ABCF->ab-angle a

Percentage Accurate: 18.7% → 58.4%

Time: 22.1s

Alternatives: 12

Speedup: 6.0×

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: 18.7% 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: 58.4% accurate, 0.3× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ t_1 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{if}\;t\_1 \leq -2 \cdot 10^{-139}:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B\_m, A - C\right)\right)}{\mathsf{fma}\left(-4, A \cdot C, {B\_m}^{2}\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{elif}\;t\_1 \leq \infty:\\ \;\;\;\;\frac{\sqrt{F \cdot \mathsf{fma}\left(A, C \cdot -4, {B\_m}^{2}\right)} \cdot \sqrt{2 \cdot \left(0.5 \cdot \frac{{B\_m}^{2}}{C - A} + 2 \cdot C\right)}}{A \cdot \left(4 \cdot C\right) - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\sqrt{2}}{B\_m} \cdot \sqrt{C + \mathsf{hypot}\left(C, B\_m\right)}\right) \cdot \left(-\sqrt{F}\right)\\ \end{array} \end{array} \]

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 (* (* 4.0 A) C))
        (t_1
         (/
          (sqrt
           (*
            (* 2.0 (* (- (pow B_m 2.0) t_0) F))
            (+ (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
          (- t_0 (pow B_m 2.0)))))
   (if (<= t_1 -2e-139)
     (*
      (sqrt
       (*
        F
        (/ (+ A (+ C (hypot B_m (- A C)))) (fma -4.0 (* A C) (pow B_m 2.0)))))
      (- (sqrt 2.0)))
     (if (<= t_1 INFINITY)
       (/
        (*
         (sqrt (* F (fma A (* C -4.0) (pow B_m 2.0))))
         (sqrt (* 2.0 (+ (* 0.5 (/ (pow B_m 2.0) (- C A))) (* 2.0 C)))))
        (- (* A (* 4.0 C)) (pow B_m 2.0)))
       (* (* (/ (sqrt 2.0) B_m) (sqrt (+ C (hypot C B_m)))) (- (sqrt F)))))))

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 t_1 = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * ((A + C) + sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / (t_0 - pow(B_m, 2.0));
	double tmp;
	if (t_1 <= -2e-139) {
		tmp = sqrt((F * ((A + (C + hypot(B_m, (A - C)))) / fma(-4.0, (A * C), pow(B_m, 2.0))))) * -sqrt(2.0);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = (sqrt((F * fma(A, (C * -4.0), pow(B_m, 2.0)))) * sqrt((2.0 * ((0.5 * (pow(B_m, 2.0) / (C - A))) + (2.0 * C))))) / ((A * (4.0 * C)) - pow(B_m, 2.0));
	} else {
		tmp = ((sqrt(2.0) / B_m) * sqrt((C + hypot(C, B_m)))) * -sqrt(F);
	}
	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)
	t_1 = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(Float64(A + C) + sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / Float64(t_0 - (B_m ^ 2.0)))
	tmp = 0.0
	if (t_1 <= -2e-139)
		tmp = Float64(sqrt(Float64(F * Float64(Float64(A + Float64(C + hypot(B_m, Float64(A - C)))) / fma(-4.0, Float64(A * C), (B_m ^ 2.0))))) * Float64(-sqrt(2.0)));
	elseif (t_1 <= Inf)
		tmp = Float64(Float64(sqrt(Float64(F * fma(A, Float64(C * -4.0), (B_m ^ 2.0)))) * sqrt(Float64(2.0 * Float64(Float64(0.5 * Float64((B_m ^ 2.0) / Float64(C - A))) + Float64(2.0 * C))))) / Float64(Float64(A * Float64(4.0 * C)) - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(Float64(sqrt(2.0) / B_m) * sqrt(Float64(C + hypot(C, B_m)))) * Float64(-sqrt(F)));
	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[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$1 = 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[(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$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, -2e-139], N[(N[Sqrt[N[(F * N[(N[(A + N[(C + N[Sqrt[B$95$m ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(-4.0 * N[(A * C), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(N[(N[Sqrt[N[(F * N[(A * N[(C * -4.0), $MachinePrecision] + N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(2.0 * N[(N[(0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / N[(C - A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(2.0 * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(N[(A * N[(4.0 * C), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[Sqrt[N[(C + N[Sqrt[C ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * (-N[Sqrt[F], $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))) < -2.00000000000000006e-139

   1. Initial program 37.6%

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

   3. Taylor expanded in F around 0 47.9%

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

      1. mul-1-neg 47.9%

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

   5. Simplified 76.2%

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

   if -2.00000000000000006e-139 < (/.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 21.1%

