Toniolo and Linder, Equation (3b), real

Percentage Accurate: 94.0% → 99.7%

Time: 21.8s

Alternatives: 20

Speedup: 1.4×

Specification

Math

\[\begin{array}{l} \\ \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))

C

double code(double kx, double ky, double th) {
	return (sin(ky) / sqrt((pow(sin(kx), 2.0) + pow(sin(ky), 2.0)))) * sin(th);
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = (sin(ky) / sqrt(((sin(kx) ** 2.0d0) + (sin(ky) ** 2.0d0)))) * sin(th)
end function

Java

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.sqrt((Math.pow(Math.sin(kx), 2.0) + Math.pow(Math.sin(ky), 2.0)))) * Math.sin(th);
}

Python

def code(kx, ky, th):
	return (math.sin(ky) / math.sqrt((math.pow(math.sin(kx), 2.0) + math.pow(math.sin(ky), 2.0)))) * math.sin(th)

Julia

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / sqrt(Float64((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / sqrt(((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th);
end

Wolfram

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[(N[Power[N[Sin[kx], $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[Sin[ky], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Initial Program: 94.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))

C

double code(double kx, double ky, double th) {
	return (sin(ky) / sqrt((pow(sin(kx), 2.0) + pow(sin(ky), 2.0)))) * sin(th);
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = (sin(ky) / sqrt(((sin(kx) ** 2.0d0) + (sin(ky) ** 2.0d0)))) * sin(th)
end function

Java

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.sqrt((Math.pow(Math.sin(kx), 2.0) + Math.pow(Math.sin(ky), 2.0)))) * Math.sin(th);
}

Python

def code(kx, ky, th):
	return (math.sin(ky) / math.sqrt((math.pow(math.sin(kx), 2.0) + math.pow(math.sin(ky), 2.0)))) * math.sin(th)

Julia

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / sqrt(Float64((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / sqrt(((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th);
end

Wolfram

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[(N[Power[N[Sin[kx], $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[Sin[ky], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.7% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (hypot (sin ky) (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	return (sin(ky) / hypot(sin(ky), sin(kx))) * sin(th);
}

Java

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx))) * Math.sin(th);
}

Python

def code(kx, ky, th):
	return (math.sin(ky) / math.hypot(math.sin(ky), math.sin(kx))) * math.sin(th)

Julia

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / hypot(sin(ky), sin(kx))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * sin(th);
end

Wolfram

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 93.6%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
2. Step-by-step derivation

   1. remove-double-neg 93.6%

      \[\leadsto \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \color{blue}{\left(-\left(-\sin th\right)\right)} \]

   2. sin-neg 93.6%

      \[\leadsto \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \left(-\color{blue}{\sin \left(-th\right)}\right) \]

   3. neg-mul-1 93.6%

      \[\leadsto \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \color{blue}{\left(-1 \cdot \sin \left(-th\right)\right)} \]

   4. *-commutative 93.6%

      \[\leadsto \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \color{blue}{\left(\sin \left(-th\right) \cdot -1\right)} \]

   5. associate-*l* 93.6%

      \[\leadsto \color{blue}{\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin \left(-th\right)\right) \cdot -1} \]

   6. associate-*l/ 91.0%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin \left(-th\right)}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \cdot -1 \]

   7. associate-/r/ 91.0%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin \left(-th\right)}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}} \]

   8. associate-*l/ 93.6%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}} \cdot \sin \left(-th\right)} \]

   9. associate-/r/ 93.5%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{\sin \left(-th\right)}}} \]

   10. sin-neg 93.5%

       \[\leadsto \frac{\sin ky}{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{\color{blue}{-\sin th}}} \]

   11. neg-mul-1 93.5%

       \[\leadsto \frac{\sin ky}{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{\color{blue}{-1 \cdot \sin th}}} \]

   12. associate-/r* 93.5%

       \[\leadsto \frac{\sin ky}{\color{blue}{\frac{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{-1}}{\sin th}}} \]

   13. associate-/r/ 93.6%

       \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{-1}} \cdot \sin th} \]

3. Simplified 99.7%

   \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]
4. Add Preprocessing

5. Final simplification 99.7%

   \[\leadsto \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th \]
6. Add Preprocessing

Alternative 2: 77.8% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;\sin ky \leq -0.01:\\ \;\;\;\;\frac{\sin ky}{t_1 \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 0.001:\\ \;\;\;\;\frac{\sin th}{t_1 \cdot \left(ky \cdot 0.16666666666666666 + \frac{1}{ky}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= (sin ky) -0.01)
     (/ (sin ky) (* t_1 (+ (/ 1.0 th) (* th 0.16666666666666666))))
     (if (<= (sin ky) 0.001)
       (/ (sin th) (* t_1 (+ (* ky 0.16666666666666666) (/ 1.0 ky))))
       (sin th)))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (sin(ky) <= -0.01) {
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (sin(ky) <= 0.001) {
		tmp = sin(th) / (t_1 * ((ky * 0.16666666666666666) + (1.0 / ky)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.hypot(Math.sin(kx), Math.sin(ky));
	double tmp;
	if (Math.sin(ky) <= -0.01) {
		tmp = Math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (Math.sin(ky) <= 0.001) {
		tmp = Math.sin(th) / (t_1 * ((ky * 0.16666666666666666) + (1.0 / ky)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if math.sin(ky) <= -0.01:
		tmp = math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)))
	elif math.sin(ky) <= 0.001:
		tmp = math.sin(th) / (t_1 * ((ky * 0.16666666666666666) + (1.0 / ky)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (sin(ky) <= -0.01)
		tmp = Float64(sin(ky) / Float64(t_1 * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))));
	elseif (sin(ky) <= 0.001)
		tmp = Float64(sin(th) / Float64(t_1 * Float64(Float64(ky * 0.16666666666666666) + Float64(1.0 / ky))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (sin(ky) <= -0.01)
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	elseif (sin(ky) <= 0.001)
		tmp = sin(th) / (t_1 * ((ky * 0.16666666666666666) + (1.0 / ky)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]}, If[LessEqual[N[Sin[ky], $MachinePrecision], -0.01], N[(N[Sin[ky], $MachinePrecision] / N[(t$95$1 * N[(N[(1.0 / th), $MachinePrecision] + N[(th * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 0.001], N[(N[Sin[th], $MachinePrecision] / N[(t$95$1 * N[(N[(ky * 0.16666666666666666), $MachinePrecision] + N[(1.0 / ky), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -0.0100000000000000002

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. associate-*l/ 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. associate-/l* 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin th}}} \]

      3. unpow2 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin th}} \]

      4. unpow2 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin th}} \]

      5. hypot-def 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin th}} \]

   4. Applied egg-rr 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin th}}} \]

   5. Taylor expanded in th around 0 56.1%

      \[\leadsto \frac{\sin ky}{\color{blue}{0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right) + \frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   6. Step-by-step derivation

      1. +-commutative 56.1%

         \[\leadsto \frac{\sin ky}{\color{blue}{\frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)}} \]

      2. unpow2 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      3. unpow2 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      4. hypot-def 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      5. associate-*r* 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \color{blue}{\left(0.16666666666666666 \cdot th\right) \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      6. unpow2 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}} \]

      7. unpow2 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}} \]

      8. hypot-def 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

      9. distribute-rgt-out 56.1%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + 0.16666666666666666 \cdot th\right)}} \]

   7. Simplified 56.1%

      \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + 0.16666666666666666 \cdot th\right)}} \]

   if -0.0100000000000000002 < (sin.f64 ky) < 1e-3

   1. Initial program 87.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 87.4%

         \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \cdot \sin th \]

      2. associate-*l/ 87.4%

         \[\leadsto \color{blue}{\frac{1 \cdot \sin th}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

      3. *-un-lft-identity 87.4%

         \[\leadsto \frac{\color{blue}{\sin th}}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}} \]

      4. unpow2 87.4%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin ky}} \]

      5. unpow2 87.4%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin ky}} \]

      6. hypot-def 99.7%

         \[\leadsto \frac{\sin th}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin ky}} \]

   4. Applied egg-rr 99.7%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}} \]
   5. Step-by-step derivation

      1. div-inv 99.5%

         \[\leadsto \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{\sin ky}}} \]

   6. Applied egg-rr 99.5%

      \[\leadsto \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{\sin ky}}} \]

   7. Taylor expanded in ky around 0 99.4%

      \[\leadsto \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \color{blue}{\left(0.16666666666666666 \cdot ky + \frac{1}{ky}\right)}} \]

   if 1e-3 < (sin.f64 ky)

