Toniolo and Linder, Equation (3b), real

Percentage Accurate: 94.3% → 99.7%

Time: 17.3s

Alternatives: 18

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 95.1%

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

   1. remove-double-neg 95.1%

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

   2. sin-neg 95.1%

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

   3. neg-mul-1 95.1%

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

   4. *-commutative 95.1%

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

   5. associate-*l* 95.1%

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

   6. associate-*l/ 93.7%

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

   7. associate-/r/ 93.7%

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

   8. associate-*l/ 95.1%

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

   9. associate-/r/ 95.0%

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

   10. sin-neg 95.0%

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

   11. neg-mul-1 95.0%

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

   12. associate-/r* 95.0%

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

   13. associate-/r/ 95.1%

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

3. Simplified 99.7%

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

5. Final simplification 99.7%

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

Alternative 2: 40.3% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.005:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\log \left(ky + 1\right)}\\ \mathbf{elif}\;\sin ky \leq 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.005)
   (* (sin th) (/ (sin ky) (log (+ ky 1.0))))
   (if (<= (sin ky) 1e-7) (* (sin th) (fabs (/ (sin ky) (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.005) {
		tmp = sin(th) * (sin(ky) / log((ky + 1.0)));
	} else if (sin(ky) <= 1e-7) {
		tmp = sin(th) * fabs((sin(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) <= (-0.005d0)) then
        tmp = sin(th) * (sin(ky) / log((ky + 1.0d0)))
    else if (sin(ky) <= 1d-7) then
        tmp = sin(th) * abs((sin(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) <= -0.005) {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.log((ky + 1.0)));
	} else if (Math.sin(ky) <= 1e-7) {
		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 math.sin(ky) <= -0.005:
		tmp = math.sin(th) * (math.sin(ky) / math.log((ky + 1.0)))
	elif math.sin(ky) <= 1e-7:
		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 (sin(ky) <= -0.005)
		tmp = Float64(sin(th) * Float64(sin(ky) / log(Float64(ky + 1.0))));
	elseif (sin(ky) <= 1e-7)
		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 (sin(ky) <= -0.005)
		tmp = sin(th) * (sin(ky) / log((ky + 1.0)));
	elseif (sin(ky) <= 1e-7)
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.005], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Log[N[(ky + 1.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-7], 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 (sin.f64 ky) < -0.0050000000000000001

   1. Initial program 99.7%

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

      1. +-commutative 99.7%

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

      2. unpow2 99.7%

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

      3. unpow2 99.7%

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

      4. hypot-udef 99.7%

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

      5. add-log-exp 99.0%

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

      6. hypot-udef 99.0%

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

      7. unpow2 99.0%

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

      8. unpow2 99.0%

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

      9. +-commutative 99.0%

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

      10. unpow2 99.0%

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

      11. unpow2 99.0%

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

      12. hypot-def 99.0%

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

   4. Applied egg-rr 99.0%

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

   5. Taylor expanded in kx around 0 2.8%

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

   6. Taylor expanded in ky around 0 6.2%

      \[\leadsto \frac{\sin ky}{\log \color{blue}{\left(1 + ky\right)}} \cdot \sin th \]
   7. Step-by-step derivation

      1. +-commutative 6.2%

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

   8. Simplified 6.2%

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

   if -0.0050000000000000001 < (sin.f64 ky) < 9.9999999999999995e-8

   1. Initial program 90.4%

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

   3. Taylor expanded in ky around 0 51.4%

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

      1. add-sqr-sqrt 19.0%

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

      2. sqrt-unprod 40.7%

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

      3. pow2 40.7%

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

   5. Applied egg-rr 40.7%

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

      1. unpow2 40.7%

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

      2. rem-sqrt-square 46.4%

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

   7. Simplified 46.4%

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

   if 9.9999999999999995e-8 < (sin.f64 ky)

   1. Initial program 99.8%

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

   3. Taylor expanded in kx around 0 65.5%

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

4. Final simplification 41.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.005:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\log \left(ky + 1\right)}\\ \mathbf{elif}\;\sin ky \leq 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 44.2% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.07:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 5 \cdot 10^{-68}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.07)
   (* (sin th) (fabs (/ ky (sin kx))))
   (if (<= (sin kx) 5e-68) (sin th) (* (sin ky) (/ (sin th) (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.07) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else if (sin(kx) <= 5e-68) {
		tmp = sin(th);
	} else {
		tmp = sin(ky) * (sin(th) / sin(kx));
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(kx) <= (-0.07d0)) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else if (sin(kx) <= 5d-68) then
        tmp = sin(th)
    else
        tmp = sin(ky) * (sin(th) / sin(kx))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.07) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else if (Math.sin(kx) <= 5e-68) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(ky) * (Math.sin(th) / Math.sin(kx));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.07:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	elif math.sin(kx) <= 5e-68:
		tmp = math.sin(th)
	else:
		tmp = math.sin(ky) * (math.sin(th) / math.sin(kx))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.07)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	elseif (sin(kx) <= 5e-68)
		tmp = sin(th);
	else
		tmp = Float64(sin(ky) * Float64(sin(th) / sin(kx)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.07)
		tmp = sin(th) * abs((ky / sin(kx)));
	elseif (sin(kx) <= 5e-68)
		tmp = sin(th);
	else
		tmp = sin(ky) * (sin(th) / sin(kx));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.07], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 5e-68], N[Sin[th], $MachinePrecision], N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.070000000000000007

