Toniolo and Linder, Equation (3b), real

Percentage Accurate: 94.3% → 99.7%

Time: 23.5s

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: 94.3% 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 91.3%

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

   1. +-commutative 91.3%

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

   2. unpow2 91.3%

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

   3. unpow2 91.3%

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

   4. hypot-def 99.7%

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

3. Simplified 99.7%

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

4. Final simplification 99.7%

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

Alternative 2: 74.9% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}\\ \mathbf{if}\;\sin ky \leq -0.635:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\sin ky \leq -0.18:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -0.005:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (/ th (/ (hypot (sin kx) (sin ky)) (sin ky)))))
   (if (<= (sin ky) -0.635)
     t_1
     (if (<= (sin ky) -0.18)
       (fabs (sin th))
       (if (<= (sin ky) -0.005)
         t_1
         (if (<= (sin ky) 1e-15)
           (* (sin ky) (/ (sin th) (hypot ky (sin kx))))
           (sin th)))))))

C

double code(double kx, double ky, double th) {
	double t_1 = th / (hypot(sin(kx), sin(ky)) / sin(ky));
	double tmp;
	if (sin(ky) <= -0.635) {
		tmp = t_1;
	} else if (sin(ky) <= -0.18) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= -0.005) {
		tmp = t_1;
	} else if (sin(ky) <= 1e-15) {
		tmp = sin(ky) * (sin(th) / hypot(ky, sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double t_1 = th / (Math.hypot(Math.sin(kx), Math.sin(ky)) / Math.sin(ky));
	double tmp;
	if (Math.sin(ky) <= -0.635) {
		tmp = t_1;
	} else if (Math.sin(ky) <= -0.18) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= -0.005) {
		tmp = t_1;
	} else if (Math.sin(ky) <= 1e-15) {
		tmp = Math.sin(ky) * (Math.sin(th) / Math.hypot(ky, Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = th / (math.hypot(math.sin(kx), math.sin(ky)) / math.sin(ky))
	tmp = 0
	if math.sin(ky) <= -0.635:
		tmp = t_1
	elif math.sin(ky) <= -0.18:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= -0.005:
		tmp = t_1
	elif math.sin(ky) <= 1e-15:
		tmp = math.sin(ky) * (math.sin(th) / math.hypot(ky, math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	t_1 = Float64(th / Float64(hypot(sin(kx), sin(ky)) / sin(ky)))
	tmp = 0.0
	if (sin(ky) <= -0.635)
		tmp = t_1;
	elseif (sin(ky) <= -0.18)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -0.005)
		tmp = t_1;
	elseif (sin(ky) <= 1e-15)
		tmp = Float64(sin(ky) * Float64(sin(th) / hypot(ky, sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = th / (hypot(sin(kx), sin(ky)) / sin(ky));
	tmp = 0.0;
	if (sin(ky) <= -0.635)
		tmp = t_1;
	elseif (sin(ky) <= -0.18)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -0.005)
		tmp = t_1;
	elseif (sin(ky) <= 1e-15)
		tmp = sin(ky) * (sin(th) / hypot(ky, sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[(th / N[(N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision] / N[Sin[ky], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Sin[ky], $MachinePrecision], -0.635], t$95$1, If[LessEqual[N[Sin[ky], $MachinePrecision], -0.18], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], -0.005], t$95$1, If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-15], N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / N[Sqrt[ky ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]]]

Derivation

1. Split input into 4 regimes

2. if (sin.f64 ky) < -0.63500000000000001 or -0.17999999999999999 < (sin.f64 ky) < -0.0050000000000000001

   1. Initial program 99.8%

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

      1. *-commutative 99.8%

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

      2. clear-num 99.7%

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

      3. +-commutative 99.7%

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

      4. unpow2 99.7%

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

      5. unpow2 99.7%

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

      6. hypot-udef 99.7%

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

      7. un-div-inv 99.6%

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

   3. Applied egg-rr 99.6%

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

   4. Taylor expanded in th around 0 56.1%

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

   if -0.63500000000000001 < (sin.f64 ky) < -0.17999999999999999

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.1%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 0.6%

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

      2. sqrt-unprod 39.8%

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

      3. pow2 39.8%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 39.8%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 39.8%

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

      2. rem-sqrt-square 47.2%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 47.2%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -0.0050000000000000001 < (sin.f64 ky) < 1.0000000000000001e-15

   1. Initial program 81.2%

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

      1. associate-*l/ 78.7%

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

      2. *-commutative 78.7%

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

      3. associate-*l/ 81.2%

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

      4. +-commutative 81.2%

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

      5. unpow2 81.2%

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

      6. unpow2 81.2%

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

      7. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 99.7%

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

   if 1.0000000000000001e-15 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 4 regimes into one program.

4. Final simplification 77.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.635:\\ \;\;\;\;\frac{th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}\\ \mathbf{elif}\;\sin ky \leq -0.18:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 3: 74.9% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.635:\\ \;\;\;\;\frac{th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}\\ \mathbf{elif}\;\sin ky \leq -0.18:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.635)
   (/ th (/ (hypot (sin kx) (sin ky)) (sin ky)))
   (if (<= (sin ky) -0.18)
     (fabs (sin th))
     (if (<= (sin ky) -0.005)
       (/ (sin ky) (/ (hypot (sin ky) (sin kx)) th))
       (if (<= (sin ky) 1e-15)
         (* (sin ky) (/ (sin th) (hypot ky (sin kx))))
         (sin th))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.635) {
		tmp = th / (hypot(sin(kx), sin(ky)) / sin(ky));
	} else if (sin(ky) <= -0.18) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= -0.005) {
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) / th);
	} else if (sin(ky) <= 1e-15) {
		tmp = sin(ky) * (sin(th) / hypot(ky, sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -0.635) {
		tmp = th / (Math.hypot(Math.sin(kx), Math.sin(ky)) / Math.sin(ky));
	} else if (Math.sin(ky) <= -0.18) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= -0.005) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(ky), Math.sin(kx)) / th);
	} else if (Math.sin(ky) <= 1e-15) {
		tmp = Math.sin(ky) * (Math.sin(th) / Math.hypot(ky, Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -0.635:
		tmp = th / (math.hypot(math.sin(kx), math.sin(ky)) / math.sin(ky))
	elif math.sin(ky) <= -0.18:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= -0.005:
		tmp = math.sin(ky) / (math.hypot(math.sin(ky), math.sin(kx)) / th)
	elif math.sin(ky) <= 1e-15:
		tmp = math.sin(ky) * (math.sin(th) / math.hypot(ky, math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -0.635)
		tmp = Float64(th / Float64(hypot(sin(kx), sin(ky)) / sin(ky)));
	elseif (sin(ky) <= -0.18)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -0.005)
		tmp = Float64(sin(ky) / Float64(hypot(sin(ky), sin(kx)) / th));
	elseif (sin(ky) <= 1e-15)
		tmp = Float64(sin(ky) * Float64(sin(th) / hypot(ky, sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -0.635)
		tmp = th / (hypot(sin(kx), sin(ky)) / sin(ky));
	elseif (sin(ky) <= -0.18)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -0.005)
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) / th);
	elseif (sin(ky) <= 1e-15)
		tmp = sin(ky) * (sin(th) / hypot(ky, sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.635], N[(th / N[(N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision] / N[Sin[ky], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], -0.18], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], -0.005], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision] / th), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-15], N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / N[Sqrt[ky ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]]

