Toniolo and Linder, Equation (3b), real

Percentage Accurate: 94.2% → 99.7%

Time: 17.8s

Alternatives: 16

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: 94.2% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 99.7% accurate, 1.4× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 93.8%

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

   1. remove-double-neg 93.8%

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

   2. sin-neg 93.8%

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

   3. neg-mul-1 93.8%

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

   4. *-commutative 93.8%

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

   5. associate-*l* 93.8%

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

   6. associate-*l/ 90.7%

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

   7. associate-/r/ 90.7%

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

   8. associate-*l/ 93.8%

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

   9. associate-/r/ 93.4%

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

   10. sin-neg 93.4%

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

   11. neg-mul-1 93.4%

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

   12. associate-/r* 93.4%

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

   13. associate-/r/ 93.8%

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

3. Simplified 99.7%

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

4. Final simplification 99.7%

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

Alternative 2: 77.5% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky}{t_1 \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 0.00027:\\ \;\;\;\;\frac{\sin th}{t_1 \cdot \left(ky \cdot 0.16666666666666666 + \frac{1}{ky}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= (sin ky) -0.005)
     (/ (sin ky) (* t_1 (+ (/ 1.0 th) (* th 0.16666666666666666))))
     (if (<= (sin ky) 0.00027)
       (/ (sin th) (* t_1 (+ (* ky 0.16666666666666666) (/ 1.0 ky))))
       (sin th)))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (sin(ky) <= -0.005) {
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (sin(ky) <= 0.00027) {
		tmp = sin(th) / (t_1 * ((ky * 0.16666666666666666) + (1.0 / ky)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.hypot(Math.sin(kx), Math.sin(ky));
	double tmp;
	if (Math.sin(ky) <= -0.005) {
		tmp = Math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (Math.sin(ky) <= 0.00027) {
		tmp = Math.sin(th) / (t_1 * ((ky * 0.16666666666666666) + (1.0 / ky)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if math.sin(ky) <= -0.005:
		tmp = math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)))
	elif math.sin(ky) <= 0.00027:
		tmp = math.sin(th) / (t_1 * ((ky * 0.16666666666666666) + (1.0 / ky)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (sin(ky) <= -0.005)
		tmp = Float64(sin(ky) / Float64(t_1 * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))));
	elseif (sin(ky) <= 0.00027)
		tmp = Float64(sin(th) / Float64(t_1 * Float64(Float64(ky * 0.16666666666666666) + Float64(1.0 / ky))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (sin(ky) <= -0.005)
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	elseif (sin(ky) <= 0.00027)
		tmp = sin(th) / (t_1 * ((ky * 0.16666666666666666) + (1.0 / ky)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]}, If[LessEqual[N[Sin[ky], $MachinePrecision], -0.005], 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], 0.00027], N[(N[Sin[th], $MachinePrecision] / N[(t$95$1 * N[(N[(ky * 0.16666666666666666), $MachinePrecision] + N[(1.0 / ky), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

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

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   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. associate-/l* 99.3%

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

      3. unpow2 99.3%

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

      4. unpow2 99.3%

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

      5. hypot-def 99.4%

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

   3. Applied egg-rr 99.4%

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

   4. Taylor expanded in th around 0 52.9%

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

      1. +-commutative 52.9%

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

      2. unpow2 52.9%

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

      3. unpow2 52.9%

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

      4. hypot-def 52.9%

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

      5. associate-*r* 52.9%

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

      6. unpow2 52.9%

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

      7. unpow2 52.9%

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

      8. hypot-def 52.9%

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

      9. distribute-rgt-out 52.9%

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

   6. Simplified 52.9%

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

   if -0.0050000000000000001 < (sin.f64 ky) < 2.70000000000000003e-4

   1. Initial program 88.8%

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

      1. *-commutative 88.8%

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

      2. clear-num 88.7%

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

      3. un-div-inv 88.7%

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

      4. unpow2 88.7%

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

      5. unpow2 88.7%

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

      6. hypot-def 99.7%

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

   3. Applied egg-rr 99.7%

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

      1. div-inv 99.6%

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

   5. Applied egg-rr 99.6%

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

   6. Taylor expanded in ky around 0 98.8%

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

   if 2.70000000000000003e-4 < (sin.f64 ky)