   1. Initial program 99.4%

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

   3. Taylor expanded in ky around 0 15.5%

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

      1. associate-/l* 15.5%

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

      2. associate-/r/ 15.5%

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

   5. Simplified 15.5%

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

      1. add-sqr-sqrt 8.0%

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

      2. sqrt-unprod 28.6%

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

      3. pow2 28.6%

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

   7. Applied egg-rr 28.6%

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

      1. unpow2 28.6%

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

      2. rem-sqrt-square 34.4%

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

   9. Simplified 34.4%

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

   if -0.070000000000000007 < (sin.f64 kx) < 4.99999999999999971e-68

   1. Initial program 89.9%

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

   3. Taylor expanded in kx around 0 41.7%

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

   if 4.99999999999999971e-68 < (sin.f64 kx)

   1. Initial program 99.6%

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

      1. expm1-log1p-u 99.5%

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

      2. expm1-udef 46.3%

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

   4. Applied egg-rr 46.3%

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

      1. expm1-def 99.5%

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

      2. expm1-log1p 99.6%

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

      3. *-commutative 99.6%

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

      4. associate-*r/ 99.1%

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

      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 99.6%

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

      8. unpow2 99.6%

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

      9. unpow2 99.6%

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

      10. +-commutative 99.6%

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

      11. unpow2 99.6%

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

      12. unpow2 99.6%

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

      13. hypot-def 99.6%

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

   6. Simplified 99.6%

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

   7. Taylor expanded in ky around 0 62.6%

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

4. Final simplification 46.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.07:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 5 \cdot 10^{-68}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 44.2% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.07:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 5 \cdot 10^{-68}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky}{\frac{\sin kx}{\sin th}}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin kx) -0.07)
   (* (sin th) (fabs (/ ky (sin kx))))
   (if (<= (sin kx) 5e-68) (sin th) (/ (sin ky) (/ (sin kx) (sin th))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(kx) <= -0.07) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else if (sin(kx) <= 5e-68) {
		tmp = sin(th);
	} else {
		tmp = sin(ky) / (sin(kx) / 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(kx) <= (-0.07d0)) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else if (sin(kx) <= 5d-68) then
        tmp = sin(th)
    else
        tmp = sin(ky) / (sin(kx) / sin(th))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(kx) <= -0.07) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else if (Math.sin(kx) <= 5e-68) {
		tmp = Math.sin(th);
	} else {
		tmp = Math.sin(ky) / (Math.sin(kx) / Math.sin(th));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(kx) <= -0.07:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	elif math.sin(kx) <= 5e-68:
		tmp = math.sin(th)
	else:
		tmp = math.sin(ky) / (math.sin(kx) / math.sin(th))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(kx) <= -0.07)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	elseif (sin(kx) <= 5e-68)
		tmp = sin(th);
	else
		tmp = Float64(sin(ky) / Float64(sin(kx) / sin(th)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(kx) <= -0.07)
		tmp = sin(th) * abs((ky / sin(kx)));
	elseif (sin(kx) <= 5e-68)
		tmp = sin(th);
	else
		tmp = sin(ky) / (sin(kx) / sin(th));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[kx], $MachinePrecision], -0.07], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 5e-68], N[Sin[th], $MachinePrecision], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sin[kx], $MachinePrecision] / N[Sin[th], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.070000000000000007

   1. Initial program 99.4%

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

   3. Taylor expanded in ky around 0 15.5%

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

      1. associate-/l* 15.5%

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

      2. associate-/r/ 15.5%

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

   5. Simplified 15.5%

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

      1. add-sqr-sqrt 8.0%

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

      2. sqrt-unprod 28.6%

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

      3. pow2 28.6%

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

   7. Applied egg-rr 28.6%

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

      1. unpow2 28.6%

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

      2. rem-sqrt-square 34.4%

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

   9. Simplified 34.4%

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

   if -0.070000000000000007 < (sin.f64 kx) < 4.99999999999999971e-68

   1. Initial program 89.9%

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

   3. Taylor expanded in kx around 0 41.7%

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

   if 4.99999999999999971e-68 < (sin.f64 kx)

   1. Initial program 99.6%

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

   3. Taylor expanded in ky around 0 62.5%

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

      1. associate-*l/ 62.0%

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

      2. associate-/l* 62.5%

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

   5. Applied egg-rr 62.5%

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

4. Final simplification 46.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.07:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 5 \cdot 10^{-68}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky}{\frac{\sin kx}{\sin th}}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 40.6% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky \cdot th}{\left|\sin kx\right|}\\ \mathbf{elif}\;\sin ky \leq 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.005)
   (/ (* (sin ky) th) (fabs (sin kx)))
   (if (<= (sin ky) 1e-7) (* (sin th) (fabs (/ ky (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.005) {
		tmp = (sin(ky) * th) / fabs(sin(kx));
	} else if (sin(ky) <= 1e-7) {
		tmp = sin(th) * fabs((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) <= (-0.005d0)) then
        tmp = (sin(ky) * th) / abs(sin(kx))
    else if (sin(ky) <= 1d-7) then
        tmp = sin(th) * abs((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) <= -0.005) {
		tmp = (Math.sin(ky) * th) / Math.abs(Math.sin(kx));
	} else if (Math.sin(ky) <= 1e-7) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -0.005:
		tmp = (math.sin(ky) * th) / math.fabs(math.sin(kx))
	elif math.sin(ky) <= 1e-7:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -0.005)
		tmp = Float64(Float64(sin(ky) * th) / abs(sin(kx)));
	elseif (sin(ky) <= 1e-7)
		tmp = Float64(sin(th) * abs(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) <= -0.005)
		tmp = (sin(ky) * th) / abs(sin(kx));
	elseif (sin(ky) <= 1e-7)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.005], N[(N[(N[Sin[ky], $MachinePrecision] * th), $MachinePrecision] / N[Abs[N[Sin[kx], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-7], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $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 99.7%

