Migdal et al, Equation (64)

Percentage Accurate: 99.6% → 99.6%

Time: 8.3s

Alternatives: 22

Speedup: 2.0×

• 43.3% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ t\_1 \cdot \left(a1 \cdot a1\right) + t\_1 \cdot \left(a2 \cdot a2\right) \end{array} \end{array} \]

FPCore

(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (/ (cos th) (sqrt 2.0))))
   (+ (* t_1 (* a1 a1)) (* t_1 (* a2 a2)))))

C

double code(double a1, double a2, double th) {
	double t_1 = cos(th) / sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}

Fortran

real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    t_1 = cos(th) / sqrt(2.0d0)
    code = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))
end function

Java

public static double code(double a1, double a2, double th) {
	double t_1 = Math.cos(th) / Math.sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}

Python

def code(a1, a2, th):
	t_1 = math.cos(th) / math.sqrt(2.0)
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))

Julia

function code(a1, a2, th)
	t_1 = Float64(cos(th) / sqrt(2.0))
	return Float64(Float64(t_1 * Float64(a1 * a1)) + Float64(t_1 * Float64(a2 * a2)))
end

MATLAB

function tmp = code(a1, a2, th)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
end

Wolfram

code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]}, N[(N[(t$95$1 * N[(a1 * a1), $MachinePrecision]), $MachinePrecision] + N[(t$95$1 * N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Initial Program: 99.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ t\_1 \cdot \left(a1 \cdot a1\right) + t\_1 \cdot \left(a2 \cdot a2\right) \end{array} \end{array} \]

FPCore

(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (/ (cos th) (sqrt 2.0))))
   (+ (* t_1 (* a1 a1)) (* t_1 (* a2 a2)))))

C

double code(double a1, double a2, double th) {
	double t_1 = cos(th) / sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}

Fortran

real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    t_1 = cos(th) / sqrt(2.0d0)
    code = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))
end function

Java

public static double code(double a1, double a2, double th) {
	double t_1 = Math.cos(th) / Math.sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}

Python

def code(a1, a2, th):
	t_1 = math.cos(th) / math.sqrt(2.0)
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))

Julia

function code(a1, a2, th)
	t_1 = Float64(cos(th) / sqrt(2.0))
	return Float64(Float64(t_1 * Float64(a1 * a1)) + Float64(t_1 * Float64(a2 * a2)))
end

MATLAB

function tmp = code(a1, a2, th)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
end

Wolfram

code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]}, N[(N[(t$95$1 * N[(a1 * a1), $MachinePrecision]), $MachinePrecision] + N[(t$95$1 * N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Alternative 1: 99.6% accurate, 1.3× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \cos th \cdot \frac{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}{\sqrt{2}} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (* (cos th) (/ (fma a2 a2 (* a1 a1)) (sqrt 2.0))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return cos(th) * (fma(a2, a2, (a1 * a1)) / sqrt(2.0));
}

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	return Float64(cos(th) * Float64(fma(a2, a2, Float64(a1 * a1)) / sqrt(2.0)))
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(N[Cos[th], $MachinePrecision] * N[(N[(a2 * a2 + N[(a1 * a1), $MachinePrecision]), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.6%

   \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
2. Step-by-step derivation

   1. distribute-lft-out 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   2. cos-neg 99.6%

      \[\leadsto \frac{\color{blue}{\cos \left(-th\right)}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   3. associate-*l/ 99.6%

      \[\leadsto \color{blue}{\frac{\cos \left(-th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

   4. associate-/l* 99.6%

      \[\leadsto \color{blue}{\cos \left(-th\right) \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]

   5. cos-neg 99.6%

      \[\leadsto \color{blue}{\cos th} \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}} \]

   6. +-commutative 99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{a2 \cdot a2 + a1 \cdot a1}}{\sqrt{2}} \]

   7. fma-define 99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}}{\sqrt{2}} \]

3. Simplified 99.6%

   \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}{\sqrt{2}}} \]
4. Add Preprocessing
5. Add Preprocessing

Alternative 2: 80.1% accurate, 1.3× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} \mathbf{if}\;\cos th \leq 0.66:\\ \;\;\;\;\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}{\sqrt{2}}\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (if (<= (cos th) 0.66)
   (* (cos th) (+ (* a1 a1) (* a2 a2)))
   (/ (fma a2 a2 (* a1 a1)) (sqrt 2.0))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double tmp;
	if (cos(th) <= 0.66) {
		tmp = cos(th) * ((a1 * a1) + (a2 * a2));
	} else {
		tmp = fma(a2, a2, (a1 * a1)) / sqrt(2.0);
	}
	return tmp;
}

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	tmp = 0.0
	if (cos(th) <= 0.66)
		tmp = Float64(cos(th) * Float64(Float64(a1 * a1) + Float64(a2 * a2)));
	else
		tmp = Float64(fma(a2, a2, Float64(a1 * a1)) / sqrt(2.0));
	end
	return tmp
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := If[LessEqual[N[Cos[th], $MachinePrecision], 0.66], N[(N[Cos[th], $MachinePrecision] * N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(a2 * a2 + N[(a1 * a1), $MachinePrecision]), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (cos.f64 th) < 0.660000000000000031

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. frac-2neg 99.6%

         \[\leadsto \color{blue}{\frac{-\cos th}{-\sqrt{2}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      2. div-inv 99.5%

         \[\leadsto \color{blue}{\left(\left(-\cos th\right) \cdot \frac{1}{-\sqrt{2}}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\left(\left(-\cos th\right) \cdot \frac{1}{-\sqrt{2}}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   8. Applied egg-rr 59.8%

      \[\leadsto \color{blue}{\left(0 + \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 0.660000000000000031 < (cos.f64 th)

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

      2. cos-neg 99.6%

         \[\leadsto \frac{\color{blue}{\cos \left(-th\right)}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      3. associate-*l/ 99.7%

         \[\leadsto \color{blue}{\frac{\cos \left(-th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

