ABCF->ab-angle b

Percentage Accurate: 19.1% → 50.5%

Time: 20.1s

Alternatives: 11

Speedup: 3.0×

Specification

Math

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

FPCore

(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)))

C

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;
}

Fortran

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

Java

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;
}

Python

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

Julia

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

MATLAB

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

Wolfram

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]]

Initial Program: 19.1% accurate, 1.0× speedup

Math

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

FPCore

(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)))

C

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;
}

Fortran

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

Java

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;
}

Python

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

Julia

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

MATLAB

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

Wolfram

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]]

Alternative 1: 50.5% accurate, 0.4× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)\\ t_1 := \left(4 \cdot A\right) \cdot C\\ t_2 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_1\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_1 - {B\_m}^{2}}\\ t_3 := \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)}\\ t_4 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;t\_2 \leq -1 \cdot 10^{-188}:\\ \;\;\;\;\frac{t\_3 \cdot \sqrt{2 \cdot t\_0}}{-t\_0}\\ \mathbf{elif}\;t\_2 \leq \infty:\\ \;\;\;\;\frac{-1}{\frac{t\_4}{\sqrt{\left(F \cdot t\_4\right) \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;t\_3 \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (fma C (* A -4.0) (pow B_m 2.0)))
        (t_1 (* (* 4.0 A) C))
        (t_2
         (/
          (sqrt
           (*
            (* 2.0 (* (- (pow B_m 2.0) t_1) F))
            (- (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
          (- t_1 (pow B_m 2.0))))
        (t_3 (sqrt (* F (- A (hypot B_m A)))))
        (t_4 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= t_2 -1e-188)
     (/ (* t_3 (sqrt (* 2.0 t_0))) (- t_0))
     (if (<= t_2 INFINITY)
       (/
        -1.0
        (/
         t_4
         (sqrt
          (* (* F t_4) (* 2.0 (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C)))))))))
       (* t_3 (/ (sqrt 2.0) (- B_m)))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = fma(C, (A * -4.0), pow(B_m, 2.0));
	double t_1 = (4.0 * A) * C;
	double t_2 = sqrt(((2.0 * ((pow(B_m, 2.0) - t_1) * F)) * ((A + C) - sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / (t_1 - pow(B_m, 2.0));
	double t_3 = sqrt((F * (A - hypot(B_m, A))));
	double t_4 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (t_2 <= -1e-188) {
		tmp = (t_3 * sqrt((2.0 * t_0))) / -t_0;
	} else if (t_2 <= ((double) INFINITY)) {
		tmp = -1.0 / (t_4 / sqrt(((F * t_4) * (2.0 * (A + (A + (-0.5 * (pow(B_m, 2.0) / C))))))));
	} else {
		tmp = t_3 * (sqrt(2.0) / -B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = fma(C, Float64(A * -4.0), (B_m ^ 2.0))
	t_1 = Float64(Float64(4.0 * A) * C)
	t_2 = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_1) * F)) * Float64(Float64(A + C) - sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / Float64(t_1 - (B_m ^ 2.0)))
	t_3 = sqrt(Float64(F * Float64(A - hypot(B_m, A))))
	t_4 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if (t_2 <= -1e-188)
		tmp = Float64(Float64(t_3 * sqrt(Float64(2.0 * t_0))) / Float64(-t_0));
	elseif (t_2 <= Inf)
		tmp = Float64(-1.0 / Float64(t_4 / sqrt(Float64(Float64(F * t_4) * Float64(2.0 * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C)))))))));
	else
		tmp = Float64(t_3 * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$2 = N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$1), $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$1 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$4 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$2, -1e-188], N[(N[(t$95$3 * N[Sqrt[N[(2.0 * t$95$0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / (-t$95$0)), $MachinePrecision], If[LessEqual[t$95$2, Infinity], N[(-1.0 / N[(t$95$4 / N[Sqrt[N[(N[(F * t$95$4), $MachinePrecision] * N[(2.0 * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(t$95$3 * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]]]]]]

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))) < -9.9999999999999995e-189