   1. Initial program 99.6%

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

   2. Taylor expanded in kx around 0 61.3%

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

4. Final simplification 79.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 0.00027:\\ \;\;\;\;\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(ky \cdot 0.16666666666666666 + \frac{1}{ky}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 3: 77.3% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;\sin ky \leq -0.004:\\ \;\;\;\;\frac{\sin ky \cdot th}{t_1}\\ \mathbf{elif}\;\sin ky \leq 0.00027:\\ \;\;\;\;\frac{\sin th}{t_1 \cdot \frac{1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= (sin ky) -0.004)
     (/ (* (sin ky) th) t_1)
     (if (<= (sin ky) 0.00027) (/ (sin th) (* t_1 (/ 1.0 ky))) (sin th)))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (sin(ky) <= -0.004) {
		tmp = (sin(ky) * th) / t_1;
	} else if (sin(ky) <= 0.00027) {
		tmp = sin(th) / (t_1 * (1.0 / ky));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.hypot(Math.sin(kx), Math.sin(ky));
	double tmp;
	if (Math.sin(ky) <= -0.004) {
		tmp = (Math.sin(ky) * th) / t_1;
	} else if (Math.sin(ky) <= 0.00027) {
		tmp = Math.sin(th) / (t_1 * (1.0 / ky));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if math.sin(ky) <= -0.004:
		tmp = (math.sin(ky) * th) / t_1
	elif math.sin(ky) <= 0.00027:
		tmp = math.sin(th) / (t_1 * (1.0 / ky))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (sin(ky) <= -0.004)
		tmp = Float64(Float64(sin(ky) * th) / t_1);
	elseif (sin(ky) <= 0.00027)
		tmp = Float64(sin(th) / Float64(t_1 * Float64(1.0 / ky)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (sin(ky) <= -0.004)
		tmp = (sin(ky) * th) / t_1;
	elseif (sin(ky) <= 0.00027)
		tmp = sin(th) / (t_1 * (1.0 / ky));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]}, If[LessEqual[N[Sin[ky], $MachinePrecision], -0.004], N[(N[(N[Sin[ky], $MachinePrecision] * th), $MachinePrecision] / t$95$1), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 0.00027], N[(N[Sin[th], $MachinePrecision] / N[(t$95$1 * N[(1.0 / ky), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -0.0040000000000000001

   1. Initial program 99.6%

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

      1. expm1-log1p-u 99.6%

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

      2. expm1-udef 51.6%

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

   3. Applied egg-rr 51.6%

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

      1. expm1-def 99.7%

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

      2. expm1-log1p 99.7%

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

      3. associate-*l/ 99.6%

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

   5. Simplified 99.6%

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

   6. Taylor expanded in th around 0 52.0%

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

   if -0.0040000000000000001 < (sin.f64 ky) < 2.70000000000000003e-4

   1. Initial program 88.7%

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

      1. *-commutative 88.7%

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

      2. clear-num 88.6%

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

      3. un-div-inv 88.6%

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

      4. unpow2 88.6%

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

      5. unpow2 88.6%

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

      6. hypot-def 99.7%

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

   3. Applied egg-rr 99.7%

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

      1. div-inv 99.6%

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

   5. Applied egg-rr 99.6%

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

   6. Taylor expanded in ky around 0 99.0%

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

   if 2.70000000000000003e-4 < (sin.f64 ky)

   1. Initial program 99.6%

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

   2. Taylor expanded in kx around 0 61.3%

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

4. Final simplification 78.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.004:\\ \;\;\;\;\frac{\sin ky \cdot th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{elif}\;\sin ky \leq 0.00027:\\ \;\;\;\;\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 4: 77.4% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;\sin ky \leq -0.004:\\ \;\;\;\;\frac{\sin ky}{t_1 \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 0.00027:\\ \;\;\;\;\frac{\sin th}{t_1 \cdot \frac{1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= (sin ky) -0.004)
     (/ (sin ky) (* t_1 (+ (/ 1.0 th) (* th 0.16666666666666666))))
     (if (<= (sin ky) 0.00027) (/ (sin th) (* t_1 (/ 1.0 ky))) (sin th)))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (sin(ky) <= -0.004) {
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (sin(ky) <= 0.00027) {
		tmp = sin(th) / (t_1 * (1.0 / ky));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.hypot(Math.sin(kx), Math.sin(ky));
	double tmp;
	if (Math.sin(ky) <= -0.004) {
		tmp = Math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (Math.sin(ky) <= 0.00027) {
		tmp = Math.sin(th) / (t_1 * (1.0 / ky));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if math.sin(ky) <= -0.004:
		tmp = math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)))
	elif math.sin(ky) <= 0.00027:
		tmp = math.sin(th) / (t_1 * (1.0 / ky))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (sin(ky) <= -0.004)
		tmp = Float64(sin(ky) / Float64(t_1 * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))));
	elseif (sin(ky) <= 0.00027)
		tmp = Float64(sin(th) / Float64(t_1 * Float64(1.0 / ky)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (sin(ky) <= -0.004)
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	elseif (sin(ky) <= 0.00027)
		tmp = sin(th) / (t_1 * (1.0 / ky));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]}, If[LessEqual[N[Sin[ky], $MachinePrecision], -0.004], 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], 0.00027], N[(N[Sin[th], $MachinePrecision] / N[(t$95$1 * N[(1.0 / ky), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -0.0040000000000000001

