2-ancestry mixing, positive discriminant

Percentage Accurate: 43.3% → 97.0%

Time: 11.1s

Alternatives: 9

Speedup: 2.6×

Specification

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{1}{2 \cdot a}\\ t_1 := \sqrt{g \cdot g - h \cdot h}\\ \sqrt[3]{t\_0 \cdot \left(\left(-g\right) + t\_1\right)} + \sqrt[3]{t\_0 \cdot \left(\left(-g\right) - t\_1\right)} \end{array} \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (let* ((t_0 (/ 1.0 (* 2.0 a))) (t_1 (sqrt (- (* g g) (* h h)))))
   (+ (cbrt (* t_0 (+ (- g) t_1))) (cbrt (* t_0 (- (- g) t_1))))))

C

double code(double g, double h, double a) {
	double t_0 = 1.0 / (2.0 * a);
	double t_1 = sqrt(((g * g) - (h * h)));
	return cbrt((t_0 * (-g + t_1))) + cbrt((t_0 * (-g - t_1)));
}

Java

public static double code(double g, double h, double a) {
	double t_0 = 1.0 / (2.0 * a);
	double t_1 = Math.sqrt(((g * g) - (h * h)));
	return Math.cbrt((t_0 * (-g + t_1))) + Math.cbrt((t_0 * (-g - t_1)));
}

Julia

function code(g, h, a)
	t_0 = Float64(1.0 / Float64(2.0 * a))
	t_1 = sqrt(Float64(Float64(g * g) - Float64(h * h)))
	return Float64(cbrt(Float64(t_0 * Float64(Float64(-g) + t_1))) + cbrt(Float64(t_0 * Float64(Float64(-g) - t_1))))
end

Wolfram

code[g_, h_, a_] := Block[{t$95$0 = N[(1.0 / N[(2.0 * a), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[Sqrt[N[(N[(g * g), $MachinePrecision] - N[(h * h), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, N[(N[Power[N[(t$95$0 * N[((-g) + t$95$1), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision] + N[Power[N[(t$95$0 * N[((-g) - t$95$1), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]]]

Initial Program: 43.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{1}{2 \cdot a}\\ t_1 := \sqrt{g \cdot g - h \cdot h}\\ \sqrt[3]{t\_0 \cdot \left(\left(-g\right) + t\_1\right)} + \sqrt[3]{t\_0 \cdot \left(\left(-g\right) - t\_1\right)} \end{array} \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (let* ((t_0 (/ 1.0 (* 2.0 a))) (t_1 (sqrt (- (* g g) (* h h)))))
   (+ (cbrt (* t_0 (+ (- g) t_1))) (cbrt (* t_0 (- (- g) t_1))))))

C

double code(double g, double h, double a) {
	double t_0 = 1.0 / (2.0 * a);
	double t_1 = sqrt(((g * g) - (h * h)));
	return cbrt((t_0 * (-g + t_1))) + cbrt((t_0 * (-g - t_1)));
}

Java

public static double code(double g, double h, double a) {
	double t_0 = 1.0 / (2.0 * a);
	double t_1 = Math.sqrt(((g * g) - (h * h)));
	return Math.cbrt((t_0 * (-g + t_1))) + Math.cbrt((t_0 * (-g - t_1)));
}

Julia

function code(g, h, a)
	t_0 = Float64(1.0 / Float64(2.0 * a))
	t_1 = sqrt(Float64(Float64(g * g) - Float64(h * h)))
	return Float64(cbrt(Float64(t_0 * Float64(Float64(-g) + t_1))) + cbrt(Float64(t_0 * Float64(Float64(-g) - t_1))))
end

Wolfram

code[g_, h_, a_] := Block[{t$95$0 = N[(1.0 / N[(2.0 * a), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[Sqrt[N[(N[(g * g), $MachinePrecision] - N[(h * h), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, N[(N[Power[N[(t$95$0 * N[((-g) + t$95$1), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision] + N[Power[N[(t$95$0 * N[((-g) - t$95$1), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]]]

Alternative 1: 97.0% accurate, 0.4× speedup

Math

\[\begin{array}{l} \\ \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \frac{\sqrt[3]{\frac{h}{g} \cdot h}}{\sqrt[3]{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (fma
  (* (/ (cbrt g) (cbrt a)) (cbrt -0.5))
  (cbrt 2.0)
  (* (/ (cbrt (* (/ h g) h)) (cbrt a)) (* (cbrt 0.5) (cbrt -0.5)))))