Derivation

1. Split input into 5 regimes

2. if (sin.f64 ky) < -0.63500000000000001

   1. Initial program 99.9%

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

      1. *-commutative 99.9%

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

      2. clear-num 99.9%

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

      3. +-commutative 99.9%

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

      4. unpow2 99.9%

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

      5. unpow2 99.9%

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

      6. hypot-udef 99.9%

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

      7. un-div-inv 99.8%

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

   3. Applied egg-rr 99.8%

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

   4. Taylor expanded in th around 0 50.0%

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

   if -0.63500000000000001 < (sin.f64 ky) < -0.17999999999999999

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.1%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 0.6%

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

      2. sqrt-unprod 39.8%

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

      3. pow2 39.8%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 39.8%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 39.8%

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

      2. rem-sqrt-square 47.2%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 47.2%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -0.17999999999999999 < (sin.f64 ky) < -0.0050000000000000001

   1. Initial program 99.3%

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

      1. associate-*l/ 99.2%

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

      2. *-commutative 99.2%

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

      3. associate-*l/ 99.3%

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

      4. +-commutative 99.3%

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

      5. unpow2 99.3%

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

      6. unpow2 99.3%

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

      7. hypot-def 99.3%

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

   3. Simplified 99.3%

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

      1. expm1-log1p-u 99.3%

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

   5. Applied egg-rr 99.3%

      \[\leadsto \frac{\sin th}{\mathsf{hypot}\left(\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin ky\right)\right)}, \sin kx\right)} \cdot \sin ky \]
   6. Step-by-step derivation

      1. *-commutative 99.3%

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

      2. clear-num 99.4%

         \[\leadsto \sin ky \cdot \color{blue}{\frac{1}{\frac{\mathsf{hypot}\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\sin ky\right)\right), \sin kx\right)}{\sin th}}} \]

      3. un-div-inv 99.3%

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

      4. expm1-log1p-u 99.3%

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

   7. Applied egg-rr 99.3%

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

   8. Taylor expanded in th around 0 85.4%

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

   if -0.0050000000000000001 < (sin.f64 ky) < 1.0000000000000001e-15

   1. Initial program 81.2%

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

      1. associate-*l/ 78.7%

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

      2. *-commutative 78.7%

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

      3. associate-*l/ 81.2%

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

      4. +-commutative 81.2%

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

      5. unpow2 81.2%

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

      6. unpow2 81.2%

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

      7. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 99.7%

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

   if 1.0000000000000001e-15 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 5 regimes into one program.

4. Final simplification 77.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.635:\\ \;\;\;\;\frac{th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}\\ \mathbf{elif}\;\sin ky \leq -0.18:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 4: 74.9% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;\sin ky \leq -0.635:\\ \;\;\;\;\frac{th}{\frac{t_1}{\sin ky}}\\ \mathbf{elif}\;\sin ky \leq -0.18:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\frac{\sin th}{\frac{t_1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= (sin ky) -0.635)
     (/ th (/ t_1 (sin ky)))
     (if (<= (sin ky) -0.18)
       (fabs (sin th))
       (if (<= (sin ky) -0.005)
         (/ (sin ky) (/ (hypot (sin ky) (sin kx)) th))
         (if (<= (sin ky) 1e-15) (/ (sin th) (/ t_1 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) <= -0.635) {
		tmp = th / (t_1 / sin(ky));
	} else if (sin(ky) <= -0.18) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= -0.005) {
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) / th);
	} else if (sin(ky) <= 1e-15) {
		tmp = sin(th) / (t_1 / 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) <= -0.635) {
		tmp = th / (t_1 / Math.sin(ky));
	} else if (Math.sin(ky) <= -0.18) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= -0.005) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(ky), Math.sin(kx)) / th);
	} else if (Math.sin(ky) <= 1e-15) {
		tmp = Math.sin(th) / (t_1 / 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) <= -0.635:
		tmp = th / (t_1 / math.sin(ky))
	elif math.sin(ky) <= -0.18:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= -0.005:
		tmp = math.sin(ky) / (math.hypot(math.sin(ky), math.sin(kx)) / th)
	elif math.sin(ky) <= 1e-15:
		tmp = math.sin(th) / (t_1 / 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) <= -0.635)
		tmp = Float64(th / Float64(t_1 / sin(ky)));
	elseif (sin(ky) <= -0.18)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -0.005)
		tmp = Float64(sin(ky) / Float64(hypot(sin(ky), sin(kx)) / th));
	elseif (sin(ky) <= 1e-15)
		tmp = Float64(sin(th) / Float64(t_1 / 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) <= -0.635)
		tmp = th / (t_1 / sin(ky));
	elseif (sin(ky) <= -0.18)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -0.005)
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) / th);
	elseif (sin(ky) <= 1e-15)
		tmp = sin(th) / (t_1 / 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], -0.635], N[(th / N[(t$95$1 / N[Sin[ky], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], -0.18], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], -0.005], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision] / th), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-15], N[(N[Sin[th], $MachinePrecision] / N[(t$95$1 / ky), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]]]

Derivation

1. Split input into 5 regimes

2. if (sin.f64 ky) < -0.63500000000000001

   1. Initial program 99.9%

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

      1. *-commutative 99.9%

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

      2. clear-num 99.9%

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

      3. +-commutative 99.9%

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

      4. unpow2 99.9%

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

      5. unpow2 99.9%

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

      6. hypot-udef 99.9%

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

      7. un-div-inv 99.8%

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

   3. Applied egg-rr 99.8%

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

   4. Taylor expanded in th around 0 50.0%

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

   if -0.63500000000000001 < (sin.f64 ky) < -0.17999999999999999

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.1%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 0.6%

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

      2. sqrt-unprod 39.8%

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

      3. pow2 39.8%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 39.8%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 39.8%

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

      2. rem-sqrt-square 47.2%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 47.2%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -0.17999999999999999 < (sin.f64 ky) < -0.0050000000000000001

   1. Initial program 99.3%

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

      1. associate-*l/ 99.2%

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

      2. *-commutative 99.2%

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

      3. associate-*l/ 99.3%

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

      4. +-commutative 99.3%

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

      5. unpow2 99.3%

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

      6. unpow2 99.3%

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

      7. hypot-def 99.3%

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

   3. Simplified 99.3%

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

      1. expm1-log1p-u 99.3%

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

   5. Applied egg-rr 99.3%

      \[\leadsto \frac{\sin th}{\mathsf{hypot}\left(\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin ky\right)\right)}, \sin kx\right)} \cdot \sin ky \]
   6. Step-by-step derivation