   1. Initial program 99.6%

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

      1. associate-*l/ 99.5%

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

      2. associate-/l* 99.3%

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

      3. unpow2 99.3%

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

      4. unpow2 99.3%

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

      5. hypot-def 99.4%

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

   3. Applied egg-rr 99.4%

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

   4. Taylor expanded in th around 0 52.1%

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

      1. +-commutative 52.1%

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

      2. unpow2 52.1%

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

      3. unpow2 52.1%

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

      4. hypot-def 52.1%

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

      5. associate-*r* 52.1%

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

      6. unpow2 52.1%

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

      7. unpow2 52.1%

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

      8. hypot-def 52.1%

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

      9. distribute-rgt-out 52.1%

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

   6. Simplified 52.1%

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

   if -0.0040000000000000001 < (sin.f64 ky) < 2.70000000000000003e-4

   1. Initial program 88.7%

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

      1. *-commutative 88.7%

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

      2. clear-num 88.6%

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

      3. un-div-inv 88.6%

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

      4. unpow2 88.6%

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

      5. unpow2 88.6%

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

      6. hypot-def 99.7%

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

   3. Applied egg-rr 99.7%

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

      1. div-inv 99.6%

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

   5. Applied egg-rr 99.6%

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

   6. Taylor expanded in ky around 0 99.0%

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

   if 2.70000000000000003e-4 < (sin.f64 ky)

   1. Initial program 99.6%

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

   2. Taylor expanded in kx around 0 61.3%

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

4. Final simplification 78.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.004:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;\sin ky \leq 0.00027:\\ \;\;\;\;\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \frac{1}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 5: 44.3% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.005:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-83}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.005)
   (* (sin th) (fabs (/ ky (sin kx))))
   (if (<= (sin kx) 4e-83) (sin th) (* (sin th) (/ (sin ky) (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.005) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else if (sin(kx) <= 4e-83) {
		tmp = sin(th);
	} else {
		tmp = sin(th) * (sin(ky) / sin(kx));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.005d0)) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else if (sin(kx) <= 4d-83) then
        tmp = sin(th)
    else
        tmp = sin(th) * (sin(ky) / sin(kx))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.005) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else if (Math.sin(kx) <= 4e-83) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.sin(kx));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.005:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	elif math.sin(kx) <= 4e-83:
		tmp = math.sin(th)
	else:
		tmp = math.sin(th) * (math.sin(ky) / math.sin(kx))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.005)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	elseif (sin(kx) <= 4e-83)
		tmp = sin(th);
	else
		tmp = Float64(sin(th) * Float64(sin(ky) / sin(kx)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.005)
		tmp = sin(th) * abs((ky / sin(kx)));
	elseif (sin(kx) <= 4e-83)
		tmp = sin(th);
	else
		tmp = sin(th) * (sin(ky) / sin(kx));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.005], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 4e-83], N[Sin[th], $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.0050000000000000001

   1. Initial program 99.4%

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

   2. Taylor expanded in ky around 0 13.4%

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

      1. associate-/l* 13.4%

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

      2. associate-/r/ 13.4%

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

   4. Simplified 13.4%

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

      1. add-sqr-sqrt 6.0%

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

      2. sqrt-unprod 26.7%

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

      3. pow2 26.7%

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

   6. Applied egg-rr 26.7%

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

      1. unpow2 26.7%

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

      2. rem-sqrt-square 38.7%

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

   8. Simplified 38.7%

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

   if -0.0050000000000000001 < (sin.f64 kx) < 4.0000000000000001e-83

   1. Initial program 86.7%

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

   2. Taylor expanded in kx around 0 40.7%

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

   if 4.0000000000000001e-83 < (sin.f64 kx)