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

   3. Simplified 25.9%

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

      1. sqrt-prod 31.3%

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

      2. *-commutative 31.3%

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

   6. Applied egg-rr 31.3%

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

   7. Taylor expanded in B around 0 23.1%

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

   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 A around 0 1.6%

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

      1. mul-1-neg 1.6%

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

      2. unpow2 1.6%

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

      3. unpow2 1.6%

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

      4. hypot-define 12.5%

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

   5. Simplified 12.5%

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

      1. pow1/2 12.6%

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

      2. *-commutative 12.6%

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

      3. hypot-undefine 1.7%

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

      4. unpow2 1.7%

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

      5. unpow2 1.7%

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

      6. unpow-prod-down 1.5%

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

   7. Applied egg-rr 23.1%

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

      1. associate-*r* 23.1%

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

   9. Applied egg-rr 23.1%

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

4. Final simplification 40.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -2 \cdot 10^{-139}:\\ \;\;\;\;\sqrt{F \cdot \frac{A + \left(C + \mathsf{hypot}\left(B, A - C\right)\right)}{\mathsf{fma}\left(-4, A \cdot C, {B}^{2}\right)}} \cdot \left(-\sqrt{2}\right)\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{F \cdot \mathsf{fma}\left(A, C \cdot -4, {B}^{2}\right)} \cdot \sqrt{2 \cdot \left(0.5 \cdot \frac{{B}^{2}}{C - A} + 2 \cdot C\right)}}{A \cdot \left(4 \cdot C\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\sqrt{2}}{B} \cdot \sqrt{C + \mathsf{hypot}\left(C, B\right)}\right) \cdot \left(-\sqrt{F}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 55.7% accurate, 1.2× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := A \cdot \left(4 \cdot C\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 10^{-125}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot \left(F \cdot \left(2 \cdot C + -0.5 \cdot \frac{{B\_m}^{2}}{A}\right)\right)\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\sqrt{2}}{B\_m} \cdot \sqrt{C + \mathsf{hypot}\left(C, B\_m\right)}\right) \cdot \left(-\sqrt{F}\right)\\ \end{array} \end{array} \]

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 (* A (* 4.0 C))))
   (if (<= (pow B_m 2.0) 1e-125)
     (/
      (sqrt
       (*
        2.0
        (*
         (- (pow B_m 2.0) t_0)
         (* F (+ (* 2.0 C) (* -0.5 (/ (pow B_m 2.0) A)))))))
      (- t_0 (pow B_m 2.0)))
     (* (* (/ (sqrt 2.0) B_m) (sqrt (+ C (hypot C B_m)))) (- (sqrt F))))))

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 = A * (4.0 * C);
	double tmp;
	if (pow(B_m, 2.0) <= 1e-125) {
		tmp = sqrt((2.0 * ((pow(B_m, 2.0) - t_0) * (F * ((2.0 * C) + (-0.5 * (pow(B_m, 2.0) / A))))))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = ((sqrt(2.0) / B_m) * sqrt((C + hypot(C, B_m)))) * -sqrt(F);
	}
	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 = A * (4.0 * C);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 1e-125) {
		tmp = Math.sqrt((2.0 * ((Math.pow(B_m, 2.0) - t_0) * (F * ((2.0 * C) + (-0.5 * (Math.pow(B_m, 2.0) / A))))))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = ((Math.sqrt(2.0) / B_m) * Math.sqrt((C + Math.hypot(C, B_m)))) * -Math.sqrt(F);
	}
	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 = A * (4.0 * C)
	tmp = 0
	if math.pow(B_m, 2.0) <= 1e-125:
		tmp = math.sqrt((2.0 * ((math.pow(B_m, 2.0) - t_0) * (F * ((2.0 * C) + (-0.5 * (math.pow(B_m, 2.0) / A))))))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = ((math.sqrt(2.0) / B_m) * math.sqrt((C + math.hypot(C, B_m)))) * -math.sqrt(F)
	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(A * Float64(4.0 * C))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 1e-125)
		tmp = Float64(sqrt(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * Float64(F * Float64(Float64(2.0 * C) + Float64(-0.5 * Float64((B_m ^ 2.0) / A))))))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(Float64(sqrt(2.0) / B_m) * sqrt(Float64(C + hypot(C, B_m)))) * Float64(-sqrt(F)));
	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 = A * (4.0 * C);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 1e-125)
		tmp = sqrt((2.0 * (((B_m ^ 2.0) - t_0) * (F * ((2.0 * C) + (-0.5 * ((B_m ^ 2.0) / A))))))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = ((sqrt(2.0) / B_m) * sqrt((C + hypot(C, B_m)))) * -sqrt(F);
	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[(A * N[(4.0 * C), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-125], N[(N[Sqrt[N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * N[(F * N[(N[(2.0 * C), $MachinePrecision] + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[Sqrt[N[(C + N[Sqrt[C ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * (-N[Sqrt[F], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 1.00000000000000001e-125

   1. Initial program 15.8%

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

      1. *-un-lft-identity 15.8%

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

   4. Applied egg-rr 28.6%

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

      1. *-lft-identity 28.6%

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

      2. distribute-frac-neg 28.6%

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

      3. distribute-frac-neg2 28.6%

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

   6. Simplified 27.7%

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

   7. Taylor expanded in A around -inf 17.1%

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

   if 1.00000000000000001e-125 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 18.3%

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

   3. Taylor expanded in A around 0 11.3%

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

      1. mul-1-neg 11.3%

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

      2. unpow2 11.3%

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

      3. unpow2 11.3%

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

      4. hypot-define 19.0%

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

   5. Simplified 19.0%

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

      1. pow1/2 19.1%

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

      2. *-commutative 19.1%

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

      3. hypot-undefine 11.4%

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

      4. unpow2 11.4%

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

      5. unpow2 11.4%

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

      6. unpow-prod-down 14.5%

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

   7. Applied egg-rr 30.4%

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

      1. associate-*r* 30.5%

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

   9. Applied egg-rr 30.5%

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

4. Final simplification 25.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 10^{-125}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(\left({B}^{2} - A \cdot \left(4 \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot C + -0.5 \cdot \frac{{B}^{2}}{A}\right)\right)\right)}}{A \cdot \left(4 \cdot C\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\sqrt{2}}{B} \cdot \sqrt{C + \mathsf{hypot}\left(C, B\right)}\right) \cdot \left(-\sqrt{F}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 55.6% accurate, 1.2× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ \mathbf{if}\;{B\_m}^{2} \leq 10^{-125}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot C + -0.5 \cdot \frac{{B\_m}^{2}}{A}\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\sqrt{2}}{B\_m} \cdot \sqrt{C + \mathsf{hypot}\left(C, B\_m\right)}\right) \cdot \left(-\sqrt{F}\right)\\ \end{array} \end{array} \]