C

double code(double g, double h, double a) {
	return fma(((cbrt(g) / cbrt(a)) * cbrt(-0.5)), cbrt(2.0), ((cbrt(((h / g) * h)) / cbrt(a)) * (cbrt(0.5) * cbrt(-0.5))));
}

Julia

function code(g, h, a)
	return fma(Float64(Float64(cbrt(g) / cbrt(a)) * cbrt(-0.5)), cbrt(2.0), Float64(Float64(cbrt(Float64(Float64(h / g) * h)) / cbrt(a)) * Float64(cbrt(0.5) * cbrt(-0.5))))
end

Wolfram

code[g_, h_, a_] := N[(N[(N[(N[Power[g, 1/3], $MachinePrecision] / N[Power[a, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[-0.5, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[2.0, 1/3], $MachinePrecision] + N[(N[(N[Power[N[(N[(h / g), $MachinePrecision] * h), $MachinePrecision], 1/3], $MachinePrecision] / N[Power[a, 1/3], $MachinePrecision]), $MachinePrecision] * N[(N[Power[0.5, 1/3], $MachinePrecision] * N[Power[-0.5, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 36.8%

   \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
2. Add Preprocessing

3. Taylor expanded in h around 0

   \[\leadsto \color{blue}{\sqrt[3]{\frac{g}{a}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{2}\right) + \sqrt[3]{\frac{{h}^{2}}{a \cdot g}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{\frac{1}{2}}\right)} \]
4. Step-by-step derivation

5. Applied rewrites 71.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right)} \]
6. Step-by-step derivation

7. Applied rewrites 90.4%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
8. Step-by-step derivation

9. Applied rewrites 95.1%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
10. Step-by-step derivation

11. Applied rewrites 97.0%

    \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \frac{\sqrt[3]{\frac{h}{g} \cdot h}}{\sqrt[3]{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
12. Add Preprocessing

Alternative 2: 95.1% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \mathsf{fma}\left(\frac{\sqrt[3]{g \cdot -0.5}}{\sqrt[3]{a}}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (fma
  (/ (cbrt (* g -0.5)) (cbrt a))
  (cbrt 2.0)
  (* (cbrt (/ (* (/ h g) h) a)) (* (cbrt 0.5) (cbrt -0.5)))))

C

double code(double g, double h, double a) {
	return fma((cbrt((g * -0.5)) / cbrt(a)), cbrt(2.0), (cbrt((((h / g) * h) / a)) * (cbrt(0.5) * cbrt(-0.5))));
}

Julia

function code(g, h, a)
	return fma(Float64(cbrt(Float64(g * -0.5)) / cbrt(a)), cbrt(2.0), Float64(cbrt(Float64(Float64(Float64(h / g) * h) / a)) * Float64(cbrt(0.5) * cbrt(-0.5))))
end

Wolfram

code[g_, h_, a_] := N[(N[(N[Power[N[(g * -0.5), $MachinePrecision], 1/3], $MachinePrecision] / N[Power[a, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[2.0, 1/3], $MachinePrecision] + N[(N[Power[N[(N[(N[(h / g), $MachinePrecision] * h), $MachinePrecision] / a), $MachinePrecision], 1/3], $MachinePrecision] * N[(N[Power[0.5, 1/3], $MachinePrecision] * N[Power[-0.5, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 36.8%

   \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
2. Add Preprocessing

3. Taylor expanded in h around 0

   \[\leadsto \color{blue}{\sqrt[3]{\frac{g}{a}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{2}\right) + \sqrt[3]{\frac{{h}^{2}}{a \cdot g}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{\frac{1}{2}}\right)} \]
4. Step-by-step derivation

5. Applied rewrites 71.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right)} \]
6. Step-by-step derivation

7. Applied rewrites 90.4%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
8. Step-by-step derivation

9. Applied rewrites 95.1%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
10. Step-by-step derivation

11. Applied rewrites 95.4%

    \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g \cdot -0.5}}{\sqrt[3]{a}}, \sqrt[3]{\color{blue}{2}}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
12. Add Preprocessing