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 64.9%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 79.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.01:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 0.001:\\ \;\;\;\;\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(ky \cdot 0.16666666666666666 + \frac{1}{ky}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 77.5% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.01:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{elif}\;\sin ky \leq 4 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.01)
   (* (/ (sin ky) (hypot (sin ky) (sin kx))) th)
   (if (<= (sin ky) 4e-5)
     (/ (sin th) (* (hypot (sin kx) (sin ky)) (/ 1.0 ky)))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.01) {
		tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * th;
	} else if (sin(ky) <= 4e-5) {
		tmp = sin(th) / (hypot(sin(kx), sin(ky)) * (1.0 / ky));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -0.01) {
		tmp = (Math.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx))) * th;
	} else if (Math.sin(ky) <= 4e-5) {
		tmp = Math.sin(th) / (Math.hypot(Math.sin(kx), Math.sin(ky)) * (1.0 / ky));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -0.01:
		tmp = (math.sin(ky) / math.hypot(math.sin(ky), math.sin(kx))) * th
	elif math.sin(ky) <= 4e-5:
		tmp = math.sin(th) / (math.hypot(math.sin(kx), math.sin(ky)) * (1.0 / ky))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -0.01)
		tmp = Float64(Float64(sin(ky) / hypot(sin(ky), sin(kx))) * th);
	elseif (sin(ky) <= 4e-5)
		tmp = Float64(sin(th) / Float64(hypot(sin(kx), sin(ky)) * Float64(1.0 / ky)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -0.01)
		tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * th;
	elseif (sin(ky) <= 4e-5)
		tmp = sin(th) / (hypot(sin(kx), sin(ky)) * (1.0 / ky));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.01], N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] * th), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 4e-5], N[(N[Sin[th], $MachinePrecision] / N[(N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision] * N[(1.0 / ky), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -0.0100000000000000002

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 55.5%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   4. Step-by-step derivation

      1. expm1-log1p-u 54.6%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}\right)\right)} \]

      2. expm1-udef 5.6%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}\right)} - 1} \]

   5. Applied egg-rr 5.6%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\right)} - 1} \]
   6. Step-by-step derivation

      1. expm1-def 54.7%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\right)\right)} \]

      2. expm1-log1p 55.6%

         \[\leadsto \color{blue}{th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

      3. hypot-def 55.5%

         \[\leadsto th \cdot \frac{\sin ky}{\color{blue}{\sqrt{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \]

      4. unpow2 55.5%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{{\sin kx}^{2}} + \sin ky \cdot \sin ky}} \]

      5. unpow2 55.5%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{{\sin kx}^{2} + \color{blue}{{\sin ky}^{2}}}} \]

      6. +-commutative 55.5%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      7. unpow2 55.5%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      8. unpow2 55.5%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      9. hypot-def 55.6%

         \[\leadsto th \cdot \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   7. Simplified 55.6%

      \[\leadsto \color{blue}{th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   if -0.0100000000000000002 < (sin.f64 ky) < 4.00000000000000033e-5

   1. Initial program 87.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 87.3%

         \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \cdot \sin th \]

      2. associate-*l/ 87.3%

         \[\leadsto \color{blue}{\frac{1 \cdot \sin th}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

      3. *-un-lft-identity 87.3%

         \[\leadsto \frac{\color{blue}{\sin th}}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}} \]

      4. unpow2 87.3%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin ky}} \]

      5. unpow2 87.3%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin ky}} \]

      6. hypot-def 99.7%

         \[\leadsto \frac{\sin th}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin ky}} \]

   4. Applied egg-rr 99.7%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}} \]
   5. Step-by-step derivation

      1. div-inv 99.5%

         \[\leadsto \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{\sin ky}}} \]

   6. Applied egg-rr 99.5%

      \[\leadsto \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{\sin ky}}} \]

   7. Taylor expanded in ky around 0 99.1%

      \[\leadsto \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \color{blue}{\frac{1}{ky}}} \]

   if 4.00000000000000033e-5 < (sin.f64 ky)

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 64.2%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 79.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.01:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{elif}\;\sin ky \leq 4 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 77.6% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;\sin ky \leq -0.01:\\ \;\;\;\;\frac{\sin ky}{t_1 \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 4 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sin th}{t_1 \cdot \frac{1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= (sin ky) -0.01)
     (/ (sin ky) (* t_1 (+ (/ 1.0 th) (* th 0.16666666666666666))))
     (if (<= (sin ky) 4e-5) (/ (sin th) (* t_1 (/ 1.0 ky))) (sin th)))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (sin(ky) <= -0.01) {
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (sin(ky) <= 4e-5) {
		tmp = sin(th) / (t_1 * (1.0 / ky));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.hypot(Math.sin(kx), Math.sin(ky));
	double tmp;
	if (Math.sin(ky) <= -0.01) {
		tmp = Math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (Math.sin(ky) <= 4e-5) {
		tmp = Math.sin(th) / (t_1 * (1.0 / ky));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if math.sin(ky) <= -0.01:
		tmp = math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)))
	elif math.sin(ky) <= 4e-5:
		tmp = math.sin(th) / (t_1 * (1.0 / ky))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (sin(ky) <= -0.01)
		tmp = Float64(sin(ky) / Float64(t_1 * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))));
	elseif (sin(ky) <= 4e-5)
		tmp = Float64(sin(th) / Float64(t_1 * Float64(1.0 / ky)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (sin(ky) <= -0.01)
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	elseif (sin(ky) <= 4e-5)
		tmp = sin(th) / (t_1 * (1.0 / ky));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]}, If[LessEqual[N[Sin[ky], $MachinePrecision], -0.01], N[(N[Sin[ky], $MachinePrecision] / N[(t$95$1 * N[(N[(1.0 / th), $MachinePrecision] + N[(th * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 4e-5], N[(N[Sin[th], $MachinePrecision] / N[(t$95$1 * N[(1.0 / ky), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -0.0100000000000000002

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. associate-*l/ 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. associate-/l* 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin th}}} \]

      3. unpow2 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin th}} \]

      4. unpow2 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin th}} \]

      5. hypot-def 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin th}} \]

   4. Applied egg-rr 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin th}}} \]

   5. Taylor expanded in th around 0 56.1%

      \[\leadsto \frac{\sin ky}{\color{blue}{0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right) + \frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   6. Step-by-step derivation

      1. +-commutative 56.1%

         \[\leadsto \frac{\sin ky}{\color{blue}{\frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)}} \]

      2. unpow2 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      3. unpow2 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      4. hypot-def 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      5. associate-*r* 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \color{blue}{\left(0.16666666666666666 \cdot th\right) \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      6. unpow2 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}} \]

      7. unpow2 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}} \]

      8. hypot-def 56.1%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

      9. distribute-rgt-out 56.1%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + 0.16666666666666666 \cdot th\right)}} \]

   7. Simplified 56.1%

      \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + 0.16666666666666666 \cdot th\right)}} \]

   if -0.0100000000000000002 < (sin.f64 ky) < 4.00000000000000033e-5

   1. Initial program 87.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 87.3%

         \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \cdot \sin th \]

      2. associate-*l/ 87.3%

         \[\leadsto \color{blue}{\frac{1 \cdot \sin th}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

      3. *-un-lft-identity 87.3%

         \[\leadsto \frac{\color{blue}{\sin th}}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}} \]

      4. unpow2 87.3%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin ky}} \]

      5. unpow2 87.3%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin ky}} \]

      6. hypot-def 99.7%

         \[\leadsto \frac{\sin th}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin ky}} \]

   4. Applied egg-rr 99.7%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}} \]
   5. Step-by-step derivation

      1. div-inv 99.5%

         \[\leadsto \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{\sin ky}}} \]

   6. Applied egg-rr 99.5%

      \[\leadsto \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{\sin ky}}} \]

   7. Taylor expanded in ky around 0 99.1%

      \[\leadsto \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \color{blue}{\frac{1}{ky}}} \]

   if 4.00000000000000033e-5 < (sin.f64 ky)

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 64.2%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 79.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.01:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 4 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 46.0% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.05:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\frac{\sin th}{1 + 0.5 \cdot {\left(\frac{kx}{\sin ky}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky \cdot \sin th}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.05)
   (* (sin th) (fabs (/ ky (sin kx))))
   (if (<= (sin kx) 0.001)
     (/ (sin th) (+ 1.0 (* 0.5 (pow (/ kx (sin ky)) 2.0))))
     (/ (* (sin ky) (sin th)) (sin kx)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.05) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else if (sin(kx) <= 0.001) {
		tmp = sin(th) / (1.0 + (0.5 * pow((kx / sin(ky)), 2.0)));
	} else {
		tmp = (sin(ky) * sin(th)) / sin(kx);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.05d0)) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else if (sin(kx) <= 0.001d0) then
        tmp = sin(th) / (1.0d0 + (0.5d0 * ((kx / sin(ky)) ** 2.0d0)))
    else
        tmp = (sin(ky) * sin(th)) / sin(kx)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.05) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else if (Math.sin(kx) <= 0.001) {
		tmp = Math.sin(th) / (1.0 + (0.5 * Math.pow((kx / Math.sin(ky)), 2.0)));
	} else {
		tmp = (Math.sin(ky) * Math.sin(th)) / Math.sin(kx);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.05:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	elif math.sin(kx) <= 0.001:
		tmp = math.sin(th) / (1.0 + (0.5 * math.pow((kx / math.sin(ky)), 2.0)))
	else:
		tmp = (math.sin(ky) * math.sin(th)) / math.sin(kx)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.05)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	elseif (sin(kx) <= 0.001)
		tmp = Float64(sin(th) / Float64(1.0 + Float64(0.5 * (Float64(kx / sin(ky)) ^ 2.0))));
	else
		tmp = Float64(Float64(sin(ky) * sin(th)) / sin(kx));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.05)
		tmp = sin(th) * abs((ky / sin(kx)));
	elseif (sin(kx) <= 0.001)
		tmp = sin(th) / (1.0 + (0.5 * ((kx / sin(ky)) ^ 2.0)));
	else
		tmp = (sin(ky) * sin(th)) / sin(kx);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.05], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 0.001], N[(N[Sin[th], $MachinePrecision] / N[(1.0 + N[(0.5 * N[Power[N[(kx / N[Sin[ky], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sin[ky], $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.050000000000000003