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 (* (* 4.0 A) C)))
   (if (<= (pow B_m 2.0) 1e-125)
     (/
      (sqrt
       (*
        (* 2.0 (* (- (pow B_m 2.0) t_0) F))
        (+ (* 2.0 C) (* -0.5 (/ (pow B_m 2.0) A)))))
      (- t_0 (pow B_m 2.0)))
     (* (* (/ (sqrt 2.0) B_m) (sqrt (+ C (hypot C B_m)))) (- (sqrt F))))))

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) <= 1e-125) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * ((2.0 * C) + (-0.5 * (pow(B_m, 2.0) / A))))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = ((sqrt(2.0) / B_m) * sqrt((C + hypot(C, B_m)))) * -sqrt(F);
	}
	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) <= 1e-125) {
		tmp = Math.sqrt(((2.0 * ((Math.pow(B_m, 2.0) - t_0) * F)) * ((2.0 * C) + (-0.5 * (Math.pow(B_m, 2.0) / A))))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = ((Math.sqrt(2.0) / B_m) * Math.sqrt((C + Math.hypot(C, B_m)))) * -Math.sqrt(F);
	}
	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) <= 1e-125:
		tmp = math.sqrt(((2.0 * ((math.pow(B_m, 2.0) - t_0) * F)) * ((2.0 * C) + (-0.5 * (math.pow(B_m, 2.0) / A))))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = ((math.sqrt(2.0) / B_m) * math.sqrt((C + math.hypot(C, B_m)))) * -math.sqrt(F)
	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) <= 1e-125)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(Float64(2.0 * C) + Float64(-0.5 * Float64((B_m ^ 2.0) / A))))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(Float64(sqrt(2.0) / B_m) * sqrt(Float64(C + hypot(C, B_m)))) * Float64(-sqrt(F)));
	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) <= 1e-125)
		tmp = sqrt(((2.0 * (((B_m ^ 2.0) - t_0) * F)) * ((2.0 * C) + (-0.5 * ((B_m ^ 2.0) / A))))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = ((sqrt(2.0) / B_m) * sqrt((C + hypot(C, B_m)))) * -sqrt(F);
	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], 1e-125], 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[(N[(2.0 * C), $MachinePrecision] + N[(-0.5 * N[(N[Power[B$95$m, 2.0], $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[Sqrt[N[(C + N[Sqrt[C ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * (-N[Sqrt[F], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 1.00000000000000001e-125

   1. Initial program 15.8%

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

   3. Taylor expanded in A around -inf 17.4%

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

   if 1.00000000000000001e-125 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 18.3%

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

   3. Taylor expanded in A around 0 11.3%

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

      1. mul-1-neg 11.3%

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

      2. unpow2 11.3%

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

      3. unpow2 11.3%

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

      4. hypot-define 19.0%

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

   5. Simplified 19.0%

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

      1. pow1/2 19.1%

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

      2. *-commutative 19.1%

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

      3. hypot-undefine 11.4%

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

      4. unpow2 11.4%

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

      5. unpow2 11.4%

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

      6. unpow-prod-down 14.5%

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

   7. Applied egg-rr 30.4%

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

      1. associate-*r* 30.5%

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

   9. Applied egg-rr 30.5%

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

4. Final simplification 25.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 10^{-125}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot C + -0.5 \cdot \frac{{B}^{2}}{A}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\sqrt{2}}{B} \cdot \sqrt{C + \mathsf{hypot}\left(C, B\right)}\right) \cdot \left(-\sqrt{F}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 44.8% accurate, 1.2× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 4 \cdot 10^{-150}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(-8 \cdot \left(A \cdot \left(F \cdot {C}^{2}\right)\right)\right)}}{A \cdot \left(4 \cdot C\right) - {B\_m}^{2}}\\ \mathbf{elif}\;{B\_m}^{2} \leq 5 \cdot 10^{+223}:\\ \;\;\;\;\frac{\sqrt{\left(C + \mathsf{hypot}\left(C, B\_m\right)\right) \cdot \left(2 \cdot F\right)}}{-B\_m}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B\_m + C}\right)\right)\\ \end{array} \end{array} \]

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) 4e-150)
   (/
    (sqrt (* 2.0 (* -8.0 (* A (* F (pow C 2.0))))))
    (- (* A (* 4.0 C)) (pow B_m 2.0)))
   (if (<= (pow B_m 2.0) 5e+223)
     (/ (sqrt (* (+ C (hypot C B_m)) (* 2.0 F))) (- B_m))
     (* (/ (sqrt 2.0) B_m) (* (sqrt F) (- (sqrt (+ B_m C))))))))