Alternative 3: 94.8% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \sqrt[3]{-0.25}\right) \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (fma
  (* (/ (cbrt g) (cbrt a)) (cbrt -0.5))
  (cbrt 2.0)
  (* (cbrt (/ (* (/ h g) h) a)) (cbrt -0.25))))

C

double code(double g, double h, double a) {
	return fma(((cbrt(g) / cbrt(a)) * cbrt(-0.5)), cbrt(2.0), (cbrt((((h / g) * h) / a)) * cbrt(-0.25)));
}

Julia

function code(g, h, a)
	return fma(Float64(Float64(cbrt(g) / cbrt(a)) * cbrt(-0.5)), cbrt(2.0), Float64(cbrt(Float64(Float64(Float64(h / g) * h) / a)) * cbrt(-0.25)))
end

Wolfram

code[g_, h_, a_] := N[(N[(N[(N[Power[g, 1/3], $MachinePrecision] / N[Power[a, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[-0.5, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[2.0, 1/3], $MachinePrecision] + N[(N[Power[N[(N[(N[(h / g), $MachinePrecision] * h), $MachinePrecision] / a), $MachinePrecision], 1/3], $MachinePrecision] * N[Power[-0.25, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 36.8%

   \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
2. Add Preprocessing

3. Taylor expanded in h around 0

   \[\leadsto \color{blue}{\sqrt[3]{\frac{g}{a}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{2}\right) + \sqrt[3]{\frac{{h}^{2}}{a \cdot g}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{\frac{1}{2}}\right)} \]
4. Step-by-step derivation

5. Applied rewrites 71.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right)} \]
6. Step-by-step derivation

7. Applied rewrites 90.4%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
8. Step-by-step derivation

9. Applied rewrites 95.1%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
10. Step-by-step derivation

11. Applied rewrites 95.1%

    \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \sqrt[3]{-0.25}\right) \]
12. Add Preprocessing

Alternative 4: 92.9% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \sqrt[3]{-0.25}\right) \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (fma
  (* (/ (cbrt g) (cbrt a)) (cbrt -0.5))
  (cbrt 2.0)
  (* (cbrt (* (/ h g) (/ h a))) (cbrt -0.25))))

C

double code(double g, double h, double a) {
	return fma(((cbrt(g) / cbrt(a)) * cbrt(-0.5)), cbrt(2.0), (cbrt(((h / g) * (h / a))) * cbrt(-0.25)));
}

Julia

function code(g, h, a)
	return fma(Float64(Float64(cbrt(g) / cbrt(a)) * cbrt(-0.5)), cbrt(2.0), Float64(cbrt(Float64(Float64(h / g) * Float64(h / a))) * cbrt(-0.25)))
end

Wolfram

code[g_, h_, a_] := N[(N[(N[(N[Power[g, 1/3], $MachinePrecision] / N[Power[a, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[-0.5, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[2.0, 1/3], $MachinePrecision] + N[(N[Power[N[(N[(h / g), $MachinePrecision] * N[(h / a), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision] * N[Power[-0.25, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 36.8%

   \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
2. Add Preprocessing

3. Taylor expanded in h around 0

   \[\leadsto \color{blue}{\sqrt[3]{\frac{g}{a}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{2}\right) + \sqrt[3]{\frac{{h}^{2}}{a \cdot g}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{\frac{1}{2}}\right)} \]
4. Step-by-step derivation

5. Applied rewrites 71.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right)} \]
6. Step-by-step derivation

7. Applied rewrites 90.4%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
8. Step-by-step derivation

9. Applied rewrites 90.4%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \sqrt[3]{-0.25}\right) \]
10. Add Preprocessing

Alternative 5: 75.0% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := a \cdot \left(g \cdot g\right)\\ \mathbf{if}\;\frac{1}{2 \cdot a} \leq 2 \cdot 10^{+258}:\\ \;\;\;\;\mathsf{fma}\left(\sqrt[3]{-0.25} \cdot \sqrt[3]{\frac{h}{g}}, \sqrt[3]{\frac{h}{a}}, \sqrt[3]{-1} \cdot \sqrt[3]{\frac{g}{a}}\right)\\ \mathbf{else}:\\ \;\;\;\;g \cdot \left(\left(-\sqrt[3]{0.5}\right) \cdot \left(\sqrt[3]{\frac{0}{t\_0}} + \sqrt[3]{\frac{2}{t\_0}}\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (let* ((t_0 (* a (* g g))))
   (if (<= (/ 1.0 (* 2.0 a)) 2e+258)
     (fma
      (* (cbrt -0.25) (cbrt (/ h g)))
      (cbrt (/ h a))
      (* (cbrt -1.0) (cbrt (/ g a))))
     (* g (* (- (cbrt 0.5)) (+ (cbrt (/ 0.0 t_0)) (cbrt (/ 2.0 t_0))))))))