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 15.4%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   4. Step-by-step derivation

      1. add-sqr-sqrt 9.9%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{\sin ky}{\sin kx}} \cdot \sqrt{\frac{\sin ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 24.5%

         \[\leadsto \color{blue}{\sqrt{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 24.5%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   5. Applied egg-rr 24.5%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   6. Step-by-step derivation

      1. unpow2 24.5%

         \[\leadsto \sqrt{\color{blue}{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 35.6%

         \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   7. Simplified 35.6%

      \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   8. Taylor expanded in ky around 0 31.3%

      \[\leadsto \left|\color{blue}{\frac{ky}{\sin kx}}\right| \cdot \sin th \]

   if -0.050000000000000003 < (sin.f64 kx) < 1e-3

   1. Initial program 88.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 88.8%

         \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \cdot \sin th \]

      2. associate-*l/ 88.8%

         \[\leadsto \color{blue}{\frac{1 \cdot \sin th}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

      3. *-un-lft-identity 88.8%

         \[\leadsto \frac{\color{blue}{\sin th}}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}} \]

      4. unpow2 88.8%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin ky}} \]

      5. unpow2 88.8%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin ky}} \]

      6. hypot-def 99.9%

         \[\leadsto \frac{\sin th}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin ky}} \]

   4. Applied egg-rr 99.9%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}} \]

   5. Taylor expanded in kx around 0 37.5%

      \[\leadsto \frac{\sin th}{\color{blue}{1 + 0.5 \cdot \frac{{kx}^{2}}{{\sin ky}^{2}}}} \]
   6. Step-by-step derivation

      1. expm1-log1p-u 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{kx}^{2}}{{\sin ky}^{2}}\right)\right)}} \]

      2. expm1-udef 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{\left(e^{\mathsf{log1p}\left(\frac{{kx}^{2}}{{\sin ky}^{2}}\right)} - 1\right)}} \]

      3. add-sqr-sqrt 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left(\color{blue}{\sqrt{\frac{{kx}^{2}}{{\sin ky}^{2}}} \cdot \sqrt{\frac{{kx}^{2}}{{\sin ky}^{2}}}}\right)} - 1\right)} \]

      4. pow2 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left(\color{blue}{{\left(\sqrt{\frac{{kx}^{2}}{{\sin ky}^{2}}}\right)}^{2}}\right)} - 1\right)} \]

      5. sqrt-div 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\color{blue}{\left(\frac{\sqrt{{kx}^{2}}}{\sqrt{{\sin ky}^{2}}}\right)}}^{2}\right)} - 1\right)} \]

      6. unpow2 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{\sqrt{\color{blue}{kx \cdot kx}}}{\sqrt{{\sin ky}^{2}}}\right)}^{2}\right)} - 1\right)} \]

      7. sqrt-prod 22.3%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{\color{blue}{\sqrt{kx} \cdot \sqrt{kx}}}{\sqrt{{\sin ky}^{2}}}\right)}^{2}\right)} - 1\right)} \]

      8. add-sqr-sqrt 38.0%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{\color{blue}{kx}}{\sqrt{{\sin ky}^{2}}}\right)}^{2}\right)} - 1\right)} \]

      9. unpow2 38.0%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{kx}{\sqrt{\color{blue}{\sin ky \cdot \sin ky}}}\right)}^{2}\right)} - 1\right)} \]

      10. sqrt-prod 39.2%

          \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{kx}{\color{blue}{\sqrt{\sin ky} \cdot \sqrt{\sin ky}}}\right)}^{2}\right)} - 1\right)} \]

      11. add-sqr-sqrt 40.7%

          \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{kx}{\color{blue}{\sin ky}}\right)}^{2}\right)} - 1\right)} \]

   7. Applied egg-rr 40.7%

      \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{\left(e^{\mathsf{log1p}\left({\left(\frac{kx}{\sin ky}\right)}^{2}\right)} - 1\right)}} \]
   8. Step-by-step derivation

      1. expm1-def 40.7%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left({\left(\frac{kx}{\sin ky}\right)}^{2}\right)\right)}} \]

      2. expm1-log1p 40.7%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{{\left(\frac{kx}{\sin ky}\right)}^{2}}} \]

   9. Simplified 40.7%

      \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{{\left(\frac{kx}{\sin ky}\right)}^{2}}} \]

   if 1e-3 < (sin.f64 kx)

   1. Initial program 99.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 63.3%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   4. Step-by-step derivation

      1. associate-*l/ 63.3%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sin kx}} \]

      2. *-commutative 63.3%

         \[\leadsto \frac{\color{blue}{\sin th \cdot \sin ky}}{\sin kx} \]

   5. Applied egg-rr 63.3%

      \[\leadsto \color{blue}{\frac{\sin th \cdot \sin ky}{\sin kx}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 43.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.05:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\frac{\sin th}{1 + 0.5 \cdot {\left(\frac{kx}{\sin ky}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky \cdot \sin th}{\sin kx}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 46.9% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.05:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\frac{\sin th}{1 + 0.5 \cdot {\left(\frac{kx}{\sin ky}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky \cdot \sin th}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.05)
   (* (sin th) (fabs (/ (sin ky) (sin kx))))
   (if (<= (sin kx) 0.001)
     (/ (sin th) (+ 1.0 (* 0.5 (pow (/ kx (sin ky)) 2.0))))
     (/ (* (sin ky) (sin th)) (sin kx)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.05) {
		tmp = sin(th) * fabs((sin(ky) / sin(kx)));
	} else if (sin(kx) <= 0.001) {
		tmp = sin(th) / (1.0 + (0.5 * pow((kx / sin(ky)), 2.0)));
	} else {
		tmp = (sin(ky) * sin(th)) / sin(kx);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.05d0)) then
        tmp = sin(th) * abs((sin(ky) / sin(kx)))
    else if (sin(kx) <= 0.001d0) then
        tmp = sin(th) / (1.0d0 + (0.5d0 * ((kx / sin(ky)) ** 2.0d0)))
    else
        tmp = (sin(ky) * sin(th)) / sin(kx)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.05) {
		tmp = Math.sin(th) * Math.abs((Math.sin(ky) / Math.sin(kx)));
	} else if (Math.sin(kx) <= 0.001) {
		tmp = Math.sin(th) / (1.0 + (0.5 * Math.pow((kx / Math.sin(ky)), 2.0)));
	} else {
		tmp = (Math.sin(ky) * Math.sin(th)) / Math.sin(kx);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.05:
		tmp = math.sin(th) * math.fabs((math.sin(ky) / math.sin(kx)))
	elif math.sin(kx) <= 0.001:
		tmp = math.sin(th) / (1.0 + (0.5 * math.pow((kx / math.sin(ky)), 2.0)))
	else:
		tmp = (math.sin(ky) * math.sin(th)) / math.sin(kx)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.05)
		tmp = Float64(sin(th) * abs(Float64(sin(ky) / sin(kx))));
	elseif (sin(kx) <= 0.001)
		tmp = Float64(sin(th) / Float64(1.0 + Float64(0.5 * (Float64(kx / sin(ky)) ^ 2.0))));
	else
		tmp = Float64(Float64(sin(ky) * sin(th)) / sin(kx));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.05)
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	elseif (sin(kx) <= 0.001)
		tmp = sin(th) / (1.0 + (0.5 * ((kx / sin(ky)) ^ 2.0)));
	else
		tmp = (sin(ky) * sin(th)) / sin(kx);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.05], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 0.001], N[(N[Sin[th], $MachinePrecision] / N[(1.0 + N[(0.5 * N[Power[N[(kx / N[Sin[ky], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sin[ky], $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.050000000000000003

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 15.4%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   4. Step-by-step derivation

      1. add-sqr-sqrt 9.9%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{\sin ky}{\sin kx}} \cdot \sqrt{\frac{\sin ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 24.5%

         \[\leadsto \color{blue}{\sqrt{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 24.5%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   5. Applied egg-rr 24.5%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   6. Step-by-step derivation

      1. unpow2 24.5%

         \[\leadsto \sqrt{\color{blue}{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 35.6%