      1. *-commutative 99.3%

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

      2. clear-num 99.4%

         \[\leadsto \sin ky \cdot \color{blue}{\frac{1}{\frac{\mathsf{hypot}\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\sin ky\right)\right), \sin kx\right)}{\sin th}}} \]

      3. un-div-inv 99.3%

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

      4. expm1-log1p-u 99.3%

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

   7. Applied egg-rr 99.3%

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

   8. Taylor expanded in th around 0 85.4%

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

   if -0.0050000000000000001 < (sin.f64 ky) < 1.0000000000000001e-15

   1. Initial program 81.2%

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

      1. *-commutative 81.2%

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

      2. clear-num 81.1%

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

      3. +-commutative 81.1%

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

      4. unpow2 81.1%

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

      5. unpow2 81.1%

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

      6. hypot-udef 99.6%

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

      7. un-div-inv 99.7%

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

   3. Applied egg-rr 99.7%

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

   4. Taylor expanded in ky around 0 99.7%

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

   if 1.0000000000000001e-15 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 5 regimes into one program.

4. Final simplification 77.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.635:\\ \;\;\;\;\frac{th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}\\ \mathbf{elif}\;\sin ky \leq -0.18:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 5: 74.9% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ t_2 := \frac{\sin ky}{t_1 \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{if}\;\sin ky \leq -0.658:\\ \;\;\;\;t_2\\ \mathbf{elif}\;\sin ky \leq -0.18:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -0.005:\\ \;\;\;\;t_2\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\frac{\sin th}{\frac{t_1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky)))
        (t_2 (/ (sin ky) (* t_1 (+ (/ 1.0 th) (* th 0.16666666666666666))))))
   (if (<= (sin ky) -0.658)
     t_2
     (if (<= (sin ky) -0.18)
       (fabs (sin th))
       (if (<= (sin ky) -0.005)
         t_2
         (if (<= (sin ky) 1e-15) (/ (sin th) (/ t_1 ky)) (sin th)))))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double t_2 = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	double tmp;
	if (sin(ky) <= -0.658) {
		tmp = t_2;
	} else if (sin(ky) <= -0.18) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= -0.005) {
		tmp = t_2;
	} else if (sin(ky) <= 1e-15) {
		tmp = sin(th) / (t_1 / 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 t_2 = Math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	double tmp;
	if (Math.sin(ky) <= -0.658) {
		tmp = t_2;
	} else if (Math.sin(ky) <= -0.18) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= -0.005) {
		tmp = t_2;
	} else if (Math.sin(ky) <= 1e-15) {
		tmp = Math.sin(th) / (t_1 / ky);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	t_2 = math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)))
	tmp = 0
	if math.sin(ky) <= -0.658:
		tmp = t_2
	elif math.sin(ky) <= -0.18:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= -0.005:
		tmp = t_2
	elif math.sin(ky) <= 1e-15:
		tmp = math.sin(th) / (t_1 / ky)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	t_2 = Float64(sin(ky) / Float64(t_1 * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))))
	tmp = 0.0
	if (sin(ky) <= -0.658)
		tmp = t_2;
	elseif (sin(ky) <= -0.18)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -0.005)
		tmp = t_2;
	elseif (sin(ky) <= 1e-15)
		tmp = Float64(sin(th) / Float64(t_1 / ky));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	t_2 = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	tmp = 0.0;
	if (sin(ky) <= -0.658)
		tmp = t_2;
	elseif (sin(ky) <= -0.18)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -0.005)
		tmp = t_2;
	elseif (sin(ky) <= 1e-15)
		tmp = sin(th) / (t_1 / 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]}, Block[{t$95$2 = 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], -0.658], t$95$2, If[LessEqual[N[Sin[ky], $MachinePrecision], -0.18], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], -0.005], t$95$2, If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-15], N[(N[Sin[th], $MachinePrecision] / N[(t$95$1 / ky), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]]]]

Derivation

1. Split input into 4 regimes

2. if (sin.f64 ky) < -0.658000000000000029 or -0.17999999999999999 < (sin.f64 ky) < -0.0050000000000000001

   1. Initial program 99.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. 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. *-commutative 99.6%

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

      3. associate-*l/ 99.6%

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

      4. +-commutative 99.6%

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

      5. unpow2 99.6%

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

      6. unpow2 99.6%

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

      7. hypot-def 99.6%

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

   3. Simplified 99.6%

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

      1. expm1-log1p-u 99.6%

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

   5. Applied egg-rr 99.6%

      \[\leadsto \frac{\sin th}{\mathsf{hypot}\left(\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin ky\right)\right)}, \sin kx\right)} \cdot \sin ky \]
   6. Step-by-step derivation

      1. *-commutative 99.6%

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

      2. clear-num 99.5%

         \[\leadsto \sin ky \cdot \color{blue}{\frac{1}{\frac{\mathsf{hypot}\left(\mathsf{expm1}\left(\mathsf{log1p}\left(\sin ky\right)\right), \sin kx\right)}{\sin th}}} \]

      3. un-div-inv 99.6%

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

      4. expm1-log1p-u 99.6%

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

   7. Applied egg-rr 99.6%

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

   8. Taylor expanded in th around 0 56.9%

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

      1. +-commutative 56.9%

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

      2. *-commutative 56.9%

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

      3. +-commutative 56.9%

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

      4. unpow2 56.9%

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

      5. unpow2 56.9%

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

      6. hypot-def 56.9%

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

      7. associate-*r/ 56.9%

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

      8. *-rgt-identity 56.9%

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

      9. *-lft-identity 56.9%

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

      10. associate-*l/ 56.9%

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

      11. associate-*r* 56.9%

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

   10. Simplified 56.9%

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

   if -0.658000000000000029 < (sin.f64 ky) < -0.17999999999999999

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.2%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 0.7%

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

      2. sqrt-unprod 40.8%

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

      3. pow2 40.8%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 40.8%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 40.8%

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

      2. rem-sqrt-square 47.6%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 47.6%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -0.0050000000000000001 < (sin.f64 ky) < 1.0000000000000001e-15

   1. Initial program 81.2%

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

      1. *-commutative 81.2%

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

      2. clear-num 81.1%

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

      3. +-commutative 81.1%

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

      4. unpow2 81.1%

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

      5. unpow2 81.1%

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

      6. hypot-udef 99.6%

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

      7. un-div-inv 99.7%

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

   3. Applied egg-rr 99.7%

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

   4. Taylor expanded in ky around 0 99.7%

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

   if 1.0000000000000001e-15 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 4 regimes into one program.