C

double code(double g, double h, double a) {
	double t_0 = a * (g * g);
	double tmp;
	if ((1.0 / (2.0 * a)) <= 2e+258) {
		tmp = fma((cbrt(-0.25) * cbrt((h / g))), cbrt((h / a)), (cbrt(-1.0) * cbrt((g / a))));
	} else {
		tmp = g * (-cbrt(0.5) * (cbrt((0.0 / t_0)) + cbrt((2.0 / t_0))));
	}
	return tmp;
}

Julia

function code(g, h, a)
	t_0 = Float64(a * Float64(g * g))
	tmp = 0.0
	if (Float64(1.0 / Float64(2.0 * a)) <= 2e+258)
		tmp = fma(Float64(cbrt(-0.25) * cbrt(Float64(h / g))), cbrt(Float64(h / a)), Float64(cbrt(-1.0) * cbrt(Float64(g / a))));
	else
		tmp = Float64(g * Float64(Float64(-cbrt(0.5)) * Float64(cbrt(Float64(0.0 / t_0)) + cbrt(Float64(2.0 / t_0)))));
	end
	return tmp
end

Wolfram

code[g_, h_, a_] := Block[{t$95$0 = N[(a * N[(g * g), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / N[(2.0 * a), $MachinePrecision]), $MachinePrecision], 2e+258], N[(N[(N[Power[-0.25, 1/3], $MachinePrecision] * N[Power[N[(h / g), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[N[(h / a), $MachinePrecision], 1/3], $MachinePrecision] + N[(N[Power[-1.0, 1/3], $MachinePrecision] * N[Power[N[(g / a), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(g * N[((-N[Power[0.5, 1/3], $MachinePrecision]) * N[(N[Power[N[(0.0 / t$95$0), $MachinePrecision], 1/3], $MachinePrecision] + N[Power[N[(2.0 / t$95$0), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 #s(literal 1 binary64) (*.f64 #s(literal 2 binary64) a)) < 2.00000000000000011e258

   1. Initial program 38.4%

      \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in h around 0

      \[\leadsto \color{blue}{\sqrt[3]{\frac{g}{a}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{2}\right) + \sqrt[3]{\frac{{h}^{2}}{a \cdot g}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{\frac{1}{2}}\right)} \]
   4. Step-by-step derivation

   5. Applied rewrites 74.0%

      \[\leadsto \color{blue}{\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right)} \]
   6. Step-by-step derivation

   7. Applied rewrites 92.4%

      \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
   8. Step-by-step derivation

   9. Applied rewrites 95.7%

      \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
   10. Step-by-step derivation

   11. Applied rewrites 74.8%

       \[\leadsto \mathsf{fma}\left(\sqrt[3]{-0.25} \cdot \sqrt[3]{\frac{h}{g}}, \color{blue}{\sqrt[3]{\frac{h}{a}}}, \sqrt[3]{-1} \cdot \sqrt[3]{\frac{g}{a}}\right) \]

   if 2.00000000000000011e258 < (/.f64 #s(literal 1 binary64) (*.f64 #s(literal 2 binary64) a))

   1. Initial program 1.9%

      \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

   4. Applied rewrites 1.7%

      \[\leadsto \color{blue}{\sqrt[3]{\mathsf{fma}\left(\sqrt{h + g}, \sqrt{g - h}, -g\right) \cdot \frac{0.5}{a}} + \sqrt[3]{\left(\left(-g\right) - \sqrt{\left(g - h\right) \cdot \left(h + g\right)}\right) \cdot \frac{0.5}{a}}} \]

   5. Taylor expanded in a around 0

      \[\leadsto \color{blue}{\sqrt[3]{\frac{\sqrt{\left(g + h\right) \cdot \left(g - h\right)} - g}{a}} \cdot \sqrt[3]{\frac{1}{2}}} + \sqrt[3]{\left(\left(-g\right) - \sqrt{\left(g - h\right) \cdot \left(h + g\right)}\right) \cdot \frac{\frac{1}{2}}{a}} \]
   6. Step-by-step derivation