      4. associate-/l* 99.6%

         \[\leadsto \color{blue}{\cos \left(-th\right) \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]

      5. cos-neg 99.6%

         \[\leadsto \color{blue}{\cos th} \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}} \]

      6. +-commutative 99.6%

         \[\leadsto \cos th \cdot \frac{\color{blue}{a2 \cdot a2 + a1 \cdot a1}}{\sqrt{2}} \]

      7. fma-define 99.6%

         \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}}{\sqrt{2}} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}{\sqrt{2}}} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 89.2%

      \[\leadsto \color{blue}{1} \cdot \frac{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}{\sqrt{2}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 78.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\cos th \leq 0.66:\\ \;\;\;\;\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}{\sqrt{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 55.3% accurate, 1.3× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} \mathbf{if}\;\cos th \leq 0.66:\\ \;\;\;\;\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{{a2}^{2}}{\sqrt{2}}\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (if (<= (cos th) 0.66)
   (* (cos th) (+ (* a1 a1) (* a2 a2)))
   (/ (pow a2 2.0) (sqrt 2.0))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double tmp;
	if (cos(th) <= 0.66) {
		tmp = cos(th) * ((a1 * a1) + (a2 * a2));
	} else {
		tmp = pow(a2, 2.0) / sqrt(2.0);
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: tmp
    if (cos(th) <= 0.66d0) then
        tmp = cos(th) * ((a1 * a1) + (a2 * a2))
    else
        tmp = (a2 ** 2.0d0) / sqrt(2.0d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double tmp;
	if (Math.cos(th) <= 0.66) {
		tmp = Math.cos(th) * ((a1 * a1) + (a2 * a2));
	} else {
		tmp = Math.pow(a2, 2.0) / Math.sqrt(2.0);
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	tmp = 0
	if math.cos(th) <= 0.66:
		tmp = math.cos(th) * ((a1 * a1) + (a2 * a2))
	else:
		tmp = math.pow(a2, 2.0) / math.sqrt(2.0)
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	tmp = 0.0
	if (cos(th) <= 0.66)
		tmp = Float64(cos(th) * Float64(Float64(a1 * a1) + Float64(a2 * a2)));
	else
		tmp = Float64((a2 ^ 2.0) / sqrt(2.0));
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	tmp = 0.0;
	if (cos(th) <= 0.66)
		tmp = cos(th) * ((a1 * a1) + (a2 * a2));
	else
		tmp = (a2 ^ 2.0) / sqrt(2.0);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := If[LessEqual[N[Cos[th], $MachinePrecision], 0.66], N[(N[Cos[th], $MachinePrecision] * N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Power[a2, 2.0], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (cos.f64 th) < 0.660000000000000031

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. frac-2neg 99.6%

         \[\leadsto \color{blue}{\frac{-\cos th}{-\sqrt{2}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      2. div-inv 99.5%

         \[\leadsto \color{blue}{\left(\left(-\cos th\right) \cdot \frac{1}{-\sqrt{2}}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\left(\left(-\cos th\right) \cdot \frac{1}{-\sqrt{2}}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   8. Applied egg-rr 59.8%

      \[\leadsto \color{blue}{\left(0 + \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 0.660000000000000031 < (cos.f64 th)

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 89.1%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   6. Taylor expanded in a1 around 0 56.3%

      \[\leadsto \color{blue}{\frac{{a2}^{2}}{\sqrt{2}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 57.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\cos th \leq 0.66:\\ \;\;\;\;\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{{a2}^{2}}{\sqrt{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 80.0% accurate, 1.9× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;\cos th \leq 0.66:\\ \;\;\;\;\cos th \cdot t\_1\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot \frac{1}{\sqrt{2}}\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= (cos th) 0.66) (* (cos th) t_1) (* t_1 (/ 1.0 (sqrt 2.0))))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (cos(th) <= 0.66) {
		tmp = cos(th) * t_1;
	} else {
		tmp = t_1 * (1.0 / sqrt(2.0));
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (cos(th) <= 0.66d0) then
        tmp = cos(th) * t_1
    else
        tmp = t_1 * (1.0d0 / sqrt(2.0d0))
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (Math.cos(th) <= 0.66) {
		tmp = Math.cos(th) * t_1;
	} else {
		tmp = t_1 * (1.0 / Math.sqrt(2.0));
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if math.cos(th) <= 0.66:
		tmp = math.cos(th) * t_1
	else:
		tmp = t_1 * (1.0 / math.sqrt(2.0))
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (cos(th) <= 0.66)
		tmp = Float64(cos(th) * t_1);
	else
		tmp = Float64(t_1 * Float64(1.0 / sqrt(2.0)));
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (cos(th) <= 0.66)
		tmp = cos(th) * t_1;
	else
		tmp = t_1 * (1.0 / sqrt(2.0));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Cos[th], $MachinePrecision], 0.66], N[(N[Cos[th], $MachinePrecision] * t$95$1), $MachinePrecision], N[(t$95$1 * N[(1.0 / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (cos.f64 th) < 0.660000000000000031

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. frac-2neg 99.6%

         \[\leadsto \color{blue}{\frac{-\cos th}{-\sqrt{2}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      2. div-inv 99.5%

         \[\leadsto \color{blue}{\left(\left(-\cos th\right) \cdot \frac{1}{-\sqrt{2}}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\left(\left(-\cos th\right) \cdot \frac{1}{-\sqrt{2}}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   8. Applied egg-rr 59.8%

      \[\leadsto \color{blue}{\left(0 + \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 0.660000000000000031 < (cos.f64 th)