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

   3. Taylor expanded in ky around 0 8.2%

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

      1. add-sqr-sqrt 1.0%

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

      2. sqrt-unprod 1.9%

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

      3. pow2 1.9%

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

   5. Applied egg-rr 1.9%

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

      1. unpow2 1.9%

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

      2. rem-sqrt-square 2.1%

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

   7. Simplified 2.1%

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

   8. Taylor expanded in th around 0 2.9%

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

      1. fabs-div 2.9%

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

      2. rem-square-sqrt 0.0%

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

      3. fabs-sqr 0.0%

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

      4. rem-square-sqrt 8.4%

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

      5. associate-*r/ 8.4%

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

   10. Simplified 8.4%

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

   if -0.0050000000000000001 < (sin.f64 ky) < 9.9999999999999995e-8

   1. Initial program 90.4%

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

   3. Taylor expanded in ky around 0 49.1%

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

      1. associate-/l* 51.4%

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

      2. associate-/r/ 51.4%

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

   5. Simplified 51.4%

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

      1. add-sqr-sqrt 19.0%

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

      2. sqrt-unprod 40.7%

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

      3. pow2 40.7%

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

   7. Applied egg-rr 40.7%

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

      1. unpow2 40.7%

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

      2. rem-sqrt-square 46.4%

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

   9. Simplified 46.4%

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

   if 9.9999999999999995e-8 < (sin.f64 ky)

   1. Initial program 99.8%

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

   3. Taylor expanded in kx around 0 65.5%

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

4. Final simplification 41.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.005:\\ \;\;\;\;\frac{\sin ky \cdot th}{\left|\sin kx\right|}\\ \mathbf{elif}\;\sin ky \leq 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 40.3% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.005:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\log \left(ky + 1\right)}\\ \mathbf{elif}\;\sin ky \leq 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.005)
   (* (sin th) (/ (sin ky) (log (+ ky 1.0))))
   (if (<= (sin ky) 1e-7) (* (sin th) (fabs (/ ky (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.005) {
		tmp = sin(th) * (sin(ky) / log((ky + 1.0)));
	} else if (sin(ky) <= 1e-7) {
		tmp = sin(th) * fabs((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) <= (-0.005d0)) then
        tmp = sin(th) * (sin(ky) / log((ky + 1.0d0)))
    else if (sin(ky) <= 1d-7) then
        tmp = sin(th) * abs((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) <= -0.005) {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.log((ky + 1.0)));
	} else if (Math.sin(ky) <= 1e-7) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -0.005:
		tmp = math.sin(th) * (math.sin(ky) / math.log((ky + 1.0)))
	elif math.sin(ky) <= 1e-7:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -0.005)
		tmp = Float64(sin(th) * Float64(sin(ky) / log(Float64(ky + 1.0))));
	elseif (sin(ky) <= 1e-7)
		tmp = Float64(sin(th) * abs(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) <= -0.005)
		tmp = sin(th) * (sin(ky) / log((ky + 1.0)));
	elseif (sin(ky) <= 1e-7)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.005], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Log[N[(ky + 1.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-7], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $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 99.7%

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

      1. +-commutative 99.7%

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

      2. unpow2 99.7%

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

      3. unpow2 99.7%

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

      4. hypot-udef 99.7%

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

      5. add-log-exp 99.0%

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

      6. hypot-udef 99.0%

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

      7. unpow2 99.0%

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

      8. unpow2 99.0%

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

      9. +-commutative 99.0%

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

      10. unpow2 99.0%

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

      11. unpow2 99.0%

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

      12. hypot-def 99.0%

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

   4. Applied egg-rr 99.0%

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

   5. Taylor expanded in kx around 0 2.8%

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

   6. Taylor expanded in ky around 0 6.2%

      \[\leadsto \frac{\sin ky}{\log \color{blue}{\left(1 + ky\right)}} \cdot \sin th \]
   7. Step-by-step derivation

      1. +-commutative 6.2%

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

   8. Simplified 6.2%

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

   if -0.0050000000000000001 < (sin.f64 ky) < 9.9999999999999995e-8

   1. Initial program 90.4%

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

   3. Taylor expanded in ky around 0 49.1%

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

      1. associate-/l* 51.4%

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

      2. associate-/r/ 51.4%

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

   5. Simplified 51.4%

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

      1. add-sqr-sqrt 19.0%

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

      2. sqrt-unprod 40.7%

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

      3. pow2 40.7%

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

   7. Applied egg-rr 40.7%

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

      1. unpow2 40.7%

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

      2. rem-sqrt-square 46.4%

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

   9. Simplified 46.4%

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

   if 9.9999999999999995e-8 < (sin.f64 ky)

   1. Initial program 99.8%

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

   3. Taylor expanded in kx around 0 65.5%

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

4. Final simplification 41.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.005:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\log \left(ky + 1\right)}\\ \mathbf{elif}\;\sin ky \leq 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