   1. Initial program 43.9%

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

   3. Simplified 44.7%

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

   5. Taylor expanded in C around 0 35.2%

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

      1. mul-1-neg 35.2%

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

      2. +-commutative 35.2%

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

      3. unpow2 35.2%

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

      4. unpow2 35.2%

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

      5. hypot-define 39.0%

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

   7. Simplified 39.0%

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

      1. pow1/2 39.0%

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

      2. associate-*r* 43.8%

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

      3. unpow-prod-down 54.5%

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

      4. unsub-neg 54.5%

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

      5. pow1/2 54.6%

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

   9. Applied egg-rr 54.6%

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

       1. unpow1/2 54.6%

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

   11. Simplified 54.6%

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

   if -9.9999999999999995e-189 < (/.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 21.2%

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

   3. Simplified 35.7%

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

   5. Taylor expanded in C around inf 41.8%

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

      1. associate-*r/ 41.8%

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

      2. mul-1-neg 41.8%

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

   7. Simplified 41.8%

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

      1. clear-num 41.8%

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

      2. inv-pow 41.8%

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

      3. associate-/l* 41.8%

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

   9. Applied egg-rr 41.8%

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

       1. unpow-1 41.8%

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

   11. Simplified 41.8%

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

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

   1. Initial program 0.0%

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

   3. Taylor expanded in C around 0 1.6%

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

      1. mul-1-neg 1.6%

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

      2. +-commutative 1.6%

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

      3. unpow2 1.6%

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

      4. unpow2 1.6%

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

      5. hypot-define 21.5%

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

   5. Simplified 21.5%

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

4. Final simplification 39.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -1 \cdot 10^{-188}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{-1}{\frac{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 46.3% accurate, 1.2× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+36}:\\ \;\;\;\;\frac{-1}{\frac{t\_0}{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B\_m}^{2}}{C}\right)\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 2e+36)
     (/
      -1.0
      (/
       t_0
       (sqrt (* (* F t_0) (* 2.0 (+ A (+ A (* -0.5 (/ (pow B_m 2.0) C)))))))))
     (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 2e+36) {
		tmp = -1.0 / (t_0 / sqrt(((F * t_0) * (2.0 * (A + (A + (-0.5 * (pow(B_m, 2.0) / C))))))));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e+36)
		tmp = Float64(-1.0 / Float64(t_0 / sqrt(Float64(Float64(F * t_0) * Float64(2.0 * Float64(A + Float64(A + Float64(-0.5 * Float64((B_m ^ 2.0) / C)))))))));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+36], N[(-1.0 / N[(t$95$0 / N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * N[(A + N[(A + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 2.00000000000000008e36

   1. Initial program 26.2%

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

   3. Simplified 39.0%

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

   5. Taylor expanded in C around inf 31.9%

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

      1. associate-*r/ 31.9%

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

      2. mul-1-neg 31.9%

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

   7. Simplified 31.9%

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

      1. clear-num 31.9%

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

      2. inv-pow 31.9%

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

      3. associate-/l* 31.9%

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

   9. Applied egg-rr 31.9%

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

       1. unpow-1 31.9%

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

   11. Simplified 31.9%

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

   if 2.00000000000000008e36 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 17.3%

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

   3. Taylor expanded in C around 0 12.7%

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

      1. mul-1-neg 12.7%

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

      2. +-commutative 12.7%

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

      3. unpow2 12.7%

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

      4. unpow2 12.7%

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

      5. hypot-define 29.4%

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

   5. Simplified 29.4%

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

4. Final simplification 30.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+36}:\\ \;\;\;\;\frac{-1}{\frac{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A + \left(A + -0.5 \cdot \frac{{B}^{2}}{C}\right)\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 46.3% accurate, 1.2× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(B\_m, B\_m, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+36}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \left(A + \left(A + \frac{{B\_m}^{2} \cdot -0.5}{C}\right)\right)\right)}}{-t\_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (fma B_m B_m (* A (* C -4.0)))))
   (if (<= (pow B_m 2.0) 2e+36)
     (/
      (sqrt (* (* F t_0) (* 2.0 (+ A (+ A (/ (* (pow B_m 2.0) -0.5) C))))))
      (- t_0))
     (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = fma(B_m, B_m, (A * (C * -4.0)));
	double tmp;
	if (pow(B_m, 2.0) <= 2e+36) {
		tmp = sqrt(((F * t_0) * (2.0 * (A + (A + ((pow(B_m, 2.0) * -0.5) / C)))))) / -t_0;
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = fma(B_m, B_m, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e+36)
		tmp = Float64(sqrt(Float64(Float64(F * t_0) * Float64(2.0 * Float64(A + Float64(A + Float64(Float64((B_m ^ 2.0) * -0.5) / C)))))) / Float64(-t_0));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(B$95$m * B$95$m + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+36], N[(N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * N[(A + N[(A + N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] * -0.5), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 2.00000000000000008e36

