\_0\right) \cdot \left(F \cdot \left(A + A\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}\\ \end{array} \end{array} \]
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 C (* A -4.0) (pow B_m 2.0))))
   (if (<= (pow B_m 2.0) 0.0001)
     (/ -1.0 (/ t_0 (sqrt (* (* 2.0 t_0) (* F (+ A A))))))
     (/ (pow (* 2.0 (* F (- A (hypot A B_m)))) 0.5) (- 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(C, (A * -4.0), pow(B_m, 2.0));
	double tmp;
	if (pow(B_m, 2.0) <= 0.0001) {
		tmp = -1.0 / (t_0 / sqrt(((2.0 * t_0) * (F * (A + A)))));
	} else {
		tmp = pow((2.0 * (F * (A - hypot(A, B_m)))), 0.5) / -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(C, Float64(A * -4.0), (B_m ^ 2.0))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 0.0001)
		tmp = Float64(-1.0 / Float64(t_0 / sqrt(Float64(Float64(2.0 * t_0) * Float64(F * Float64(A + A))))));
	else
		tmp = Float64((Float64(2.0 * Float64(F * Float64(A - hypot(A, B_m)))) ^ 0.5) / 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[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 0.0001], N[(-1.0 / N[(t$95$0 / N[Sqrt[N[(N[(2.0 * t$95$0), $MachinePrecision] * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Power[N[(2.0 * N[(F * N[(A - N[Sqrt[A ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 0.5], $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}
t_0 := \mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 0.0001:\\
\;\;\;\;\frac{-1}{\frac{t\_0}{\sqrt{\left(2 \cdot t\_0\right) \cdot \left(F \cdot \left(A + A\right)\right)}}}\\

\mathbf{else}:\\
\;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B 2) < 1.00000000000000005e-4

    1. Initial program 23.2%

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

      \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
    3. Add Preprocessing
    4. Step-by-step derivation
      1. clear-num28.9%

        \[\leadsto \color{blue}{\frac{1}{\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}}} \]
      2. inv-pow28.9%

        \[\leadsto \color{blue}{{\left(\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}\right)}^{-1}} \]
    5. Applied egg-rr28.0%

      \[\leadsto \color{blue}{{\left(\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}}\right)}^{-1}} \]
    6. Step-by-step derivation
      1. unpow-128.0%

        \[\leadsto \color{blue}{\frac{1}{\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}}}} \]
    7. Simplified28.0%

      \[\leadsto \color{blue}{\frac{1}{\frac{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(F \cdot \left(\left(A + C\right) - \mathsf{hypot}\left(A - C, B\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)}}}} \]
    8. Taylor expanded in C around inf 30.5%

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

    if 1.00000000000000005e-4 < (pow.f64 B 2)

    1. Initial program 14.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 14.2%

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

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    5. Simplified14.2%

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

        \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]
      2. pow1/214.2%

        \[\leadsto -\frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B} \]
      3. pow1/214.2%

        \[\leadsto -\frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)}^{0.5}}}{B} \]
      4. pow-prod-down14.2%

        \[\leadsto -\frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)\right)}^{0.5}}}{B} \]
      5. unpow214.2%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)\right)}^{0.5}}{B} \]
      6. unpow214.2%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)\right)}^{0.5}}{B} \]
      7. hypot-define28.8%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)\right)}^{0.5}}{B} \]
    7. Applied egg-rr28.8%

      \[\leadsto -\color{blue}{\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{B}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification29.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 0.0001:\\ \;\;\;\;\frac{-1}{\frac{\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}{\sqrt{\left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right) \cdot \left(F \cdot \left(A + A\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 46.2% accurate, 1.5× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := C \cdot \left(A \cdot 4\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 0.0001:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B\_m}^{2} - t\_0\right)\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}\\ \end{array} \end{array} \]
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 (* A 4.0))))
   (if (<= (pow B_m 2.0) 0.0001)
     (/
      (sqrt (* (* 2.0 (* F (- (pow B_m 2.0) t_0))) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (/ (pow (* 2.0 (* F (- A (hypot A B_m)))) 0.5) (- 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 * (A * 4.0);
	double tmp;
	if (pow(B_m, 2.0) <= 0.0001) {
		tmp = sqrt(((2.0 * (F * (pow(B_m, 2.0) - t_0))) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = pow((2.0 * (F * (A - hypot(A, B_m)))), 0.5) / -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 * (A * 4.0);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 0.0001) {
		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.pow((2.0 * (F * (A - Math.hypot(A, B_m)))), 0.5) / -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 * (A * 4.0)
	tmp = 0
	if math.pow(B_m, 2.0) <= 0.0001:
		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.pow((2.0 * (F * (A - math.hypot(A, B_m)))), 0.5) / -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(A * 4.0))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 0.0001)
		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((Float64(2.0 * Float64(F * Float64(A - hypot(A, B_m)))) ^ 0.5) / 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 * (A * 4.0);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 0.0001)
		tmp = sqrt(((2.0 * (F * ((B_m ^ 2.0) - t_0))) * (2.0 * A))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = ((2.0 * (F * (A - hypot(A, B_m)))) ^ 0.5) / -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[(A * 4.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 0.0001], 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[Power[N[(2.0 * N[(F * N[(A - N[Sqrt[A ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 0.5], $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}
t_0 := C \cdot \left(A \cdot 4\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 0.0001:\\
\;\;\;\;\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}:\\
\;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B 2) < 1.00000000000000005e-4