4. Final simplification 77.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.658:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq -0.18:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 6: 72.0% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.01:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.01)
   (fabs (sin th))
   (if (<= (sin ky) 1e-15)
     (* (sin th) (/ ky (hypot (sin ky) (sin kx))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.01) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= 1e-15) {
		tmp = sin(th) * (ky / hypot(sin(ky), sin(kx)));
	} 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.abs(Math.sin(th));
	} else if (Math.sin(ky) <= 1e-15) {
		tmp = Math.sin(th) * (ky / Math.hypot(Math.sin(ky), Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -0.01:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= 1e-15:
		tmp = math.sin(th) * (ky / math.hypot(math.sin(ky), math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -0.01)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-15)
		tmp = Float64(sin(th) * Float64(ky / hypot(sin(ky), sin(kx))));
	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 = abs(sin(th));
	elseif (sin(ky) <= 1e-15)
		tmp = sin(th) * (ky / hypot(sin(ky), sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.01], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-15], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

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

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.3%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.1%

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

      2. sqrt-unprod 22.6%

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

      3. pow2 22.6%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 22.6%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 22.6%

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

      2. rem-sqrt-square 31.5%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 31.5%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -0.0100000000000000002 < (sin.f64 ky) < 1.0000000000000001e-15

   1. Initial program 81.3%

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

      1. +-commutative 81.3%

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

      2. unpow2 81.3%

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

      3. unpow2 81.3%

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

      4. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 99.2%

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

   if 1.0000000000000001e-15 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.5%

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

4. Final simplification 71.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.01:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 7: 72.0% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.02:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\mathsf{hypot}\left(ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.02)
   (fabs (sin th))
   (if (<= (sin ky) 1e-15)
     (* (sin th) (/ (sin ky) (hypot ky (sin kx))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.02) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= 1e-15) {
		tmp = sin(th) * (sin(ky) / hypot(ky, sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -0.02) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= 1e-15) {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.hypot(ky, Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -0.02:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= 1e-15:
		tmp = math.sin(th) * (math.sin(ky) / math.hypot(ky, math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -0.02)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-15)
		tmp = Float64(sin(th) * Float64(sin(ky) / hypot(ky, sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -0.02)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-15)
		tmp = sin(th) * (sin(ky) / hypot(ky, sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.02], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-15], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[ky ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -0.0200000000000000004

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.3%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.1%

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

      2. sqrt-unprod 22.9%

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

      3. pow2 22.9%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 22.9%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 22.9%

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

      2. rem-sqrt-square 31.9%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 31.9%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -0.0200000000000000004 < (sin.f64 ky) < 1.0000000000000001e-15

   1. Initial program 81.5%

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

      1. +-commutative 81.5%

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

      2. unpow2 81.5%

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

      3. unpow2 81.5%

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

      4. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 98.6%

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

   if 1.0000000000000001e-15 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.5%

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

4. Final simplification 71.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.02:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\mathsf{hypot}\left(ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 8: 72.0% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.02:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.02)
   (fabs (sin th))
   (if (<= (sin ky) 1e-15)
     (* (sin ky) (/ (sin th) (hypot ky (sin kx))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.02) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= 1e-15) {
		tmp = sin(ky) * (sin(th) / hypot(ky, sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -0.02) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= 1e-15) {
		tmp = Math.sin(ky) * (Math.sin(th) / Math.hypot(ky, Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -0.02:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= 1e-15:
		tmp = math.sin(ky) * (math.sin(th) / math.hypot(ky, math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -0.02)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-15)
		tmp = Float64(sin(ky) * Float64(sin(th) / hypot(ky, sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -0.02)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-15)
		tmp = sin(ky) * (sin(th) / hypot(ky, sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.02], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-15], N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / N[Sqrt[ky ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -0.0200000000000000004

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.3%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.1%

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

      2. sqrt-unprod 22.9%

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

      3. pow2 22.9%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 22.9%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 22.9%

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

      2. rem-sqrt-square 31.9%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 31.9%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -0.0200000000000000004 < (sin.f64 ky) < 1.0000000000000001e-15

   1. Initial program 81.5%

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

      1. associate-*l/ 79.1%

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

      2. *-commutative 79.1%

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

      3. associate-*l/ 81.5%

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

      4. +-commutative 81.5%

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

      5. unpow2 81.5%

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

      6. unpow2 81.5%

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

      7. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 98.6%

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

   if 1.0000000000000001e-15 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.5%

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

4. Final simplification 71.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.02:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-15}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(ky, \sin kx\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 9: 60.6% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-11}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin kx\right|}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -1e-11)
   (fabs (sin th))
   (if (<= (sin ky) 1e-35)
     (* (sin th) (/ (sin ky) (fabs (sin kx))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -1e-11) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= 1e-35) {
		tmp = sin(th) * (sin(ky) / fabs(sin(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 (sin(ky) <= (-1d-11)) then
        tmp = abs(sin(th))
    else if (sin(ky) <= 1d-35) then
        tmp = sin(th) * (sin(ky) / abs(sin(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 (Math.sin(ky) <= -1e-11) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= 1e-35) {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.abs(Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -1e-11:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= 1e-35:
		tmp = math.sin(th) * (math.sin(ky) / math.fabs(math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -1e-11)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-35)
		tmp = Float64(sin(th) * Float64(sin(ky) / abs(sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -1e-11)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-35)
		tmp = sin(th) * (sin(ky) / abs(sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -1e-11], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-35], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Abs[N[Sin[kx], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -9.99999999999999939e-12

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.3%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.1%

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

      2. sqrt-unprod 23.7%

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

      3. pow2 23.7%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 23.7%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 23.7%

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

      2. rem-sqrt-square 32.3%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 32.3%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -9.99999999999999939e-12 < (sin.f64 ky) < 1.00000000000000001e-35

   1. Initial program 80.3%

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

      1. unpow2 80.3%

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

      2. sin-mult 65.7%

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

   3. Applied egg-rr 65.7%

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

      1. div-sub 65.7%

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

      2. +-inverses 65.7%

         \[\leadsto \frac{\sin ky}{\sqrt{\left(\frac{\cos \color{blue}{0}}{2} - \frac{\cos \left(kx + kx\right)}{2}\right) + {\sin ky}^{2}}} \cdot \sin th \]

      3. +-inverses 65.7%

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

      4. +-inverses 65.7%

         \[\leadsto \frac{\sin ky}{\sqrt{\left(\frac{\cos \color{blue}{0}}{2} - \frac{\cos \left(kx + kx\right)}{2}\right) + {\sin ky}^{2}}} \cdot \sin th \]

      5. cos-0 65.7%

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

      6. metadata-eval 65.7%

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

      7. count-2 65.7%

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

      8. *-commutative 65.7%

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

   5. Simplified 65.7%

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

   6. Taylor expanded in ky around 0 46.2%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sqrt{0.5 - 0.5 \cdot \cos \left(2 \cdot kx\right)}}} \cdot \sin th \]
   7. Step-by-step derivation