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 95.1%

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

   1. expm1-log1p-u 95.0%

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

   2. expm1-udef 41.3%

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

4. Applied egg-rr 43.0%

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

   1. expm1-def 99.6%

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

   2. expm1-log1p 99.7%

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

   3. *-commutative 99.7%

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

   4. associate-*r/ 96.5%

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

   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 95.0%

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

   8. unpow2 95.0%

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

   9. unpow2 95.0%

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

   10. +-commutative 95.0%

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

   11. unpow2 95.0%

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

   12. unpow2 95.0%

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

   13. hypot-def 99.6%

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

6. Simplified 99.6%

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

7. Final simplification 99.6%

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

Alternative 8: 62.3% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;th \leq 1250:\\ \;\;\;\;\frac{\sin ky}{t_1 \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;th \leq 2.65 \cdot 10^{+199}:\\ \;\;\;\;\frac{ky \cdot \sin th}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin ky\right|}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= th 1250.0)
     (/ (sin ky) (* t_1 (+ (/ 1.0 th) (* th 0.16666666666666666))))
     (if (<= th 2.65e+199)
       (/ (* ky (sin th)) t_1)
       (* (sin th) (/ (sin ky) (fabs (sin ky))))))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (th <= 1250.0) {
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (th <= 2.65e+199) {
		tmp = (ky * sin(th)) / t_1;
	} else {
		tmp = sin(th) * (sin(ky) / fabs(sin(ky)));
	}
	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 <= 1250.0) {
		tmp = Math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else if (th <= 2.65e+199) {
		tmp = (ky * Math.sin(th)) / t_1;
	} else {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.abs(Math.sin(ky)));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if th <= 1250.0:
		tmp = math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)))
	elif th <= 2.65e+199:
		tmp = (ky * math.sin(th)) / t_1
	else:
		tmp = math.sin(th) * (math.sin(ky) / math.fabs(math.sin(ky)))
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (th <= 1250.0)
		tmp = Float64(sin(ky) / Float64(t_1 * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))));
	elseif (th <= 2.65e+199)
		tmp = Float64(Float64(ky * sin(th)) / t_1);
	else
		tmp = Float64(sin(th) * Float64(sin(ky) / abs(sin(ky))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (th <= 1250.0)
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	elseif (th <= 2.65e+199)
		tmp = (ky * sin(th)) / t_1;
	else
		tmp = sin(th) * (sin(ky) / abs(sin(ky)));
	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, 1250.0], 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[th, 2.65e+199], N[(N[(ky * N[Sin[th], $MachinePrecision]), $MachinePrecision] / t$95$1), $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Abs[N[Sin[ky], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if th < 1250

   1. Initial program 96.4%

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

      1. expm1-log1p-u 96.3%

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

      2. expm1-udef 39.2%

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

   4. Applied egg-rr 39.3%

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

      1. expm1-def 99.6%

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

      2. expm1-log1p 99.7%

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

      3. *-commutative 99.7%

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

      4. associate-*r/ 95.8%

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

      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 96.3%

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

      8. unpow2 96.3%

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

      9. unpow2 96.3%

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

      10. +-commutative 96.3%

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

      11. unpow2 96.3%

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

      12. unpow2 96.3%

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

      13. hypot-def 99.6%

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

   6. Simplified 99.6%

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

      1. clear-num 99.4%

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

      2. un-div-inv 99.5%

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

   8. Applied egg-rr 99.5%

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

   9. Taylor expanded in th around 0 66.3%

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

       1. +-commutative 66.3%

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

       2. +-commutative 66.3%

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

       3. unpow2 66.3%

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

       4. unpow2 66.3%

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

       5. hypot-def 69.1%

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

       6. associate-*r* 69.1%

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

       7. +-commutative 69.1%

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

       8. unpow2 69.1%

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

       9. unpow2 69.1%

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

       10. hypot-def 69.1%

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

       11. distribute-rgt-out 69.1%

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

   11. Simplified 69.1%

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

   if 1250 < th < 2.6499999999999998e199

   1. Initial program 92.4%

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

      1. expm1-log1p-u 92.0%

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

      2. expm1-udef 41.6%

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

   4. Applied egg-rr 49.0%

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

      1. expm1-def 99.4%

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

      2. expm1-log1p 99.6%

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

      3. associate-*l/ 99.7%

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

   6. Simplified 99.7%

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

   7. Taylor expanded in ky around 0 67.8%

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

   if 2.6499999999999998e199 < th

   1. Initial program 86.4%

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

      1. +-commutative 86.4%

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

      2. unpow2 86.4%

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

      3. unpow2 86.4%

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

      4. hypot-udef 99.7%

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

      5. add-log-exp 77.0%

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

      6. hypot-udef 77.0%

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

      7. unpow2 77.0%

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

      8. unpow2 77.0%

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

      9. +-commutative 77.0%

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

      10. unpow2 77.0%

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

      11. unpow2 77.0%

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

      12. hypot-def 77.0%

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

   4. Applied egg-rr 77.0%

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

   5. Taylor expanded in kx around 0 18.1%

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

      1. add-sqr-sqrt 17.5%

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

      2. fabs-sqr 17.5%

         \[\leadsto \frac{\sin ky}{\color{blue}{\left|\sqrt{\log \left(e^{\sin ky}\right)} \cdot \sqrt{\log \left(e^{\sin ky}\right)}\right|}} \cdot \sin th \]