    1. Initial program 23.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 28.8%

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

    if 1.00000000000000005e-4 < (pow.f64 B 2)

    1. Initial program 14.1%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 14.2%

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

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    5. Simplified14.2%

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

        \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]
      2. pow1/214.2%

        \[\leadsto -\frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B} \]
      3. pow1/214.2%

        \[\leadsto -\frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)}^{0.5}}}{B} \]
      4. pow-prod-down14.2%

        \[\leadsto -\frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)\right)}^{0.5}}}{B} \]
      5. unpow214.2%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)\right)}^{0.5}}{B} \]
      6. unpow214.2%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)\right)}^{0.5}}{B} \]
      7. hypot-define28.8%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)\right)}^{0.5}}{B} \]
    7. Applied egg-rr28.8%

      \[\leadsto -\color{blue}{\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{B}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification28.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 0.0001:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right)\right) \cdot \left(2 \cdot A\right)}}{C \cdot \left(A \cdot 4\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 4: 44.3% accurate, 2.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;B\_m \leq 0.0092:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}\\ \end{array} \end{array} \]
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 0.0092)
   (/
    (sqrt (* -8.0 (* A (* C (* F (+ A A))))))
    (- (fma C (* A -4.0) (pow B_m 2.0))))
   (/ (pow (* 2.0 (* F (- A (hypot A B_m)))) 0.5) (- 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 <= 0.0092) {
		tmp = sqrt((-8.0 * (A * (C * (F * (A + A)))))) / -fma(C, (A * -4.0), pow(B_m, 2.0));
	} else {
		tmp = pow((2.0 * (F * (A - hypot(A, B_m)))), 0.5) / -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 <= 0.0092)
		tmp = Float64(sqrt(Float64(-8.0 * Float64(A * Float64(C * Float64(F * Float64(A + A)))))) / Float64(-fma(C, Float64(A * -4.0), (B_m ^ 2.0))));
	else
		tmp = Float64((Float64(2.0 * Float64(F * Float64(A - hypot(A, B_m)))) ^ 0.5) / Float64(-B_m));
	end
	return tmp
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[B$95$m, 0.0092], N[(N[Sqrt[N[(-8.0 * N[(A * N[(C * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-N[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision])), $MachinePrecision], N[(N[Power[N[(2.0 * N[(F * N[(A - N[Sqrt[A ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 0.5], $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 0.0092:\\
\;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if B < 0.0091999999999999998

    1. Initial program 18.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Simplified21.5%

      \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
    3. Add Preprocessing
    4. Taylor expanded in C around inf 20.2%

      \[\leadsto \frac{\sqrt{\color{blue}{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right)\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    5. Step-by-step derivation
      1. mul-1-neg20.2%

        \[\leadsto \frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A - \color{blue}{\left(-A\right)}\right)\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
    6. Simplified20.2%

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

    if 0.0091999999999999998 < B

    1. Initial program 21.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 26.3%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg26.3%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    5. Simplified26.3%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    6. Step-by-step derivation
      1. associate-*l/26.3%

        \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]
      2. pow1/226.3%

        \[\leadsto -\frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B} \]
      3. pow1/226.3%

        \[\leadsto -\frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)}^{0.5}}}{B} \]
      4. pow-prod-down26.4%

        \[\leadsto -\frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)\right)}^{0.5}}}{B} \]
      5. unpow226.4%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)\right)}^{0.5}}{B} \]
      6. unpow226.4%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)\right)}^{0.5}}{B} \]
      7. hypot-define53.5%

        \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)\right)}^{0.5}}{B} \]
    7. Applied egg-rr53.5%

      \[\leadsto -\color{blue}{\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{B}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification28.3%