   1. Initial program 26.2%

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

   3. Simplified 39.0%

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

   5. Taylor expanded in C around inf 31.9%

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

      1. associate-*r/ 31.9%

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

      2. mul-1-neg 31.9%

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

   7. Simplified 31.9%

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

   if 2.00000000000000008e36 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 17.3%

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

   3. Taylor expanded in C around 0 12.7%

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

      1. mul-1-neg 12.7%

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

      2. +-commutative 12.7%

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

      3. unpow2 12.7%

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

      4. unpow2 12.7%

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

      5. hypot-define 29.4%

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

   5. Simplified 29.4%

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

4. Final simplification 30.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+36}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right) \cdot \left(2 \cdot \left(A + \left(A + \frac{{B}^{2} \cdot -0.5}{C}\right)\right)\right)}}{-\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 46.5% accurate, 1.5× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* (* 4.0 A) C)))
   (if (<= (pow B_m 2.0) 2e-41)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 A)))
      (- t_0 (pow B_m 2.0)))
     (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double tmp;
	if (pow(B_m, 2.0) <= 2e-41) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}

Java

B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e-41) {
		tmp = Math.sqrt(((2.0 * ((Math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
	}
	return tmp;
}

Python

B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = (4.0 * A) * C
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e-41:
		tmp = math.sqrt(((2.0 * ((math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
	return tmp

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-41)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * A))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end

MATLAB

B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = (4.0 * A) * C;
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e-41)
		tmp = sqrt(((2.0 * (((B_m ^ 2.0) - t_0) * F)) * (2.0 * A))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	end
	tmp_2 = tmp;
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-41], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 2.00000000000000001e-41

   1. Initial program 26.3%

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

   3. Taylor expanded in A around -inf 32.1%

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

   if 2.00000000000000001e-41 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 18.1%

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

   3. Taylor expanded in C around 0 14.0%

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

      1. mul-1-neg 14.0%

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

      2. +-commutative 14.0%

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

      3. unpow2 14.0%

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

      4. unpow2 14.0%

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

      5. hypot-define 29.0%

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

   5. Simplified 29.0%

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

4. Final simplification 30.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 45.4% accurate, 1.5× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B\_m}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 2e-41)
   (/
    (sqrt (* -8.0 (* (* A C) (* F (+ A A)))))
    (- (fma C (* A -4.0) (pow B_m 2.0))))
   (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m)))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 2e-41) {
		tmp = sqrt((-8.0 * ((A * C) * (F * (A + A))))) / -fma(C, (A * -4.0), pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-41)
		tmp = Float64(sqrt(Float64(-8.0 * Float64(Float64(A * C) * Float64(F * Float64(A + A))))) / Float64(-fma(C, Float64(A * -4.0), (B_m ^ 2.0))));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-41], N[(N[Sqrt[N[(-8.0 * N[(N[(A * C), $MachinePrecision] * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-N[(C * N[(A * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision])), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 2.00000000000000001e-41

   1. Initial program 26.3%

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

   3. Simplified 31.2%

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

   5. Taylor expanded in C around inf 30.0%

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

      1. associate-*r* 30.3%

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

      2. *-commutative 30.3%

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

      3. mul-1-neg 30.3%

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

   7. Simplified 30.3%

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

   if 2.00000000000000001e-41 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 18.1%