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 89.1%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 78.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\cos th \leq 0.66:\\ \;\;\;\;\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \frac{1}{\sqrt{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 80.1% accurate, 1.9× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;\cos th \leq 0.66:\\ \;\;\;\;\cos th \cdot t\_1\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot \sqrt{0.5}\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= (cos th) 0.66) (* (cos th) t_1) (* t_1 (sqrt 0.5)))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (cos(th) <= 0.66) {
		tmp = cos(th) * t_1;
	} else {
		tmp = t_1 * sqrt(0.5);
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (cos(th) <= 0.66d0) then
        tmp = cos(th) * t_1
    else
        tmp = t_1 * sqrt(0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (Math.cos(th) <= 0.66) {
		tmp = Math.cos(th) * t_1;
	} else {
		tmp = t_1 * Math.sqrt(0.5);
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if math.cos(th) <= 0.66:
		tmp = math.cos(th) * t_1
	else:
		tmp = t_1 * math.sqrt(0.5)
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (cos(th) <= 0.66)
		tmp = Float64(cos(th) * t_1);
	else
		tmp = Float64(t_1 * sqrt(0.5));
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (cos(th) <= 0.66)
		tmp = cos(th) * t_1;
	else
		tmp = t_1 * sqrt(0.5);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Cos[th], $MachinePrecision], 0.66], N[(N[Cos[th], $MachinePrecision] * t$95$1), $MachinePrecision], N[(t$95$1 * N[Sqrt[0.5], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (cos.f64 th) < 0.660000000000000031

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. frac-2neg 99.6%

         \[\leadsto \color{blue}{\frac{-\cos th}{-\sqrt{2}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      2. div-inv 99.5%

         \[\leadsto \color{blue}{\left(\left(-\cos th\right) \cdot \frac{1}{-\sqrt{2}}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\left(\left(-\cos th\right) \cdot \frac{1}{-\sqrt{2}}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   8. Applied egg-rr 59.8%

      \[\leadsto \color{blue}{\left(0 + \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 0.660000000000000031 < (cos.f64 th)

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. clear-num 99.6%

         \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{2}}{\cos th}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      2. associate-/r/ 99.6%

         \[\leadsto \color{blue}{\left(\frac{1}{\sqrt{2}} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      3. pow1/2 99.6%

         \[\leadsto \left(\frac{1}{\color{blue}{{2}^{0.5}}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      4. pow-flip 99.5%

         \[\leadsto \left(\color{blue}{{2}^{\left(-0.5\right)}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      5. metadata-eval 99.5%

         \[\leadsto \left({2}^{\color{blue}{-0.5}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   6. Applied egg-rr 99.5%

      \[\leadsto \color{blue}{\left({2}^{-0.5} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Taylor expanded in th around 0 89.1%

      \[\leadsto \color{blue}{\sqrt{0.5}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 78.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\cos th \leq 0.66:\\ \;\;\;\;\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 99.6% accurate, 2.0× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (* (/ (cos th) (sqrt 2.0)) (+ (* a1 a1) (* a2 a2))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return (cos(th) / sqrt(2.0)) * ((a1 * a1) + (a2 * a2));
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = (cos(th) / sqrt(2.0d0)) * ((a1 * a1) + (a2 * a2))
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	return (Math.cos(th) / Math.sqrt(2.0)) * ((a1 * a1) + (a2 * a2));
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	return (math.cos(th) / math.sqrt(2.0)) * ((a1 * a1) + (a2 * a2))

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	return Float64(Float64(cos(th) / sqrt(2.0)) * Float64(Float64(a1 * a1) + Float64(a2 * a2)))
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = (cos(th) / sqrt(2.0)) * ((a1 * a1) + (a2 * a2));
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision] * N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.6%

   \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
2. Step-by-step derivation

   1. distribute-lft-out 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

3. Simplified 99.6%

   \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
4. Add Preprocessing
5. Add Preprocessing

Alternative 7: 99.6% accurate, 2.0× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \left(\cos th \cdot \sqrt{0.5}\right) \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (* (+ (* a1 a1) (* a2 a2)) (* (cos th) (sqrt 0.5))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return ((a1 * a1) + (a2 * a2)) * (cos(th) * sqrt(0.5));
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = ((a1 * a1) + (a2 * a2)) * (cos(th) * sqrt(0.5d0))
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	return ((a1 * a1) + (a2 * a2)) * (Math.cos(th) * Math.sqrt(0.5));
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	return ((a1 * a1) + (a2 * a2)) * (math.cos(th) * math.sqrt(0.5))

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	return Float64(Float64(Float64(a1 * a1) + Float64(a2 * a2)) * Float64(cos(th) * sqrt(0.5)))
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = ((a1 * a1) + (a2 * a2)) * (cos(th) * sqrt(0.5));
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision] * N[(N[Cos[th], $MachinePrecision] * N[Sqrt[0.5], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.6%

   \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
2. Step-by-step derivation

   1. distribute-lft-out 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

3. Simplified 99.6%

   \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
4. Add Preprocessing
5. Step-by-step derivation

   1. clear-num 99.6%

      \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{2}}{\cos th}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   2. associate-/r/ 99.5%

      \[\leadsto \color{blue}{\left(\frac{1}{\sqrt{2}} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   3. pow1/2 99.5%

      \[\leadsto \left(\frac{1}{\color{blue}{{2}^{0.5}}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   4. pow-flip 99.6%

      \[\leadsto \left(\color{blue}{{2}^{\left(-0.5\right)}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   5. metadata-eval 99.6%

      \[\leadsto \left({2}^{\color{blue}{-0.5}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

6. Applied egg-rr 99.6%

   \[\leadsto \color{blue}{\left({2}^{-0.5} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

7. Taylor expanded in th around inf 99.6%

   \[\leadsto \color{blue}{\left(\cos th \cdot \sqrt{0.5}\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
8. Step-by-step derivation

   1. *-commutative 99.6%

      \[\leadsto \color{blue}{\left(\sqrt{0.5} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

9. Simplified 99.6%

   \[\leadsto \color{blue}{\left(\sqrt{0.5} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

10. Final simplification 99.6%

    \[\leadsto \left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \left(\cos th \cdot \sqrt{0.5}\right) \]
11. Add Preprocessing