         \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   7. Simplified 35.6%

      \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   if -0.050000000000000003 < (sin.f64 kx) < 1e-3

   1. Initial program 88.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 88.8%

         \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \cdot \sin th \]

      2. associate-*l/ 88.8%

         \[\leadsto \color{blue}{\frac{1 \cdot \sin th}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

      3. *-un-lft-identity 88.8%

         \[\leadsto \frac{\color{blue}{\sin th}}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}} \]

      4. unpow2 88.8%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin ky}} \]

      5. unpow2 88.8%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin ky}} \]

      6. hypot-def 99.9%

         \[\leadsto \frac{\sin th}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin ky}} \]

   4. Applied egg-rr 99.9%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}} \]

   5. Taylor expanded in kx around 0 37.5%

      \[\leadsto \frac{\sin th}{\color{blue}{1 + 0.5 \cdot \frac{{kx}^{2}}{{\sin ky}^{2}}}} \]
   6. Step-by-step derivation

      1. expm1-log1p-u 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{kx}^{2}}{{\sin ky}^{2}}\right)\right)}} \]

      2. expm1-udef 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{\left(e^{\mathsf{log1p}\left(\frac{{kx}^{2}}{{\sin ky}^{2}}\right)} - 1\right)}} \]

      3. add-sqr-sqrt 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left(\color{blue}{\sqrt{\frac{{kx}^{2}}{{\sin ky}^{2}}} \cdot \sqrt{\frac{{kx}^{2}}{{\sin ky}^{2}}}}\right)} - 1\right)} \]

      4. pow2 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left(\color{blue}{{\left(\sqrt{\frac{{kx}^{2}}{{\sin ky}^{2}}}\right)}^{2}}\right)} - 1\right)} \]

      5. sqrt-div 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\color{blue}{\left(\frac{\sqrt{{kx}^{2}}}{\sqrt{{\sin ky}^{2}}}\right)}}^{2}\right)} - 1\right)} \]

      6. unpow2 37.5%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{\sqrt{\color{blue}{kx \cdot kx}}}{\sqrt{{\sin ky}^{2}}}\right)}^{2}\right)} - 1\right)} \]

      7. sqrt-prod 22.3%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{\color{blue}{\sqrt{kx} \cdot \sqrt{kx}}}{\sqrt{{\sin ky}^{2}}}\right)}^{2}\right)} - 1\right)} \]

      8. add-sqr-sqrt 38.0%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{\color{blue}{kx}}{\sqrt{{\sin ky}^{2}}}\right)}^{2}\right)} - 1\right)} \]

      9. unpow2 38.0%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{kx}{\sqrt{\color{blue}{\sin ky \cdot \sin ky}}}\right)}^{2}\right)} - 1\right)} \]

      10. sqrt-prod 39.2%

          \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{kx}{\color{blue}{\sqrt{\sin ky} \cdot \sqrt{\sin ky}}}\right)}^{2}\right)} - 1\right)} \]

      11. add-sqr-sqrt 40.7%

          \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \left(e^{\mathsf{log1p}\left({\left(\frac{kx}{\color{blue}{\sin ky}}\right)}^{2}\right)} - 1\right)} \]

   7. Applied egg-rr 40.7%

      \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{\left(e^{\mathsf{log1p}\left({\left(\frac{kx}{\sin ky}\right)}^{2}\right)} - 1\right)}} \]
   8. Step-by-step derivation

      1. expm1-def 40.7%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left({\left(\frac{kx}{\sin ky}\right)}^{2}\right)\right)}} \]

      2. expm1-log1p 40.7%

         \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{{\left(\frac{kx}{\sin ky}\right)}^{2}}} \]

   9. Simplified 40.7%

      \[\leadsto \frac{\sin th}{1 + 0.5 \cdot \color{blue}{{\left(\frac{kx}{\sin ky}\right)}^{2}}} \]

   if 1e-3 < (sin.f64 kx)

   1. Initial program 99.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 63.3%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   4. Step-by-step derivation

      1. associate-*l/ 63.3%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sin kx}} \]

      2. *-commutative 63.3%

         \[\leadsto \frac{\color{blue}{\sin th \cdot \sin ky}}{\sin kx} \]

   5. Applied egg-rr 63.3%

      \[\leadsto \color{blue}{\frac{\sin th \cdot \sin ky}{\sin kx}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 44.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.05:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\frac{\sin th}{1 + 0.5 \cdot {\left(\frac{kx}{\sin ky}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky \cdot \sin th}{\sin kx}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 43.9% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.05:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 2 \cdot 10^{-11}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.05)
   (* (sin th) (fabs (/ ky (sin kx))))
   (if (<= (sin kx) 2e-11) (sin th) (* (sin th) (/ (sin ky) (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.05) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else if (sin(kx) <= 2e-11) {
		tmp = sin(th);
	} else {
		tmp = sin(th) * (sin(ky) / sin(kx));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.05d0)) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else if (sin(kx) <= 2d-11) then
        tmp = sin(th)
    else
        tmp = sin(th) * (sin(ky) / sin(kx))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.05) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else if (Math.sin(kx) <= 2e-11) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.sin(kx));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.05:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	elif math.sin(kx) <= 2e-11:
		tmp = math.sin(th)
	else:
		tmp = math.sin(th) * (math.sin(ky) / math.sin(kx))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.05)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	elseif (sin(kx) <= 2e-11)
		tmp = sin(th);
	else
		tmp = Float64(sin(th) * Float64(sin(ky) / sin(kx)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.05)
		tmp = sin(th) * abs((ky / sin(kx)));
	elseif (sin(kx) <= 2e-11)
		tmp = sin(th);
	else
		tmp = sin(th) * (sin(ky) / sin(kx));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.05], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 2e-11], N[Sin[th], $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.050000000000000003

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 15.4%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   4. Step-by-step derivation

      1. add-sqr-sqrt 9.9%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{\sin ky}{\sin kx}} \cdot \sqrt{\frac{\sin ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 24.5%

         \[\leadsto \color{blue}{\sqrt{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 24.5%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   5. Applied egg-rr 24.5%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   6. Step-by-step derivation

      1. unpow2 24.5%

         \[\leadsto \sqrt{\color{blue}{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 35.6%

         \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   7. Simplified 35.6%

      \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   8. Taylor expanded in ky around 0 31.3%

      \[\leadsto \left|\color{blue}{\frac{ky}{\sin kx}}\right| \cdot \sin th \]

   if -0.050000000000000003 < (sin.f64 kx) < 1.99999999999999988e-11

   1. Initial program 88.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 37.4%

      \[\leadsto \color{blue}{\sin th} \]

   if 1.99999999999999988e-11 < (sin.f64 kx)

   1. Initial program 99.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 60.8%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
3. Recombined 3 regimes into one program.

4. Final simplification 41.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.05:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 2 \cdot 10^{-11}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 99.6% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (sin ky) (/ (sin th) (hypot (sin ky) (sin kx)))))

C

double code(double kx, double ky, double th) {
	return sin(ky) * (sin(th) / hypot(sin(ky), sin(kx)));
}

Java

public static double code(double kx, double ky, double th) {
	return Math.sin(ky) * (Math.sin(th) / Math.hypot(Math.sin(ky), Math.sin(kx)));
}

Python

def code(kx, ky, th):
	return math.sin(ky) * (math.sin(th) / math.hypot(math.sin(ky), math.sin(kx)))

Julia

function code(kx, ky, th)
	return Float64(sin(ky) * Float64(sin(th) / hypot(sin(ky), sin(kx))))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = sin(ky) * (sin(th) / hypot(sin(ky), sin(kx)));
end

Wolfram

code[kx_, ky_, th_] := N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 93.6%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
2. Add Preprocessing
3. Step-by-step derivation

   1. expm1-log1p-u 93.4%

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)\right)} \]

   2. expm1-udef 41.6%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)} - 1} \]

4. Applied egg-rr 43.4%

   \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)} - 1} \]
5. Step-by-step derivation

   1. expm1-def 99.6%

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)\right)} \]

   2. expm1-log1p 99.7%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th} \]

   3. *-commutative 99.7%

      \[\leadsto \color{blue}{\sin th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   4. associate-*r/ 95.0%

      \[\leadsto \color{blue}{\frac{\sin th \cdot \sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   5. associate-*l/ 99.6%

      \[\leadsto \color{blue}{\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin ky} \]

   6. *-commutative 99.6%

      \[\leadsto \color{blue}{\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   7. hypot-def 93.5%

      \[\leadsto \sin ky \cdot \frac{\sin th}{\color{blue}{\sqrt{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \]

   8. unpow2 93.5%

      \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{\color{blue}{{\sin kx}^{2}} + \sin ky \cdot \sin ky}} \]

   9. unpow2 93.5%

      \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{{\sin kx}^{2} + \color{blue}{{\sin ky}^{2}}}} \]