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) <= 4e-150) {
		tmp = sqrt((2.0 * (-8.0 * (A * (F * pow(C, 2.0)))))) / ((A * (4.0 * C)) - pow(B_m, 2.0));
	} else if (pow(B_m, 2.0) <= 5e+223) {
		tmp = sqrt(((C + hypot(C, B_m)) * (2.0 * F))) / -B_m;
	} else {
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	}
	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 tmp;
	if (Math.pow(B_m, 2.0) <= 4e-150) {
		tmp = Math.sqrt((2.0 * (-8.0 * (A * (F * Math.pow(C, 2.0)))))) / ((A * (4.0 * C)) - Math.pow(B_m, 2.0));
	} else if (Math.pow(B_m, 2.0) <= 5e+223) {
		tmp = Math.sqrt(((C + Math.hypot(C, B_m)) * (2.0 * F))) / -B_m;
	} else {
		tmp = (Math.sqrt(2.0) / B_m) * (Math.sqrt(F) * -Math.sqrt((B_m + C)));
	}
	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) <= 4e-150:
		tmp = math.sqrt((2.0 * (-8.0 * (A * (F * math.pow(C, 2.0)))))) / ((A * (4.0 * C)) - math.pow(B_m, 2.0))
	elif math.pow(B_m, 2.0) <= 5e+223:
		tmp = math.sqrt(((C + math.hypot(C, B_m)) * (2.0 * F))) / -B_m
	else:
		tmp = (math.sqrt(2.0) / B_m) * (math.sqrt(F) * -math.sqrt((B_m + C)))
	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) <= 4e-150)
		tmp = Float64(sqrt(Float64(2.0 * Float64(-8.0 * Float64(A * Float64(F * (C ^ 2.0)))))) / Float64(Float64(A * Float64(4.0 * C)) - (B_m ^ 2.0)));
	elseif ((B_m ^ 2.0) <= 5e+223)
		tmp = Float64(sqrt(Float64(Float64(C + hypot(C, B_m)) * Float64(2.0 * F))) / Float64(-B_m));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(sqrt(F) * Float64(-sqrt(Float64(B_m + C)))));
	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) <= 4e-150)
		tmp = sqrt((2.0 * (-8.0 * (A * (F * (C ^ 2.0)))))) / ((A * (4.0 * C)) - (B_m ^ 2.0));
	elseif ((B_m ^ 2.0) <= 5e+223)
		tmp = sqrt(((C + hypot(C, B_m)) * (2.0 * F))) / -B_m;
	else
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	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], 4e-150], N[(N[Sqrt[N[(2.0 * N[(-8.0 * N[(A * N[(F * N[Power[C, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(A * N[(4.0 * C), $MachinePrecision]), $MachinePrecision] - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e+223], N[(N[Sqrt[N[(N[(C + N[Sqrt[C ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision] * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-B$95$m)), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[(N[Sqrt[F], $MachinePrecision] * (-N[Sqrt[N[(B$95$m + C), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 4.00000000000000003e-150

   1. Initial program 16.2%

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

      1. *-un-lft-identity 16.2%

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

   4. Applied egg-rr 28.3%

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

      1. *-lft-identity 28.3%

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

      2. distribute-frac-neg 28.3%

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

      3. distribute-frac-neg2 28.3%

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

   6. Simplified 27.5%

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

   7. Taylor expanded in A around -inf 11.3%

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

   if 4.00000000000000003e-150 < (pow.f64 B #s(literal 2 binary64)) < 4.99999999999999985e223

   1. Initial program 39.5%

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

   3. Taylor expanded in A around 0 19.3%

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

      1. mul-1-neg 19.3%

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

      2. unpow2 19.3%

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

      3. unpow2 19.3%

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

      4. hypot-define 19.7%

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

   5. Simplified 19.7%

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

      1. associate-*l/ 19.7%

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

      2. pow1/2 19.7%

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

      3. pow1/2 19.9%

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

      4. hypot-undefine 19.6%

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

      5. unpow2 19.6%

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

      6. unpow2 19.6%

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

      7. pow-prod-down 19.6%

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

      8. +-commutative 19.6%

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

      9. unpow2 19.6%

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

      10. unpow2 19.6%

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

      11. hypot-define 20.0%

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

   7. Applied egg-rr 20.0%

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

      1. unpow1/2 19.8%

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

      2. associate-*r* 19.8%

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

   9. Simplified 19.8%

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

   if 4.99999999999999985e223 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 1.8%

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

   3. Taylor expanded in A around 0 5.0%

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

      1. mul-1-neg 5.0%

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

      2. unpow2 5.0%

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

      3. unpow2 5.0%

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

      4. hypot-define 17.9%

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

   5. Simplified 17.9%

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

      1. pow1/2 17.9%

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

      2. *-commutative 17.9%

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

      3. hypot-undefine 5.0%

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

      4. unpow2 5.0%

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

      5. unpow2 5.0%

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

      6. unpow-prod-down 9.3%

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

   7. Applied egg-rr 36.4%

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

   8. Taylor expanded in C around 0 32.9%

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

4. Final simplification 21.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 4 \cdot 10^{-150}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(-8 \cdot \left(A \cdot \left(F \cdot {C}^{2}\right)\right)\right)}}{A \cdot \left(4 \cdot C\right) - {B}^{2}}\\ \mathbf{elif}\;{B}^{2} \leq 5 \cdot 10^{+223}:\\ \;\;\;\;\frac{\sqrt{\left(C + \mathsf{hypot}\left(C, B\right)\right) \cdot \left(2 \cdot F\right)}}{-B}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B + C}\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 56.1% accurate, 1.2× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-102}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot C\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\sqrt{2}}{B\_m} \cdot \sqrt{C + \mathsf{hypot}\left(C, B\_m\right)}\right) \cdot \left(-\sqrt{F}\right)\\ \end{array} \end{array} \]

