ABCF->ab-angle a

Percentage Accurate: 19.0% → 59.4%
Time: 17.9s
Alternatives: 15
Speedup: 16.9×

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 15 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: 19.0% accurate, 1.0× speedup?

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

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

Alternative 1: 59.4% accurate, 0.2× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(A, C \cdot -4, B\_m \cdot B\_m\right)\\ t_1 := 2 \cdot t\_0\\ t_2 := t\_0 \cdot \left(2 \cdot F\right)\\ t_3 := \mathsf{fma}\left(B\_m, \frac{B\_m}{A} \cdot -0.5, 2 \cdot C\right)\\ t_4 := \left(4 \cdot A\right) \cdot C\\ t_5 := t\_4 - {B\_m}^{2}\\ t_6 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_4\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_5}\\ \mathbf{if}\;t\_6 \leq -\infty:\\ \;\;\;\;\frac{\sqrt{t\_1} \cdot \sqrt{F \cdot t\_3}}{t\_5}\\ \mathbf{elif}\;t\_6 \leq -1 \cdot 10^{-186}:\\ \;\;\;\;\frac{\sqrt{t\_2 \cdot \left(C + \sqrt{\mathsf{fma}\left(B\_m, B\_m, C \cdot C\right)}\right)}}{-t\_0}\\ \mathbf{elif}\;t\_6 \leq 0:\\ \;\;\;\;\frac{\sqrt{F} \cdot \sqrt{t\_1 \cdot t\_3}}{t\_5}\\ \mathbf{elif}\;t\_6 \leq \infty:\\ \;\;\;\;\left(\sqrt{t\_2} \cdot \frac{-1}{t\_0}\right) \cdot \sqrt{2 \cdot C}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{F}}{\sqrt{B\_m \cdot 0.5}}\\ \end{array} \end{array} \]
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 A (* C -4.0) (* B_m B_m)))
        (t_1 (* 2.0 t_0))
        (t_2 (* t_0 (* 2.0 F)))
        (t_3 (fma B_m (* (/ B_m A) -0.5) (* 2.0 C)))
        (t_4 (* (* 4.0 A) C))
        (t_5 (- t_4 (pow B_m 2.0)))
        (t_6
         (/
          (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_5)))
   (if (<= t_6 (- INFINITY))
     (/ (* (sqrt t_1) (sqrt (* F t_3))) t_5)
     (if (<= t_6 -1e-186)
       (/ (sqrt (* t_2 (+ C (sqrt (fma B_m B_m (* C C)))))) (- t_0))
       (if (<= t_6 0.0)
         (/ (* (sqrt F) (sqrt (* t_1 t_3))) t_5)
         (if (<= t_6 INFINITY)
           (* (* (sqrt t_2) (/ -1.0 t_0)) (sqrt (* 2.0 C)))
           (- (/ (sqrt F) (sqrt (* B_m 0.5))))))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = fma(A, (C * -4.0), (B_m * B_m));
	double t_1 = 2.0 * t_0;
	double t_2 = t_0 * (2.0 * F);
	double t_3 = fma(B_m, ((B_m / A) * -0.5), (2.0 * C));
	double t_4 = (4.0 * A) * C;
	double t_5 = t_4 - pow(B_m, 2.0);
	double t_6 = 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_5;
	double tmp;
	if (t_6 <= -((double) INFINITY)) {
		tmp = (sqrt(t_1) * sqrt((F * t_3))) / t_5;
	} else if (t_6 <= -1e-186) {
		tmp = sqrt((t_2 * (C + sqrt(fma(B_m, B_m, (C * C)))))) / -t_0;
	} else if (t_6 <= 0.0) {
		tmp = (sqrt(F) * sqrt((t_1 * t_3))) / t_5;
	} else if (t_6 <= ((double) INFINITY)) {
		tmp = (sqrt(t_2) * (-1.0 / t_0)) * sqrt((2.0 * C));
	} else {
		tmp = -(sqrt(F) / sqrt((B_m * 0.5)));
	}
	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(A, Float64(C * -4.0), Float64(B_m * B_m))
	t_1 = Float64(2.0 * t_0)
	t_2 = Float64(t_0 * Float64(2.0 * F))
	t_3 = fma(B_m, Float64(Float64(B_m / A) * -0.5), Float64(2.0 * C))
	t_4 = Float64(Float64(4.0 * A) * C)
	t_5 = Float64(t_4 - (B_m ^ 2.0))
	t_6 = 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)))))) / t_5)
	tmp = 0.0
	if (t_6 <= Float64(-Inf))
		tmp = Float64(Float64(sqrt(t_1) * sqrt(Float64(F * t_3))) / t_5);
	elseif (t_6 <= -1e-186)
		tmp = Float64(sqrt(Float64(t_2 * Float64(C + sqrt(fma(B_m, B_m, Float64(C * C)))))) / Float64(-t_0));
	elseif (t_6 <= 0.0)
		tmp = Float64(Float64(sqrt(F) * sqrt(Float64(t_1 * t_3))) / t_5);
	elseif (t_6 <= Inf)
		tmp = Float64(Float64(sqrt(t_2) * Float64(-1.0 / t_0)) * sqrt(Float64(2.0 * C)));
	else
		tmp = Float64(-Float64(sqrt(F) / sqrt(Float64(B_m * 0.5))));
	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[(A * N[(C * -4.0), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(2.0 * t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(t$95$0 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(B$95$m * N[(N[(B$95$m / A), $MachinePrecision] * -0.5), $MachinePrecision] + N[(2.0 * C), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$4 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$5 = N[(t$95$4 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$6 = 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] / t$95$5), $MachinePrecision]}, If[LessEqual[t$95$6, (-Infinity)], N[(N[(N[Sqrt[t$95$1], $MachinePrecision] * N[Sqrt[N[(F * t$95$3), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / t$95$5), $MachinePrecision], If[LessEqual[t$95$6, -1e-186], N[(N[Sqrt[N[(t$95$2 * N[(C + N[Sqrt[N[(B$95$m * B$95$m + N[(C * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], If[LessEqual[t$95$6, 0.0], N[(N[(N[Sqrt[F], $MachinePrecision] * N[Sqrt[N[(t$95$1 * t$95$3), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / t$95$5), $MachinePrecision], If[LessEqual[t$95$6, Infinity], N[(N[(N[Sqrt[t$95$2], $MachinePrecision] * N[(-1.0 / t$95$0), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(2.0 * C), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], (-N[(N[Sqrt[F], $MachinePrecision] / N[Sqrt[N[(B$95$m * 0.5), $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 := \mathsf{fma}\left(A, C \cdot -4, B\_m \cdot B\_m\right)\\
t_1 := 2 \cdot t\_0\\
t_2 := t\_0 \cdot \left(2 \cdot F\right)\\
t_3 := \mathsf{fma}\left(B\_m, \frac{B\_m}{A} \cdot -0.5, 2 \cdot C\right)\\
t_4 := \left(4 \cdot A\right) \cdot C\\
t_5 := t\_4 - {B\_m}^{2}\\
t_6 := \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\_5}\\
\mathbf{if}\;t\_6 \leq -\infty:\\
\;\;\;\;\frac{\sqrt{t\_1} \cdot \sqrt{F \cdot t\_3}}{t\_5}\\