      1. sqr-sin-a 61.5%

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

      2. rem-sqrt-square 72.0%

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

   8. Applied egg-rr 72.0%

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

   if 1.00000000000000001e-35 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.0%

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

4. Final simplification 59.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-11}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin kx\right|}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 10: 49.1% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-11}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -4 \cdot 10^{-163}:\\ \;\;\;\;ky \cdot \left(\sin th \cdot \sqrt{\frac{1}{0.5 + -0.5 \cdot \cos \left(kx \cdot 2\right)}}\right)\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -1e-11)
   (fabs (sin th))
   (if (<= (sin ky) -4e-163)
     (* ky (* (sin th) (sqrt (/ 1.0 (+ 0.5 (* -0.5 (cos (* kx 2.0))))))))
     (if (<= (sin ky) 1e-35) (* ky (/ (sin th) (sin kx))) (sin th)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -1e-11) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= -4e-163) {
		tmp = ky * (sin(th) * sqrt((1.0 / (0.5 + (-0.5 * cos((kx * 2.0)))))));
	} else if (sin(ky) <= 1e-35) {
		tmp = ky * (sin(th) / sin(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 (sin(ky) <= (-1d-11)) then
        tmp = abs(sin(th))
    else if (sin(ky) <= (-4d-163)) then
        tmp = ky * (sin(th) * sqrt((1.0d0 / (0.5d0 + ((-0.5d0) * cos((kx * 2.0d0)))))))
    else if (sin(ky) <= 1d-35) then
        tmp = ky * (sin(th) / sin(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 (Math.sin(ky) <= -1e-11) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= -4e-163) {
		tmp = ky * (Math.sin(th) * Math.sqrt((1.0 / (0.5 + (-0.5 * Math.cos((kx * 2.0)))))));
	} else if (Math.sin(ky) <= 1e-35) {
		tmp = ky * (Math.sin(th) / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -1e-11:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= -4e-163:
		tmp = ky * (math.sin(th) * math.sqrt((1.0 / (0.5 + (-0.5 * math.cos((kx * 2.0)))))))
	elif math.sin(ky) <= 1e-35:
		tmp = ky * (math.sin(th) / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -1e-11)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -4e-163)
		tmp = Float64(ky * Float64(sin(th) * sqrt(Float64(1.0 / Float64(0.5 + Float64(-0.5 * cos(Float64(kx * 2.0))))))));
	elseif (sin(ky) <= 1e-35)
		tmp = Float64(ky * Float64(sin(th) / sin(kx)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -1e-11)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -4e-163)
		tmp = ky * (sin(th) * sqrt((1.0 / (0.5 + (-0.5 * cos((kx * 2.0)))))));
	elseif (sin(ky) <= 1e-35)
		tmp = ky * (sin(th) / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -1e-11], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], -4e-163], N[(ky * N[(N[Sin[th], $MachinePrecision] * N[Sqrt[N[(1.0 / N[(0.5 + N[(-0.5 * N[Cos[N[(kx * 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-35], N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]

Derivation

1. Split input into 4 regimes

2. if (sin.f64 ky) < -9.99999999999999939e-12

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.3%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.1%

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

      2. sqrt-unprod 23.7%

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

      3. pow2 23.7%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 23.7%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 23.7%

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

      2. rem-sqrt-square 32.3%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 32.3%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -9.99999999999999939e-12 < (sin.f64 ky) < -3.99999999999999969e-163

   1. Initial program 96.6%

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

      1. unpow2 96.6%

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

      2. sin-mult 91.3%

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

   3. Applied egg-rr 91.3%

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

      1. div-sub 91.3%

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

      2. +-inverses 91.3%

         \[\leadsto \frac{\sin ky}{\sqrt{\left(\frac{\cos \color{blue}{0}}{2} - \frac{\cos \left(kx + kx\right)}{2}\right) + {\sin ky}^{2}}} \cdot \sin th \]

      3. +-inverses 91.3%

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

      4. +-inverses 91.3%

         \[\leadsto \frac{\sin ky}{\sqrt{\left(\frac{\cos \color{blue}{0}}{2} - \frac{\cos \left(kx + kx\right)}{2}\right) + {\sin ky}^{2}}} \cdot \sin th \]

      5. cos-0 91.3%

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

      6. metadata-eval 91.3%

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

      7. count-2 91.3%

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

      8. *-commutative 91.3%

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

   5. Simplified 91.3%

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

   6. Taylor expanded in ky around 0 48.6%

      \[\leadsto \color{blue}{\left(ky \cdot \sin th\right) \cdot \sqrt{\frac{1}{0.5 - 0.5 \cdot \cos \left(2 \cdot kx\right)}}} \]
   7. Step-by-step derivation

      1. associate-*l* 48.8%

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

      2. cancel-sign-sub-inv 48.8%

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

      3. metadata-eval 48.8%

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

      4. *-commutative 48.8%

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

   8. Simplified 48.8%

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

   if -3.99999999999999969e-163 < (sin.f64 ky) < 1.00000000000000001e-35

   1. Initial program 73.7%

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

      1. associate-*l/ 71.4%

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

      2. *-commutative 71.4%

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

      3. associate-*l/ 73.8%

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

      4. +-commutative 73.8%

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

      5. unpow2 73.8%

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

      6. unpow2 73.8%

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

      7. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 55.7%

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

   5. Taylor expanded in ky around 0 55.7%

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

   if 1.00000000000000001e-35 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.0%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 4 regimes into one program.

4. Final simplification 51.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-11}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -4 \cdot 10^{-163}:\\ \;\;\;\;ky \cdot \left(\sin th \cdot \sqrt{\frac{1}{0.5 + -0.5 \cdot \cos \left(kx \cdot 2\right)}}\right)\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 11: 49.1% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-11}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -4 \cdot 10^{-163}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sqrt{0.5 - 0.5 \cdot \cos \left(kx \cdot 2\right)}}\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -1e-11)
   (fabs (sin th))
   (if (<= (sin ky) -4e-163)
     (* (sin th) (/ ky (sqrt (- 0.5 (* 0.5 (cos (* kx 2.0)))))))
     (if (<= (sin ky) 1e-35) (* ky (/ (sin th) (sin kx))) (sin th)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -1e-11) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= -4e-163) {
		tmp = sin(th) * (ky / sqrt((0.5 - (0.5 * cos((kx * 2.0))))));
	} else if (sin(ky) <= 1e-35) {
		tmp = ky * (sin(th) / sin(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 (sin(ky) <= (-1d-11)) then
        tmp = abs(sin(th))
    else if (sin(ky) <= (-4d-163)) then
        tmp = sin(th) * (ky / sqrt((0.5d0 - (0.5d0 * cos((kx * 2.0d0))))))
    else if (sin(ky) <= 1d-35) then
        tmp = ky * (sin(th) / sin(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 (Math.sin(ky) <= -1e-11) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= -4e-163) {
		tmp = Math.sin(th) * (ky / Math.sqrt((0.5 - (0.5 * Math.cos((kx * 2.0))))));
	} else if (Math.sin(ky) <= 1e-35) {
		tmp = ky * (Math.sin(th) / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -1e-11:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= -4e-163:
		tmp = math.sin(th) * (ky / math.sqrt((0.5 - (0.5 * math.cos((kx * 2.0))))))
	elif math.sin(ky) <= 1e-35:
		tmp = ky * (math.sin(th) / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -1e-11)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -4e-163)
		tmp = Float64(sin(th) * Float64(ky / sqrt(Float64(0.5 - Float64(0.5 * cos(Float64(kx * 2.0)))))));
	elseif (sin(ky) <= 1e-35)
		tmp = Float64(ky * Float64(sin(th) / sin(kx)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -1e-11)
		tmp = abs(sin(th));
	elseif (sin(ky) <= -4e-163)
		tmp = sin(th) * (ky / sqrt((0.5 - (0.5 * cos((kx * 2.0))))));
	elseif (sin(ky) <= 1e-35)
		tmp = ky * (sin(th) / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -1e-11], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], -4e-163], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sqrt[N[(0.5 - N[(0.5 * N[Cos[N[(kx * 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-35], N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]