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 (* (* 4.0 A) C)))
   (if (<= (pow B_m 2.0) 5e-102)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 C)))
      (- t_0 (pow B_m 2.0)))
     (* (* (/ (sqrt 2.0) B_m) (sqrt (+ C (hypot C B_m)))) (- (sqrt F))))))

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) <= 5e-102) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * C))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = ((sqrt(2.0) / B_m) * sqrt((C + hypot(C, B_m)))) * -sqrt(F);
	}
	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) <= 5e-102) {
		tmp = Math.sqrt(((2.0 * ((Math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * C))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = ((Math.sqrt(2.0) / B_m) * Math.sqrt((C + Math.hypot(C, B_m)))) * -Math.sqrt(F);
	}
	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) <= 5e-102:
		tmp = math.sqrt(((2.0 * ((math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * C))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = ((math.sqrt(2.0) / B_m) * math.sqrt((C + math.hypot(C, B_m)))) * -math.sqrt(F)
	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) <= 5e-102)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * C))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(Float64(sqrt(2.0) / B_m) * sqrt(Float64(C + hypot(C, B_m)))) * Float64(-sqrt(F)));
	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) <= 5e-102)
		tmp = sqrt(((2.0 * (((B_m ^ 2.0) - t_0) * F)) * (2.0 * C))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = ((sqrt(2.0) / B_m) * sqrt((C + hypot(C, B_m)))) * -sqrt(F);
	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], 5e-102], 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 * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[Sqrt[N[(C + N[Sqrt[C ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * (-N[Sqrt[F], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 5.00000000000000026e-102

   1. Initial program 16.0%

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

   3. Taylor expanded in A around -inf 17.5%

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

   if 5.00000000000000026e-102 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 18.2%

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

   3. Taylor expanded in A around 0 11.0%

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

      1. mul-1-neg 11.0%

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

      2. unpow2 11.0%

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

      3. unpow2 11.0%

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

      4. hypot-define 18.9%

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

   5. Simplified 18.9%

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

      1. pow1/2 19.0%

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

      2. *-commutative 19.0%

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

      3. hypot-undefine 11.1%

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

      4. unpow2 11.1%

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

      5. unpow2 11.1%

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

      6. unpow-prod-down 14.2%

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

   7. Applied egg-rr 30.7%

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

      1. associate-*r* 30.7%

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

   9. Applied egg-rr 30.7%

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

4. Final simplification 25.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-102}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot C\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\sqrt{2}}{B} \cdot \sqrt{C + \mathsf{hypot}\left(C, B\right)}\right) \cdot \left(-\sqrt{F}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 50.7% accurate, 1.5× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-102}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(2 \cdot C\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B\_m + C}\right)\right)\\ \end{array} \end{array} \]

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 (* (* 4.0 A) C)))
   (if (<= (pow B_m 2.0) 5e-102)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (* 2.0 C)))
      (- t_0 (pow B_m 2.0)))
     (* (/ (sqrt 2.0) B_m) (* (sqrt F) (- (sqrt (+ B_m C))))))))

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) <= 5e-102) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (2.0 * C))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	}
	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 = (4.0d0 * a) * c
    if ((b_m ** 2.0d0) <= 5d-102) then
        tmp = sqrt(((2.0d0 * (((b_m ** 2.0d0) - t_0) * f)) * (2.0d0 * c))) / (t_0 - (b_m ** 2.0d0))
    else
        tmp = (sqrt(2.0d0) / b_m) * (sqrt(f) * -sqrt((b_m + c)))
    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 = (4.0 * A) * C;
	double tmp;
	if (Math.pow(B_m, 2.0) <= 5e-102) {
		tmp = Math.sqrt(((2.0 * ((Math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * C))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = (Math.sqrt(2.0) / B_m) * (Math.sqrt(F) * -Math.sqrt((B_m + C)));
	}
	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) <= 5e-102:
		tmp = math.sqrt(((2.0 * ((math.pow(B_m, 2.0) - t_0) * F)) * (2.0 * C))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = (math.sqrt(2.0) / B_m) * (math.sqrt(F) * -math.sqrt((B_m + C)))
	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) <= 5e-102)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(2.0 * C))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(sqrt(F) * Float64(-sqrt(Float64(B_m + C)))));
	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) <= 5e-102)
		tmp = sqrt(((2.0 * (((B_m ^ 2.0) - t_0) * F)) * (2.0 * C))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	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], 5e-102], 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 * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[(N[Sqrt[F], $MachinePrecision] * (-N[Sqrt[N[(B$95$m + C), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 5.00000000000000026e-102

   1. Initial program 16.0%

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

   3. Taylor expanded in A around -inf 17.5%

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

   if 5.00000000000000026e-102 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 18.2%