\mathbf{elif}\;t\_6 \leq -1 \cdot 10^{-186}:\\
\;\;\;\;\frac{\sqrt{t\_2 \cdot \left(C + \sqrt{\mathsf{fma}\left(B\_m, B\_m, C \cdot C\right)}\right)}}{-t\_0}\\

\mathbf{elif}\;t\_6 \leq 0:\\
\;\;\;\;\frac{\sqrt{F} \cdot \sqrt{t\_1 \cdot t\_3}}{t\_5}\\

\mathbf{elif}\;t\_6 \leq \infty:\\
\;\;\;\;\left(\sqrt{t\_2} \cdot \frac{-1}{t\_0}\right) \cdot \sqrt{2 \cdot C}\\

\mathbf{else}:\\
\;\;\;\;-\frac{\sqrt{F}}{\sqrt{B\_m \cdot 0.5}}\\


\end{array}
\end{array}
Derivation
  1. Split input into 5 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 A around -inf

      \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(\frac{-1}{2} \cdot \frac{{B}^{2}}{A} + 2 \cdot C\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Step-by-step derivation
      1. *-commutativeN/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\color{blue}{\frac{{B}^{2}}{A} \cdot \frac{-1}{2}} + 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. lower-fma.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{{B}^{2}}{A}, \frac{-1}{2}, 2 \cdot C\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      3. lower-/.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\color{blue}{\frac{{B}^{2}}{A}}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      4. unpow2N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{\color{blue}{B \cdot B}}{A}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      5. lower-*.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{\color{blue}{B \cdot B}}{A}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      6. lower-*.f6431.3