   1. Initial program 99.6%

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

   2. Taylor expanded in ky around 0 60.2%

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

4. Final simplification 47.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.005:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-83}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \]

Alternative 6: 45.3% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{\sin ky}{\sin kx}\\ \mathbf{if}\;\sin kx \leq -0.005:\\ \;\;\;\;\sin th \cdot \left|t_1\right|\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-83}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot t_1\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (/ (sin ky) (sin kx))))
   (if (<= (sin kx) -0.005)
     (* (sin th) (fabs t_1))
     (if (<= (sin kx) 4e-83) (sin th) (* (sin th) t_1)))))

C

double code(double kx, double ky, double th) {
	double t_1 = sin(ky) / sin(kx);
	double tmp;
	if (sin(kx) <= -0.005) {
		tmp = sin(th) * fabs(t_1);
	} else if (sin(kx) <= 4e-83) {
		tmp = sin(th);
	} else {
		tmp = sin(th) * t_1;
	}
	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) :: t_1
    real(8) :: tmp
    t_1 = sin(ky) / sin(kx)
    if (sin(kx) <= (-0.005d0)) then
        tmp = sin(th) * abs(t_1)
    else if (sin(kx) <= 4d-83) then
        tmp = sin(th)
    else
        tmp = sin(th) * t_1
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.sin(ky) / Math.sin(kx);
	double tmp;
	if (Math.sin(kx) <= -0.005) {
		tmp = Math.sin(th) * Math.abs(t_1);
	} else if (Math.sin(kx) <= 4e-83) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(th) * t_1;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.sin(ky) / math.sin(kx)
	tmp = 0
	if math.sin(kx) <= -0.005:
		tmp = math.sin(th) * math.fabs(t_1)
	elif math.sin(kx) <= 4e-83:
		tmp = math.sin(th)
	else:
		tmp = math.sin(th) * t_1
	return tmp

Julia

function code(kx, ky, th)
	t_1 = Float64(sin(ky) / sin(kx))
	tmp = 0.0
	if (sin(kx) <= -0.005)
		tmp = Float64(sin(th) * abs(t_1));
	elseif (sin(kx) <= 4e-83)
		tmp = sin(th);
	else
		tmp = Float64(sin(th) * t_1);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = sin(ky) / sin(kx);
	tmp = 0.0;
	if (sin(kx) <= -0.005)
		tmp = sin(th) * abs(t_1);
	elseif (sin(kx) <= 4e-83)
		tmp = sin(th);
	else
		tmp = sin(th) * t_1;
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Sin[kx], $MachinePrecision], -0.005], N[(N[Sin[th], $MachinePrecision] * N[Abs[t$95$1], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 4e-83], N[Sin[th], $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * t$95$1), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.0050000000000000001

   1. Initial program 99.4%

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

   2. Taylor expanded in ky around 0 13.6%

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

      1. add-sqr-sqrt 6.1%

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

      2. sqrt-unprod 31.0%

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

      3. pow2 31.0%

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

   4. Applied egg-rr 31.0%

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

      1. unpow2 31.0%

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

      2. rem-sqrt-square 42.7%

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

   6. Simplified 42.7%

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

   if -0.0050000000000000001 < (sin.f64 kx) < 4.0000000000000001e-83

   1. Initial program 86.7%

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

   2. Taylor expanded in kx around 0 40.7%

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

   if 4.0000000000000001e-83 < (sin.f64 kx)

   1. Initial program 99.6%

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

   2. Taylor expanded in ky around 0 60.2%

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

4. Final simplification 48.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.005:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-83}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \]

Alternative 7: 99.6% accurate, 1.4× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 93.8%