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

   3. Taylor expanded in A around 0 11.0%

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

      1. mul-1-neg 11.0%

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

      2. unpow2 11.0%

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

      3. unpow2 11.0%

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

      4. hypot-define 18.9%

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

   5. Simplified 18.9%

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

      1. pow1/2 19.0%

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

      2. *-commutative 19.0%

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

      3. hypot-undefine 11.1%

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

      4. unpow2 11.1%

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

      5. unpow2 11.1%

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

      6. unpow-prod-down 14.2%

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

   7. Applied egg-rr 30.7%

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

   8. Taylor expanded in C around 0 26.9%

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

4. Final simplification 23.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-102}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(2 \cdot C\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B + C}\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 38.5% accurate, 2.0× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;F \leq 1.75 \cdot 10^{+21}:\\ \;\;\;\;\frac{\sqrt{\left(C + \mathsf{hypot}\left(C, B\_m\right)\right) \cdot \left(2 \cdot F\right)}}{-B\_m}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B\_m + C}\right)\right)\\ \end{array} \end{array} \]

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 (<= F 1.75e+21)
   (/ (sqrt (* (+ C (hypot C B_m)) (* 2.0 F))) (- B_m))
   (* (/ (sqrt 2.0) B_m) (* (sqrt F) (- (sqrt (+ B_m C)))))))

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 (F <= 1.75e+21) {
		tmp = sqrt(((C + hypot(C, B_m)) * (2.0 * F))) / -B_m;
	} else {
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	}
	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 tmp;
	if (F <= 1.75e+21) {
		tmp = Math.sqrt(((C + Math.hypot(C, B_m)) * (2.0 * F))) / -B_m;
	} else {
		tmp = (Math.sqrt(2.0) / B_m) * (Math.sqrt(F) * -Math.sqrt((B_m + C)));
	}
	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 F <= 1.75e+21:
		tmp = math.sqrt(((C + math.hypot(C, B_m)) * (2.0 * F))) / -B_m
	else:
		tmp = (math.sqrt(2.0) / B_m) * (math.sqrt(F) * -math.sqrt((B_m + C)))
	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 (F <= 1.75e+21)
		tmp = Float64(sqrt(Float64(Float64(C + hypot(C, B_m)) * Float64(2.0 * F))) / Float64(-B_m));
	else
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(sqrt(F) * Float64(-sqrt(Float64(B_m + C)))));
	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 (F <= 1.75e+21)
		tmp = sqrt(((C + hypot(C, B_m)) * (2.0 * F))) / -B_m;
	else
		tmp = (sqrt(2.0) / B_m) * (sqrt(F) * -sqrt((B_m + C)));
	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[F, 1.75e+21], N[(N[Sqrt[N[(N[(C + N[Sqrt[C ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision] * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-B$95$m)), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * N[(N[Sqrt[F], $MachinePrecision] * (-N[Sqrt[N[(B$95$m + C), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if F < 1.75e21

   1. Initial program 20.1%

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

   3. Taylor expanded in A around 0 8.0%

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

      1. mul-1-neg 8.0%

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

      2. unpow2 8.0%

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

      3. unpow2 8.0%

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

      4. hypot-define 16.5%

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

   5. Simplified 16.5%

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

      1. associate-*l/ 16.5%

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

      2. pow1/2 16.5%

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

      3. pow1/2 16.7%

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

      4. hypot-undefine 8.3%

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

      5. unpow2 8.3%

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

      6. unpow2 8.3%

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

      7. pow-prod-down 8.3%

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

      8. +-commutative 8.3%

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

      9. unpow2 8.3%

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

      10. unpow2 8.3%

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

      11. hypot-define 16.8%

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

   7. Applied egg-rr 16.8%

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

      1. unpow1/2 16.6%

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

      2. associate-*r* 16.6%

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

   9. Simplified 16.6%

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

   if 1.75e21 < F

   1. Initial program 13.8%

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

   3. Taylor expanded in A around 0 7.4%

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

      1. mul-1-neg 7.4%

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

      2. unpow2 7.4%

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

      3. unpow2 7.4%

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

      4. hypot-define 7.7%

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

   5. Simplified 7.7%

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

      1. pow1/2 7.7%

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

      2. *-commutative 7.7%

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

      3. hypot-undefine 7.4%

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

      4. unpow2 7.4%

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

      5. unpow2 7.4%

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

      6. unpow-prod-down 11.7%

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

   7. Applied egg-rr 23.2%

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

   8. Taylor expanded in C around 0 19.6%

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

4. Final simplification 17.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;F \leq 1.75 \cdot 10^{+21}:\\ \;\;\;\;\frac{\sqrt{\left(C + \mathsf{hypot}\left(C, B\right)\right) \cdot \left(2 \cdot F\right)}}{-B}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(\sqrt{F} \cdot \left(-\sqrt{B + C}\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 37.7% accurate, 2.9× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;F \leq 1.15 \cdot 10^{+22}:\\ \;\;\;\;\frac{\sqrt{\left(C + \mathsf{hypot}\left(C, B\_m\right)\right) \cdot \left(2 \cdot F\right)}}{-B\_m}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{2 \cdot \frac{F}{B\_m}}\\ \end{array} \end{array} \]