   10. +-commutative 93.5%

       \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

   11. unpow2 93.5%

       \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

   12. unpow2 93.5%

       \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

   13. hypot-def 99.6%

       \[\leadsto \sin ky \cdot \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

6. Simplified 99.6%

   \[\leadsto \color{blue}{\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

7. Final simplification 99.6%

   \[\leadsto \sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]
8. Add Preprocessing

Alternative 9: 56.6% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;th \leq 0.00043:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{elif}\;th \leq 1.45 \cdot 10^{+41} \lor \neg \left(th \leq 1.14 \cdot 10^{+101}\right):\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= th 0.00043)
   (* (/ (sin ky) (hypot (sin ky) (sin kx))) th)
   (if (or (<= th 1.45e+41) (not (<= th 1.14e+101)))
     (* (sin th) (fabs (/ (sin ky) (sin kx))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.00043) {
		tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * th;
	} else if ((th <= 1.45e+41) || !(th <= 1.14e+101)) {
		tmp = sin(th) * fabs((sin(ky) / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.00043) {
		tmp = (Math.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx))) * th;
	} else if ((th <= 1.45e+41) || !(th <= 1.14e+101)) {
		tmp = Math.sin(th) * Math.abs((Math.sin(ky) / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if th <= 0.00043:
		tmp = (math.sin(ky) / math.hypot(math.sin(ky), math.sin(kx))) * th
	elif (th <= 1.45e+41) or not (th <= 1.14e+101):
		tmp = math.sin(th) * math.fabs((math.sin(ky) / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (th <= 0.00043)
		tmp = Float64(Float64(sin(ky) / hypot(sin(ky), sin(kx))) * th);
	elseif ((th <= 1.45e+41) || !(th <= 1.14e+101))
		tmp = Float64(sin(th) * abs(Float64(sin(ky) / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (th <= 0.00043)
		tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * th;
	elseif ((th <= 1.45e+41) || ~((th <= 1.14e+101)))
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[th, 0.00043], N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] * th), $MachinePrecision], If[Or[LessEqual[th, 1.45e+41], N[Not[LessEqual[th, 1.14e+101]], $MachinePrecision]], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if th < 4.29999999999999989e-4

   1. Initial program 92.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 59.5%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   4. Step-by-step derivation

      1. expm1-log1p-u 58.9%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}\right)\right)} \]

      2. expm1-udef 13.0%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}\right)} - 1} \]

   5. Applied egg-rr 13.2%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\right)} - 1} \]
   6. Step-by-step derivation

      1. expm1-def 67.4%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\right)\right)} \]

      2. expm1-log1p 67.9%

         \[\leadsto \color{blue}{th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

      3. hypot-def 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\color{blue}{\sqrt{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \]

      4. unpow2 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{{\sin kx}^{2}} + \sin ky \cdot \sin ky}} \]

      5. unpow2 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{{\sin kx}^{2} + \color{blue}{{\sin ky}^{2}}}} \]

      6. +-commutative 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      7. unpow2 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      8. unpow2 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      9. hypot-def 67.9%

         \[\leadsto th \cdot \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   7. Simplified 67.9%

      \[\leadsto \color{blue}{th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   if 4.29999999999999989e-4 < th < 1.44999999999999994e41 or 1.13999999999999997e101 < th

   1. Initial program 94.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 30.2%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   4. Step-by-step derivation

      1. add-sqr-sqrt 12.7%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{\sin ky}{\sin kx}} \cdot \sqrt{\frac{\sin ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 9.8%

         \[\leadsto \color{blue}{\sqrt{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 9.8%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   5. Applied egg-rr 9.8%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   6. Step-by-step derivation

      1. unpow2 9.8%

         \[\leadsto \sqrt{\color{blue}{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 22.4%

         \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   7. Simplified 22.4%

      \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   if 1.44999999999999994e41 < th < 1.13999999999999997e101

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 33.3%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 55.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 0.00043:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{elif}\;th \leq 1.45 \cdot 10^{+41} \lor \neg \left(th \leq 1.14 \cdot 10^{+101}\right):\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 38.2% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.1586:\\ \;\;\;\;\left|ky \cdot \frac{th}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.1586)
   (fabs (* ky (/ th (sin kx))))
   (if (<= (sin kx) 0.001) (sin th) (* (sin th) (/ ky (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.1586) {
		tmp = fabs((ky * (th / sin(kx))));
	} else if (sin(kx) <= 0.001) {
		tmp = sin(th);
	} else {
		tmp = sin(th) * (ky / sin(kx));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.1586d0)) then
        tmp = abs((ky * (th / sin(kx))))
    else if (sin(kx) <= 0.001d0) then
        tmp = sin(th)
    else
        tmp = sin(th) * (ky / sin(kx))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.1586) {
		tmp = Math.abs((ky * (th / Math.sin(kx))));
	} else if (Math.sin(kx) <= 0.001) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(th) * (ky / Math.sin(kx));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.1586:
		tmp = math.fabs((ky * (th / math.sin(kx))))
	elif math.sin(kx) <= 0.001:
		tmp = math.sin(th)
	else:
		tmp = math.sin(th) * (ky / math.sin(kx))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.1586)
		tmp = abs(Float64(ky * Float64(th / sin(kx))));
	elseif (sin(kx) <= 0.001)
		tmp = sin(th);
	else
		tmp = Float64(sin(th) * Float64(ky / sin(kx)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.1586)
		tmp = abs((ky * (th / sin(kx))));
	elseif (sin(kx) <= 0.001)
		tmp = sin(th);
	else
		tmp = sin(th) * (ky / sin(kx));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.1586], N[Abs[N[(ky * N[(th / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 0.001], N[Sin[th], $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.158599999999999991

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 57.9%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   4. Taylor expanded in ky around 0 16.7%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{\sin kx}} \]
   5. Step-by-step derivation

      1. associate-/l* 16.7%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   6. Simplified 16.7%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 16.7%

         \[\leadsto \frac{ky}{\color{blue}{\left(\sqrt[3]{\frac{\sin kx}{th}} \cdot \sqrt[3]{\frac{\sin kx}{th}}\right) \cdot \sqrt[3]{\frac{\sin kx}{th}}}} \]

      2. pow3 16.7%

         \[\leadsto \frac{ky}{\color{blue}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \]

   8. Applied egg-rr 16.7%

      \[\leadsto \frac{ky}{\color{blue}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \]
   9. Step-by-step derivation

      1. rem-square-sqrt 12.5%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \cdot \sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}}} \]

      2. sqrt-unprod 18.6%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}} \cdot \frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}}} \]

      3. pow2 18.6%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}\right)}^{2}}} \]

      4. rem-cube-cbrt 18.6%

         \[\leadsto \sqrt{{\left(\frac{ky}{\color{blue}{\frac{\sin kx}{th}}}\right)}^{2}} \]

      5. div-inv 18.6%

         \[\leadsto \sqrt{{\color{blue}{\left(ky \cdot \frac{1}{\frac{\sin kx}{th}}\right)}}^{2}} \]

      6. clear-num 18.6%

         \[\leadsto \sqrt{{\left(ky \cdot \color{blue}{\frac{th}{\sin kx}}\right)}^{2}} \]

   10. Applied egg-rr 18.6%

       \[\leadsto \color{blue}{\sqrt{{\left(ky \cdot \frac{th}{\sin kx}\right)}^{2}}} \]
   11. Step-by-step derivation

       1. unpow2 18.6%

          \[\leadsto \sqrt{\color{blue}{\left(ky \cdot \frac{th}{\sin kx}\right) \cdot \left(ky \cdot \frac{th}{\sin kx}\right)}} \]

       2. rem-sqrt-square 26.4%

          \[\leadsto \color{blue}{\left|ky \cdot \frac{th}{\sin kx}\right|} \]

   12. Simplified 26.4%

       \[\leadsto \color{blue}{\left|ky \cdot \frac{th}{\sin kx}\right|} \]

   if -0.158599999999999991 < (sin.f64 kx) < 1e-3

   1. Initial program 89.0%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 37.2%

      \[\leadsto \color{blue}{\sin th} \]

   if 1e-3 < (sin.f64 kx)

   1. Initial program 99.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 55.2%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
   4. Step-by-step derivation

