ABCF->ab-angle a

Percentage Accurate: 18.8% → 57.1%
Time: 20.8s
Alternatives: 12
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 12 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: 18.8% 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: 57.1% accurate, 0.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\ t_1 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\ t_2 := \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\\ t_3 := \left(4 \cdot A\right) \cdot C\\ t_4 := t\_3 - {B\_m}^{2}\\ t_5 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_3\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_4}\\ \mathbf{if}\;t\_5 \leq -\infty:\\ \;\;\;\;\frac{\sqrt{2 \cdot t\_0} \cdot \sqrt{F \cdot t\_2}}{t\_4}\\ \mathbf{elif}\;t\_5 \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}}}{t\_1}\\ \mathbf{elif}\;t\_5 \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot t\_2\right)}}{-t\_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
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) (* B_m B_m)))
        (t_1 (fma B_m B_m (* -4.0 (* A C))))
        (t_2 (fma 2.0 C (/ (* (* B_m B_m) -0.5) A)))
        (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 -2e-194)
       (*
        (/ (sqrt (* t_1 (* 2.0 F))) -1.0)
        (/ (sqrt (+ (+ A C) (sqrt (fma (- A C) (- A C) (* B_m B_m))))) t_1))
       (if (<= t_5 INFINITY)
         (/ (sqrt (* (* F t_0) (* 2.0 t_2))) (- t_0))
         (/ (sqrt (* 2.0 F)) (- (sqrt 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), (B_m * B_m));
	double t_1 = fma(B_m, B_m, (-4.0 * (A * C)));
	double t_2 = fma(2.0, C, (((B_m * B_m) * -0.5) / A));
	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 <= -2e-194) {
		tmp = (sqrt((t_1 * (2.0 * F))) / -1.0) * (sqrt(((A + C) + sqrt(fma((A - C), (A - C), (B_m * B_m))))) / t_1);
	} else if (t_5 <= ((double) INFINITY)) {
		tmp = sqrt(((F * t_0) * (2.0 * t_2))) / -t_0;
	} else {
		tmp = sqrt((2.0 * F)) / -sqrt(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), Float64(B_m * B_m))
	t_1 = fma(B_m, B_m, Float64(-4.0 * Float64(A * C)))
	t_2 = fma(2.0, C, Float64(Float64(Float64(B_m * B_m) * -0.5) / A))
	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 <= -2e-194)
		tmp = Float64(Float64(sqrt(Float64(t_1 * Float64(2.0 * F))) / -1.0) * Float64(sqrt(Float64(Float64(A + C) + sqrt(fma(Float64(A - C), Float64(A - C), Float64(B_m * B_m))))) / t_1));
	elseif (t_5 <= Inf)
		tmp = Float64(sqrt(Float64(Float64(F * t_0) * Float64(2.0 * t_2))) / Float64(-t_0));
	else
		tmp = Float64(sqrt(Float64(2.0 * F)) / Float64(-sqrt(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[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(B$95$m * B$95$m + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(2.0 * C + N[(N[(N[(B$95$m * B$95$m), $MachinePrecision] * -0.5), $MachinePrecision] / A), $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, -2e-194], N[(N[(N[Sqrt[N[(t$95$1 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / -1.0), $MachinePrecision] * N[(N[Sqrt[N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[(A - C), $MachinePrecision] * N[(A - C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$5, Infinity], N[(N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * t$95$2), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], N[(N[Sqrt[N[(2.0 * F), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]]]]]]]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\
t_1 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\
t_2 := \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\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 -2 \cdot 10^{-194}:\\
\;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}}}{t\_1}\\

\mathbf{elif}\;t\_5 \leq \infty:\\
\;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot t\_2\right)}}{-t\_0}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -inf.0

    1. Initial program 2.9%

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

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

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

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

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

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

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      7. lower-*.f6444.2

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{-0.5 \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites44.2%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(2, C, \frac{-0.5 \cdot \left(B \cdot B\right)}{A}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Step-by-step derivation
      1. lift-sqrt.f64N/A

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

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

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

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

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

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

        \[\leadsto \frac{\mathsf{neg}\left(\color{blue}{\sqrt{2 \cdot \left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)} \cdot \sqrt{F \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \left(B \cdot B\right)}{A}\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    7. Applied rewrites62.9%

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

    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))) < -2.00000000000000004e-194

    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. Applied rewrites99.3%

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

    if -2.00000000000000004e-194 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < +inf.0

    1. Initial program 14.2%

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

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

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

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

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

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

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      7. lower-*.f6456.6

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{-0.5 \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites56.6%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(2, C, \frac{-0.5 \cdot \left(B \cdot B\right)}{A}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Step-by-step derivation
      1. lift-/.f64N/A

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

        \[\leadsto \color{blue}{\frac{\mathsf{neg}\left(\left(\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \left(B \cdot B\right)}{A}\right)}\right)\right)\right)}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
    7. Applied rewrites56.6%

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

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

    1. Initial program 0.0%

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

      \[\leadsto \color{blue}{-1 \cdot \left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
    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-/.f6432.6

        \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
    5. Applied rewrites32.6%

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

        \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
    7. Recombined 4 regimes into one program.
    8. Final simplification60.0%

      \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -\infty:\\ \;\;\;\;\frac{\sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)} \cdot \sqrt{F \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)}}}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B}}\\ \end{array} \]
    9. Add Preprocessing

    Alternative 2: 54.2% accurate, 0.3× speedup?