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 (<= F 1.15e+22)
   (/ (sqrt (* (+ C (hypot C B_m)) (* 2.0 F))) (- B_m))
   (- (sqrt (* 2.0 (/ 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) {
	double tmp;
	if (F <= 1.15e+22) {
		tmp = sqrt(((C + hypot(C, B_m)) * (2.0 * F))) / -B_m;
	} else {
		tmp = -sqrt((2.0 * (F / 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 tmp;
	if (F <= 1.15e+22) {
		tmp = Math.sqrt(((C + Math.hypot(C, B_m)) * (2.0 * F))) / -B_m;
	} else {
		tmp = -Math.sqrt((2.0 * (F / 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 F <= 1.15e+22:
		tmp = math.sqrt(((C + math.hypot(C, B_m)) * (2.0 * F))) / -B_m
	else:
		tmp = -math.sqrt((2.0 * (F / 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 (F <= 1.15e+22)
		tmp = Float64(sqrt(Float64(Float64(C + hypot(C, B_m)) * Float64(2.0 * F))) / Float64(-B_m));
	else
		tmp = Float64(-sqrt(Float64(2.0 * Float64(F / 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 (F <= 1.15e+22)
		tmp = sqrt(((C + hypot(C, B_m)) * (2.0 * F))) / -B_m;
	else
		tmp = -sqrt((2.0 * (F / 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[F, 1.15e+22], N[(N[Sqrt[N[(N[(C + N[Sqrt[C ^ 2 + B$95$m ^ 2], $MachinePrecision]), $MachinePrecision] * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-B$95$m)), $MachinePrecision], (-N[Sqrt[N[(2.0 * N[(F / B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])]

Derivation

1. Split input into 2 regimes

2. if F < 1.1500000000000001e22

   1. Initial program 20.7%

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

   3. Taylor expanded in A around 0 8.0%

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

      1. mul-1-neg 8.0%

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

      2. unpow2 8.0%

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

      3. unpow2 8.0%

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

      4. hypot-define 16.5%

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

   5. Simplified 16.5%

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

      1. associate-*l/ 16.5%

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

      2. pow1/2 16.5%

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

      3. pow1/2 16.7%

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

      4. hypot-undefine 8.3%

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

      5. unpow2 8.3%

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

      6. unpow2 8.3%

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

      7. pow-prod-down 8.3%

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

      8. +-commutative 8.3%

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

      9. unpow2 8.3%

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

      10. unpow2 8.3%

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

      11. hypot-define 16.7%

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

   7. Applied egg-rr 16.7%

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

      1. unpow1/2 16.5%

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

      2. associate-*r* 16.5%

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

   9. Simplified 16.5%

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

   if 1.1500000000000001e22 < F

   1. Initial program 13.0%

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

   3. Taylor expanded in B around inf 19.9%

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

      1. mul-1-neg 19.9%

         \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]

   5. Simplified 19.9%

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

      1. *-un-lft-identity 19.9%

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

      2. sqrt-unprod 20.1%

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

   7. Applied egg-rr 20.1%

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

      1. *-lft-identity 20.1%

         \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]

   9. Simplified 20.1%

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

4. Final simplification 18.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;F \leq 1.15 \cdot 10^{+22}:\\ \;\;\;\;\frac{\sqrt{\left(C + \mathsf{hypot}\left(C, B\right)\right) \cdot \left(2 \cdot F\right)}}{-B}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{2 \cdot \frac{F}{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 34.3% accurate, 2.9× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;F \leq 3400000:\\ \;\;\;\;\frac{\sqrt{2}}{B\_m} \cdot \left(-\sqrt{F \cdot \left(B\_m + C\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{2 \cdot \frac{F}{B\_m}}\\ \end{array} \end{array} \]

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 (<= F 3400000.0)
   (* (/ (sqrt 2.0) B_m) (- (sqrt (* F (+ B_m C)))))
   (- (sqrt (* 2.0 (/ 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) {
	double tmp;
	if (F <= 3400000.0) {
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (B_m + C)));
	} else {
		tmp = -sqrt((2.0 * (F / 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 (f <= 3400000.0d0) then
        tmp = (sqrt(2.0d0) / b_m) * -sqrt((f * (b_m + c)))
    else
        tmp = -sqrt((2.0d0 * (f / 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 (F <= 3400000.0) {
		tmp = (Math.sqrt(2.0) / B_m) * -Math.sqrt((F * (B_m + C)));
	} else {
		tmp = -Math.sqrt((2.0 * (F / 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 F <= 3400000.0:
		tmp = (math.sqrt(2.0) / B_m) * -math.sqrt((F * (B_m + C)))
	else:
		tmp = -math.sqrt((2.0 * (F / 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 (F <= 3400000.0)
		tmp = Float64(Float64(sqrt(2.0) / B_m) * Float64(-sqrt(Float64(F * Float64(B_m + C)))));
	else
		tmp = Float64(-sqrt(Float64(2.0 * Float64(F / 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 (F <= 3400000.0)
		tmp = (sqrt(2.0) / B_m) * -sqrt((F * (B_m + C)));
	else
		tmp = -sqrt((2.0 * (F / 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[F, 3400000.0], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B$95$m), $MachinePrecision] * (-N[Sqrt[N[(F * N[(B$95$m + C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision], (-N[Sqrt[N[(2.0 * N[(F / B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])]

Derivation

1. Split input into 2 regimes

2. if F < 3.4e6

   1. Initial program 20.9%

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

   3. Taylor expanded in A around 0 7.6%

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

      1. mul-1-neg 7.6%

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

      2. unpow2 7.6%

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

      3. unpow2 7.6%

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

      4. hypot-define 16.5%

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

   5. Simplified 16.5%

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

   6. Taylor expanded in C around 0 13.7%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{B \cdot F + C \cdot F}} \]
   7. Step-by-step derivation