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

   3. Taylor expanded in C around 0 14.0%

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

      1. mul-1-neg 14.0%

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

      2. +-commutative 14.0%

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

      3. unpow2 14.0%

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

      4. unpow2 14.0%

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

      5. hypot-define 29.0%

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

   5. Simplified 29.0%

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

4. Final simplification 29.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(\left(A \cdot C\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, {B}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 30.6% accurate, 1.5× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := F \cdot \left(A + C\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-103}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot t\_0\right)\right) + B\_m \cdot \left(8 \cdot \left(A \cdot \left(C \cdot F\right)\right) + B\_m \cdot \left(-2 \cdot \left(B\_m \cdot F\right) + 2 \cdot t\_0\right)\right)}}{\left(4 \cdot A\right) \cdot C - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B\_m, A\right)\right)} \cdot \frac{\sqrt{2}}{-B\_m}\\ \end{array} \end{array} \]

FPCore

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 (* F (+ A C))))
   (if (<= (pow B_m 2.0) 2e-103)
     (/
      (sqrt
       (+
        (* -8.0 (* A (* C t_0)))
        (*
         B_m
         (+
          (* 8.0 (* A (* C F)))
          (* B_m (+ (* -2.0 (* B_m F)) (* 2.0 t_0)))))))
      (- (* (* 4.0 A) C) (pow B_m 2.0)))
     (* (sqrt (* F (- A (hypot B_m A)))) (/ (sqrt 2.0) (- B_m))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = F * (A + C);
	double tmp;
	if (pow(B_m, 2.0) <= 2e-103) {
		tmp = sqrt(((-8.0 * (A * (C * t_0))) + (B_m * ((8.0 * (A * (C * F))) + (B_m * ((-2.0 * (B_m * F)) + (2.0 * t_0))))))) / (((4.0 * A) * C) - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	}
	return tmp;
}

Java

B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = F * (A + C);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e-103) {
		tmp = Math.sqrt(((-8.0 * (A * (C * t_0))) + (B_m * ((8.0 * (A * (C * F))) + (B_m * ((-2.0 * (B_m * F)) + (2.0 * t_0))))))) / (((4.0 * A) * C) - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(B_m, A)))) * (Math.sqrt(2.0) / -B_m);
	}
	return tmp;
}

Python

B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = F * (A + C)
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e-103:
		tmp = math.sqrt(((-8.0 * (A * (C * t_0))) + (B_m * ((8.0 * (A * (C * F))) + (B_m * ((-2.0 * (B_m * F)) + (2.0 * t_0))))))) / (((4.0 * A) * C) - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * (A - math.hypot(B_m, A)))) * (math.sqrt(2.0) / -B_m)
	return tmp

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(F * Float64(A + C))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-103)
		tmp = Float64(sqrt(Float64(Float64(-8.0 * Float64(A * Float64(C * t_0))) + Float64(B_m * Float64(Float64(8.0 * Float64(A * Float64(C * F))) + Float64(B_m * Float64(Float64(-2.0 * Float64(B_m * F)) + Float64(2.0 * t_0))))))) / Float64(Float64(Float64(4.0 * A) * C) - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(B_m, A)))) * Float64(sqrt(2.0) / Float64(-B_m)));
	end
	return tmp
end

MATLAB

B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = F * (A + C);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e-103)
		tmp = sqrt(((-8.0 * (A * (C * t_0))) + (B_m * ((8.0 * (A * (C * F))) + (B_m * ((-2.0 * (B_m * F)) + (2.0 * t_0))))))) / (((4.0 * A) * C) - (B_m ^ 2.0));
	else
		tmp = sqrt((F * (A - hypot(B_m, A)))) * (sqrt(2.0) / -B_m);
	end
	tmp_2 = tmp;
end

Wolfram

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[(F * N[(A + C), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-103], N[(N[Sqrt[N[(N[(-8.0 * N[(A * N[(C * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(B$95$m * N[(N[(8.0 * N[(A * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(B$95$m * N[(N[(-2.0 * N[(B$95$m * F), $MachinePrecision]), $MachinePrecision] + N[(2.0 * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[B$95$m ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 1.99999999999999992e-103

   1. Initial program 25.0%

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

   3. Taylor expanded in B around inf 9.9%

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

   4. Taylor expanded in B around 0 9.2%

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

   if 1.99999999999999992e-103 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 20.2%