      3. add-sqr-sqrt 38.2%

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

      4. rem-log-exp 52.5%

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

   7. Applied egg-rr 52.5%

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

4. Final simplification 67.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 1250:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{elif}\;th \leq 2.65 \cdot 10^{+199}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin ky\right|}\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 61.9% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;th \leq 1250:\\ \;\;\;\;\frac{\sin ky}{\frac{t_1}{th}}\\ \mathbf{elif}\;th \leq 1.06 \cdot 10^{+200}:\\ \;\;\;\;\frac{ky \cdot \sin th}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin ky\right|}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= th 1250.0)
     (/ (sin ky) (/ t_1 th))
     (if (<= th 1.06e+200)
       (/ (* ky (sin th)) t_1)
       (* (sin th) (/ (sin ky) (fabs (sin ky))))))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (th <= 1250.0) {
		tmp = sin(ky) / (t_1 / th);
	} else if (th <= 1.06e+200) {
		tmp = (ky * sin(th)) / t_1;
	} else {
		tmp = sin(th) * (sin(ky) / fabs(sin(ky)));
	}
	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 <= 1250.0) {
		tmp = Math.sin(ky) / (t_1 / th);
	} else if (th <= 1.06e+200) {
		tmp = (ky * Math.sin(th)) / t_1;
	} else {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.abs(Math.sin(ky)));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if th <= 1250.0:
		tmp = math.sin(ky) / (t_1 / th)
	elif th <= 1.06e+200:
		tmp = (ky * math.sin(th)) / t_1
	else:
		tmp = math.sin(th) * (math.sin(ky) / math.fabs(math.sin(ky)))
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (th <= 1250.0)
		tmp = Float64(sin(ky) / Float64(t_1 / th));
	elseif (th <= 1.06e+200)
		tmp = Float64(Float64(ky * sin(th)) / t_1);
	else
		tmp = Float64(sin(th) * Float64(sin(ky) / abs(sin(ky))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (th <= 1250.0)
		tmp = sin(ky) / (t_1 / th);
	elseif (th <= 1.06e+200)
		tmp = (ky * sin(th)) / t_1;
	else
		tmp = sin(th) * (sin(ky) / abs(sin(ky)));
	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, 1250.0], N[(N[Sin[ky], $MachinePrecision] / N[(t$95$1 / th), $MachinePrecision]), $MachinePrecision], If[LessEqual[th, 1.06e+200], N[(N[(ky * N[Sin[th], $MachinePrecision]), $MachinePrecision] / t$95$1), $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Abs[N[Sin[ky], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if th < 1250

   1. Initial program 96.4%

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

      1. expm1-log1p-u 96.3%

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

      2. expm1-udef 39.2%

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

   4. Applied egg-rr 39.3%

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

      1. expm1-def 99.6%

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

      2. expm1-log1p 99.7%

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

      3. *-commutative 99.7%

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

      4. associate-*r/ 95.8%

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

      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 96.3%

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

      8. unpow2 96.3%

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

      9. unpow2 96.3%

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

      10. +-commutative 96.3%

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

      11. unpow2 96.3%

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

      12. unpow2 96.3%

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

      13. hypot-def 99.6%

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

   6. Simplified 99.6%

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

      1. clear-num 99.4%

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

      2. un-div-inv 99.5%

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

   8. Applied egg-rr 99.5%

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

   9. Taylor expanded in th around 0 65.8%

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

       1. associate-*l/ 65.9%

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

       2. +-commutative 65.9%

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

       3. unpow2 65.9%

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

       4. unpow2 65.9%

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

       5. hypot-def 68.7%

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

       6. *-lft-identity 68.7%

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

       7. hypot-def 65.9%

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

       8. unpow2 65.9%

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

       9. unpow2 65.9%

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

       10. +-commutative 65.9%

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

       11. unpow2 65.9%

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

       12. unpow2 65.9%

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

       13. hypot-def 68.7%

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

   11. Simplified 68.7%

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

   if 1250 < th < 1.0599999999999999e200

   1. Initial program 92.4%

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

      1. expm1-log1p-u 92.0%

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

      2. expm1-udef 41.6%

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

   4. Applied egg-rr 49.0%

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

      1. expm1-def 99.4%

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

      2. expm1-log1p 99.6%

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

      3. associate-*l/ 99.7%

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

   6. Simplified 99.7%

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

   7. Taylor expanded in ky around 0 67.8%

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

   if 1.0599999999999999e200 < th

   1. Initial program 86.4%

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

      1. +-commutative 86.4%

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

      2. unpow2 86.4%

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

      3. unpow2 86.4%

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

      4. hypot-udef 99.7%

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

      5. add-log-exp 77.0%

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

      6. hypot-udef 77.0%

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

      7. unpow2 77.0%

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

      8. unpow2 77.0%

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

      9. +-commutative 77.0%

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

      10. unpow2 77.0%

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

      11. unpow2 77.0%

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

      12. hypot-def 77.0%

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

   4. Applied egg-rr 77.0%

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

   5. Taylor expanded in kx around 0 18.1%

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

      1. add-sqr-sqrt 17.5%

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

      2. fabs-sqr 17.5%

         \[\leadsto \frac{\sin ky}{\color{blue}{\left|\sqrt{\log \left(e^{\sin ky}\right)} \cdot \sqrt{\log \left(e^{\sin ky}\right)}\right|}} \cdot \sin th \]