    \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 0.0092:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{-B}\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 27.0% accurate, 2.9× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \frac{\sqrt{2}}{B\_m}\\ \mathbf{if}\;A \leq -7.2 \cdot 10^{+226}:\\ \;\;\;\;t\_0 \cdot \left(-\sqrt{F \cdot \left(2 \cdot A\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;t\_0 \cdot \left(-\sqrt{F \cdot \left(A - B\_m\right)}\right)\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (/ (sqrt 2.0) B_m)))
   (if (<= A -7.2e+226)
     (* t_0 (- (sqrt (* F (* 2.0 A)))))
     (* t_0 (- (sqrt (* F (- A 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 = sqrt(2.0) / B_m;
	double tmp;
	if (A <= -7.2e+226) {
		tmp = t_0 * -sqrt((F * (2.0 * A)));
	} else {
		tmp = t_0 * -sqrt((F * (A - B_m)));
	}
	return tmp;
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: tmp
    t_0 = sqrt(2.0d0) / b_m
    if (a <= (-7.2d+226)) then
        tmp = t_0 * -sqrt((f * (2.0d0 * a)))
    else
        tmp = t_0 * -sqrt((f * (a - b_m)))
    end if
    code = tmp
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = Math.sqrt(2.0) / B_m;
	double tmp;
	if (A <= -7.2e+226) {
		tmp = t_0 * -Math.sqrt((F * (2.0 * A)));
	} else {
		tmp = t_0 * -Math.sqrt((F * (A - 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 = math.sqrt(2.0) / B_m
	tmp = 0
	if A <= -7.2e+226:
		tmp = t_0 * -math.sqrt((F * (2.0 * A)))
	else:
		tmp = t_0 * -math.sqrt((F * (A - 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(sqrt(2.0) / B_m)
	tmp = 0.0
	if (A <= -7.2e+226)
		tmp = Float64(t_0 * Float64(-sqrt(Float64(F * Float64(2.0 * A)))));
	else
		tmp = Float64(t_0 * Float64(-sqrt(Float64(F * Float64(A - 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 = sqrt(2.0) / B_m;
	tmp = 0.0;
	if (A <= -7.2e+226)
		tmp = t_0 * -sqrt((F * (2.0 * A)));
	else
		tmp = t_0 * -sqrt((F * (A - 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[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision]}, If[LessEqual[A, -7.2e+226], N[(t$95$0 * (-N[Sqrt[N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision], N[(t$95$0 * (-N[Sqrt[N[(F * N[(A - B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \frac{\sqrt{2}}{B\_m}\\
\mathbf{if}\;A \leq -7.2 \cdot 10^{+226}:\\
\;\;\;\;t\_0 \cdot \left(-\sqrt{F \cdot \left(2 \cdot A\right)}\right)\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if A < -7.19999999999999962e226

    1. Initial program 1.4%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 0.9%

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

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    5. Simplified0.9%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    6. Taylor expanded in A around -inf 10.4%

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

    if -7.19999999999999962e226 < A

    1. Initial program 20.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around 0 11.0%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
    4. Step-by-step derivation
      1. mul-1-neg11.0%

        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    5. Simplified11.0%

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
    6. Taylor expanded in A around 0 15.1%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{B}\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification14.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -7.2 \cdot 10^{+226}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(2 \cdot A\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 6: 30.9% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m} \end{array} \]
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
 (/ (pow (* 2.0 (* F (- A (hypot A B_m)))) 0.5) (- 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 pow((2.0 * (F * (A - hypot(A, B_m)))), 0.5) / -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.pow((2.0 * (F * (A - Math.hypot(A, B_m)))), 0.5) / -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.pow((2.0 * (F * (A - math.hypot(A, B_m)))), 0.5) / -B_m
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64((Float64(2.0 * Float64(F * Float64(A - hypot(A, B_m)))) ^ 0.5) / 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 = ((2.0 * (F * (A - hypot(A, B_m)))) ^ 0.5) / -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[Power[N[(2.0 * N[(F * N[(A - N[Sqrt[A ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 0.5], $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{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\_m\right)\right)\right)\right)}^{0.5}}{-B\_m}
\end{array}
Derivation
  1. Initial program 18.9%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in C around 0 10.1%

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

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
  5. Simplified10.1%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
  6. Step-by-step derivation
    1. associate-*l/10.1%

      \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]
    2. pow1/210.1%

      \[\leadsto -\frac{\color{blue}{{2}^{0.5}} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B} \]
    3. pow1/210.2%

      \[\leadsto -\frac{{2}^{0.5} \cdot \color{blue}{{\left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)}^{0.5}}}{B} \]
    4. pow-prod-down10.2%

      \[\leadsto -\frac{\color{blue}{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)\right)\right)}^{0.5}}}{B} \]
    5. unpow210.2%

      \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)\right)}^{0.5}}{B} \]
    6. unpow210.2%

      \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)\right)}^{0.5}}{B} \]
    7. hypot-define17.2%

      \[\leadsto -\frac{{\left(2 \cdot \left(F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)\right)}^{0.5}}{B} \]
  7. Applied egg-rr17.2%

    \[\leadsto -\color{blue}{\frac{{\left(2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}^{0.5}}{B}} \]
  8. Final simplification17.2%