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

   3. Taylor expanded in C around 0 14.2%

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

      1. mul-1-neg 14.2%

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

      2. +-commutative 14.2%

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

      3. unpow2 14.2%

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

      4. unpow2 14.2%

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

      5. hypot-define 27.3%

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

   5. Simplified 27.3%

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

4. Final simplification 19.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-103}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + C\right)\right)\right)\right) + B \cdot \left(8 \cdot \left(A \cdot \left(C \cdot F\right)\right) + B \cdot \left(-2 \cdot \left(B \cdot F\right) + 2 \cdot \left(F \cdot \left(A + C\right)\right)\right)\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)} \cdot \frac{\sqrt{2}}{-B}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 26.9% accurate, 1.8× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := F \cdot \left(A + C\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-94}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot t\_0\right)\right) + B\_m \cdot \left(8 \cdot \left(A \cdot \left(C \cdot F\right)\right) + B\_m \cdot \left(-2 \cdot \left(B\_m \cdot F\right) + 2 \cdot t\_0\right)\right)}}{\left(4 \cdot A\right) \cdot C - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(A - B\_m\right)}\right)\\ \end{array} \end{array} \]

FPCore

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 (* F (+ A C))))
   (if (<= (pow B_m 2.0) 2e-94)
     (/
      (sqrt
       (+
        (* -8.0 (* A (* C t_0)))
        (*
         B_m
         (+
          (* 8.0 (* A (* C F)))
          (* B_m (+ (* -2.0 (* B_m F)) (* 2.0 t_0)))))))
      (- (* (* 4.0 A) C) (pow B_m 2.0)))
     (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (- A B_m))))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = F * (A + C);
	double tmp;
	if (pow(B_m, 2.0) <= 2e-94) {
		tmp = sqrt(((-8.0 * (A * (C * t_0))) + (B_m * ((8.0 * (A * (C * F))) + (B_m * ((-2.0 * (B_m * F)) + (2.0 * t_0))))))) / (((4.0 * A) * C) - pow(B_m, 2.0));
	} else {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - B_m)));
	}
	return tmp;
}

Fortran

B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: tmp
    t_0 = f * (a + c)
    if ((b_m ** 2.0d0) <= 2d-94) then
        tmp = sqrt((((-8.0d0) * (a * (c * t_0))) + (b_m * ((8.0d0 * (a * (c * f))) + (b_m * (((-2.0d0) * (b_m * f)) + (2.0d0 * t_0))))))) / (((4.0d0 * a) * c) - (b_m ** 2.0d0))
    else
        tmp = (sqrt(2.0d0) / b_m) * -sqrt((f * (a - b_m)))
    end if
    code = tmp
end function

Java

B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = F * (A + C);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e-94) {
		tmp = Math.sqrt(((-8.0 * (A * (C * t_0))) + (B_m * ((8.0 * (A * (C * F))) + (B_m * ((-2.0 * (B_m * F)) + (2.0 * t_0))))))) / (((4.0 * A) * C) - Math.pow(B_m, 2.0));
	} else {
		tmp = (Math.sqrt(2.0) / B_m) * -Math.sqrt((F * (A - B_m)));
	}
	return tmp;
}

Python

B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = F * (A + C)
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e-94:
		tmp = math.sqrt(((-8.0 * (A * (C * t_0))) + (B_m * ((8.0 * (A * (C * F))) + (B_m * ((-2.0 * (B_m * F)) + (2.0 * t_0))))))) / (((4.0 * A) * C) - math.pow(B_m, 2.0))
	else:
		tmp = (math.sqrt(2.0) / B_m) * -math.sqrt((F * (A - B_m)))
	return tmp

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(F * Float64(A + C))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-94)
		tmp = Float64(sqrt(Float64(Float64(-8.0 * Float64(A * Float64(C * t_0))) + Float64(B_m * Float64(Float64(8.0 * Float64(A * Float64(C * F))) + Float64(B_m * Float64(Float64(-2.0 * Float64(B_m * F)) + Float64(2.0 * t_0))))))) / Float64(Float64(Float64(4.0 * A) * C) - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(A - B_m)))));
	end
	return tmp
end

MATLAB

B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = F * (A + C);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e-94)
		tmp = sqrt(((-8.0 * (A * (C * t_0))) + (B_m * ((8.0 * (A * (C * F))) + (B_m * ((-2.0 * (B_m * F)) + (2.0 * t_0))))))) / (((4.0 * A) * C) - (B_m ^ 2.0));
	else
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - B_m)));
	end
	tmp_2 = tmp;
end