Alternative 8: 67.0% accurate, 3.6× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;t\_1 \cdot \sqrt{0.5}\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= th 3.9e+97) (* t_1 (sqrt 0.5)) (* t_1 -0.5))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * sqrt(0.5);
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (th <= 3.9d+97) then
        tmp = t_1 * sqrt(0.5d0)
    else
        tmp = t_1 * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * Math.sqrt(0.5);
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if th <= 3.9e+97:
		tmp = t_1 * math.sqrt(0.5)
	else:
		tmp = t_1 * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (th <= 3.9e+97)
		tmp = Float64(t_1 * sqrt(0.5));
	else
		tmp = Float64(t_1 * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (th <= 3.9e+97)
		tmp = t_1 * sqrt(0.5);
	else
		tmp = t_1 * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 3.9e+97], N[(t$95$1 * N[Sqrt[0.5], $MachinePrecision]), $MachinePrecision], N[(t$95$1 * -0.5), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 3.8999999999999999e97

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. clear-num 99.6%

         \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{2}}{\cos th}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      2. associate-/r/ 99.5%

         \[\leadsto \color{blue}{\left(\frac{1}{\sqrt{2}} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      3. pow1/2 99.5%

         \[\leadsto \left(\frac{1}{\color{blue}{{2}^{0.5}}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      4. pow-flip 99.6%

         \[\leadsto \left(\color{blue}{{2}^{\left(-0.5\right)}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

      5. metadata-eval 99.6%

         \[\leadsto \left({2}^{\color{blue}{-0.5}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   6. Applied egg-rr 99.6%

      \[\leadsto \color{blue}{\left({2}^{-0.5} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Taylor expanded in th around 0 71.4%

      \[\leadsto \color{blue}{\sqrt{0.5}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 3.8999999999999999e97 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.6%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.4%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 65.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \sqrt{0.5}\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 32.8% accurate, 3.8× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;{a2}^{2} \cdot 0.6666666666666666\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (if (<= th 3.9e+97)
   (* (pow a2 2.0) 0.6666666666666666)
   (* (+ (* a1 a1) (* a2 a2)) -0.5)))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double tmp;
	if (th <= 3.9e+97) {
		tmp = pow(a2, 2.0) * 0.6666666666666666;
	} else {
		tmp = ((a1 * a1) + (a2 * a2)) * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: tmp
    if (th <= 3.9d+97) then
        tmp = (a2 ** 2.0d0) * 0.6666666666666666d0
    else
        tmp = ((a1 * a1) + (a2 * a2)) * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double tmp;
	if (th <= 3.9e+97) {
		tmp = Math.pow(a2, 2.0) * 0.6666666666666666;
	} else {
		tmp = ((a1 * a1) + (a2 * a2)) * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	tmp = 0
	if th <= 3.9e+97:
		tmp = math.pow(a2, 2.0) * 0.6666666666666666
	else:
		tmp = ((a1 * a1) + (a2 * a2)) * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	tmp = 0.0
	if (th <= 3.9e+97)
		tmp = Float64((a2 ^ 2.0) * 0.6666666666666666);
	else
		tmp = Float64(Float64(Float64(a1 * a1) + Float64(a2 * a2)) * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	tmp = 0.0;
	if (th <= 3.9e+97)
		tmp = (a2 ^ 2.0) * 0.6666666666666666;
	else
		tmp = ((a1 * a1) + (a2 * a2)) * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := If[LessEqual[th, 3.9e+97], N[(N[Power[a2, 2.0], $MachinePrecision] * 0.6666666666666666), $MachinePrecision], N[(N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision] * -0.5), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 3.8999999999999999e97

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 71.4%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 49.3%

      \[\leadsto \color{blue}{0.6666666666666666} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   8. Taylor expanded in a1 around 0 32.1%

      \[\leadsto \color{blue}{0.6666666666666666 \cdot {a2}^{2}} \]
   9. Step-by-step derivation

      1. *-commutative 32.1%

         \[\leadsto \color{blue}{{a2}^{2} \cdot 0.6666666666666666} \]

   10. Simplified 32.1%

       \[\leadsto \color{blue}{{a2}^{2} \cdot 0.6666666666666666} \]

   if 3.8999999999999999e97 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.6%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.4%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 33.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;{a2}^{2} \cdot 0.6666666666666666\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 48.0% accurate, 29.6× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;t\_1 \cdot 0.6666666666666666\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= th 3.9e+97) (* t_1 0.6666666666666666) (* t_1 -0.5))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.6666666666666666;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (th <= 3.9d+97) then
        tmp = t_1 * 0.6666666666666666d0
    else
        tmp = t_1 * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.6666666666666666;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if th <= 3.9e+97:
		tmp = t_1 * 0.6666666666666666
	else:
		tmp = t_1 * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (th <= 3.9e+97)
		tmp = Float64(t_1 * 0.6666666666666666);
	else
		tmp = Float64(t_1 * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (th <= 3.9e+97)
		tmp = t_1 * 0.6666666666666666;
	else
		tmp = t_1 * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 3.9e+97], N[(t$95$1 * 0.6666666666666666), $MachinePrecision], N[(t$95$1 * -0.5), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 3.8999999999999999e97

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 71.4%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 49.3%

      \[\leadsto \color{blue}{0.6666666666666666} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 3.8999999999999999e97 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.6%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.4%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 47.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot 0.6666666666666666\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 46.9% accurate, 29.6× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;t\_1 \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= th 3.9e+97) (* t_1 0.5) (* t_1 -0.5))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.5;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (th <= 3.9d+97) then
        tmp = t_1 * 0.5d0
    else
        tmp = t_1 * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.5;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if th <= 3.9e+97:
		tmp = t_1 * 0.5
	else:
		tmp = t_1 * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (th <= 3.9e+97)
		tmp = Float64(t_1 * 0.5);
	else
		tmp = Float64(t_1 * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (th <= 3.9e+97)
		tmp = t_1 * 0.5;
	else
		tmp = t_1 * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 3.9e+97], N[(t$95$1 * 0.5), $MachinePrecision], N[(t$95$1 * -0.5), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 3.8999999999999999e97