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{B \cdot B}{A}, -0.5, \color{blue}{2 \cdot C}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites31.3%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{B \cdot B}{A}, -0.5, 2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Applied rewrites59.4%

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

    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))) < -9.9999999999999991e-187

    1. Initial program 97.9%

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

      \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Step-by-step derivation
      1. lower-+.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. lower-sqrt.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \color{blue}{\sqrt{{B}^{2} + {C}^{2}}}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      3. +-commutativeN/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\color{blue}{{C}^{2} + {B}^{2}}}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      4. unpow2N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\color{blue}{C \cdot C} + {B}^{2}}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      5. lower-fma.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\color{blue}{\mathsf{fma}\left(C, C, {B}^{2}\right)}}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      6. unpow2N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\mathsf{fma}\left(C, C, \color{blue}{B \cdot B}\right)}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      7. lower-*.f6498.1

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\mathsf{fma}\left(C, C, \color{blue}{B \cdot B}\right)}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites98.1%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C + \sqrt{\mathsf{fma}\left(C, C, B \cdot B\right)}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Applied rewrites98.1%

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

    if -9.9999999999999991e-187 < (/.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))) < -0.0

    1. Initial program 3.4%

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

      \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(\frac{-1}{2} \cdot \frac{{B}^{2}}{A} + 2 \cdot C\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    4. Step-by-step derivation
      1. *-commutativeN/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\color{blue}{\frac{{B}^{2}}{A} \cdot \frac{-1}{2}} + 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. lower-fma.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{{B}^{2}}{A}, \frac{-1}{2}, 2 \cdot C\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      3. lower-/.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\color{blue}{\frac{{B}^{2}}{A}}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      4. unpow2N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{\color{blue}{B \cdot B}}{A}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      5. lower-*.f64N/A

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{\color{blue}{B \cdot B}}{A}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      6. lower-*.f6465.8

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{B \cdot B}{A}, -0.5, \color{blue}{2 \cdot C}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites65.8%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{B \cdot B}{A}, -0.5, 2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Applied rewrites70.2%

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

    if -0.0 < (/.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))) < +inf.0

    1. Initial program 43.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied rewrites43.9%

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

      \[\leadsto \sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \color{blue}{\left(C \cdot \left(2 + \left(-1 \cdot \frac{A}{C} + \frac{A}{C}\right)\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
    5. Step-by-step derivation
      1. distribute-lft1-inN/A

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

        \[\leadsto \sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot \left(2 + \color{blue}{0} \cdot \frac{A}{C}\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
      3. mul0-lftN/A

        \[\leadsto \sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot \left(2 + \color{blue}{0}\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
      4. metadata-evalN/A

        \[\leadsto \sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot \color{blue}{2}\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
      5. lower-*.f6469.3

        \[\leadsto \sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \color{blue}{\left(C \cdot 2\right)}} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
    6. Applied rewrites69.3%

      \[\leadsto \sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \color{blue}{\left(C \cdot 2\right)}} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
    7. Applied rewrites79.1%

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

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

    1. Initial program 0.0%

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

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

        \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
      2. lower-neg.f64N/A

        \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
      3. *-commutativeN/A

        \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
      4. lower-*.f64N/A

        \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
      5. lower-sqrt.f64N/A

        \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2}} \cdot \sqrt{\frac{F}{B}}\right) \]
      6. lower-sqrt.f64N/A

        \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
      7. lower-/.f6427.3

        \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
    5. Applied rewrites27.3%

      \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    6. Step-by-step derivation
      1. Applied rewrites27.5%

        \[\leadsto \color{blue}{-\sqrt{\frac{2 \cdot F}{B}}} \]
      2. Step-by-step derivation
        1. Applied rewrites27.5%

          \[\leadsto -\sqrt{F \cdot \frac{2}{B}} \]
        2. Step-by-step derivation
          1. Applied rewrites39.3%

            \[\leadsto -\frac{\sqrt{F}}{\sqrt{B \cdot 0.5}} \]
        3. Recombined 5 regimes into one program.
        4. Final simplification58.2%