Wolfram

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[(F * N[(A + C), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-94], N[(N[Sqrt[N[(N[(-8.0 * N[(A * N[(C * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(B$95$m * N[(N[(8.0 * N[(A * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(B$95$m * N[(N[(-2.0 * N[(B$95$m * F), $MachinePrecision]), $MachinePrecision] + N[(2.0 * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 1.9999999999999999e-94

   1. Initial program 26.5%

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

   3. Taylor expanded in B around inf 9.8%

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

   4. Taylor expanded in B around 0 9.1%

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

   if 1.9999999999999999e-94 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 18.8%

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

   3. Taylor expanded in B around inf 8.7%

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

   4. Taylor expanded in C around 0 26.5%

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

      1. mul-1-neg 26.5%

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

   6. Simplified 26.5%

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

4. Final simplification 18.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-94}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + C\right)\right)\right)\right) + B \cdot \left(8 \cdot \left(A \cdot \left(C \cdot F\right)\right) + B \cdot \left(-2 \cdot \left(B \cdot F\right) + 2 \cdot \left(F \cdot \left(A + C\right)\right)\right)\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 27.0% accurate, 1.9× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-245}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + C\right)\right)\right)\right)}}{\left(4 \cdot A\right) \cdot C - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(A - B\_m\right)}\right)\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 5e-245)
   (/
    (sqrt (* -8.0 (* A (* C (* F (+ A C))))))
    (- (* (* 4.0 A) C) (pow B_m 2.0)))
   (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (- A B_m)))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 5e-245) {
		tmp = sqrt((-8.0 * (A * (C * (F * (A + C)))))) / (((4.0 * A) * C) - pow(B_m, 2.0));
	} else {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - B_m)));
	}
	return tmp;
}

Fortran

B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: tmp
    if ((b_m ** 2.0d0) <= 5d-245) then
        tmp = sqrt(((-8.0d0) * (a * (c * (f * (a + c)))))) / (((4.0d0 * a) * c) - (b_m ** 2.0d0))
    else
        tmp = (sqrt(2.0d0) / b_m) * -sqrt((f * (a - b_m)))
    end if
    code = tmp
end function

Java

B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double tmp;
	if (Math.pow(B_m, 2.0) <= 5e-245) {
		tmp = Math.sqrt((-8.0 * (A * (C * (F * (A + C)))))) / (((4.0 * A) * C) - Math.pow(B_m, 2.0));
	} else {
		tmp = (Math.sqrt(2.0) / B_m) * -Math.sqrt((F * (A - B_m)));
	}
	return tmp;
}

Python

B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	tmp = 0
	if math.pow(B_m, 2.0) <= 5e-245:
		tmp = math.sqrt((-8.0 * (A * (C * (F * (A + C)))))) / (((4.0 * A) * C) - math.pow(B_m, 2.0))
	else:
		tmp = (math.sqrt(2.0) / B_m) * -math.sqrt((F * (A - B_m)))
	return tmp

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-245)
		tmp = Float64(sqrt(Float64(-8.0 * Float64(A * Float64(C * Float64(F * Float64(A + C)))))) / Float64(Float64(Float64(4.0 * A) * C) - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(A - B_m)))));
	end
	return tmp
end

MATLAB

B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 5e-245)
		tmp = sqrt((-8.0 * (A * (C * (F * (A + C)))))) / (((4.0 * A) * C) - (B_m ^ 2.0));
	else
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - B_m)));
	end
	tmp_2 = tmp;
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-245], N[(N[Sqrt[N[(-8.0 * N[(A * N[(C * N[(F * N[(A + C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 4.9999999999999997e-245

   1. Initial program 22.6%

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

   3. Taylor expanded in B around inf 8.0%

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

   4. Taylor expanded in B around 0 7.8%

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

   if 4.9999999999999997e-245 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 22.3%

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

   3. Taylor expanded in B around inf 9.9%

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

   4. Taylor expanded in C around 0 23.4%

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

      1. mul-1-neg 23.4%

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

   6. Simplified 23.4%

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

4. Final simplification 17.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-245}:\\ \;\;\;\;\frac{\sqrt{-8 \cdot \left(A \cdot \left(C \cdot \left(F \cdot \left(A + C\right)\right)\right)\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 26.7% accurate, 3.0× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(A - B\_m\right)}\right) \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (- A B_m))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return (sqrt(2.0) / B_m) * -sqrt((F * (A - B_m)));
}