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 71.4%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 47.8%

      \[\leadsto \color{blue}{0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 3.8999999999999999e97 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.6%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.4%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 46.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 46.5% accurate, 29.6× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;t\_1 \cdot 0.375\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= th 3.9e+97) (* t_1 0.375) (* t_1 -0.5))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.375;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (th <= 3.9d+97) then
        tmp = t_1 * 0.375d0
    else
        tmp = t_1 * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.375;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if th <= 3.9e+97:
		tmp = t_1 * 0.375
	else:
		tmp = t_1 * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (th <= 3.9e+97)
		tmp = Float64(t_1 * 0.375);
	else
		tmp = Float64(t_1 * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (th <= 3.9e+97)
		tmp = t_1 * 0.375;
	else
		tmp = t_1 * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 3.9e+97], N[(t$95$1 * 0.375), $MachinePrecision], N[(t$95$1 * -0.5), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 3.8999999999999999e97

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 71.4%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 47.3%

      \[\leadsto \color{blue}{0.375} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 3.8999999999999999e97 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.6%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.4%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 45.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot 0.375\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 46.4% accurate, 29.6× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;t\_1 \cdot 0.3333333333333333\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= th 3.9e+97) (* t_1 0.3333333333333333) (* t_1 -0.5))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.3333333333333333;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (th <= 3.9d+97) then
        tmp = t_1 * 0.3333333333333333d0
    else
        tmp = t_1 * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.3333333333333333;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if th <= 3.9e+97:
		tmp = t_1 * 0.3333333333333333
	else:
		tmp = t_1 * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (th <= 3.9e+97)
		tmp = Float64(t_1 * 0.3333333333333333);
	else
		tmp = Float64(t_1 * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (th <= 3.9e+97)
		tmp = t_1 * 0.3333333333333333;
	else
		tmp = t_1 * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 3.9e+97], N[(t$95$1 * 0.3333333333333333), $MachinePrecision], N[(t$95$1 * -0.5), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 3.8999999999999999e97

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 71.4%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 47.1%

      \[\leadsto \color{blue}{0.3333333333333333} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 3.8999999999999999e97 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.6%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.4%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 45.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot 0.3333333333333333\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 46.2% accurate, 29.6× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;t\_1 \cdot 0.25\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= th 3.9e+97) (* t_1 0.25) (* t_1 -0.5))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.25;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (th <= 3.9d+97) then
        tmp = t_1 * 0.25d0
    else
        tmp = t_1 * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.25;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if th <= 3.9e+97:
		tmp = t_1 * 0.25
	else:
		tmp = t_1 * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (th <= 3.9e+97)
		tmp = Float64(t_1 * 0.25);
	else
		tmp = Float64(t_1 * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (th <= 3.9e+97)
		tmp = t_1 * 0.25;
	else
		tmp = t_1 * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 3.9e+97], N[(t$95$1 * 0.25), $MachinePrecision], N[(t$95$1 * -0.5), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 3.8999999999999999e97

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 71.4%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 46.9%

      \[\leadsto \color{blue}{0.25} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 3.8999999999999999e97 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.6%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.4%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 45.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot 0.25\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 15: 45.9% accurate, 29.6× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a1 \cdot a1 + a2 \cdot a2\\ \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;t\_1 \cdot 0.16666666666666666\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ (* a1 a1) (* a2 a2))))
   (if (<= th 3.9e+97) (* t_1 0.16666666666666666) (* t_1 -0.5))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.16666666666666666;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = (a1 * a1) + (a2 * a2)
    if (th <= 3.9d+97) then
        tmp = t_1 * 0.16666666666666666d0
    else
        tmp = t_1 * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = (a1 * a1) + (a2 * a2);
	double tmp;
	if (th <= 3.9e+97) {
		tmp = t_1 * 0.16666666666666666;
	} else {
		tmp = t_1 * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = (a1 * a1) + (a2 * a2)
	tmp = 0
	if th <= 3.9e+97:
		tmp = t_1 * 0.16666666666666666
	else:
		tmp = t_1 * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(Float64(a1 * a1) + Float64(a2 * a2))
	tmp = 0.0
	if (th <= 3.9e+97)
		tmp = Float64(t_1 * 0.16666666666666666);
	else
		tmp = Float64(t_1 * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = (a1 * a1) + (a2 * a2);
	tmp = 0.0;
	if (th <= 3.9e+97)
		tmp = t_1 * 0.16666666666666666;
	else
		tmp = t_1 * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 3.9e+97], N[(t$95$1 * 0.16666666666666666), $MachinePrecision], N[(t$95$1 * -0.5), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 3.8999999999999999e97

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 71.4%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 46.6%