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

   1. expm1-log1p-u 93.8%

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

   2. expm1-udef 39.3%

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

3. Applied egg-rr 42.3%

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

   1. expm1-def 99.7%

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

   2. expm1-log1p 99.7%

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

   3. *-commutative 99.7%

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

   4. associate-*r/ 95.0%

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

   5. associate-*l/ 99.6%

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

   6. *-commutative 99.6%

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

   7. hypot-def 93.8%

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

   8. unpow2 93.8%

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

   9. unpow2 93.8%

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

   10. +-commutative 93.8%

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

   11. unpow2 93.8%

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

   12. unpow2 93.8%

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

   13. hypot-def 99.6%

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

5. Simplified 99.6%

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

6. Final simplification 99.6%

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

Alternative 8: 55.6% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;th \leq 1.02 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{elif}\;th \leq 1.4 \cdot 10^{+83} \lor \neg \left(th \leq 2.5 \cdot 10^{+178}\right) \land \left(th \leq 1.55 \cdot 10^{+259} \lor \neg \left(th \leq 5.8 \cdot 10^{+281}\right)\right):\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= th 1.02e-8)
   (/ (sin ky) (/ (hypot (sin kx) (sin ky)) th))
   (if (or (<= th 1.4e+83)
           (and (not (<= th 2.5e+178))
                (or (<= th 1.55e+259) (not (<= th 5.8e+281)))))
     (* (sin th) (fabs (/ (sin ky) (sin kx))))
     (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 1.02e-8) {
		tmp = sin(ky) / (hypot(sin(kx), sin(ky)) / th);
	} else if ((th <= 1.4e+83) || (!(th <= 2.5e+178) && ((th <= 1.55e+259) || !(th <= 5.8e+281)))) {
		tmp = sin(th) * fabs((sin(ky) / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 1.02e-8) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(kx), Math.sin(ky)) / th);
	} else if ((th <= 1.4e+83) || (!(th <= 2.5e+178) && ((th <= 1.55e+259) || !(th <= 5.8e+281)))) {
		tmp = Math.sin(th) * Math.abs((Math.sin(ky) / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if th <= 1.02e-8:
		tmp = math.sin(ky) / (math.hypot(math.sin(kx), math.sin(ky)) / th)
	elif (th <= 1.4e+83) or (not (th <= 2.5e+178) and ((th <= 1.55e+259) or not (th <= 5.8e+281))):
		tmp = math.sin(th) * math.fabs((math.sin(ky) / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (th <= 1.02e-8)
		tmp = Float64(sin(ky) / Float64(hypot(sin(kx), sin(ky)) / th));
	elseif ((th <= 1.4e+83) || (!(th <= 2.5e+178) && ((th <= 1.55e+259) || !(th <= 5.8e+281))))
		tmp = Float64(sin(th) * abs(Float64(sin(ky) / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (th <= 1.02e-8)
		tmp = sin(ky) / (hypot(sin(kx), sin(ky)) / th);
	elseif ((th <= 1.4e+83) || (~((th <= 2.5e+178)) && ((th <= 1.55e+259) || ~((th <= 5.8e+281)))))
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[th, 1.02e-8], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision] / th), $MachinePrecision]), $MachinePrecision], If[Or[LessEqual[th, 1.4e+83], And[N[Not[LessEqual[th, 2.5e+178]], $MachinePrecision], Or[LessEqual[th, 1.55e+259], N[Not[LessEqual[th, 5.8e+281]], $MachinePrecision]]]], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if th < 1.02000000000000003e-8

   1. Initial program 95.0%

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

      1. associate-*l/ 90.9%

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

      2. associate-/l* 94.5%

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

      3. unpow2 94.5%

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

      4. unpow2 94.5%

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

      5. hypot-def 99.2%

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

   3. Applied egg-rr 99.2%

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

   4. Taylor expanded in th around 0 64.6%

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

      1. associate-*l/ 64.7%

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

      2. unpow2 64.7%

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

      3. unpow2 64.7%

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

      4. hypot-def 67.9%

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

      5. *-lft-identity 67.9%

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

   6. Simplified 67.9%

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

   if 1.02000000000000003e-8 < th < 1.4e83 or 2.49999999999999995e178 < th < 1.5500000000000002e259 or 5.80000000000000019e281 < th

   1. Initial program 89.2%

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

   2. Taylor expanded in ky around 0 39.4%

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

      1. add-sqr-sqrt 7.1%

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

      2. sqrt-unprod 12.0%

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

      3. pow2 12.0%

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

   4. Applied egg-rr 12.0%

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

      1. unpow2 12.0%

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

      2. rem-sqrt-square 16.9%

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

   6. Simplified 16.9%

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

   if 1.4e83 < th < 2.49999999999999995e178 or 1.5500000000000002e259 < th < 5.80000000000000019e281