      1. associate-/l* 55.1%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

      2. associate-/r/ 55.2%

         \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   5. Simplified 55.2%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 39.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.1586:\\ \;\;\;\;\left|ky \cdot \frac{th}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 38.2% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.1586:\\ \;\;\;\;\left|ky \cdot \frac{th}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.1586)
   (fabs (* ky (/ th (sin kx))))
   (if (<= (sin kx) 0.001) (sin th) (/ (* ky (sin th)) (sin kx)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.1586) {
		tmp = fabs((ky * (th / sin(kx))));
	} else if (sin(kx) <= 0.001) {
		tmp = sin(th);
	} else {
		tmp = (ky * sin(th)) / sin(kx);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.1586d0)) then
        tmp = abs((ky * (th / sin(kx))))
    else if (sin(kx) <= 0.001d0) then
        tmp = sin(th)
    else
        tmp = (ky * sin(th)) / sin(kx)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.1586) {
		tmp = Math.abs((ky * (th / Math.sin(kx))));
	} else if (Math.sin(kx) <= 0.001) {
		tmp = Math.sin(th);
	} else {
		tmp = (ky * Math.sin(th)) / Math.sin(kx);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.1586:
		tmp = math.fabs((ky * (th / math.sin(kx))))
	elif math.sin(kx) <= 0.001:
		tmp = math.sin(th)
	else:
		tmp = (ky * math.sin(th)) / math.sin(kx)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.1586)
		tmp = abs(Float64(ky * Float64(th / sin(kx))));
	elseif (sin(kx) <= 0.001)
		tmp = sin(th);
	else
		tmp = Float64(Float64(ky * sin(th)) / sin(kx));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.1586)
		tmp = abs((ky * (th / sin(kx))));
	elseif (sin(kx) <= 0.001)
		tmp = sin(th);
	else
		tmp = (ky * sin(th)) / sin(kx);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.1586], N[Abs[N[(ky * N[(th / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 0.001], N[Sin[th], $MachinePrecision], N[(N[(ky * N[Sin[th], $MachinePrecision]), $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.158599999999999991

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 57.9%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   4. Taylor expanded in ky around 0 16.7%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{\sin kx}} \]
   5. Step-by-step derivation

      1. associate-/l* 16.7%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   6. Simplified 16.7%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 16.7%

         \[\leadsto \frac{ky}{\color{blue}{\left(\sqrt[3]{\frac{\sin kx}{th}} \cdot \sqrt[3]{\frac{\sin kx}{th}}\right) \cdot \sqrt[3]{\frac{\sin kx}{th}}}} \]

      2. pow3 16.7%

         \[\leadsto \frac{ky}{\color{blue}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \]

   8. Applied egg-rr 16.7%

      \[\leadsto \frac{ky}{\color{blue}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \]
   9. Step-by-step derivation

      1. rem-square-sqrt 12.5%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \cdot \sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}}} \]

      2. sqrt-unprod 18.6%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}} \cdot \frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}}} \]

      3. pow2 18.6%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}\right)}^{2}}} \]

      4. rem-cube-cbrt 18.6%

         \[\leadsto \sqrt{{\left(\frac{ky}{\color{blue}{\frac{\sin kx}{th}}}\right)}^{2}} \]

      5. div-inv 18.6%

         \[\leadsto \sqrt{{\color{blue}{\left(ky \cdot \frac{1}{\frac{\sin kx}{th}}\right)}}^{2}} \]

      6. clear-num 18.6%

         \[\leadsto \sqrt{{\left(ky \cdot \color{blue}{\frac{th}{\sin kx}}\right)}^{2}} \]

   10. Applied egg-rr 18.6%

       \[\leadsto \color{blue}{\sqrt{{\left(ky \cdot \frac{th}{\sin kx}\right)}^{2}}} \]
   11. Step-by-step derivation

       1. unpow2 18.6%

          \[\leadsto \sqrt{\color{blue}{\left(ky \cdot \frac{th}{\sin kx}\right) \cdot \left(ky \cdot \frac{th}{\sin kx}\right)}} \]

       2. rem-sqrt-square 26.4%

          \[\leadsto \color{blue}{\left|ky \cdot \frac{th}{\sin kx}\right|} \]

   12. Simplified 26.4%

       \[\leadsto \color{blue}{\left|ky \cdot \frac{th}{\sin kx}\right|} \]

   if -0.158599999999999991 < (sin.f64 kx) < 1e-3

   1. Initial program 89.0%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 37.2%

      \[\leadsto \color{blue}{\sin th} \]

   if 1e-3 < (sin.f64 kx)

   1. Initial program 99.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 55.2%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 39.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.1586:\\ \;\;\;\;\left|ky \cdot \frac{th}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\sin kx}\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 38.8% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.1586:\\ \;\;\;\;\left|\sin ky \cdot \frac{th}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.1586)
   (fabs (* (sin ky) (/ th (sin kx))))
   (if (<= (sin kx) 0.001) (sin th) (/ (* ky (sin th)) (sin kx)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.1586) {
		tmp = fabs((sin(ky) * (th / sin(kx))));
	} else if (sin(kx) <= 0.001) {
		tmp = sin(th);
	} else {
		tmp = (ky * sin(th)) / sin(kx);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.1586d0)) then
        tmp = abs((sin(ky) * (th / sin(kx))))
    else if (sin(kx) <= 0.001d0) then
        tmp = sin(th)
    else
        tmp = (ky * sin(th)) / sin(kx)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.1586) {
		tmp = Math.abs((Math.sin(ky) * (th / Math.sin(kx))));
	} else if (Math.sin(kx) <= 0.001) {
		tmp = Math.sin(th);
	} else {
		tmp = (ky * Math.sin(th)) / Math.sin(kx);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.1586:
		tmp = math.fabs((math.sin(ky) * (th / math.sin(kx))))
	elif math.sin(kx) <= 0.001:
		tmp = math.sin(th)
	else:
		tmp = (ky * math.sin(th)) / math.sin(kx)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.1586)
		tmp = abs(Float64(sin(ky) * Float64(th / sin(kx))));
	elseif (sin(kx) <= 0.001)
		tmp = sin(th);
	else
		tmp = Float64(Float64(ky * sin(th)) / sin(kx));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.1586)
		tmp = abs((sin(ky) * (th / sin(kx))));
	elseif (sin(kx) <= 0.001)
		tmp = sin(th);
	else
		tmp = (ky * sin(th)) / sin(kx);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.1586], N[Abs[N[(N[Sin[ky], $MachinePrecision] * N[(th / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 0.001], N[Sin[th], $MachinePrecision], N[(N[(ky * N[Sin[th], $MachinePrecision]), $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.158599999999999991

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 16.3%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]

   4. Taylor expanded in th around 0 16.5%

      \[\leadsto \color{blue}{\frac{th \cdot \sin ky}{\sin kx}} \]
   5. Step-by-step derivation

      1. associate-/l* 16.5%

         \[\leadsto \color{blue}{\frac{th}{\frac{\sin kx}{\sin ky}}} \]

   6. Simplified 16.5%

      \[\leadsto \color{blue}{\frac{th}{\frac{\sin kx}{\sin ky}}} \]
   7. Step-by-step derivation

      1. add-sqr-sqrt 12.7%

         \[\leadsto \color{blue}{\sqrt{\frac{th}{\frac{\sin kx}{\sin ky}}} \cdot \sqrt{\frac{th}{\frac{\sin kx}{\sin ky}}}} \]

      2. sqrt-unprod 20.6%

         \[\leadsto \color{blue}{\sqrt{\frac{th}{\frac{\sin kx}{\sin ky}} \cdot \frac{th}{\frac{\sin kx}{\sin ky}}}} \]

      3. pow2 20.6%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{th}{\frac{\sin kx}{\sin ky}}\right)}^{2}}} \]

      4. associate-/r/ 20.6%

         \[\leadsto \sqrt{{\color{blue}{\left(\frac{th}{\sin kx} \cdot \sin ky\right)}}^{2}} \]

      5. *-commutative 20.6%

         \[\leadsto \sqrt{{\color{blue}{\left(\sin ky \cdot \frac{th}{\sin kx}\right)}}^{2}} \]

   8. Applied egg-rr 20.6%

      \[\leadsto \color{blue}{\sqrt{{\left(\sin ky \cdot \frac{th}{\sin kx}\right)}^{2}}} \]
   9. Step-by-step derivation

      1. unpow2 20.6%

         \[\leadsto \sqrt{\color{blue}{\left(\sin ky \cdot \frac{th}{\sin kx}\right) \cdot \left(\sin ky \cdot \frac{th}{\sin kx}\right)}} \]

      2. rem-sqrt-square 28.5%

         \[\leadsto \color{blue}{\left|\sin ky \cdot \frac{th}{\sin kx}\right|} \]

   10. Simplified 28.5%

       \[\leadsto \color{blue}{\left|\sin ky \cdot \frac{th}{\sin kx}\right|} \]

   if -0.158599999999999991 < (sin.f64 kx) < 1e-3

   1. Initial program 89.0%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 37.2%

      \[\leadsto \color{blue}{\sin th} \]

   if 1e-3 < (sin.f64 kx)