    \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\ t_1 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\ t_2 := \frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\right)}}{-t\_0}\\ t_3 := \left(4 \cdot A\right) \cdot C\\ t_4 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_3\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_3 - {B\_m}^{2}}\\ \mathbf{if}\;t\_4 \leq -\infty:\\ \;\;\;\;t\_2\\ \mathbf{elif}\;t\_4 \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}}}{t\_1}\\ \mathbf{elif}\;t\_4 \leq \infty:\\ \;\;\;\;t\_2\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
    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) (* B_m B_m)))
            (t_1 (fma B_m B_m (* -4.0 (* A C))))
            (t_2
             (/
              (sqrt (* (* F t_0) (* 2.0 (fma 2.0 C (/ (* (* B_m B_m) -0.5) A)))))
              (- t_0)))
            (t_3 (* (* 4.0 A) C))
            (t_4
             (/
              (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_3 (pow B_m 2.0)))))
       (if (<= t_4 (- INFINITY))
         t_2
         (if (<= t_4 -2e-194)
           (*
            (/ (sqrt (* t_1 (* 2.0 F))) -1.0)
            (/ (sqrt (+ (+ A C) (sqrt (fma (- A C) (- A C) (* B_m B_m))))) t_1))
           (if (<= t_4 INFINITY) t_2 (/ (sqrt (* 2.0 F)) (- (sqrt 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), (B_m * B_m));
    	double t_1 = fma(B_m, B_m, (-4.0 * (A * C)));
    	double t_2 = sqrt(((F * t_0) * (2.0 * fma(2.0, C, (((B_m * B_m) * -0.5) / A))))) / -t_0;
    	double t_3 = (4.0 * A) * C;
    	double t_4 = 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_3 - pow(B_m, 2.0));
    	double tmp;
    	if (t_4 <= -((double) INFINITY)) {
    		tmp = t_2;
    	} else if (t_4 <= -2e-194) {
    		tmp = (sqrt((t_1 * (2.0 * F))) / -1.0) * (sqrt(((A + C) + sqrt(fma((A - C), (A - C), (B_m * B_m))))) / t_1);
    	} else if (t_4 <= ((double) INFINITY)) {
    		tmp = t_2;
    	} else {
    		tmp = sqrt((2.0 * F)) / -sqrt(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), Float64(B_m * B_m))
    	t_1 = fma(B_m, B_m, Float64(-4.0 * Float64(A * C)))
    	t_2 = Float64(sqrt(Float64(Float64(F * t_0) * Float64(2.0 * fma(2.0, C, Float64(Float64(Float64(B_m * B_m) * -0.5) / A))))) / Float64(-t_0))
    	t_3 = Float64(Float64(4.0 * A) * C)
    	t_4 = 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)))))) / Float64(t_3 - (B_m ^ 2.0)))
    	tmp = 0.0
    	if (t_4 <= Float64(-Inf))
    		tmp = t_2;
    	elseif (t_4 <= -2e-194)
    		tmp = Float64(Float64(sqrt(Float64(t_1 * Float64(2.0 * F))) / -1.0) * Float64(sqrt(Float64(Float64(A + C) + sqrt(fma(Float64(A - C), Float64(A - C), Float64(B_m * B_m))))) / t_1));
    	elseif (t_4 <= Inf)
    		tmp = t_2;
    	else
    		tmp = Float64(sqrt(Float64(2.0 * F)) / Float64(-sqrt(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[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(B$95$m * B$95$m + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * N[(2.0 * C + N[(N[(N[(B$95$m * B$95$m), $MachinePrecision] * -0.5), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision]}, Block[{t$95$3 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$4 = 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] / N[(t$95$3 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$4, (-Infinity)], t$95$2, If[LessEqual[t$95$4, -2e-194], N[(N[(N[Sqrt[N[(t$95$1 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / -1.0), $MachinePrecision] * N[(N[Sqrt[N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[(A - C), $MachinePrecision] * N[(A - C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$4, Infinity], t$95$2, N[(N[Sqrt[N[(2.0 * F), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]]]]]]]]
    
    \begin{array}{l}
    B_m = \left|B\right|
    \\
    [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
    \\
    \begin{array}{l}
    t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\
    t_1 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\
    t_2 := \frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\right)}}{-t\_0}\\
    t_3 := \left(4 \cdot A\right) \cdot C\\
    t_4 := \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\_3 - {B\_m}^{2}}\\
    \mathbf{if}\;t\_4 \leq -\infty:\\
    \;\;\;\;t\_2\\
    
    \mathbf{elif}\;t\_4 \leq -2 \cdot 10^{-194}:\\
    \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}}}{t\_1}\\
    
    \mathbf{elif}\;t\_4 \leq \infty:\\
    \;\;\;\;t\_2\\
    
    \mathbf{else}:\\
    \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B\_m}}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 3 regimes
    2. if (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -inf.0 or -2.00000000000000004e-194 < (/.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 11.6%

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

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

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

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

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

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

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

          \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        7. lower-*.f6446.6

          \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{-0.5 \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      5. Applied rewrites46.6%

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(2, C, \frac{-0.5 \cdot \left(B \cdot B\right)}{A}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      6. Step-by-step derivation
        1. lift-/.f64N/A

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

          \[\leadsto \color{blue}{\frac{\mathsf{neg}\left(\left(\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \left(B \cdot B\right)}{A}\right)}\right)\right)\right)}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
      7. Applied rewrites46.6%

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

      if -inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -2.00000000000000004e-194

      1. Initial program 97.8%

        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. Add Preprocessing
      3. Applied rewrites98.0%

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

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

      1. Initial program 0.0%

        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. Add Preprocessing
      3. Taylor expanded in 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-/.f6431.2

          \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
      5. Applied rewrites31.2%

        \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
      6. Step-by-step derivation
        1. Applied rewrites47.4%

          \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
      7. Recombined 3 regimes into one program.
      8. Final simplification54.2%

        \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -\infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)}}}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B}}\\ \end{array} \]
      9. Add Preprocessing

      Reproduce

      ?
      herbie shell --seed 2024219 
      (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))))