Fortran

B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = (sqrt(2.0d0) / b_m) * -sqrt((f * (a - b_m)))
end function

Java

B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return (Math.sqrt(2.0) / B_m) * -Math.sqrt((F * (A - B_m)));
}

Python

B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return (math.sqrt(2.0) / B_m) * -math.sqrt((F * (A - B_m)))

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(A - B_m)))))
end

MATLAB

B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = (sqrt(2.0) / B_m) * -sqrt((F * (A - B_m)));
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]

Derivation

1. Initial program 22.4%

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

3. Taylor expanded in B around inf 9.2%

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

4. Taylor expanded in C around 0 15.6%

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

   1. mul-1-neg 15.6%

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

6. Simplified 15.6%

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

7. Final simplification 15.6%

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

Alternative 10: 1.7% accurate, 3.1× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{2} \cdot {\left(\frac{F}{B\_m}\right)}^{0.5} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (* (sqrt 2.0) (pow (/ F B_m) 0.5)))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt(2.0) * pow((F / B_m), 0.5);
}

Fortran

B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = sqrt(2.0d0) * ((f / b_m) ** 0.5d0)
end function

Java

B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt(2.0) * Math.pow((F / B_m), 0.5);
}

Python

B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt(2.0) * math.pow((F / B_m), 0.5)

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(2.0) * (Float64(F / B_m) ^ 0.5))
end

MATLAB

B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt(2.0) * ((F / B_m) ^ 0.5);
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[2.0], $MachinePrecision] * N[Power[N[(F / B$95$m), $MachinePrecision], 0.5], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 22.4%

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

3. Taylor expanded in B around -inf 0.0%

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

   1. mul-1-neg 0.0%

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

   2. unpow2 0.0%

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

   3. rem-square-sqrt 1.9%

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

5. Simplified 1.9%

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

   1. pow1/2 2.0%

      \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \left(-1 \cdot \sqrt{2}\right) \]

7. Applied egg-rr 2.0%

   \[\leadsto -\color{blue}{{\left(\frac{F}{B}\right)}^{0.5}} \cdot \left(-1 \cdot \sqrt{2}\right) \]

8. Final simplification 2.0%

   \[\leadsto \sqrt{2} \cdot {\left(\frac{F}{B}\right)}^{0.5} \]
9. Add Preprocessing

Alternative 11: 1.5% accurate, 3.1× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{2} \cdot \sqrt{\frac{F}{B\_m}} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (* (sqrt 2.0) (sqrt (/ F B_m))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	return sqrt(2.0) * sqrt((F / B_m));
}

Fortran

B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = sqrt(2.0d0) * sqrt((f / b_m))
end function

Java

B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	return Math.sqrt(2.0) * Math.sqrt((F / B_m));
}

Python

B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	return math.sqrt(2.0) * math.sqrt((F / B_m))

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	return Float64(sqrt(2.0) * sqrt(Float64(F / B_m)))
end

MATLAB

B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp = code(A, B_m, C, F)
	tmp = sqrt(2.0) * sqrt((F / B_m));
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[2.0], $MachinePrecision] * N[Sqrt[N[(F / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 22.4%

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

3. Taylor expanded in B around -inf 0.0%

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

   1. mul-1-neg 0.0%

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

   2. unpow2 0.0%

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

   3. rem-square-sqrt 1.9%

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

5. Simplified 1.9%

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

6. Taylor expanded in F around 0 1.9%

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

   1. mul-1-neg 1.9%

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

   2. distribute-rgt-neg-out 1.9%

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

8. Simplified 1.9%

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

9. Final simplification 1.9%

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

Reproduce

herbie shell --seed 2024108 
(FPCore (A B C F)
  :name "ABCF->ab-angle b"
  :precision binary64
  (/ (- (sqrt (* (* 2.0 (* (- (pow B 2.0) (* (* 4.0 A) C)) F)) (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))) (- (pow B 2.0) (* (* 4.0 A) C))))