   1. Initial program 91.9%

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

   2. Taylor expanded in kx around 0 15.6%

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

4. Final simplification 55.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 1.02 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{elif}\;th \leq 1.4 \cdot 10^{+83} \lor \neg \left(th \leq 2.5 \cdot 10^{+178}\right) \land \left(th \leq 1.55 \cdot 10^{+259} \lor \neg \left(th \leq 5.8 \cdot 10^{+281}\right)\right):\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 9: 42.1% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.005:\\ \;\;\;\;\left|ky \cdot \frac{\sin th}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-83}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.005)
   (fabs (* ky (/ (sin th) (sin kx))))
   (if (<= (sin kx) 4e-83) (sin th) (* (sin th) (/ ky (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.005) {
		tmp = fabs((ky * (sin(th) / sin(kx))));
	} else if (sin(kx) <= 4e-83) {
		tmp = sin(th);
	} else {
		tmp = sin(th) * (ky / sin(kx));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.005d0)) then
        tmp = abs((ky * (sin(th) / sin(kx))))
    else if (sin(kx) <= 4d-83) then
        tmp = sin(th)
    else
        tmp = sin(th) * (ky / sin(kx))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.005) {
		tmp = Math.abs((ky * (Math.sin(th) / Math.sin(kx))));
	} else if (Math.sin(kx) <= 4e-83) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(th) * (ky / Math.sin(kx));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.005:
		tmp = math.fabs((ky * (math.sin(th) / math.sin(kx))))
	elif math.sin(kx) <= 4e-83:
		tmp = math.sin(th)
	else:
		tmp = math.sin(th) * (ky / math.sin(kx))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.005)
		tmp = abs(Float64(ky * Float64(sin(th) / sin(kx))));
	elseif (sin(kx) <= 4e-83)
		tmp = sin(th);
	else
		tmp = Float64(sin(th) * Float64(ky / sin(kx)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.005)
		tmp = abs((ky * (sin(th) / sin(kx))));
	elseif (sin(kx) <= 4e-83)
		tmp = sin(th);
	else
		tmp = sin(th) * (ky / sin(kx));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.005], N[Abs[N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 4e-83], N[Sin[th], $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.0050000000000000001

   1. Initial program 99.4%

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

   2. Taylor expanded in ky around 0 13.4%

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

      1. associate-/l* 13.4%

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

      2. associate-/r/ 13.4%

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

   4. Simplified 13.4%

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

      1. add-sqr-sqrt 10.0%

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

      2. sqrt-unprod 19.9%

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

      3. pow2 19.9%

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

      4. div-inv 19.9%

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

      5. associate-*l* 19.9%

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

      6. associate-/r/ 20.0%

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

      7. clear-num 20.0%

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

   6. Applied egg-rr 20.0%

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

      1. unpow2 20.0%

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

      2. rem-sqrt-square 36.9%

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

   8. Simplified 36.9%

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

   if -0.0050000000000000001 < (sin.f64 kx) < 4.0000000000000001e-83

   1. Initial program 86.7%

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

   2. Taylor expanded in kx around 0 40.7%

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

   if 4.0000000000000001e-83 < (sin.f64 kx)

   1. Initial program 99.6%

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

   2. Taylor expanded in ky around 0 51.2%

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

      1. associate-/l* 51.7%

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

      2. associate-/r/ 51.8%

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

   4. Simplified 51.8%

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

4. Final simplification 44.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.005:\\ \;\;\;\;\left|ky \cdot \frac{\sin th}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 4 \cdot 10^{-83}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \end{array} \]

Alternative 10: 63.0% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;th \leq 3.1 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sin ky}{\frac{t_1}{th}}\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{t_1}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= th 3.1e-8) (/ (sin ky) (/ t_1 th)) (/ (* ky (sin th)) t_1))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (th <= 3.1e-8) {
		tmp = sin(ky) / (t_1 / th);
	} else {
		tmp = (ky * sin(th)) / t_1;
	}
	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 (th <= 3.1e-8) {
		tmp = Math.sin(ky) / (t_1 / th);
	} else {
		tmp = (ky * Math.sin(th)) / t_1;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if th <= 3.1e-8:
		tmp = math.sin(ky) / (t_1 / th)
	else:
		tmp = (ky * math.sin(th)) / t_1
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (th <= 3.1e-8)
		tmp = Float64(sin(ky) / Float64(t_1 / th));
	else
		tmp = Float64(Float64(ky * sin(th)) / t_1);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (th <= 3.1e-8)
		tmp = sin(ky) / (t_1 / th);
	else
		tmp = (ky * sin(th)) / t_1;
	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[th, 3.1e-8], N[(N[Sin[ky], $MachinePrecision] / N[(t$95$1 / th), $MachinePrecision]), $MachinePrecision], N[(N[(ky * N[Sin[th], $MachinePrecision]), $MachinePrecision] / t$95$1), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 3.1e-8