      \[\leadsto \color{blue}{0.16666666666666666} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   if 3.8999999999999999e97 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.6%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.4%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 45.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 3.9 \cdot 10^{+97}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot 0.16666666666666666\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 29.4% accurate, 29.6× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} \mathbf{if}\;th \leq 5.8 \cdot 10^{-18}:\\ \;\;\;\;0.75 \cdot \left(a2 + a1 \cdot a1\right)\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (if (<= th 5.8e-18)
   (* 0.75 (+ a2 (* a1 a1)))
   (* (+ (* a1 a1) (* a2 a2)) -0.5)))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double tmp;
	if (th <= 5.8e-18) {
		tmp = 0.75 * (a2 + (a1 * a1));
	} else {
		tmp = ((a1 * a1) + (a2 * a2)) * -0.5;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: tmp
    if (th <= 5.8d-18) then
        tmp = 0.75d0 * (a2 + (a1 * a1))
    else
        tmp = ((a1 * a1) + (a2 * a2)) * (-0.5d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double tmp;
	if (th <= 5.8e-18) {
		tmp = 0.75 * (a2 + (a1 * a1));
	} else {
		tmp = ((a1 * a1) + (a2 * a2)) * -0.5;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	tmp = 0
	if th <= 5.8e-18:
		tmp = 0.75 * (a2 + (a1 * a1))
	else:
		tmp = ((a1 * a1) + (a2 * a2)) * -0.5
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	tmp = 0.0
	if (th <= 5.8e-18)
		tmp = Float64(0.75 * Float64(a2 + Float64(a1 * a1)));
	else
		tmp = Float64(Float64(Float64(a1 * a1) + Float64(a2 * a2)) * -0.5);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	tmp = 0.0;
	if (th <= 5.8e-18)
		tmp = 0.75 * (a2 + (a1 * a1));
	else
		tmp = ((a1 * a1) + (a2 * a2)) * -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := If[LessEqual[th, 5.8e-18], N[(0.75 * N[(a2 + N[(a1 * a1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision] * -0.5), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 5.8e-18

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 76.2%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 50.4%

      \[\leadsto \color{blue}{0.75} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 37.6%

      \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 25.9%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 25.9%

       \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   if 5.8e-18 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 32.0%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 35.7%

      \[\leadsto \color{blue}{-0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 28.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 5.8 \cdot 10^{-18}:\\ \;\;\;\;0.75 \cdot \left(a2 + a1 \cdot a1\right)\\ \mathbf{else}:\\ \;\;\;\;\left(a1 \cdot a1 + a2 \cdot a2\right) \cdot -0.5\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 25.6% accurate, 34.5× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a2 + a1 \cdot a1\\ \mathbf{if}\;th \leq 2.25 \cdot 10^{+170}:\\ \;\;\;\;0.75 \cdot t\_1\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.125\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ a2 (* a1 a1))))
   (if (<= th 2.25e+170) (* 0.75 t_1) (* t_1 -0.125))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = a2 + (a1 * a1);
	double tmp;
	if (th <= 2.25e+170) {
		tmp = 0.75 * t_1;
	} else {
		tmp = t_1 * -0.125;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = a2 + (a1 * a1)
    if (th <= 2.25d+170) then
        tmp = 0.75d0 * t_1
    else
        tmp = t_1 * (-0.125d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = a2 + (a1 * a1);
	double tmp;
	if (th <= 2.25e+170) {
		tmp = 0.75 * t_1;
	} else {
		tmp = t_1 * -0.125;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = a2 + (a1 * a1)
	tmp = 0
	if th <= 2.25e+170:
		tmp = 0.75 * t_1
	else:
		tmp = t_1 * -0.125
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(a2 + Float64(a1 * a1))
	tmp = 0.0
	if (th <= 2.25e+170)
		tmp = Float64(0.75 * t_1);
	else
		tmp = Float64(t_1 * -0.125);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = a2 + (a1 * a1);
	tmp = 0.0;
	if (th <= 2.25e+170)
		tmp = 0.75 * t_1;
	else
		tmp = t_1 * -0.125;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(a2 + N[(a1 * a1), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 2.25e+170], N[(0.75 * t$95$1), $MachinePrecision], N[(t$95$1 * -0.125), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 2.25000000000000011e170

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 68.8%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 47.6%

      \[\leadsto \color{blue}{0.75} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 35.1%

      \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 24.2%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 24.2%

       \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   if 2.25000000000000011e170 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 27.3%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.3%

      \[\leadsto \color{blue}{-0.125} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 31.7%

      \[\leadsto -0.125 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 7.4%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 20.0%

       \[\leadsto -0.125 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 23.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 2.25 \cdot 10^{+170}:\\ \;\;\;\;0.75 \cdot \left(a2 + a1 \cdot a1\right)\\ \mathbf{else}:\\ \;\;\;\;\left(a2 + a1 \cdot a1\right) \cdot -0.125\\ \end{array} \]
5. Add Preprocessing

Alternative 18: 25.6% accurate, 34.5× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a2 + a1 \cdot a1\\ \mathbf{if}\;th \leq 2.25 \cdot 10^{+170}:\\ \;\;\;\;0.6666666666666666 \cdot t\_1\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.125\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ a2 (* a1 a1))))
   (if (<= th 2.25e+170) (* 0.6666666666666666 t_1) (* t_1 -0.125))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = a2 + (a1 * a1);
	double tmp;
	if (th <= 2.25e+170) {
		tmp = 0.6666666666666666 * t_1;
	} else {
		tmp = t_1 * -0.125;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = a2 + (a1 * a1)
    if (th <= 2.25d+170) then
        tmp = 0.6666666666666666d0 * t_1
    else
        tmp = t_1 * (-0.125d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = a2 + (a1 * a1);
	double tmp;
	if (th <= 2.25e+170) {
		tmp = 0.6666666666666666 * t_1;
	} else {
		tmp = t_1 * -0.125;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = a2 + (a1 * a1)
	tmp = 0
	if th <= 2.25e+170:
		tmp = 0.6666666666666666 * t_1
	else:
		tmp = t_1 * -0.125
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(a2 + Float64(a1 * a1))
	tmp = 0.0
	if (th <= 2.25e+170)
		tmp = Float64(0.6666666666666666 * t_1);
	else
		tmp = Float64(t_1 * -0.125);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = a2 + (a1 * a1);
	tmp = 0.0;
	if (th <= 2.25e+170)
		tmp = 0.6666666666666666 * t_1;
	else
		tmp = t_1 * -0.125;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(a2 + N[(a1 * a1), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 2.25e+170], N[(0.6666666666666666 * t$95$1), $MachinePrecision], N[(t$95$1 * -0.125), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 2.25000000000000011e170