      3. add-sqr-sqrt 38.2%

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

      4. rem-log-exp 52.5%

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

   7. Applied egg-rr 52.5%

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

4. Final simplification 67.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 1250:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{elif}\;th \leq 1.06 \cdot 10^{+200}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin ky\right|}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 61.0% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;th \leq 2.9 \cdot 10^{-6}:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin ky\right|}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= th 2.9e-6)
   (/ (sin ky) (/ (hypot (sin kx) (sin ky)) th))
   (* (sin th) (/ (sin ky) (fabs (sin ky))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 2.9e-6) {
		tmp = sin(ky) / (hypot(sin(kx), sin(ky)) / th);
	} else {
		tmp = sin(th) * (sin(ky) / fabs(sin(ky)));
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 2.9e-6) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(kx), Math.sin(ky)) / th);
	} else {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.abs(Math.sin(ky)));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if th <= 2.9e-6:
		tmp = math.sin(ky) / (math.hypot(math.sin(kx), math.sin(ky)) / th)
	else:
		tmp = math.sin(th) * (math.sin(ky) / math.fabs(math.sin(ky)))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (th <= 2.9e-6)
		tmp = Float64(sin(ky) / Float64(hypot(sin(kx), sin(ky)) / th));
	else
		tmp = Float64(sin(th) * Float64(sin(ky) / abs(sin(ky))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (th <= 2.9e-6)
		tmp = sin(ky) / (hypot(sin(kx), sin(ky)) / th);
	else
		tmp = sin(th) * (sin(ky) / abs(sin(ky)));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[th, 2.9e-6], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision] / th), $MachinePrecision]), $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Abs[N[Sin[ky], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 2.9000000000000002e-6

   1. Initial program 96.4%

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

      1. expm1-log1p-u 96.3%

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

      2. expm1-udef 38.7%

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

   4. Applied egg-rr 38.9%

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

      1. expm1-def 99.6%

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

      2. expm1-log1p 99.7%

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

      3. *-commutative 99.7%

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

      4. associate-*r/ 95.7%

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

      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 96.3%

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

      8. unpow2 96.3%

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

      9. unpow2 96.3%

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

      10. +-commutative 96.3%

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

      11. unpow2 96.3%

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

      12. unpow2 96.3%

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

      13. hypot-def 99.6%

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

   6. Simplified 99.6%

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

      1. clear-num 99.4%

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

      2. un-div-inv 99.5%

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

   8. Applied egg-rr 99.5%

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

   9. Taylor expanded in th around 0 66.0%

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

       1. associate-*l/ 66.0%

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

       2. +-commutative 66.0%

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

       3. unpow2 66.0%

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

       4. unpow2 66.0%

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

       5. hypot-def 68.9%

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

       6. *-lft-identity 68.9%

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

       7. hypot-def 66.0%

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

       8. unpow2 66.0%

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

       9. unpow2 66.0%

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

       10. +-commutative 66.0%

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

       11. unpow2 66.0%

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

       12. unpow2 66.0%

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

       13. hypot-def 68.9%

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

   11. Simplified 68.9%

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

   if 2.9000000000000002e-6 < th

   1. Initial program 90.1%

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

      1. +-commutative 90.1%

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

      2. unpow2 90.1%

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

      3. unpow2 90.1%

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

      4. hypot-udef 99.6%

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

      5. add-log-exp 74.3%

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

      6. hypot-udef 74.3%

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

      7. unpow2 74.3%

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

      8. unpow2 74.3%

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

      9. +-commutative 74.3%

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

      10. unpow2 74.3%

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

      11. unpow2 74.3%

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

      12. hypot-def 74.3%

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

   4. Applied egg-rr 74.3%

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

   5. Taylor expanded in kx around 0 13.7%

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

      1. add-sqr-sqrt 13.2%

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

      2. fabs-sqr 13.2%

         \[\leadsto \frac{\sin ky}{\color{blue}{\left|\sqrt{\log \left(e^{\sin ky}\right)} \cdot \sqrt{\log \left(e^{\sin ky}\right)}\right|}} \cdot \sin th \]

      3. add-sqr-sqrt 31.4%

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

      4. rem-log-exp 42.5%

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

   7. Applied egg-rr 42.5%

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

4. Final simplification 63.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 2.9 \cdot 10^{-6}:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin ky\right|}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 49.6% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-53}:\\ \;\;\;\;\sin th \cdot \mathsf{expm1}\left(\frac{ky}{\sin kx}\right)\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 1e-53) (* (sin th) (expm1 (/ ky (sin kx)))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 1e-53) {
		tmp = sin(th) * expm1((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) <= 1e-53) {
		tmp = Math.sin(th) * Math.expm1((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 1e-53:
		tmp = math.sin(th) * math.expm1((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 1e-53)
		tmp = Float64(sin(th) * expm1(Float64(ky / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-53], N[(N[Sin[th], $MachinePrecision] * N[(Exp[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 1.00000000000000003e-53