   1. Initial program 95.0%

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

      1. associate-*l/ 91.0%

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

      2. associate-/l* 94.5%

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

      3. unpow2 94.5%

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

      4. unpow2 94.5%

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

      5. hypot-def 99.2%

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

   3. Applied egg-rr 99.2%

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

   4. Taylor expanded in th around 0 64.8%

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

      1. associate-*l/ 64.9%

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

      2. unpow2 64.9%

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

      3. unpow2 64.9%

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

      4. hypot-def 68.1%

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

      5. *-lft-identity 68.1%

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

   6. Simplified 68.1%

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

   if 3.1e-8 < th

   1. Initial program 90.1%

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

      1. expm1-log1p-u 90.0%

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

      2. expm1-udef 53.4%

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

   3. Applied egg-rr 60.7%

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

      1. expm1-def 99.6%

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

      2. expm1-log1p 99.7%

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

      3. associate-*l/ 99.4%

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

   5. Simplified 99.4%

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

   6. Taylor expanded in ky around 0 61.5%

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

4. Final simplification 66.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.1 \cdot 10^{-8}:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \]

Alternative 11: 30.6% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;kx \leq 3.8 \cdot 10^{-83}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= kx 3.8e-83) (sin th) (* (sin th) (fabs (/ ky (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 3.8e-83) {
		tmp = sin(th);
	} else {
		tmp = sin(th) * fabs((ky / sin(kx)));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (kx <= 3.8d-83) then
        tmp = sin(th)
    else
        tmp = sin(th) * abs((ky / sin(kx)))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 3.8e-83) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if kx <= 3.8e-83:
		tmp = math.sin(th)
	else:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (kx <= 3.8e-83)
		tmp = sin(th);
	else
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (kx <= 3.8e-83)
		tmp = sin(th);
	else
		tmp = sin(th) * abs((ky / sin(kx)));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[kx, 3.8e-83], N[Sin[th], $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if kx < 3.79999999999999977e-83

   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 28.9%

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

   if 3.79999999999999977e-83 < kx

   1. Initial program 99.5%

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

   2. Taylor expanded in ky around 0 37.1%

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

      1. associate-/l* 37.6%

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

      2. associate-/r/ 37.7%

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

   4. Simplified 37.7%

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

      1. add-sqr-sqrt 14.1%

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

      2. sqrt-unprod 26.8%

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

      3. pow2 26.8%

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

   6. Applied egg-rr 26.8%

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

      1. unpow2 26.8%

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

      2. rem-sqrt-square 36.2%

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

   8. Simplified 36.2%

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

4. Final simplification 31.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;kx \leq 3.8 \cdot 10^{-83}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \end{array} \]

Alternative 12: 33.4% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 5.2 \cdot 10^{-80}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 5.2e-80) (* ky (/ (sin th) (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 5.2e-80) {
		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 (ky <= 5.2d-80) 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 (ky <= 5.2e-80) {
		tmp = ky * (Math.sin(th) / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 5.2e-80:
		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 (ky <= 5.2e-80)
		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 (ky <= 5.2e-80)
		tmp = ky * (sin(th) / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 5.2e-80], N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 5.2000000000000002e-80

   1. Initial program 91.8%

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

      1. *-commutative 91.8%

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

      2. clear-num 91.8%

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

      3. un-div-inv 91.8%

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

      4. unpow2 91.8%

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

      5. unpow2 91.8%

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

      6. hypot-def 99.6%

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

   3. Applied egg-rr 99.6%

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

      1. associate-/r/ 99.6%

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

   5. Applied egg-rr 99.6%

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

   6. Taylor expanded in ky around 0 31.2%

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

      1. associate-*r/ 32.7%

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

   8. Simplified 32.7%

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

   if 5.2000000000000002e-80 < 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 41.2%