Derivation

1. Split input into 4 regimes

2. if (sin.f64 ky) < -9.99999999999999939e-12

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.3%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.1%

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

      2. sqrt-unprod 23.7%

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

      3. pow2 23.7%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 23.7%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 23.7%

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

      2. rem-sqrt-square 32.3%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 32.3%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -9.99999999999999939e-12 < (sin.f64 ky) < -3.99999999999999969e-163

   1. Initial program 96.6%

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

      1. unpow2 96.6%

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

      2. sin-mult 91.3%

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

   3. Applied egg-rr 91.3%

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

      1. div-sub 91.3%

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

      2. +-inverses 91.3%

         \[\leadsto \frac{\sin ky}{\sqrt{\left(\frac{\cos \color{blue}{0}}{2} - \frac{\cos \left(kx + kx\right)}{2}\right) + {\sin ky}^{2}}} \cdot \sin th \]

      3. +-inverses 91.3%

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

      4. +-inverses 91.3%

         \[\leadsto \frac{\sin ky}{\sqrt{\left(\frac{\cos \color{blue}{0}}{2} - \frac{\cos \left(kx + kx\right)}{2}\right) + {\sin ky}^{2}}} \cdot \sin th \]

      5. cos-0 91.3%

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

      6. metadata-eval 91.3%

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

      7. count-2 91.3%

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

      8. *-commutative 91.3%

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

   5. Simplified 91.3%

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

   6. Taylor expanded in ky around 0 48.6%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sqrt{0.5 - 0.5 \cdot \cos \left(2 \cdot kx\right)}}} \cdot \sin th \]

   7. Taylor expanded in ky around 0 48.6%

      \[\leadsto \frac{\color{blue}{ky}}{\sqrt{0.5 - 0.5 \cdot \cos \left(2 \cdot kx\right)}} \cdot \sin th \]

   if -3.99999999999999969e-163 < (sin.f64 ky) < 1.00000000000000001e-35

   1. Initial program 73.7%

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

      1. associate-*l/ 71.4%

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

      2. *-commutative 71.4%

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

      3. associate-*l/ 73.8%

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

      4. +-commutative 73.8%

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

      5. unpow2 73.8%

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

      6. unpow2 73.8%

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

      7. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 55.7%

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

   5. Taylor expanded in ky around 0 55.7%

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

   if 1.00000000000000001e-35 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.0%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 4 regimes into one program.

4. Final simplification 51.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -1 \cdot 10^{-11}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq -4 \cdot 10^{-163}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sqrt{0.5 - 0.5 \cdot \cos \left(kx \cdot 2\right)}}\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 12: 47.0% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-94}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -5e-94)
   (fabs (sin th))
   (if (<= (sin ky) 1e-35) (* (sin th) (/ ky (sin kx))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -5e-94) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= 1e-35) {
		tmp = sin(th) * (ky / sin(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 (sin(ky) <= (-5d-94)) then
        tmp = abs(sin(th))
    else if (sin(ky) <= 1d-35) then
        tmp = sin(th) * (ky / sin(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 (Math.sin(ky) <= -5e-94) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= 1e-35) {
		tmp = Math.sin(th) * (ky / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -5e-94:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= 1e-35:
		tmp = math.sin(th) * (ky / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -5e-94)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-35)
		tmp = Float64(sin(th) * Float64(ky / sin(kx)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -5e-94)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-35)
		tmp = sin(th) * (ky / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -5e-94], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-35], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -4.9999999999999995e-94

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.4%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.2%

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

      2. sqrt-unprod 24.3%

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

      3. pow2 24.3%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 24.3%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 24.3%

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

      2. rem-sqrt-square 32.2%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 32.2%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -4.9999999999999995e-94 < (sin.f64 ky) < 1.00000000000000001e-35

   1. Initial program 76.8%

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

      1. +-commutative 76.8%

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

      2. unpow2 76.8%

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

      3. unpow2 76.8%

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

      4. hypot-def 99.6%

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

   3. Simplified 99.6%

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

   4. Taylor expanded in ky around 0 52.5%

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

   if 1.00000000000000001e-35 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.0%

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

4. Final simplification 49.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-94}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 13: 47.0% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-94}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -5e-94)
   (fabs (sin th))
   (if (<= (sin ky) 1e-35) (* ky (/ (sin th) (sin kx))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -5e-94) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= 1e-35) {
		tmp = ky * (sin(th) / sin(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 (sin(ky) <= (-5d-94)) then
        tmp = abs(sin(th))
    else if (sin(ky) <= 1d-35) then
        tmp = ky * (sin(th) / sin(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 (Math.sin(ky) <= -5e-94) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= 1e-35) {
		tmp = ky * (Math.sin(th) / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -5e-94:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= 1e-35:
		tmp = ky * (math.sin(th) / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -5e-94)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-35)
		tmp = Float64(ky * Float64(sin(th) / sin(kx)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -5e-94)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 1e-35)
		tmp = ky * (sin(th) / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -5e-94], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-35], N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -4.9999999999999995e-94

   1. Initial program 99.8%

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

   2. Taylor expanded in kx around 0 2.4%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.2%

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

      2. sqrt-unprod 24.3%

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

      3. pow2 24.3%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 24.3%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 24.3%

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

      2. rem-sqrt-square 32.2%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 32.2%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -4.9999999999999995e-94 < (sin.f64 ky) < 1.00000000000000001e-35

   1. Initial program 76.8%

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

      1. associate-*l/ 73.8%

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

      2. *-commutative 73.8%

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

      3. associate-*l/ 76.9%

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

      4. +-commutative 76.9%

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

      5. unpow2 76.9%

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

      6. unpow2 76.9%

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

      7. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 52.5%

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

   5. Taylor expanded in ky around 0 52.5%

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

   if 1.00000000000000001e-35 < (sin.f64 ky)