   1. Initial program 93.3%

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

      1. remove-double-neg 93.3%

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

      2. sin-neg 93.3%

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

      3. neg-mul-1 93.3%

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

      4. *-commutative 93.3%

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

      5. associate-*l* 93.3%

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

      6. associate-*l/ 91.7%

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

      7. associate-/r/ 91.7%

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

      8. associate-*l/ 93.3%

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

      9. associate-/r/ 93.2%

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

      10. sin-neg 93.2%

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

      11. neg-mul-1 93.2%

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

      12. associate-/r* 93.2%

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

      13. associate-/r/ 93.3%

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

   3. Simplified 99.7%

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

      1. expm1-log1p-u 99.7%

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

      2. hypot-udef 93.2%

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

      3. unpow2 93.2%

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

      4. unpow2 93.2%

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

      5. +-commutative 93.2%

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

      6. unpow2 93.2%

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

      7. unpow2 93.2%

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

      8. hypot-def 99.7%

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

   6. Applied egg-rr 99.7%

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

   7. Taylor expanded in ky around 0 48.5%

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

   if 1.00000000000000003e-53 < (sin.f64 ky)

   1. Initial program 99.7%

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

   3. Taylor expanded in kx around 0 60.6%

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

4. Final simplification 52.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-53}:\\ \;\;\;\;\sin th \cdot \mathsf{expm1}\left(\frac{ky}{\sin kx}\right)\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 39.5% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 9 \cdot 10^{-8}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin ky\right|}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 9e-8)
   (* (sin th) (fabs (/ (sin ky) (sin kx))))
   (* (sin th) (/ (sin ky) (fabs (sin ky))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 9e-8) {
		tmp = sin(th) * fabs((sin(ky) / sin(kx)));
	} else {
		tmp = sin(th) * (sin(ky) / fabs(sin(ky)));
	}
	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 <= 9d-8) then
        tmp = sin(th) * abs((sin(ky) / sin(kx)))
    else
        tmp = sin(th) * (sin(ky) / abs(sin(ky)))
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 9e-8) {
		tmp = Math.sin(th) * Math.abs((Math.sin(ky) / Math.sin(kx)));
	} else {
		tmp = Math.sin(th) * (Math.sin(ky) / Math.abs(Math.sin(ky)));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 9e-8:
		tmp = math.sin(th) * math.fabs((math.sin(ky) / math.sin(kx)))
	else:
		tmp = math.sin(th) * (math.sin(ky) / math.fabs(math.sin(ky)))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 9e-8)
		tmp = Float64(sin(th) * abs(Float64(sin(ky) / sin(kx))));
	else
		tmp = Float64(sin(th) * Float64(sin(ky) / abs(sin(ky))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 9e-8)
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	else
		tmp = sin(th) * (sin(ky) / abs(sin(ky)));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 9e-8], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Abs[N[Sin[ky], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 8.99999999999999986e-8

   1. Initial program 93.5%

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

   3. Taylor expanded in ky around 0 36.3%

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

      1. add-sqr-sqrt 13.5%

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

      2. sqrt-unprod 29.4%

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

      3. pow2 29.4%

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

   5. Applied egg-rr 29.4%

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

      1. unpow2 29.4%

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

      2. rem-sqrt-square 33.3%

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

   7. Simplified 33.3%

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

   if 8.99999999999999986e-8 < ky

   1. Initial program 99.7%

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

      1. +-commutative 99.7%

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

      2. unpow2 99.7%

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

      3. unpow2 99.7%

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

      4. hypot-udef 99.7%

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

      5. add-log-exp 98.6%

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

      6. hypot-udef 98.6%

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

      7. unpow2 98.6%

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

      8. unpow2 98.6%

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

      9. +-commutative 98.6%

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

      10. unpow2 98.6%

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

      11. unpow2 98.6%

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

      12. hypot-def 98.6%

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

   4. Applied egg-rr 98.6%

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

   5. Taylor expanded in kx around 0 40.4%

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

      1. add-sqr-sqrt 39.2%

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

      2. fabs-sqr 39.2%

         \[\leadsto \frac{\sin ky}{\color{blue}{\left|\sqrt{\log \left(e^{\sin ky}\right)} \cdot \sqrt{\log \left(e^{\sin ky}\right)}\right|}} \cdot \sin th \]

      3. add-sqr-sqrt 64.3%

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

      4. rem-log-exp 65.3%

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

   7. Applied egg-rr 65.3%

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

4. Final simplification 41.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 9 \cdot 10^{-8}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\left|\sin ky\right|}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 31.9% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 8.4 \cdot 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 8.4e-7) (* (sin th) (fabs (/ ky (sin kx)))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 8.4e-7) {
		tmp = sin(th) * fabs((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 <= 8.4d-7) then
        tmp = sin(th) * abs((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 <= 8.4e-7) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 8.4e-7:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 8.4e-7)
		tmp = Float64(sin(th) * abs(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 <= 8.4e-7)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 8.4e-7], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 8.4e-7

   1. Initial program 93.5%

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

   3. Taylor expanded in ky around 0 33.8%

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

      1. associate-/l* 35.3%

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

      2. associate-/r/ 35.2%

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

   5. Simplified 35.2%

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

      1. add-sqr-sqrt 13.2%

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

      2. sqrt-unprod 27.8%

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

      3. pow2 27.8%

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

   7. Applied egg-rr 27.8%

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

      1. unpow2 27.8%

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

      2. rem-sqrt-square 31.8%

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

   9. Simplified 31.8%

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

   if 8.4e-7 < ky

   1. Initial program 99.7%

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

   3. Taylor expanded in kx around 0 41.2%

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

4. Final simplification 34.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 8.4 \cdot 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 33.1% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 3 \cdot 10^{-51}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 3e-51) (* (sin th) (/ ky (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 3e-51) {
		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 <= 3d-51) 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 <= 3e-51) {
		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 <= 3e-51:
		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 <= 3e-51)
		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 <= 3e-51)
		tmp = sin(th) * (ky / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 3e-51], 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 < 3.00000000000000002e-51