   7. Applied rewrites 1.9%

      \[\leadsto \color{blue}{\sqrt[3]{\frac{\sqrt{\left(g + h\right) \cdot \left(g - h\right)} - g}{a}} \cdot \sqrt[3]{0.5}} + \sqrt[3]{\left(\left(-g\right) - \sqrt{\left(g - h\right) \cdot \left(h + g\right)}\right) \cdot \frac{0.5}{a}} \]

   8. Taylor expanded in g around -inf

      \[\leadsto \color{blue}{-1 \cdot \left(g \cdot \left(\sqrt[3]{\frac{1 + {\left(\sqrt{-1}\right)}^{2}}{a \cdot {g}^{2}}} \cdot \sqrt[3]{\frac{1}{2}} + \sqrt[3]{\frac{1 - {\left(\sqrt{-1}\right)}^{2}}{a \cdot {g}^{2}}} \cdot \sqrt[3]{\frac{1}{2}}\right)\right)} \]
   9. Step-by-step derivation

   10. Applied rewrites 46.3%

       \[\leadsto \color{blue}{\left(-g\right) \cdot \left(\sqrt[3]{0.5} \cdot \left(\sqrt[3]{\frac{0}{a \cdot \left(g \cdot g\right)}} + \sqrt[3]{\frac{2}{a \cdot \left(g \cdot g\right)}}\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 73.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{2 \cdot a} \leq 2 \cdot 10^{+258}:\\ \;\;\;\;\mathsf{fma}\left(\sqrt[3]{-0.25} \cdot \sqrt[3]{\frac{h}{g}}, \sqrt[3]{\frac{h}{a}}, \sqrt[3]{-1} \cdot \sqrt[3]{\frac{g}{a}}\right)\\ \mathbf{else}:\\ \;\;\;\;g \cdot \left(\left(-\sqrt[3]{0.5}\right) \cdot \left(\sqrt[3]{\frac{0}{a \cdot \left(g \cdot g\right)}} + \sqrt[3]{\frac{2}{a \cdot \left(g \cdot g\right)}}\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 75.0% accurate, 0.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := a \cdot \left(g \cdot g\right)\\ \mathbf{if}\;\frac{1}{2 \cdot a} \leq 2 \cdot 10^{+258}:\\ \;\;\;\;\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}}, \sqrt[3]{-1}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a} \cdot -0.25}\right)\\ \mathbf{else}:\\ \;\;\;\;g \cdot \left(\left(-\sqrt[3]{0.5}\right) \cdot \left(\sqrt[3]{\frac{0}{t\_0}} + \sqrt[3]{\frac{2}{t\_0}}\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (let* ((t_0 (* a (* g g))))
   (if (<= (/ 1.0 (* 2.0 a)) 2e+258)
     (fma (cbrt (/ g a)) (cbrt -1.0) (cbrt (* (/ (* (/ h g) h) a) -0.25)))
     (* g (* (- (cbrt 0.5)) (+ (cbrt (/ 0.0 t_0)) (cbrt (/ 2.0 t_0))))))))

C

double code(double g, double h, double a) {
	double t_0 = a * (g * g);
	double tmp;
	if ((1.0 / (2.0 * a)) <= 2e+258) {
		tmp = fma(cbrt((g / a)), cbrt(-1.0), cbrt(((((h / g) * h) / a) * -0.25)));
	} else {
		tmp = g * (-cbrt(0.5) * (cbrt((0.0 / t_0)) + cbrt((2.0 / t_0))));
	}
	return tmp;
}

Julia

function code(g, h, a)
	t_0 = Float64(a * Float64(g * g))
	tmp = 0.0
	if (Float64(1.0 / Float64(2.0 * a)) <= 2e+258)
		tmp = fma(cbrt(Float64(g / a)), cbrt(-1.0), cbrt(Float64(Float64(Float64(Float64(h / g) * h) / a) * -0.25)));
	else
		tmp = Float64(g * Float64(Float64(-cbrt(0.5)) * Float64(cbrt(Float64(0.0 / t_0)) + cbrt(Float64(2.0 / t_0)))));
	end
	return tmp
end

