Migdal et al, Equation (64)

Percentage Accurate: 99.5% → 99.6%

Time: 9.4s

Alternatives: 9

Speedup: 2.0×

• 42.7% 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: 99.5% 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, 1.0× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \left(\sqrt{0.5} \cdot \cos th\right) \cdot \mathsf{fma}\left(a2, a2, {a1}^{2}\right) \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (* (* (sqrt 0.5) (cos th)) (fma a2 a2 (pow a1 2.0))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return (sqrt(0.5) * cos(th)) * fma(a2, a2, pow(a1, 2.0));
}

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	return Float64(Float64(sqrt(0.5) * cos(th)) * fma(a2, a2, (a1 ^ 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[(N[Sqrt[0.5], $MachinePrecision] * N[Cos[th], $MachinePrecision]), $MachinePrecision] * N[(a2 * a2 + N[Power[a1, 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%

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

3. Simplified 99.6%

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

   1. clear-num 99.5%

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

   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%

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

5. Applied egg-rr 99.6%

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

6. Taylor expanded in th around inf 99.6%

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

   1. *-commutative 99.6%

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

8. Simplified 99.6%

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

   1. +-commutative 99.6%

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

   2. fma-def 99.6%

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

   3. pow2 99.6%

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

10. Applied egg-rr 99.6%

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

11. Final simplification 99.6%

    \[\leadsto \left(\sqrt{0.5} \cdot \cos th\right) \cdot \mathsf{fma}\left(a2, a2, {a1}^{2}\right) \]

Alternative 2: 72.0% accurate, 2.0× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} \mathbf{if}\;\cos th \leq 0.7:\\ \;\;\;\;a2 \cdot \left(\cos th \cdot a2\right)\\ \mathbf{else}:\\ \;\;\;\;\sqrt{0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)\\ \end{array} \end{array} \]

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) 0.7)
   (* a2 (* (cos th) a2))
   (* (sqrt 0.5) (+ (* a1 a1) (* a2 a2)))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double tmp;
	if (cos(th) <= 0.7) {
		tmp = a2 * (cos(th) * a2);
	} else {
		tmp = sqrt(0.5) * ((a1 * a1) + (a2 * a2));
	}
	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) <= 0.7d0) then
        tmp = a2 * (cos(th) * a2)
    else
        tmp = sqrt(0.5d0) * ((a1 * a1) + (a2 * a2))
    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) <= 0.7) {
		tmp = a2 * (Math.cos(th) * a2);
	} else {
		tmp = Math.sqrt(0.5) * ((a1 * a1) + (a2 * a2));
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	tmp = 0
	if math.cos(th) <= 0.7:
		tmp = a2 * (math.cos(th) * a2)
	else:
		tmp = math.sqrt(0.5) * ((a1 * a1) + (a2 * a2))
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	tmp = 0.0
	if (cos(th) <= 0.7)
		tmp = Float64(a2 * Float64(cos(th) * a2));
	else
		tmp = Float64(sqrt(0.5) * Float64(Float64(a1 * a1) + Float64(a2 * a2)));
	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) <= 0.7)
		tmp = a2 * (cos(th) * a2);
	else
		tmp = sqrt(0.5) * ((a1 * a1) + (a2 * a2));
	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], 0.7], N[(a2 * N[(N[Cos[th], $MachinePrecision] * a2), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[0.5], $MachinePrecision] * N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (cos.f64 th) < 0.69999999999999996

   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. Taylor expanded in a1 around 0 44.1%

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

   6. Applied egg-rr 32.3%

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

   if 0.69999999999999996 < (cos.f64 th)

   1. Initial program 99.5%

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

      1. distribute-lft-out 99.5%

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

   3. Simplified 99.5%

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

      1. clear-num 99.5%

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

      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%

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

   5. Applied egg-rr 99.6%

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

   6. Taylor expanded in th around 0 90.3%

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

4. Final simplification 65.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\cos th \leq 0.7:\\ \;\;\;\;a2 \cdot \left(\cos th \cdot a2\right)\\ \mathbf{else}:\\ \;\;\;\;\sqrt{0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)\\ \end{array} \]

Alternative 3: 99.6% accurate, 2.0× speedup

Math

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

FPCore

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

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return (sqrt(0.5) * cos(th)) * ((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 = (sqrt(0.5d0) * cos(th)) * ((a1 * a1) + (a2 * a2))
end function

Java

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

Python

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

Julia

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

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = (sqrt(0.5) * cos(th)) * ((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[Sqrt[0.5], $MachinePrecision] * N[Cos[th], $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. Step-by-step derivation

   1. clear-num 99.5%

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

   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%

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

5. Applied egg-rr 99.6%

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

6. Taylor expanded in th around inf 99.6%

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

   1. *-commutative 99.6%

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

8. Simplified 99.6%

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

9. Final simplification 99.6%

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

Alternative 4: 56.8% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

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 = (cos(th) * (a2 * a2)) / sqrt(2.0d0)
end function