   1. Initial program 93.3%

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

   3. Taylor expanded in ky around 0 33.0%

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

      1. associate-/l* 34.6%

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

      2. associate-/r/ 34.6%

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

   5. Simplified 34.6%

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

   if 3.00000000000000002e-51 < ky

   1. Initial program 99.7%

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

   3. Taylor expanded in kx around 0 39.1%

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

4. Final simplification 35.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 3 \cdot 10^{-51}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 15: 26.8% accurate, 6.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 1.25 \cdot 10^{-149}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 1.25e-149) (* (sin th) (/ ky kx)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 1.25e-149) {
		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 <= 1.25d-149) 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 <= 1.25e-149) {
		tmp = Math.sin(th) * (ky / kx);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 1.25e-149:
		tmp = math.sin(th) * (ky / kx)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 1.25e-149)
		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 <= 1.25e-149)
		tmp = sin(th) * (ky / kx);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 1.25e-149], N[(N[Sin[th], $MachinePrecision] * N[(ky / kx), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 1.24999999999999992e-149

   1. Initial program 92.6%

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

   3. Taylor expanded in ky around 0 32.7%

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

      1. associate-/l* 34.5%

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

      2. associate-/r/ 34.5%

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

   5. Simplified 34.5%

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

   6. Taylor expanded in kx around 0 25.1%

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

   if 1.24999999999999992e-149 < ky

   1. Initial program 99.7%

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

   3. Taylor expanded in kx around 0 40.8%

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

4. Final simplification 30.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 1.25 \cdot 10^{-149}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 26.9% accurate, 6.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 2.6 \cdot 10^{-150}:\\ \;\;\;\;\frac{\sin th}{\frac{kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 2.6e-150) (/ (sin th) (/ kx ky)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 2.6e-150) {
		tmp = sin(th) / (kx / ky);
	} 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 <= 2.6d-150) then
        tmp = sin(th) / (kx / ky)
    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 <= 2.6e-150) {
		tmp = Math.sin(th) / (kx / ky);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 2.6e-150:
		tmp = math.sin(th) / (kx / ky)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 2.6e-150)
		tmp = Float64(sin(th) / Float64(kx / ky));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 2.6e-150)
		tmp = sin(th) / (kx / ky);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 2.6e-150], N[(N[Sin[th], $MachinePrecision] / N[(kx / ky), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 2.5999999999999998e-150

   1. Initial program 92.6%

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

   3. Taylor expanded in ky around 0 32.7%

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

      1. associate-/l* 34.5%

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

      2. associate-/r/ 34.5%

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

   5. Simplified 34.5%

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

   6. Taylor expanded in kx around 0 25.1%

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

      1. *-commutative 25.1%

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

      2. clear-num 25.1%

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

      3. un-div-inv 25.1%

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

   8. Applied egg-rr 25.1%

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

   if 2.5999999999999998e-150 < ky

   1. Initial program 99.7%

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

   3. Taylor expanded in kx around 0 40.8%

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

4. Final simplification 30.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 2.6 \cdot 10^{-150}:\\ \;\;\;\;\frac{\sin th}{\frac{kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 24.1% accurate, 6.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 4.2 \cdot 10^{-160}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 4.2e-160) (/ ky (/ kx th)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 4.2e-160) {
		tmp = ky / (kx / th);
	} 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 <= 4.2d-160) then
        tmp = ky / (kx / th)
    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 <= 4.2e-160) {
		tmp = ky / (kx / th);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 4.2e-160:
		tmp = ky / (kx / th)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 4.2e-160)
		tmp = Float64(ky / Float64(kx / th));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 4.2e-160)
		tmp = ky / (kx / th);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 4.2e-160], N[(ky / N[(kx / th), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 4.2000000000000001e-160

   1. Initial program 92.5%

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

   3. Taylor expanded in ky around 0 33.1%

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

      1. associate-/l* 34.9%

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

      2. associate-/r/ 34.8%

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

   5. Simplified 34.8%

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

   6. Taylor expanded in kx around 0 25.4%

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

   7. Taylor expanded in th around 0 19.7%

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

      1. associate-/l* 21.5%

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

   9. Simplified 21.5%

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

   if 4.2000000000000001e-160 < ky

   1. Initial program 99.7%

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

   3. Taylor expanded in kx around 0 40.1%

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

4. Final simplification 28.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 4.2 \cdot 10^{-160}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]
5. Add Preprocessing

Alternative 18: 13.5% 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 95.1%

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

3. Taylor expanded in ky around 0 25.7%

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

   1. associate-/l* 26.9%

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

   2. associate-/r/ 26.8%

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

5. Simplified 26.8%

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

6. Taylor expanded in kx around 0 18.9%

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

7. Taylor expanded in th around 0 15.3%

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

   1. associate-/l* 16.5%

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

9. Simplified 16.5%

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

10. Final simplification 16.5%

    \[\leadsto \frac{ky}{\frac{kx}{th}} \]
11. Add Preprocessing

Reproduce

herbie shell --seed 2024010 
(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)))