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

Alternative 7: 8.8% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(2 \cdot A\right)}\right) \end{array} \]
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 (* F (* 2.0 A))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return (sqrt(2.0) / B_m) * -sqrt((F * (2.0 * A)));
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = (sqrt(2.0d0) / b_m) * -sqrt((f * (2.0d0 * a)))
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return (Math.sqrt(2.0) / B_m) * -Math.sqrt((F * (2.0 * A)));
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return (math.sqrt(2.0) / B_m) * -math.sqrt((F * (2.0 * A)))
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(F * Float64(2.0 * A)))))
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = (sqrt(2.0) / B_m) * -sqrt((F * (2.0 * A)));
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(2.0 * A), $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{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(2 \cdot A\right)}\right)
\end{array}
Derivation
  1. Initial program 18.9%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Add Preprocessing
  3. Taylor expanded in C around 0 10.1%

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

      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
  5. Simplified10.1%

    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]
  6. Taylor expanded in A around -inf 3.5%

    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(2 \cdot A\right)}} \]
  7. Final simplification3.5%

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

Alternative 8: 4.6% accurate, 3.0× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{A \cdot F} \cdot \frac{\sqrt{8}}{-B\_m} \end{array} \]
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 (* A F)) (/ (sqrt 8.0) (- 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((A * F)) * (sqrt(8.0) / -B_m);
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = sqrt((a * f)) * (sqrt(8.0d0) / -b_m)
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt((A * F)) * (Math.sqrt(8.0) / -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((A * F)) * (math.sqrt(8.0) / -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(A * F)) * Float64(sqrt(8.0) / 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((A * F)) * (sqrt(8.0) / -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[(A * F), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[8.0], $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{A \cdot F} \cdot \frac{\sqrt{8}}{-B\_m}
\end{array}
Derivation
  1. Initial program 18.9%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Simplified21.3%

    \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
  3. Add Preprocessing
  4. Taylor expanded in C around inf 14.1%

    \[\leadsto \frac{\sqrt{\color{blue}{C \cdot \left(-8 \cdot \left(A \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right) + 2 \cdot \frac{F \cdot \left(2 \cdot \left(A \cdot {B}^{2}\right) + {B}^{2} \cdot \left(A - -1 \cdot A\right)\right)}{C}\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
  5. Taylor expanded in A around 0 2.3%

    \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{A \cdot F} \cdot \frac{\sqrt{8}}{B}\right)} \]
  6. Final simplification2.3%

    \[\leadsto \sqrt{A \cdot F} \cdot \frac{\sqrt{8}}{-B} \]
  7. Add Preprocessing

Alternative 9: 0.0% accurate, 3.1× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ 0.25 \cdot \left(\sqrt{\frac{F}{C}} \cdot \sqrt{-16}\right) \end{array} \]
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
 (* 0.25 (* (sqrt (/ F C)) (sqrt -16.0))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return 0.25 * (sqrt((F / C)) * sqrt(-16.0));
}
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 = 0.25d0 * (sqrt((f / c)) * sqrt((-16.0d0)))
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 0.25 * (Math.sqrt((F / C)) * Math.sqrt(-16.0));
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return 0.25 * (math.sqrt((F / C)) * math.sqrt(-16.0))
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(0.25 * Float64(sqrt(Float64(F / C)) * sqrt(-16.0)))
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 = 0.25 * (sqrt((F / C)) * sqrt(-16.0));
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[(0.25 * N[(N[Sqrt[N[(F / C), $MachinePrecision]], $MachinePrecision] * N[Sqrt[-16.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
0.25 \cdot \left(\sqrt{\frac{F}{C}} \cdot \sqrt{-16}\right)
\end{array}
Derivation
  1. Initial program 18.9%

    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
  2. Simplified21.3%

    \[\leadsto \color{blue}{\frac{\sqrt{F \cdot \left(\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}} \]
  3. Add Preprocessing
  4. Taylor expanded in C around inf 14.1%

    \[\leadsto \frac{\sqrt{\color{blue}{C \cdot \left(-8 \cdot \left(A \cdot \left(F \cdot \left(A - -1 \cdot A\right)\right)\right) + 2 \cdot \frac{F \cdot \left(2 \cdot \left(A \cdot {B}^{2}\right) + {B}^{2} \cdot \left(A - -1 \cdot A\right)\right)}{C}\right)}}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)} \]
  5. Taylor expanded in A around inf 0.0%

    \[\leadsto \color{blue}{0.25 \cdot \left(\sqrt{\frac{F}{C}} \cdot \sqrt{-16}\right)} \]
  6. Final simplification0.0%

    \[\leadsto 0.25 \cdot \left(\sqrt{\frac{F}{C}} \cdot \sqrt{-16}\right) \]
  7. Add Preprocessing

Reproduce

?
herbie shell --seed 2024055 
(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))))