   1. Initial program 99.7%

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

   2. Taylor expanded in kx around 0 63.0%

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

4. Final simplification 49.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -5 \cdot 10^{-94}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 10^{-35}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 14: 38.8% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -2.35 \cdot 10^{-222}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 5.2 \cdot 10^{-204}:\\ \;\;\;\;0\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -2.35e-222)
   (fabs (sin th))
   (if (<= (sin ky) 5.2e-204) 0.0 (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -2.35e-222) {
		tmp = fabs(sin(th));
	} else if (sin(ky) <= 5.2e-204) {
		tmp = 0.0;
	} 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 (sin(ky) <= (-2.35d-222)) then
        tmp = abs(sin(th))
    else if (sin(ky) <= 5.2d-204) then
        tmp = 0.0d0
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -2.35e-222) {
		tmp = Math.abs(Math.sin(th));
	} else if (Math.sin(ky) <= 5.2e-204) {
		tmp = 0.0;
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -2.35e-222:
		tmp = math.fabs(math.sin(th))
	elif math.sin(ky) <= 5.2e-204:
		tmp = 0.0
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -2.35e-222)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 5.2e-204)
		tmp = 0.0;
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -2.35e-222)
		tmp = abs(sin(th));
	elseif (sin(ky) <= 5.2e-204)
		tmp = 0.0;
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -2.35e-222], N[Abs[N[Sin[th], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 5.2e-204], 0.0, N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -2.3499999999999999e-222

   1. Initial program 92.2%

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

   2. Taylor expanded in kx around 0 2.6%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 1.2%

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

      2. sqrt-unprod 23.8%

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

      3. pow2 23.8%

         \[\leadsto \sqrt{\color{blue}{{\sin th}^{2}}} \]

   4. Applied egg-rr 23.8%

      \[\leadsto \color{blue}{\sqrt{{\sin th}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 23.8%

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

      2. rem-sqrt-square 28.9%

         \[\leadsto \color{blue}{\left|\sin th\right|} \]

   6. Simplified 28.9%

      \[\leadsto \color{blue}{\left|\sin th\right|} \]

   if -2.3499999999999999e-222 < (sin.f64 ky) < 5.19999999999999965e-204

   1. Initial program 71.1%

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

   2. Taylor expanded in kx around 0 7.4%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-log-exp 28.3%

         \[\leadsto \color{blue}{\log \left(e^{\sin th}\right)} \]

   4. Applied egg-rr 28.3%

      \[\leadsto \color{blue}{\log \left(e^{\sin th}\right)} \]

   5. Taylor expanded in th around 0 29.3%

      \[\leadsto \log \color{blue}{1} \]

   if 5.19999999999999965e-204 < (sin.f64 ky)

   1. Initial program 96.2%

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

   2. Taylor expanded in kx around 0 55.2%

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

4. Final simplification 40.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -2.35 \cdot 10^{-222}:\\ \;\;\;\;\left|\sin th\right|\\ \mathbf{elif}\;\sin ky \leq 5.2 \cdot 10^{-204}:\\ \;\;\;\;0\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 15: 28.4% accurate, 3.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 4.4 \cdot 10^{-202}:\\ \;\;\;\;th \cdot \left(\sin ky \cdot \left(kx \cdot 0.16666666666666666 + \frac{1}{kx}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 4.4e-202)
   (* th (* (sin ky) (+ (* kx 0.16666666666666666) (/ 1.0 kx))))
   (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 4.4e-202) {
		tmp = th * (sin(ky) * ((kx * 0.16666666666666666) + (1.0 / 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 (sin(ky) <= 4.4d-202) then
        tmp = th * (sin(ky) * ((kx * 0.16666666666666666d0) + (1.0d0 / 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 (Math.sin(ky) <= 4.4e-202) {
		tmp = th * (Math.sin(ky) * ((kx * 0.16666666666666666) + (1.0 / kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 4.4e-202:
		tmp = th * (math.sin(ky) * ((kx * 0.16666666666666666) + (1.0 / kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 4.4e-202)
		tmp = Float64(th * Float64(sin(ky) * Float64(Float64(kx * 0.16666666666666666) + Float64(1.0 / kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 4.4e-202)
		tmp = th * (sin(ky) * ((kx * 0.16666666666666666) + (1.0 / kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 4.4e-202], N[(th * N[(N[Sin[ky], $MachinePrecision] * N[(N[(kx * 0.16666666666666666), $MachinePrecision] + N[(1.0 / kx), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 4.40000000000000016e-202

   1. Initial program 86.9%

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

      1. associate-*l/ 85.5%

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

      2. *-commutative 85.5%

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

      3. associate-*l/ 86.8%

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

      4. +-commutative 86.8%

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

      5. unpow2 86.8%

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

      6. unpow2 86.8%

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

      7. hypot-def 99.7%

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

   3. Simplified 99.7%

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

   4. Taylor expanded in ky around 0 29.5%

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

   5. Taylor expanded in kx around 0 14.4%

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

   6. Taylor expanded in th around 0 9.5%

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

   if 4.40000000000000016e-202 < (sin.f64 ky)

   1. Initial program 97.1%

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

   2. Taylor expanded in kx around 0 55.6%

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

4. Final simplification 29.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 4.4 \cdot 10^{-202}:\\ \;\;\;\;th \cdot \left(\sin ky \cdot \left(kx \cdot 0.16666666666666666 + \frac{1}{kx}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 16: 26.9% accurate, 6.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;kx \leq 1.35 \cdot 10^{+14}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\left(\sin th + 1\right) + -1\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= kx 1.35e+14) (sin th) (+ (+ (sin th) 1.0) -1.0)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.35e+14) {
		tmp = sin(th);
	} else {
		tmp = (sin(th) + 1.0) + -1.0;
	}
	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 <= 1.35d+14) then
        tmp = sin(th)
    else
        tmp = (sin(th) + 1.0d0) + (-1.0d0)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.35e+14) {
		tmp = Math.sin(th);
	} else {
		tmp = (Math.sin(th) + 1.0) + -1.0;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if kx <= 1.35e+14:
		tmp = math.sin(th)
	else:
		tmp = (math.sin(th) + 1.0) + -1.0
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (kx <= 1.35e+14)
		tmp = sin(th);
	else
		tmp = Float64(Float64(sin(th) + 1.0) + -1.0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (kx <= 1.35e+14)
		tmp = sin(th);
	else
		tmp = (sin(th) + 1.0) + -1.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[kx, 1.35e+14], N[Sin[th], $MachinePrecision], N[(N[(N[Sin[th], $MachinePrecision] + 1.0), $MachinePrecision] + -1.0), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if kx < 1.35e14

   1. Initial program 88.6%

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

   2. Taylor expanded in kx around 0 32.1%

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

   if 1.35e14 < kx

   1. Initial program 99.5%

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

   2. Taylor expanded in kx around 0 9.2%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. expm1-log1p-u 9.2%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin th\right)\right)} \]

   4. Applied egg-rr 9.2%

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin th\right)\right)} \]
   5. Step-by-step derivation

      1. expm1-udef 18.6%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\sin th\right)} - 1} \]

      2. log1p-udef 18.6%

         \[\leadsto e^{\color{blue}{\log \left(1 + \sin th\right)}} - 1 \]

      3. rem-exp-log 18.6%

         \[\leadsto \color{blue}{\left(1 + \sin th\right)} - 1 \]

      4. +-commutative 18.6%

         \[\leadsto \color{blue}{\left(\sin th + 1\right)} - 1 \]

   6. Applied egg-rr 18.6%

      \[\leadsto \color{blue}{\left(\sin th + 1\right) - 1} \]
3. Recombined 2 regimes into one program.