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 68.8%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 47.9%

      \[\leadsto \color{blue}{0.6666666666666666} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 35.2%

      \[\leadsto 0.6666666666666666 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 24.2%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 24.3%

       \[\leadsto 0.6666666666666666 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   if 2.25000000000000011e170 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 27.3%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.3%

      \[\leadsto \color{blue}{-0.125} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 31.7%

      \[\leadsto -0.125 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 7.4%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 20.0%

       \[\leadsto -0.125 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 23.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 2.25 \cdot 10^{+170}:\\ \;\;\;\;0.6666666666666666 \cdot \left(a2 + a1 \cdot a1\right)\\ \mathbf{else}:\\ \;\;\;\;\left(a2 + a1 \cdot a1\right) \cdot -0.125\\ \end{array} \]
5. Add Preprocessing

Alternative 19: 24.7% accurate, 34.5× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} t_1 := a2 + a1 \cdot a1\\ \mathbf{if}\;th \leq 2.25 \cdot 10^{+170}:\\ \;\;\;\;0.16666666666666666 \cdot t\_1\\ \mathbf{else}:\\ \;\;\;\;t\_1 \cdot -0.125\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (+ a2 (* a1 a1))))
   (if (<= th 2.25e+170) (* 0.16666666666666666 t_1) (* t_1 -0.125))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double t_1 = a2 + (a1 * a1);
	double tmp;
	if (th <= 2.25e+170) {
		tmp = 0.16666666666666666 * t_1;
	} else {
		tmp = t_1 * -0.125;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    real(8) :: tmp
    t_1 = a2 + (a1 * a1)
    if (th <= 2.25d+170) then
        tmp = 0.16666666666666666d0 * t_1
    else
        tmp = t_1 * (-0.125d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double t_1 = a2 + (a1 * a1);
	double tmp;
	if (th <= 2.25e+170) {
		tmp = 0.16666666666666666 * t_1;
	} else {
		tmp = t_1 * -0.125;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	t_1 = a2 + (a1 * a1)
	tmp = 0
	if th <= 2.25e+170:
		tmp = 0.16666666666666666 * t_1
	else:
		tmp = t_1 * -0.125
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	t_1 = Float64(a2 + Float64(a1 * a1))
	tmp = 0.0
	if (th <= 2.25e+170)
		tmp = Float64(0.16666666666666666 * t_1);
	else
		tmp = Float64(t_1 * -0.125);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	t_1 = a2 + (a1 * a1);
	tmp = 0.0;
	if (th <= 2.25e+170)
		tmp = 0.16666666666666666 * t_1;
	else
		tmp = t_1 * -0.125;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := Block[{t$95$1 = N[(a2 + N[(a1 * a1), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[th, 2.25e+170], N[(0.16666666666666666 * t$95$1), $MachinePrecision], N[(t$95$1 * -0.125), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 2.25000000000000011e170

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 68.8%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 45.4%

      \[\leadsto \color{blue}{0.16666666666666666} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 34.3%

      \[\leadsto 0.16666666666666666 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 24.2%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 23.4%

       \[\leadsto 0.16666666666666666 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   if 2.25000000000000011e170 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 27.3%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 39.3%

      \[\leadsto \color{blue}{-0.125} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 31.7%

      \[\leadsto -0.125 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 7.4%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 20.0%

       \[\leadsto -0.125 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 23.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 2.25 \cdot 10^{+170}:\\ \;\;\;\;0.16666666666666666 \cdot \left(a2 + a1 \cdot a1\right)\\ \mathbf{else}:\\ \;\;\;\;\left(a2 + a1 \cdot a1\right) \cdot -0.125\\ \end{array} \]
5. Add Preprocessing

Alternative 20: 7.5% accurate, 34.5× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} \mathbf{if}\;th \leq 11200:\\ \;\;\;\;a2 \cdot 0.16666666666666666\\ \mathbf{else}:\\ \;\;\;\;\left(a2 + a1 \cdot a1\right) \cdot -0.125\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (if (<= th 11200.0) (* a2 0.16666666666666666) (* (+ a2 (* a1 a1)) -0.125)))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double tmp;
	if (th <= 11200.0) {
		tmp = a2 * 0.16666666666666666;
	} else {
		tmp = (a2 + (a1 * a1)) * -0.125;
	}
	return tmp;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: tmp
    if (th <= 11200.0d0) then
        tmp = a2 * 0.16666666666666666d0
    else
        tmp = (a2 + (a1 * a1)) * (-0.125d0)
    end if
    code = tmp
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	double tmp;
	if (th <= 11200.0) {
		tmp = a2 * 0.16666666666666666;
	} else {
		tmp = (a2 + (a1 * a1)) * -0.125;
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	tmp = 0
	if th <= 11200.0:
		tmp = a2 * 0.16666666666666666
	else:
		tmp = (a2 + (a1 * a1)) * -0.125
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	tmp = 0.0
	if (th <= 11200.0)
		tmp = Float64(a2 * 0.16666666666666666);
	else
		tmp = Float64(Float64(a2 + Float64(a1 * a1)) * -0.125);
	end
	return tmp
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp_2 = code(a1, a2, th)
	tmp = 0.0;
	if (th <= 11200.0)
		tmp = a2 * 0.16666666666666666;
	else
		tmp = (a2 + (a1 * a1)) * -0.125;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := If[LessEqual[th, 11200.0], N[(a2 * 0.16666666666666666), $MachinePrecision], N[(N[(a2 + N[(a1 * a1), $MachinePrecision]), $MachinePrecision] * -0.125), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 11200

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 76.3%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 48.8%

      \[\leadsto \color{blue}{0.16666666666666666} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 37.5%

      \[\leadsto 0.16666666666666666 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 26.2%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 25.3%

       \[\leadsto 0.16666666666666666 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   12. Taylor expanded in a1 around 0 3.9%

       \[\leadsto \color{blue}{0.16666666666666666 \cdot a2} \]

   if 11200 < th

   1. Initial program 99.6%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
   2. Step-by-step derivation

      1. distribute-lft-out 99.6%

         \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   3. Simplified 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in th around 0 26.9%