          \[\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:\\ \;\;\;\;\frac{\sqrt{2 \cdot \mathsf{fma}\left(A, C \cdot -4, B \cdot B\right)} \cdot \sqrt{F \cdot \mathsf{fma}\left(B, \frac{B}{A} \cdot -0.5, 2 \cdot C\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \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 -1 \cdot 10^{-186}:\\ \;\;\;\;\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, B \cdot B\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(C + \sqrt{\mathsf{fma}\left(B, B, C \cdot C\right)}\right)}}{-\mathsf{fma}\left(A, C \cdot -4, B \cdot B\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 0:\\ \;\;\;\;\frac{\sqrt{F} \cdot \sqrt{\left(2 \cdot \mathsf{fma}\left(A, C \cdot -4, B \cdot B\right)\right) \cdot \mathsf{fma}\left(B, \frac{B}{A} \cdot -0.5, 2 \cdot C\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \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:\\ \;\;\;\;\left(\sqrt{\mathsf{fma}\left(A, C \cdot -4, B \cdot B\right) \cdot \left(2 \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(A, C \cdot -4, B \cdot B\right)}\right) \cdot \sqrt{2 \cdot C}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{F}}{\sqrt{B \cdot 0.5}}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 2: 56.9% accurate, 0.3× speedup?

        \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(A, C \cdot -4, B\_m \cdot B\_m\right)\\ t_1 := t\_0 \cdot \left(2 \cdot F\right)\\ t_2 := \mathsf{fma}\left(B\_m, \frac{B\_m}{A} \cdot -0.5, 2 \cdot C\right)\\ t_3 := \left(4 \cdot A\right) \cdot C\\ t_4 := t\_3 - {B\_m}^{2}\\ t_5 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_3\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_4}\\ \mathbf{if}\;t\_5 \leq -\infty:\\ \;\;\;\;\frac{\sqrt{2 \cdot t\_0} \cdot \sqrt{F \cdot t\_2}}{t\_4}\\ \mathbf{elif}\;t\_5 \leq -1 \cdot 10^{-186}:\\ \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(C + \sqrt{\mathsf{fma}\left(B\_m, B\_m, C \cdot C\right)}\right)}}{-t\_0}\\ \mathbf{elif}\;t\_5 \leq \infty:\\ \;\;\;\;\frac{\sqrt{t\_1} \cdot \sqrt{t\_2}}{t\_4}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{F}}{\sqrt{B\_m \cdot 0.5}}\\ \end{array} \end{array} \]
        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 A (* C -4.0) (* B_m B_m)))
                (t_1 (* t_0 (* 2.0 F)))
                (t_2 (fma B_m (* (/ B_m A) -0.5) (* 2.0 C)))
                (t_3 (* (* 4.0 A) C))
                (t_4 (- t_3 (pow B_m 2.0)))
                (t_5
                 (/
                  (sqrt
                   (*
                    (* 2.0 (* (- (pow B_m 2.0) t_3) F))
                    (+ (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
                  t_4)))
           (if (<= t_5 (- INFINITY))
             (/ (* (sqrt (* 2.0 t_0)) (sqrt (* F t_2))) t_4)
             (if (<= t_5 -1e-186)
               (/ (sqrt (* t_1 (+ C (sqrt (fma B_m B_m (* C C)))))) (- t_0))
               (if (<= t_5 INFINITY)
                 (/ (* (sqrt t_1) (sqrt t_2)) t_4)
                 (- (/ (sqrt F) (sqrt (* B_m 0.5)))))))))
        B_m = fabs(B);
        assert(A < B_m && B_m < C && C < F);
        double code(double A, double B_m, double C, double F) {
        	double t_0 = fma(A, (C * -4.0), (B_m * B_m));
        	double t_1 = t_0 * (2.0 * F);
        	double t_2 = fma(B_m, ((B_m / A) * -0.5), (2.0 * C));
        	double t_3 = (4.0 * A) * C;
        	double t_4 = t_3 - pow(B_m, 2.0);
        	double t_5 = sqrt(((2.0 * ((pow(B_m, 2.0) - t_3) * F)) * ((A + C) + sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / t_4;
        	double tmp;
        	if (t_5 <= -((double) INFINITY)) {
        		tmp = (sqrt((2.0 * t_0)) * sqrt((F * t_2))) / t_4;
        	} else if (t_5 <= -1e-186) {
        		tmp = sqrt((t_1 * (C + sqrt(fma(B_m, B_m, (C * C)))))) / -t_0;
        	} else if (t_5 <= ((double) INFINITY)) {
        		tmp = (sqrt(t_1) * sqrt(t_2)) / t_4;
        	} else {
        		tmp = -(sqrt(F) / sqrt((B_m * 0.5)));
        	}
        	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(A, Float64(C * -4.0), Float64(B_m * B_m))
        	t_1 = Float64(t_0 * Float64(2.0 * F))
        	t_2 = fma(B_m, Float64(Float64(B_m / A) * -0.5), Float64(2.0 * C))
        	t_3 = Float64(Float64(4.0 * A) * C)
        	t_4 = Float64(t_3 - (B_m ^ 2.0))
        	t_5 = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_3) * F)) * Float64(Float64(A + C) + sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / t_4)
        	tmp = 0.0
        	if (t_5 <= Float64(-Inf))
        		tmp = Float64(Float64(sqrt(Float64(2.0 * t_0)) * sqrt(Float64(F * t_2))) / t_4);
        	elseif (t_5 <= -1e-186)
        		tmp = Float64(sqrt(Float64(t_1 * Float64(C + sqrt(fma(B_m, B_m, Float64(C * C)))))) / Float64(-t_0));
        	elseif (t_5 <= Inf)
        		tmp = Float64(Float64(sqrt(t_1) * sqrt(t_2)) / t_4);
        	else
        		tmp = Float64(-Float64(sqrt(F) / sqrt(Float64(B_m * 0.5))));
        	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[(A * N[(C * -4.0), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(B$95$m * N[(N[(B$95$m / A), $MachinePrecision] * -0.5), $MachinePrecision] + N[(2.0 * C), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$4 = N[(t$95$3 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$5 = N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$3), $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] / t$95$4), $MachinePrecision]}, If[LessEqual[t$95$5, (-Infinity)], N[(N[(N[Sqrt[N[(2.0 * t$95$0), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(F * t$95$2), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / t$95$4), $MachinePrecision], If[LessEqual[t$95$5, -1e-186], N[(N[Sqrt[N[(t$95$1 * N[(C + N[Sqrt[N[(B$95$m * B$95$m + N[(C * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], If[LessEqual[t$95$5, Infinity], N[(N[(N[Sqrt[t$95$1], $MachinePrecision] * N[Sqrt[t$95$2], $MachinePrecision]), $MachinePrecision] / t$95$4), $MachinePrecision], (-N[(N[Sqrt[F], $MachinePrecision] / N[Sqrt[N[(B$95$m * 0.5), $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 := \mathsf{fma}\left(A, C \cdot -4, B\_m \cdot B\_m\right)\\
        t_1 := t\_0 \cdot \left(2 \cdot F\right)\\
        t_2 := \mathsf{fma}\left(B\_m, \frac{B\_m}{A} \cdot -0.5, 2 \cdot C\right)\\
        t_3 := \left(4 \cdot A\right) \cdot C\\
        t_4 := t\_3 - {B\_m}^{2}\\
        t_5 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_3\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_4}\\
        \mathbf{if}\;t\_5 \leq -\infty:\\
        \;\;\;\;\frac{\sqrt{2 \cdot t\_0} \cdot \sqrt{F \cdot t\_2}}{t\_4}\\
        