   1. Initial program 99.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 55.2%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 39.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.1586:\\ \;\;\;\;\left|\sin ky \cdot \frac{th}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 0.001:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\sin kx}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 63.7% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;th \leq 0.000475:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= th 0.000475)
   (* (/ (sin ky) (hypot (sin ky) (sin kx))) th)
   (/ (* ky (sin th)) (hypot (sin kx) (sin ky)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.000475) {
		tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * th;
	} else {
		tmp = (ky * sin(th)) / hypot(sin(kx), sin(ky));
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.000475) {
		tmp = (Math.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx))) * th;
	} else {
		tmp = (ky * Math.sin(th)) / Math.hypot(Math.sin(kx), Math.sin(ky));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if th <= 0.000475:
		tmp = (math.sin(ky) / math.hypot(math.sin(ky), math.sin(kx))) * th
	else:
		tmp = (ky * math.sin(th)) / math.hypot(math.sin(kx), math.sin(ky))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (th <= 0.000475)
		tmp = Float64(Float64(sin(ky) / hypot(sin(ky), sin(kx))) * th);
	else
		tmp = Float64(Float64(ky * sin(th)) / hypot(sin(kx), sin(ky)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (th <= 0.000475)
		tmp = (sin(ky) / hypot(sin(ky), sin(kx))) * th;
	else
		tmp = (ky * sin(th)) / hypot(sin(kx), sin(ky));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[th, 0.000475], N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] * th), $MachinePrecision], N[(N[(ky * N[Sin[th], $MachinePrecision]), $MachinePrecision] / N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 4.74999999999999999e-4

   1. Initial program 92.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 59.5%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   4. Step-by-step derivation

      1. expm1-log1p-u 58.9%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}\right)\right)} \]

      2. expm1-udef 13.0%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}\right)} - 1} \]

   5. Applied egg-rr 13.2%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\right)} - 1} \]
   6. Step-by-step derivation

      1. expm1-def 67.4%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\right)\right)} \]

      2. expm1-log1p 67.9%

         \[\leadsto \color{blue}{th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

      3. hypot-def 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\color{blue}{\sqrt{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \]

      4. unpow2 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{{\sin kx}^{2}} + \sin ky \cdot \sin ky}} \]

      5. unpow2 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{{\sin kx}^{2} + \color{blue}{{\sin ky}^{2}}}} \]

      6. +-commutative 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

      7. unpow2 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

      8. unpow2 63.1%

         \[\leadsto th \cdot \frac{\sin ky}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

      9. hypot-def 67.9%

         \[\leadsto th \cdot \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   7. Simplified 67.9%

      \[\leadsto \color{blue}{th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

   if 4.74999999999999999e-4 < th

   1. Initial program 95.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. expm1-log1p-u 95.5%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)\right)} \]

      2. expm1-udef 59.4%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)} - 1} \]

   4. Applied egg-rr 62.0%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)} - 1} \]
   5. Step-by-step derivation

      1. expm1-def 99.5%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)\right)} \]

      2. expm1-log1p 99.7%

         \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th} \]

      3. associate-*l/ 99.6%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   6. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   7. Taylor expanded in ky around 0 52.8%

      \[\leadsto \frac{\color{blue}{ky \cdot \sin th}}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 63.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 0.000475:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot th\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 30.9% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;kx \leq 0.0008:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= kx 0.0008) (sin th) (* (sin th) (fabs (/ ky (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 0.0008) {
		tmp = sin(th);
	} else {
		tmp = sin(th) * fabs((ky / sin(kx)));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (kx <= 0.0008d0) then
        tmp = sin(th)
    else
        tmp = sin(th) * abs((ky / sin(kx)))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 0.0008) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if kx <= 0.0008:
		tmp = math.sin(th)
	else:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (kx <= 0.0008)
		tmp = sin(th);
	else
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (kx <= 0.0008)
		tmp = sin(th);
	else
		tmp = sin(th) * abs((ky / sin(kx)));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[kx, 0.0008], N[Sin[th], $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if kx < 8.00000000000000038e-4

   1. Initial program 91.9%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 28.7%

      \[\leadsto \color{blue}{\sin th} \]

   if 8.00000000000000038e-4 < kx

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in ky around 0 31.2%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   4. Step-by-step derivation

      1. add-sqr-sqrt 11.1%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{\sin ky}{\sin kx}} \cdot \sqrt{\frac{\sin ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 20.6%

         \[\leadsto \color{blue}{\sqrt{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 20.6%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   5. Applied egg-rr 20.6%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   6. Step-by-step derivation

      1. unpow2 20.6%

         \[\leadsto \sqrt{\color{blue}{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 28.4%

         \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   7. Simplified 28.4%

      \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   8. Taylor expanded in ky around 0 23.4%

      \[\leadsto \left|\color{blue}{\frac{ky}{\sin kx}}\right| \cdot \sin th \]
3. Recombined 2 regimes into one program.

4. Final simplification 27.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;kx \leq 0.0008:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \end{array} \]
5. Add Preprocessing

Alternative 15: 27.6% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;kx \leq 1.5 \cdot 10^{+18}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\left|ky \cdot \frac{th}{\sin kx}\right|\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= kx 1.5e+18) (sin th) (fabs (* ky (/ th (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.5e+18) {
		tmp = sin(th);
	} else {
		tmp = fabs((ky * (th / sin(kx))));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (kx <= 1.5d+18) then
        tmp = sin(th)
    else
        tmp = abs((ky * (th / sin(kx))))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.5e+18) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.abs((ky * (th / Math.sin(kx))));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if kx <= 1.5e+18:
		tmp = math.sin(th)
	else:
		tmp = math.fabs((ky * (th / math.sin(kx))))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (kx <= 1.5e+18)
		tmp = sin(th);
	else
		tmp = abs(Float64(ky * Float64(th / sin(kx))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (kx <= 1.5e+18)
		tmp = sin(th);
	else
		tmp = abs((ky * (th / sin(kx))));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[kx, 1.5e+18], N[Sin[th], $MachinePrecision], N[Abs[N[(ky * N[(th / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if kx < 1.5e18

   1. Initial program 92.2%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 28.2%

      \[\leadsto \color{blue}{\sin th} \]

   if 1.5e18 < kx

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 46.7%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   4. Taylor expanded in ky around 0 14.9%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{\sin kx}} \]
   5. Step-by-step derivation

      1. associate-/l* 14.9%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   6. Simplified 14.9%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 14.8%

         \[\leadsto \frac{ky}{\color{blue}{\left(\sqrt[3]{\frac{\sin kx}{th}} \cdot \sqrt[3]{\frac{\sin kx}{th}}\right) \cdot \sqrt[3]{\frac{\sin kx}{th}}}} \]

      2. pow3 14.9%

         \[\leadsto \frac{ky}{\color{blue}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \]

   8. Applied egg-rr 14.9%

      \[\leadsto \frac{ky}{\color{blue}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \]
   9. Step-by-step derivation

      1. rem-square-sqrt 13.3%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}} \cdot \sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}}} \]

      2. sqrt-unprod 14.8%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}} \cdot \frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}}} \]

      3. pow2 14.8%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{{\left(\sqrt[3]{\frac{\sin kx}{th}}\right)}^{3}}\right)}^{2}}} \]

      4. rem-cube-cbrt 14.8%

         \[\leadsto \sqrt{{\left(\frac{ky}{\color{blue}{\frac{\sin kx}{th}}}\right)}^{2}} \]

      5. div-inv 14.8%

         \[\leadsto \sqrt{{\color{blue}{\left(ky \cdot \frac{1}{\frac{\sin kx}{th}}\right)}}^{2}} \]

      6. clear-num 14.8%

         \[\leadsto \sqrt{{\left(ky \cdot \color{blue}{\frac{th}{\sin kx}}\right)}^{2}} \]

   10. Applied egg-rr 14.8%

       \[\leadsto \color{blue}{\sqrt{{\left(ky \cdot \frac{th}{\sin kx}\right)}^{2}}} \]
   11. Step-by-step derivation

       1. unpow2 14.8%

          \[\leadsto \sqrt{\color{blue}{\left(ky \cdot \frac{th}{\sin kx}\right) \cdot \left(ky \cdot \frac{th}{\sin kx}\right)}} \]

       2. rem-sqrt-square 18.9%

          \[\leadsto \color{blue}{\left|ky \cdot \frac{th}{\sin kx}\right|} \]

   12. Simplified 18.9%

       \[\leadsto \color{blue}{\left|ky \cdot \frac{th}{\sin kx}\right|} \]
3. Recombined 2 regimes into one program.