      \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   7. Applied egg-rr 36.2%

      \[\leadsto \color{blue}{-0.125} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   9. Applied egg-rr 33.5%

      \[\leadsto -0.125 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
   10. Step-by-step derivation

       1. rem-log-exp 10.4%

          \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

   11. Simplified 24.0%

       \[\leadsto -0.125 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 9.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 11200:\\ \;\;\;\;a2 \cdot 0.16666666666666666\\ \mathbf{else}:\\ \;\;\;\;\left(a2 + a1 \cdot a1\right) \cdot -0.125\\ \end{array} \]
5. Add Preprocessing

Alternative 21: 4.4% accurate, 138.3× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ a2 \cdot 0.16666666666666666 \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (* a2 0.16666666666666666))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return a2 * 0.16666666666666666;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = a2 * 0.16666666666666666d0
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	return a2 * 0.16666666666666666;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	return a2 * 0.16666666666666666

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	return Float64(a2 * 0.16666666666666666)
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = a2 * 0.16666666666666666;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(a2 * 0.16666666666666666), $MachinePrecision]

Derivation

1. Initial program 99.6%

   \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
2. Step-by-step derivation

   1. distribute-lft-out 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

3. Simplified 99.6%

   \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
4. Add Preprocessing

5. Taylor expanded in th around 0 63.7%

   \[\leadsto \frac{\color{blue}{1}}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

7. Applied egg-rr 43.0%

   \[\leadsto \color{blue}{0.16666666666666666} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

9. Applied egg-rr 33.1%

   \[\leadsto 0.16666666666666666 \cdot \left(a1 \cdot a1 + \color{blue}{\log \left(e^{a2}\right)}\right) \]
10. Step-by-step derivation

    1. rem-log-exp 22.2%

       \[\leadsto 0.75 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

11. Simplified 21.4%

    \[\leadsto 0.16666666666666666 \cdot \left(a1 \cdot a1 + \color{blue}{a2}\right) \]

12. Taylor expanded in a1 around 0 3.8%

    \[\leadsto \color{blue}{0.16666666666666666 \cdot a2} \]

13. Final simplification 3.8%

    \[\leadsto a2 \cdot 0.16666666666666666 \]
14. Add Preprocessing

Alternative 22: 3.5% accurate, 415.0× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ 1 \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 1.0)

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return 1.0;
}

Fortran

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = 1.0d0
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	return 1.0;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	return 1.0

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	return 1.0
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = 1.0;
end

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := 1.0

Derivation

1. Initial program 99.6%

   \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. distribute-lft-in 99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   2. associate-*l/ 99.6%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

   3. +-commutative 99.6%

      \[\leadsto \frac{\cos th \cdot \color{blue}{\left(a2 \cdot a2 + a1 \cdot a1\right)}}{\sqrt{2}} \]

   4. fma-undefine 99.6%

      \[\leadsto \frac{\cos th \cdot \color{blue}{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}}{\sqrt{2}} \]

   5. div-inv 99.5%

      \[\leadsto \color{blue}{\left(\cos th \cdot \mathsf{fma}\left(a2, a2, a1 \cdot a1\right)\right) \cdot \frac{1}{\sqrt{2}}} \]

   6. add-sqr-sqrt 99.5%

      \[\leadsto \left(\cos th \cdot \color{blue}{\left(\sqrt{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)} \cdot \sqrt{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}\right)}\right) \cdot \frac{1}{\sqrt{2}} \]

   7. pow2 99.5%

      \[\leadsto \left(\cos th \cdot \color{blue}{{\left(\sqrt{\mathsf{fma}\left(a2, a2, a1 \cdot a1\right)}\right)}^{2}}\right) \cdot \frac{1}{\sqrt{2}} \]

   8. fma-undefine 99.5%

      \[\leadsto \left(\cos th \cdot {\left(\sqrt{\color{blue}{a2 \cdot a2 + a1 \cdot a1}}\right)}^{2}\right) \cdot \frac{1}{\sqrt{2}} \]

   9. hypot-define 99.5%

      \[\leadsto \left(\cos th \cdot {\color{blue}{\left(\mathsf{hypot}\left(a2, a1\right)\right)}}^{2}\right) \cdot \frac{1}{\sqrt{2}} \]

   10. pow1/2 99.5%

       \[\leadsto \left(\cos th \cdot {\left(\mathsf{hypot}\left(a2, a1\right)\right)}^{2}\right) \cdot \frac{1}{\color{blue}{{2}^{0.5}}} \]

   11. pow-flip 99.6%

       \[\leadsto \left(\cos th \cdot {\left(\mathsf{hypot}\left(a2, a1\right)\right)}^{2}\right) \cdot \color{blue}{{2}^{\left(-0.5\right)}} \]

   12. metadata-eval 99.6%

       \[\leadsto \left(\cos th \cdot {\left(\mathsf{hypot}\left(a2, a1\right)\right)}^{2}\right) \cdot {2}^{\color{blue}{-0.5}} \]

4. Applied egg-rr 99.6%

   \[\leadsto \color{blue}{\left(\cos th \cdot {\left(\mathsf{hypot}\left(a2, a1\right)\right)}^{2}\right) \cdot {2}^{-0.5}} \]

6. Applied egg-rr 3.4%

   \[\leadsto \color{blue}{\frac{\cos th \cdot a1 - \cos th \cdot a2}{\cos th \cdot a1 - \cos th \cdot a2}} \]
7. Step-by-step derivation

   1. *-inverses 3.4%

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

8. Simplified 3.4%

   \[\leadsto \color{blue}{1} \]
9. Add Preprocessing

Reproduce

herbie shell --seed 2024145 
(FPCore (a1 a2 th)
  :name "Migdal et al, Equation (64)"
  :precision binary64
  (+ (* (/ (cos th) (sqrt 2.0)) (* a1 a1)) (* (/ (cos th) (sqrt 2.0)) (* a2 a2))))