        \mathbf{elif}\;t\_5 \leq -1 \cdot 10^{-186}:\\
        \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(C + \sqrt{\mathsf{fma}\left(B\_m, B\_m, C \cdot C\right)}\right)}}{-t\_0}\\
        
        \mathbf{elif}\;t\_5 \leq \infty:\\
        \;\;\;\;\frac{\sqrt{t\_1} \cdot \sqrt{t\_2}}{t\_4}\\
        
        \mathbf{else}:\\
        \;\;\;\;-\frac{\sqrt{F}}{\sqrt{B\_m \cdot 0.5}}\\
        
        
        \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 3.2%

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

            \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(\frac{-1}{2} \cdot \frac{{B}^{2}}{A} + 2 \cdot C\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          4. Step-by-step derivation
            1. *-commutativeN/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\color{blue}{\frac{{B}^{2}}{A} \cdot \frac{-1}{2}} + 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. lower-fma.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{{B}^{2}}{A}, \frac{-1}{2}, 2 \cdot C\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            3. lower-/.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\color{blue}{\frac{{B}^{2}}{A}}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            4. unpow2N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{\color{blue}{B \cdot B}}{A}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            5. lower-*.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{\color{blue}{B \cdot B}}{A}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            6. lower-*.f6433.8

              \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{B \cdot B}{A}, -0.5, \color{blue}{2 \cdot C}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          5. Applied rewrites33.8%

            \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{B \cdot B}{A}, -0.5, 2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          6. Applied rewrites47.4%