4. Final simplification 26.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;kx \leq 1.5 \cdot 10^{+18}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\left|ky \cdot \frac{th}{\sin kx}\right|\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 27.6% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;kx \leq 1.7 \cdot 10^{+18}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\left|th \cdot \frac{ky}{\sin kx}\right|\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= kx 1.7e+18) (sin th) (fabs (* th (/ ky (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.7e+18) {
		tmp = sin(th);
	} else {
		tmp = fabs((th * (ky / sin(kx))));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (kx <= 1.7d+18) then
        tmp = sin(th)
    else
        tmp = abs((th * (ky / sin(kx))))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.7e+18) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.abs((th * (ky / Math.sin(kx))));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if kx <= 1.7e+18:
		tmp = math.sin(th)
	else:
		tmp = math.fabs((th * (ky / math.sin(kx))))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (kx <= 1.7e+18)
		tmp = sin(th);
	else
		tmp = abs(Float64(th * Float64(ky / sin(kx))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (kx <= 1.7e+18)
		tmp = sin(th);
	else
		tmp = abs((th * (ky / sin(kx))));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[kx, 1.7e+18], N[Sin[th], $MachinePrecision], N[Abs[N[(th * N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if kx < 1.7e18

   1. Initial program 92.2%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 28.2%

      \[\leadsto \color{blue}{\sin th} \]

   if 1.7e18 < kx

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 46.7%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   4. Taylor expanded in ky around 0 14.9%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{\sin kx}} \]
   5. Step-by-step derivation

      1. associate-/l* 14.9%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   6. Simplified 14.9%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]
   7. Step-by-step derivation

      1. add-sqr-sqrt 13.3%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{\frac{\sin kx}{th}}} \cdot \sqrt{\frac{ky}{\frac{\sin kx}{th}}}} \]

      2. sqrt-unprod 14.8%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{\frac{\sin kx}{th}} \cdot \frac{ky}{\frac{\sin kx}{th}}}} \]

      3. pow2 14.8%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{\frac{\sin kx}{th}}\right)}^{2}}} \]

      4. div-inv 14.8%

         \[\leadsto \sqrt{{\color{blue}{\left(ky \cdot \frac{1}{\frac{\sin kx}{th}}\right)}}^{2}} \]

      5. clear-num 14.8%

         \[\leadsto \sqrt{{\left(ky \cdot \color{blue}{\frac{th}{\sin kx}}\right)}^{2}} \]

   8. Applied egg-rr 14.8%

      \[\leadsto \color{blue}{\sqrt{{\left(ky \cdot \frac{th}{\sin kx}\right)}^{2}}} \]
   9. Step-by-step derivation

      1. unpow2 14.8%

         \[\leadsto \sqrt{\color{blue}{\left(ky \cdot \frac{th}{\sin kx}\right) \cdot \left(ky \cdot \frac{th}{\sin kx}\right)}} \]

      2. rem-sqrt-square 18.9%

         \[\leadsto \color{blue}{\left|ky \cdot \frac{th}{\sin kx}\right|} \]

      3. *-commutative 18.9%

         \[\leadsto \left|\color{blue}{\frac{th}{\sin kx} \cdot ky}\right| \]

      4. associate-*l/ 18.9%

         \[\leadsto \left|\color{blue}{\frac{th \cdot ky}{\sin kx}}\right| \]

      5. associate-*r/ 18.9%

         \[\leadsto \left|\color{blue}{th \cdot \frac{ky}{\sin kx}}\right| \]

   10. Simplified 18.9%

       \[\leadsto \color{blue}{\left|th \cdot \frac{ky}{\sin kx}\right|} \]
3. Recombined 2 regimes into one program.

4. Final simplification 26.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;kx \leq 1.7 \cdot 10^{+18}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\left|th \cdot \frac{ky}{\sin kx}\right|\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 27.8% accurate, 6.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;kx \leq 1.75 \cdot 10^{+18}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;th \cdot \frac{ky}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= kx 1.75e+18) (sin th) (* th (/ ky (sin kx)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.75e+18) {
		tmp = sin(th);
	} else {
		tmp = th * (ky / sin(kx));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (kx <= 1.75d+18) then
        tmp = sin(th)
    else
        tmp = th * (ky / sin(kx))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.75e+18) {
		tmp = Math.sin(th);
	} else {
		tmp = th * (ky / Math.sin(kx));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if kx <= 1.75e+18:
		tmp = math.sin(th)
	else:
		tmp = th * (ky / math.sin(kx))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (kx <= 1.75e+18)
		tmp = sin(th);
	else
		tmp = Float64(th * Float64(ky / sin(kx)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (kx <= 1.75e+18)
		tmp = sin(th);
	else
		tmp = th * (ky / sin(kx));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[kx, 1.75e+18], N[Sin[th], $MachinePrecision], N[(th * N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if kx < 1.75e18

   1. Initial program 92.2%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 28.2%

      \[\leadsto \color{blue}{\sin th} \]

   if 1.75e18 < kx

   1. Initial program 99.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 46.7%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   4. Taylor expanded in ky around 0 14.9%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{\sin kx}} \]
   5. Step-by-step derivation

      1. associate-/l* 14.9%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   6. Simplified 14.9%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]
   7. Step-by-step derivation

      1. associate-/r/ 14.9%

         \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot th} \]

   8. Applied egg-rr 14.9%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 25.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;kx \leq 1.75 \cdot 10^{+18}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;th \cdot \frac{ky}{\sin kx}\\ \end{array} \]
5. Add Preprocessing

Alternative 18: 24.4% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 5.5 \cdot 10^{-174}:\\ \;\;\;\;th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 5.5e-174) (* th (/ ky kx)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 5.5e-174) {
		tmp = th * (ky / kx);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 5.5d-174) then
        tmp = th * (ky / kx)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 5.5e-174) {
		tmp = th * (ky / kx);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 5.5e-174:
		tmp = th * (ky / kx)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 5.5e-174)
		tmp = Float64(th * Float64(ky / kx));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 5.5e-174)
		tmp = th * (ky / kx);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 5.5e-174], N[(th * N[(ky / kx), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 5.4999999999999999e-174

   1. Initial program 89.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 40.4%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   4. Taylor expanded in ky around 0 15.9%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{\sin kx}} \]
   5. Step-by-step derivation

      1. associate-/l* 16.3%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   6. Simplified 16.3%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   7. Taylor expanded in kx around 0 13.2%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{kx}} \]
   8. Step-by-step derivation

      1. associate-/l* 13.6%

         \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]

   9. Simplified 13.6%

      \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]
   10. Step-by-step derivation

       1. associate-/r/ 13.7%

          \[\leadsto \color{blue}{\frac{ky}{kx} \cdot th} \]

   11. Applied egg-rr 13.7%

       \[\leadsto \color{blue}{\frac{ky}{kx} \cdot th} \]

   if 5.4999999999999999e-174 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in kx around 0 36.6%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 22.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 5.5 \cdot 10^{-174}:\\ \;\;\;\;th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 19: 17.9% accurate, 70.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 2.2 \cdot 10^{-168}:\\ \;\;\;\;th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 2.2e-168) (* th (/ ky kx)) th))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 2.2e-168) {
		tmp = th * (ky / kx);
	} else {
		tmp = th;
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 2.2d-168) then
        tmp = th * (ky / kx)
    else
        tmp = th
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 2.2e-168) {
		tmp = th * (ky / kx);
	} else {
		tmp = th;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 2.2e-168:
		tmp = th * (ky / kx)
	else:
		tmp = th
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 2.2e-168)
		tmp = Float64(th * Float64(ky / kx));
	else
		tmp = th;
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 2.2e-168)
		tmp = th * (ky / kx);
	else
		tmp = th;
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 2.2e-168], N[(th * N[(ky / kx), $MachinePrecision]), $MachinePrecision], th]

Derivation

1. Split input into 2 regimes

2. if ky < 2.1999999999999998e-168

   1. Initial program 89.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 40.1%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   4. Taylor expanded in ky around 0 15.8%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{\sin kx}} \]
   5. Step-by-step derivation

      1. associate-/l* 16.2%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   6. Simplified 16.2%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{th}}} \]

   7. Taylor expanded in kx around 0 13.2%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{kx}} \]
   8. Step-by-step derivation

      1. associate-/l* 13.6%

         \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]

   9. Simplified 13.6%

      \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]
   10. Step-by-step derivation

       1. associate-/r/ 13.6%

          \[\leadsto \color{blue}{\frac{ky}{kx} \cdot th} \]

   11. Applied egg-rr 13.6%

       \[\leadsto \color{blue}{\frac{ky}{kx} \cdot th} \]

   if 2.1999999999999998e-168 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Add Preprocessing

   3. Taylor expanded in th around 0 49.7%

      \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   4. Taylor expanded in kx around 0 17.5%

      \[\leadsto \color{blue}{th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 15.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 2.2 \cdot 10^{-168}:\\ \;\;\;\;th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;th\\ \end{array} \]
5. Add Preprocessing

Alternative 20: 14.1% accurate, 709.0× speedup

Math

\[\begin{array}{l} \\ th \end{array} \]

FPCore

(FPCore (kx ky th) :precision binary64 th)

C

double code(double kx, double ky, double th) {
	return th;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = th
end function

Java

public static double code(double kx, double ky, double th) {
	return th;
}

Python

def code(kx, ky, th):
	return th

Julia

function code(kx, ky, th)
	return th
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = th;
end

Wolfram

code[kx_, ky_, th_] := th

Derivation

1. Initial program 93.6%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
2. Add Preprocessing

3. Taylor expanded in th around 0 43.9%

   \[\leadsto \color{blue}{\left(th \cdot \sin ky\right) \cdot \sqrt{\frac{1}{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

4. Taylor expanded in kx around 0 11.7%

   \[\leadsto \color{blue}{th} \]

5. Final simplification 11.7%

   \[\leadsto th \]
6. Add Preprocessing

Reproduce

herbie shell --seed 2024016 
(FPCore (kx ky th)
  :name "Toniolo and Linder, Equation (3b), real"
  :precision binary64
  (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))