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

4. Final simplification 34.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 5.2 \cdot 10^{-80}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 13: 33.4% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 6.4 \cdot 10^{-80}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 6.4e-80) (* (sin th) (/ ky (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 6.4e-80) {
		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 (ky <= 6.4d-80) 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 (ky <= 6.4e-80) {
		tmp = Math.sin(th) * (ky / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 6.4e-80:
		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 (ky <= 6.4e-80)
		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 (ky <= 6.4e-80)
		tmp = sin(th) * (ky / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 6.4e-80], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 6.3999999999999998e-80

   1. Initial program 91.8%

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

   2. Taylor expanded in ky around 0 31.2%

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

      1. associate-/l* 32.8%

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

      2. associate-/r/ 32.8%

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

   4. Simplified 32.8%

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

   if 6.3999999999999998e-80 < 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 41.2%

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

4. Final simplification 35.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 6.4 \cdot 10^{-80}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 14: 27.7% accurate, 6.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 6.2 \cdot 10^{-134}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 6.2e-134) (* (sin th) (/ ky kx)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 6.2e-134) {
		tmp = sin(th) * (ky / kx);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 6.2d-134) then
        tmp = sin(th) * (ky / kx)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 6.2e-134) {
		tmp = Math.sin(th) * (ky / kx);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 6.2e-134:
		tmp = math.sin(th) * (ky / kx)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 6.2e-134)
		tmp = Float64(sin(th) * Float64(ky / kx));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 6.2e-134)
		tmp = sin(th) * (ky / kx);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 6.2e-134], N[(N[Sin[th], $MachinePrecision] * N[(ky / kx), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 6.20000000000000012e-134

   1. Initial program 91.1%

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

   2. Taylor expanded in ky around 0 30.8%

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

      1. associate-/l* 32.5%

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

      2. associate-/r/ 32.6%

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

   4. Simplified 32.6%

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

   5. Taylor expanded in kx around 0 22.1%

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

   if 6.20000000000000012e-134 < 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 41.7%

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

4. Final simplification 28.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 6.2 \cdot 10^{-134}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 15: 26.4% accurate, 6.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;kx \leq 9.6 \cdot 10^{-39}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= kx 9.6e-39) (sin th) (/ ky (/ kx th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 9.6e-39) {
		tmp = sin(th);
	} else {
		tmp = ky / (kx / 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 (kx <= 9.6d-39) then
        tmp = sin(th)
    else
        tmp = ky / (kx / th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (kx <= 9.6e-39) {
		tmp = Math.sin(th);
	} else {
		tmp = ky / (kx / th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if kx <= 9.6e-39:
		tmp = math.sin(th)
	else:
		tmp = ky / (kx / th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (kx <= 9.6e-39)
		tmp = sin(th);
	else
		tmp = Float64(ky / Float64(kx / th));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (kx <= 9.6e-39)
		tmp = sin(th);
	else
		tmp = ky / (kx / th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[kx, 9.6e-39], N[Sin[th], $MachinePrecision], N[(ky / N[(kx / th), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if kx < 9.60000000000000063e-39

   1. Initial program 91.9%

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

   2. Taylor expanded in kx around 0 29.7%

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

   if 9.60000000000000063e-39 < kx

   1. Initial program 99.4%

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

   2. Taylor expanded in ky around 0 33.6%

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

      1. associate-/l* 34.2%

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

      2. associate-/r/ 34.3%

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

   4. Simplified 34.3%

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

   5. Taylor expanded in kx around 0 22.7%

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

   6. Taylor expanded in th around 0 19.3%

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

      1. associate-/l* 19.9%

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

   8. Simplified 19.9%

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

4. Final simplification 27.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;kx \leq 9.6 \cdot 10^{-39}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \end{array} \]

Alternative 16: 13.7% accurate, 141.8× speedup

Math

\[\begin{array}{l} \\ \frac{ky}{\frac{kx}{th}} \end{array} \]

FPCore

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

C

double code(double kx, double ky, double th) {
	return ky / (kx / 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 = ky / (kx / th)
end function

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[kx_, ky_, th_] := N[(ky / N[(kx / th), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 93.8%

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

2. Taylor expanded in ky around 0 26.2%

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

   1. associate-/l* 27.3%

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

   2. associate-/r/ 27.3%

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

4. Simplified 27.3%

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

5. Taylor expanded in kx around 0 17.6%

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

6. Taylor expanded in th around 0 13.3%

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

   1. associate-/l* 14.6%

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

8. Simplified 14.6%

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

9. Final simplification 14.6%

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

Reproduce

herbie shell --seed 2023334 
(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)))