Wolfram

code[g_, h_, a_] := Block[{t$95$0 = N[(a * N[(g * g), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / N[(2.0 * a), $MachinePrecision]), $MachinePrecision], 2e+258], N[(N[Power[N[(g / a), $MachinePrecision], 1/3], $MachinePrecision] * N[Power[-1.0, 1/3], $MachinePrecision] + N[Power[N[(N[(N[(N[(h / g), $MachinePrecision] * h), $MachinePrecision] / a), $MachinePrecision] * -0.25), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision], N[(g * N[((-N[Power[0.5, 1/3], $MachinePrecision]) * N[(N[Power[N[(0.0 / t$95$0), $MachinePrecision], 1/3], $MachinePrecision] + N[Power[N[(2.0 / t$95$0), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 #s(literal 1 binary64) (*.f64 #s(literal 2 binary64) a)) < 2.00000000000000011e258

   1. Initial program 38.4%

      \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in h around 0

      \[\leadsto \color{blue}{\sqrt[3]{\frac{g}{a}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{2}\right) + \sqrt[3]{\frac{{h}^{2}}{a \cdot g}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{\frac{1}{2}}\right)} \]
   4. Step-by-step derivation

   5. Applied rewrites 74.0%

      \[\leadsto \color{blue}{\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right)} \]
   6. Step-by-step derivation

   7. Applied rewrites 92.4%

      \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
   8. Step-by-step derivation

   9. Applied rewrites 95.7%

      \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]

   11. Applied rewrites 74.8%

       \[\leadsto \mathsf{fma}\left(\sqrt[3]{\frac{g}{a}}, \color{blue}{\sqrt[3]{-1}}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a} \cdot -0.25}\right) \]

   if 2.00000000000000011e258 < (/.f64 #s(literal 1 binary64) (*.f64 #s(literal 2 binary64) a))

   1. Initial program 1.9%

      \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

   4. Applied rewrites 1.7%

      \[\leadsto \color{blue}{\sqrt[3]{\mathsf{fma}\left(\sqrt{h + g}, \sqrt{g - h}, -g\right) \cdot \frac{0.5}{a}} + \sqrt[3]{\left(\left(-g\right) - \sqrt{\left(g - h\right) \cdot \left(h + g\right)}\right) \cdot \frac{0.5}{a}}} \]

   5. Taylor expanded in a around 0

      \[\leadsto \color{blue}{\sqrt[3]{\frac{\sqrt{\left(g + h\right) \cdot \left(g - h\right)} - g}{a}} \cdot \sqrt[3]{\frac{1}{2}}} + \sqrt[3]{\left(\left(-g\right) - \sqrt{\left(g - h\right) \cdot \left(h + g\right)}\right) \cdot \frac{\frac{1}{2}}{a}} \]
   6. Step-by-step derivation

   7. Applied rewrites 1.9%

      \[\leadsto \color{blue}{\sqrt[3]{\frac{\sqrt{\left(g + h\right) \cdot \left(g - h\right)} - g}{a}} \cdot \sqrt[3]{0.5}} + \sqrt[3]{\left(\left(-g\right) - \sqrt{\left(g - h\right) \cdot \left(h + g\right)}\right) \cdot \frac{0.5}{a}} \]

   8. Taylor expanded in g around -inf

      \[\leadsto \color{blue}{-1 \cdot \left(g \cdot \left(\sqrt[3]{\frac{1 + {\left(\sqrt{-1}\right)}^{2}}{a \cdot {g}^{2}}} \cdot \sqrt[3]{\frac{1}{2}} + \sqrt[3]{\frac{1 - {\left(\sqrt{-1}\right)}^{2}}{a \cdot {g}^{2}}} \cdot \sqrt[3]{\frac{1}{2}}\right)\right)} \]
   9. Step-by-step derivation