4. Final simplification 28.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;kx \leq 1.35 \cdot 10^{+14}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\left(\sin th + 1\right) + -1\\ \end{array} \]

Alternative 17: 30.1% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq -8500000000 \lor \neg \left(ky \leq 1.35 \cdot 10^{-204}\right):\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (or (<= ky -8500000000.0) (not (<= ky 1.35e-204))) (sin th) 0.0))

C

double code(double kx, double ky, double th) {
	double tmp;
	if ((ky <= -8500000000.0) || !(ky <= 1.35e-204)) {
		tmp = sin(th);
	} else {
		tmp = 0.0;
	}
	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 <= (-8500000000.0d0)) .or. (.not. (ky <= 1.35d-204))) then
        tmp = sin(th)
    else
        tmp = 0.0d0
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if ((ky <= -8500000000.0) || !(ky <= 1.35e-204)) {
		tmp = Math.sin(th);
	} else {
		tmp = 0.0;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if (ky <= -8500000000.0) or not (ky <= 1.35e-204):
		tmp = math.sin(th)
	else:
		tmp = 0.0
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if ((ky <= -8500000000.0) || !(ky <= 1.35e-204))
		tmp = sin(th);
	else
		tmp = 0.0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if ((ky <= -8500000000.0) || ~((ky <= 1.35e-204)))
		tmp = sin(th);
	else
		tmp = 0.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[Or[LessEqual[ky, -8500000000.0], N[Not[LessEqual[ky, 1.35e-204]], $MachinePrecision]], N[Sin[th], $MachinePrecision], 0.0]

Derivation

1. Split input into 2 regimes

2. if ky < -8.5e9 or 1.34999999999999996e-204 < ky

   1. Initial program 97.5%

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

   2. Taylor expanded in kx around 0 36.4%

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

   if -8.5e9 < ky < 1.34999999999999996e-204

   1. Initial program 78.0%

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

   2. Taylor expanded in kx around 0 4.7%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-log-exp 16.8%

         \[\leadsto \color{blue}{\log \left(e^{\sin th}\right)} \]

   4. Applied egg-rr 16.8%

      \[\leadsto \color{blue}{\log \left(e^{\sin th}\right)} \]

   5. Taylor expanded in th around 0 17.8%

      \[\leadsto \log \color{blue}{1} \]
3. Recombined 2 regimes into one program.

4. Final simplification 30.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq -8500000000 \lor \neg \left(ky \leq 1.35 \cdot 10^{-204}\right):\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 18: 16.5% accurate, 6.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;kx \leq 1.26 \cdot 10^{+86}:\\ \;\;\;\;\mathsf{expm1}\left(th\right)\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th) :precision binary64 (if (<= kx 1.26e+86) (expm1 th) 0.0))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.26e+86) {
		tmp = expm1(th);
	} else {
		tmp = 0.0;
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 1.26e+86) {
		tmp = Math.expm1(th);
	} else {
		tmp = 0.0;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if kx <= 1.26e+86:
		tmp = math.expm1(th)
	else:
		tmp = 0.0
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (kx <= 1.26e+86)
		tmp = expm1(th);
	else
		tmp = 0.0;
	end
	return tmp
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[kx, 1.26e+86], N[(Exp[th] - 1), $MachinePrecision], 0.0]

Derivation

1. Split input into 2 regimes

2. if kx < 1.25999999999999999e86

   1. Initial program 89.4%

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

   2. Taylor expanded in kx around 0 30.3%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. expm1-log1p-u 30.3%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin th\right)\right)} \]

   4. Applied egg-rr 30.3%

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin th\right)\right)} \]

   5. Taylor expanded in th around 0 19.2%

      \[\leadsto \mathsf{expm1}\left(\color{blue}{th}\right) \]

   if 1.25999999999999999e86 < kx

   1. Initial program 99.5%

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

   2. Taylor expanded in kx around 0 9.4%

      \[\leadsto \color{blue}{\sin th} \]
   3. Step-by-step derivation

      1. add-log-exp 22.6%

         \[\leadsto \color{blue}{\log \left(e^{\sin th}\right)} \]

   4. Applied egg-rr 22.6%

      \[\leadsto \color{blue}{\log \left(e^{\sin th}\right)} \]

   5. Taylor expanded in th around 0 19.3%

      \[\leadsto \log \color{blue}{1} \]
3. Recombined 2 regimes into one program.

4. Final simplification 19.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;kx \leq 1.26 \cdot 10^{+86}:\\ \;\;\;\;\mathsf{expm1}\left(th\right)\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 19: 14.6% accurate, 7.0× speedup

Math

\[\begin{array}{l} \\ \mathsf{expm1}\left(th\right) \end{array} \]

FPCore

(FPCore (kx ky th) :precision binary64 (expm1 th))

C

double code(double kx, double ky, double th) {
	return expm1(th);
}

Java

public static double code(double kx, double ky, double th) {
	return Math.expm1(th);
}

Python

def code(kx, ky, th):
	return math.expm1(th)

Julia

function code(kx, ky, th)
	return expm1(th)
end

Wolfram

code[kx_, ky_, th_] := N[(Exp[th] - 1), $MachinePrecision]

Derivation

1. Initial program 91.3%

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

2. Taylor expanded in kx around 0 26.4%

   \[\leadsto \color{blue}{\sin th} \]
3. Step-by-step derivation

   1. expm1-log1p-u 26.4%

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin th\right)\right)} \]

4. Applied egg-rr 26.4%

   \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sin th\right)\right)} \]

5. Taylor expanded in th around 0 16.6%

   \[\leadsto \mathsf{expm1}\left(\color{blue}{th}\right) \]

6. Final simplification 16.6%

   \[\leadsto \mathsf{expm1}\left(th\right) \]

Alternative 20: 13.6% 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 91.3%

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

2. Taylor expanded in kx around 0 26.4%

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

3. Taylor expanded in th around 0 15.9%

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

4. Final simplification 15.9%

   \[\leadsto th \]

Reproduce

herbie shell --seed 2023311 
(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)))