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

          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))) < -9.9999999999999991e-187

          1. Initial program 97.6%

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

            \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          4. Step-by-step derivation
            1. lower-+.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. lower-sqrt.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \color{blue}{\sqrt{{B}^{2} + {C}^{2}}}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            3. +-commutativeN/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\color{blue}{{C}^{2} + {B}^{2}}}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            4. unpow2N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\color{blue}{C \cdot C} + {B}^{2}}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            5. lower-fma.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\color{blue}{\mathsf{fma}\left(C, C, {B}^{2}\right)}}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            6. unpow2N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\mathsf{fma}\left(C, C, \color{blue}{B \cdot B}\right)}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            7. lower-*.f6496.3

              \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C + \sqrt{\mathsf{fma}\left(C, C, \color{blue}{B \cdot B}\right)}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          5. Applied rewrites96.3%

            \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C + \sqrt{\mathsf{fma}\left(C, C, B \cdot B\right)}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          6. Applied rewrites96.2%

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

          if -9.9999999999999991e-187 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < +inf.0

          1. Initial program 20.4%

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

            \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(\frac{-1}{2} \cdot \frac{{B}^{2}}{A} + 2 \cdot C\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          4. Step-by-step derivation
            1. *-commutativeN/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\color{blue}{\frac{{B}^{2}}{A} \cdot \frac{-1}{2}} + 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. lower-fma.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{{B}^{2}}{A}, \frac{-1}{2}, 2 \cdot C\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            3. lower-/.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\color{blue}{\frac{{B}^{2}}{A}}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            4. unpow2N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{\color{blue}{B \cdot B}}{A}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            5. lower-*.f64N/A

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{\color{blue}{B \cdot B}}{A}, \frac{-1}{2}, 2 \cdot C\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            6. lower-*.f6455.1

              \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(\frac{B \cdot B}{A}, -0.5, \color{blue}{2 \cdot C}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          5. Applied rewrites55.1%

            \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{B \cdot B}{A}, -0.5, 2 \cdot C\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          6. Applied rewrites57.0%

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

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

          1. Initial program 0.0%

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

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

              \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
            2. lower-neg.f64N/A

              \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
            3. *-commutativeN/A

              \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
            4. lower-*.f64N/A

              \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
            5. lower-sqrt.f64N/A

              \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2}} \cdot \sqrt{\frac{F}{B}}\right) \]
            6. lower-sqrt.f64N/A

              \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
            7. lower-/.f6432.0

              \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
          5. Applied rewrites32.0%

            \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
          6. Step-by-step derivation
            1. Applied rewrites32.2%

              \[\leadsto \color{blue}{-\sqrt{\frac{2 \cdot F}{B}}} \]
            2. Step-by-step derivation
              1. Applied rewrites32.2%

                \[\leadsto -\sqrt{F \cdot \frac{2}{B}} \]
              2. Step-by-step derivation
                1. Applied rewrites48.6%

                  \[\leadsto -\frac{\sqrt{F}}{\sqrt{B \cdot 0.5}} \]
              3. Recombined 4 regimes into one program.
              4. Final simplification56.9%

                \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -\infty:\\ \;\;\;\;\frac{\sqrt{2 \cdot \mathsf{fma}\left(A, C \cdot -4, B \cdot B\right)} \cdot \sqrt{F \cdot \mathsf{fma}\left(B, \frac{B}{A} \cdot -0.5, 2 \cdot C\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -1 \cdot 10^{-186}:\\ \;\;\;\;\frac{\sqrt{\left(\mathsf{fma}\left(A, C \cdot -4, B \cdot B\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(C + \sqrt{\mathsf{fma}\left(B, B, C \cdot C\right)}\right)}}{-\mathsf{fma}\left(A, C \cdot -4, B \cdot B\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(A, C \cdot -4, B \cdot B\right) \cdot \left(2 \cdot F\right)} \cdot \sqrt{\mathsf{fma}\left(B, \frac{B}{A} \cdot -0.5, 2 \cdot C\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{F}}{\sqrt{B \cdot 0.5}}\\ \end{array} \]
              5. Add Preprocessing

              Reproduce

              ?
              herbie shell --seed 2024228 
              (FPCore (A B C F)
                :name "ABCF->ab-angle a"
                :precision binary64
                (/ (- (sqrt (* (* 2.0 (* (- (pow B 2.0) (* (* 4.0 A) C)) F)) (+ (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))) (- (pow B 2.0) (* (* 4.0 A) C))))