   10. Applied rewrites 46.3%

       \[\leadsto \color{blue}{\left(-g\right) \cdot \left(\sqrt[3]{0.5} \cdot \left(\sqrt[3]{\frac{0}{a \cdot \left(g \cdot g\right)}} + \sqrt[3]{\frac{2}{a \cdot \left(g \cdot g\right)}}\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 73.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{2 \cdot a} \leq 2 \cdot 10^{+258}:\\ \;\;\;\;\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}}, \sqrt[3]{-1}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a} \cdot -0.25}\right)\\ \mathbf{else}:\\ \;\;\;\;g \cdot \left(\left(-\sqrt[3]{0.5}\right) \cdot \left(\sqrt[3]{\frac{0}{a \cdot \left(g \cdot g\right)}} + \sqrt[3]{\frac{2}{a \cdot \left(g \cdot g\right)}}\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 75.1% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \mathsf{fma}\left(\sqrt[3]{\frac{g}{a}}, \sqrt[3]{-1}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a} \cdot -0.25}\right) \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (fma (cbrt (/ g a)) (cbrt -1.0) (cbrt (* (/ (* (/ h g) h) a) -0.25))))

C

double code(double g, double h, double a) {
	return fma(cbrt((g / a)), cbrt(-1.0), cbrt(((((h / g) * h) / a) * -0.25)));
}

Julia

function code(g, h, a)
	return fma(cbrt(Float64(g / a)), cbrt(-1.0), cbrt(Float64(Float64(Float64(Float64(h / g) * h) / a) * -0.25)))
end

Wolfram

code[g_, h_, a_] := N[(N[Power[N[(g / a), $MachinePrecision], 1/3], $MachinePrecision] * N[Power[-1.0, 1/3], $MachinePrecision] + N[Power[N[(N[(N[(N[(h / g), $MachinePrecision] * h), $MachinePrecision] / a), $MachinePrecision] * -0.25), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 36.8%

   \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
2. Add Preprocessing

3. Taylor expanded in h around 0

   \[\leadsto \color{blue}{\sqrt[3]{\frac{g}{a}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{2}\right) + \sqrt[3]{\frac{{h}^{2}}{a \cdot g}} \cdot \left(\sqrt[3]{\frac{-1}{2}} \cdot \sqrt[3]{\frac{1}{2}}\right)} \]
4. Step-by-step derivation

5. Applied rewrites 71.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(\sqrt[3]{\frac{g}{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right)} \]
6. Step-by-step derivation

7. Applied rewrites 90.4%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{h}{g} \cdot \frac{h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]
8. Step-by-step derivation

9. Applied rewrites 95.1%

   \[\leadsto \mathsf{fma}\left(\frac{\sqrt[3]{g}}{\sqrt[3]{a}} \cdot \sqrt[3]{-0.5}, \sqrt[3]{2}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a}} \cdot \left(\sqrt[3]{0.5} \cdot \sqrt[3]{-0.5}\right)\right) \]

11. Applied rewrites 71.7%

    \[\leadsto \mathsf{fma}\left(\sqrt[3]{\frac{g}{a}}, \color{blue}{\sqrt[3]{-1}}, \sqrt[3]{\frac{\frac{h}{g} \cdot h}{a} \cdot -0.25}\right) \]
12. Add Preprocessing

Alternative 8: 75.1% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(h \cdot \frac{h}{g}\right) \cdot -0.5\right)} + \sqrt[3]{\frac{-g}{a}} \end{array} \]

FPCore

(FPCore (g h a)
 :precision binary64
 (+ (cbrt (* (/ 1.0 (* 2.0 a)) (* (* h (/ h g)) -0.5))) (cbrt (/ (- g) a))))

C

double code(double g, double h, double a) {
	return cbrt(((1.0 / (2.0 * a)) * ((h * (h / g)) * -0.5))) + cbrt((-g / a));
}

Java

public static double code(double g, double h, double a) {
	return Math.cbrt(((1.0 / (2.0 * a)) * ((h * (h / g)) * -0.5))) + Math.cbrt((-g / a));
}

Julia

function code(g, h, a)
	return Float64(cbrt(Float64(Float64(1.0 / Float64(2.0 * a)) * Float64(Float64(h * Float64(h / g)) * -0.5))) + cbrt(Float64(Float64(-g) / a)))
end

Wolfram

code[g_, h_, a_] := N[(N[Power[N[(N[(1.0 / N[(2.0 * a), $MachinePrecision]), $MachinePrecision] * N[(N[(h * N[(h / g), $MachinePrecision]), $MachinePrecision] * -0.5), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision] + N[Power[N[((-g) / a), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 36.8%

   \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
2. Add Preprocessing

3. Taylor expanded in g around inf

   \[\leadsto \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\color{blue}{-1 \cdot \frac{g}{a}}} \]
4. Step-by-step derivation