      1. distribute-rgt-out 13.7%

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

   8. Simplified 13.7%

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

   if 3.4e6 < F

   1. Initial program 13.2%

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

   3. Taylor expanded in B around inf 19.7%

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

      1. mul-1-neg 19.7%

         \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]

   5. Simplified 19.7%

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

      1. *-un-lft-identity 19.7%

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

      2. sqrt-unprod 19.8%

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

   7. Applied egg-rr 19.8%

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

      1. *-lft-identity 19.8%

         \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]

   9. Simplified 19.8%

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

4. Final simplification 16.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;F \leq 3400000:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(B + C\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{2 \cdot \frac{F}{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 33.8% accurate, 3.0× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;F \leq 6.4 \cdot 10^{-63}:\\ \;\;\;\;\sqrt{B\_m \cdot F} \cdot \frac{\sqrt{2}}{-B\_m}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{2 \cdot \frac{F}{B\_m}}\\ \end{array} \end{array} \]

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 (<= F 6.4e-63)
   (* (sqrt (* B_m F)) (/ (sqrt 2.0) (- B_m)))
   (- (sqrt (* 2.0 (/ 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) {
	double tmp;
	if (F <= 6.4e-63) {
		tmp = sqrt((B_m * F)) * (sqrt(2.0) / -B_m);
	} else {
		tmp = -sqrt((2.0 * (F / 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 (f <= 6.4d-63) then
        tmp = sqrt((b_m * f)) * (sqrt(2.0d0) / -b_m)
    else
        tmp = -sqrt((2.0d0 * (f / 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 (F <= 6.4e-63) {
		tmp = Math.sqrt((B_m * F)) * (Math.sqrt(2.0) / -B_m);
	} else {
		tmp = -Math.sqrt((2.0 * (F / 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 F <= 6.4e-63:
		tmp = math.sqrt((B_m * F)) * (math.sqrt(2.0) / -B_m)
	else:
		tmp = -math.sqrt((2.0 * (F / 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 (F <= 6.4e-63)
		tmp = Float64(sqrt(Float64(B_m * F)) * Float64(sqrt(2.0) / Float64(-B_m)));
	else
		tmp = Float64(-sqrt(Float64(2.0 * Float64(F / 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 (F <= 6.4e-63)
		tmp = sqrt((B_m * F)) * (sqrt(2.0) / -B_m);
	else
		tmp = -sqrt((2.0 * (F / 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[F, 6.4e-63], N[(N[Sqrt[N[(B$95$m * F), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / (-B$95$m)), $MachinePrecision]), $MachinePrecision], (-N[Sqrt[N[(2.0 * N[(F / B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])]

Derivation

1. Split input into 2 regimes

2. if F < 6.39999999999999978e-63

   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 A around 0 7.9%

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

      1. mul-1-neg 7.9%

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

      2. unpow2 7.9%

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

      3. unpow2 7.9%

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

      4. hypot-define 16.5%

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

   5. Simplified 16.5%

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

   6. Taylor expanded in C around 0 15.0%

      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \color{blue}{\sqrt{B \cdot F}} \]

   if 6.39999999999999978e-63 < F

   1. Initial program 14.9%

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

   3. Taylor expanded in B around inf 18.9%

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

      1. mul-1-neg 18.9%

         \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]

   5. Simplified 18.9%

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

      1. *-un-lft-identity 18.9%

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

      2. sqrt-unprod 19.0%

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

   7. Applied egg-rr 19.0%

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

      1. *-lft-identity 19.0%

         \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]

   9. Simplified 19.0%

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

4. Final simplification 17.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;F \leq 6.4 \cdot 10^{-63}:\\ \;\;\;\;\sqrt{B \cdot F} \cdot \frac{\sqrt{2}}{-B}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{2 \cdot \frac{F}{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 26.7% accurate, 5.9× speedup

Math

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

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 (- (pow (* 2.0 (/ 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 -pow((2.0 * (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 = -((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.pow((2.0 * (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.pow((2.0 * (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(-(Float64(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 = -((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[Power[N[(2.0 * N[(F / B$95$m), $MachinePrecision]), $MachinePrecision], 0.5], $MachinePrecision])

Derivation

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 B around inf 13.0%

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

   1. mul-1-neg 13.0%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]

5. Simplified 13.0%

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

   1. sqrt-unprod 13.0%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]

   2. pow1/2 13.1%

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

7. Applied egg-rr 13.1%

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

8. Final simplification 13.1%

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

Alternative 12: 26.7% accurate, 6.0× speedup

Math

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

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 (/ 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 * (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 * (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 * (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 * (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(Float64(2.0 * 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 * (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[Sqrt[N[(2.0 * N[(F / B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])

Derivation

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 B around inf 13.0%

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

   1. mul-1-neg 13.0%

      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \sqrt{2}} \]

5. Simplified 13.0%

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

   1. *-un-lft-identity 13.0%

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

   2. sqrt-unprod 13.0%

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

7. Applied egg-rr 13.0%

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

   1. *-lft-identity 13.0%

      \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]

9. Simplified 13.0%

   \[\leadsto -\color{blue}{\sqrt{\frac{F}{B} \cdot 2}} \]

10. Final simplification 13.0%

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

Reproduce

herbie shell --seed 2024096 
(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))))