5. Applied rewrites 25.7%

   \[\leadsto \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\color{blue}{\frac{-g}{a}}} \]

6. Taylor expanded in g around inf

   \[\leadsto \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \color{blue}{g}\right)} + \sqrt[3]{\frac{-g}{a}} \]
7. Step-by-step derivation

8. Applied rewrites 69.7%

   \[\leadsto \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \color{blue}{g}\right)} + \sqrt[3]{\frac{-g}{a}} \]

9. Taylor expanded in g around inf

   \[\leadsto \sqrt[3]{\frac{1}{2 \cdot a} \cdot \color{blue}{\left(\frac{-1}{2} \cdot \frac{{h}^{2}}{g}\right)}} + \sqrt[3]{\frac{-g}{a}} \]
10. Step-by-step derivation

11. Applied rewrites 71.7%

    \[\leadsto \sqrt[3]{\frac{1}{2 \cdot a} \cdot \color{blue}{\left(\left(h \cdot \frac{h}{g}\right) \cdot -0.5\right)}} + \sqrt[3]{\frac{-g}{a}} \]
12. Add Preprocessing

Alternative 9: 73.5% accurate, 2.6× speedup

Math

\[\begin{array}{l} \\ -\sqrt[3]{\frac{g}{a}} \end{array} \]

FPCore

(FPCore (g h a) :precision binary64 (- (cbrt (/ g a))))

C

double code(double g, double h, double a) {
	return -cbrt((g / a));
}

Java

public static double code(double g, double h, double a) {
	return -Math.cbrt((g / a));
}

Julia

function code(g, h, a)
	return Float64(-cbrt(Float64(g / a)))
end

Wolfram

code[g_, h_, a_] := (-N[Power[N[(g / a), $MachinePrecision], 1/3], $MachinePrecision])

Derivation

1. Initial program 36.8%

   \[\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) + \sqrt{g \cdot g - h \cdot h}\right)} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
2. Add Preprocessing

3. Taylor expanded in g around inf

   \[\leadsto \sqrt[3]{\frac{1}{2 \cdot a} \cdot \color{blue}{\left(\frac{-1}{2} \cdot \frac{{h}^{2}}{g}\right)}} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
4. Step-by-step derivation

5. Applied rewrites 24.4%

   \[\leadsto \sqrt[3]{\frac{1}{2 \cdot a} \cdot \color{blue}{\left(\frac{h \cdot h}{g} \cdot -0.5\right)}} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]
6. Step-by-step derivation

   1. lift-cbrt.f64 N/A

      \[\leadsto \color{blue}{\sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\frac{h \cdot h}{g} \cdot \frac{-1}{2}\right)}} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]

   2. pow1/3 N/A

      \[\leadsto \color{blue}{{\left(\frac{1}{2 \cdot a} \cdot \left(\frac{h \cdot h}{g} \cdot \frac{-1}{2}\right)\right)}^{\frac{1}{3}}} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]

   3. lower-pow.f64 20.3

      \[\leadsto \color{blue}{{\left(\frac{1}{2 \cdot a} \cdot \left(\frac{h \cdot h}{g} \cdot -0.5\right)\right)}^{0.3333333333333333}} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]

7. Applied rewrites 18.8%

   \[\leadsto \color{blue}{{\left(\left(\left(h \cdot \frac{h}{g}\right) \cdot -0.5\right) \cdot \frac{0.5}{a}\right)}^{0.3333333333333333}} + \sqrt[3]{\frac{1}{2 \cdot a} \cdot \left(\left(-g\right) - \sqrt{g \cdot g - h \cdot h}\right)} \]

8. Taylor expanded in g around -inf

   \[\leadsto \color{blue}{-1 \cdot \sqrt[3]{\frac{g}{a}}} \]
9. Step-by-step derivation

10. Applied rewrites 69.7%

    \[\leadsto \color{blue}{-\sqrt[3]{\frac{g}{a}}} \]
11. Add Preprocessing

Reproduce

herbie shell --seed 2025019 
(FPCore (g h a)
  :name "2-ancestry mixing, positive discriminant"
  :precision binary64
  (+ (cbrt (* (/ 1.0 (* 2.0 a)) (+ (- g) (sqrt (- (* g g) (* h h)))))) (cbrt (* (/ 1.0 (* 2.0 a)) (- (- g) (sqrt (- (* g g) (* h h))))))))