Java

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

Python

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

Julia

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

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = (cos(th) * (a2 * a2)) / 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[(N[Cos[th], $MachinePrecision] * N[(a2 * a2), $MachinePrecision]), $MachinePrecision] / N[Sqrt[2.0], $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. Taylor expanded in a1 around 0 53.0%

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

6. Applied egg-rr 53.0%

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

7. Final simplification 53.0%

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

Alternative 5: 37.0% accurate, 4.0× speedup

Math

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

FPCore

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

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return a2 * (cos(th) * 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 = a2 * (cos(th) * a2)
end function

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(a2 * N[(N[Cos[th], $MachinePrecision] * a2), $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. Taylor expanded in a1 around 0 53.0%

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

6. Applied egg-rr 36.0%

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

7. Final simplification 36.0%

   \[\leadsto a2 \cdot \left(\cos th \cdot a2\right) \]

Alternative 6: 46.4% accurate, 46.1× speedup

Math

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

FPCore

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

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return 0.5 * ((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 = 0.5d0 * ((a1 * a1) + (a2 * a2))
end function

Java

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

Python

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

Julia

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

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = 0.5 * ((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[(0.5 * 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. Taylor expanded in th around 0 61.8%

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

6. Applied egg-rr 42.0%

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

7. Final simplification 42.0%

   \[\leadsto 0.5 \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

Alternative 7: 6.8% accurate, 58.8× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \begin{array}{l} \mathbf{if}\;th \leq 5500000:\\ \;\;\;\;a2 + a1\\ \mathbf{else}:\\ \;\;\;\;a1 - a2 \cdot a2\\ \end{array} \end{array} \]

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (if (<= th 5500000.0) (+ a2 a1) (- a1 (* a2 a2))))

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	double tmp;
	if (th <= 5500000.0) {
		tmp = a2 + a1;
	} else {
		tmp = a1 - (a2 * a2);
	}
	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 <= 5500000.0d0) then
        tmp = a2 + a1
    else
        tmp = a1 - (a2 * a2)
    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 <= 5500000.0) {
		tmp = a2 + a1;
	} else {
		tmp = a1 - (a2 * a2);
	}
	return tmp;
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	tmp = 0
	if th <= 5500000.0:
		tmp = a2 + a1
	else:
		tmp = a1 - (a2 * a2)
	return tmp

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	tmp = 0.0
	if (th <= 5500000.0)
		tmp = Float64(a2 + a1);
	else
		tmp = Float64(a1 - Float64(a2 * a2));
	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 <= 5500000.0)
		tmp = a2 + a1;
	else
		tmp = a1 - (a2 * a2);
	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, 5500000.0], N[(a2 + a1), $MachinePrecision], N[(a1 - N[(a2 * a2), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 5.5e6

   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. Taylor expanded in th around 0 74.5%

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

   6. Applied egg-rr 4.1%

      \[\leadsto \color{blue}{a2 + a1} \]

   if 5.5e6 < th

   1. Initial program 99.5%

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

      1. distribute-lft-out 99.5%

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

   3. Simplified 99.5%

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

   4. Taylor expanded in th around 0 28.6%

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

   6. Applied egg-rr 13.9%

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

4. Final simplification 6.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 5500000:\\ \;\;\;\;a2 + a1\\ \mathbf{else}:\\ \;\;\;\;a1 - a2 \cdot a2\\ \end{array} \]

Alternative 8: 30.4% accurate, 59.3× speedup

Math

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

FPCore

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

C

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

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 = a2 * (a2 + (a1 * 2.0d0))
end function

Java

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

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	return a2 * (a2 + (a1 * 2.0))

Julia

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

MATLAB

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

Wolfram

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

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

3. Simplified 99.6%

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

4. Taylor expanded in th around 0 61.8%

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

6. Applied egg-rr 35.9%

   \[\leadsto \color{blue}{\left(a1 + a2\right) \cdot a1 + \left(a1 + a2\right) \cdot a2} \]
7. Step-by-step derivation

   1. distribute-lft-in 41.8%

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

8. Simplified 41.8%

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

9. Taylor expanded in a1 around 0 25.1%

   \[\leadsto \color{blue}{2 \cdot \left(a1 \cdot a2\right) + {a2}^{2}} \]
10. Step-by-step derivation

    1. +-commutative 25.1%

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

    2. unpow2 25.1%

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

    3. associate-*r* 25.1%

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

    4. distribute-rgt-out 29.8%

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

11. Simplified 29.8%

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

12. Final simplification 29.8%

    \[\leadsto a2 \cdot \left(a2 + a1 \cdot 2\right) \]

Alternative 9: 4.0% accurate, 138.3× speedup

Math

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

FPCore

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

C

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

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 + a1
end function

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(a2 + a1), $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. Taylor expanded in th around 0 61.8%

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

6. Applied egg-rr 4.2%

   \[\leadsto \color{blue}{a2 + a1} \]

7. Final simplification 4.2%

   \[\leadsto a2 + a1 \]

Reproduce

herbie shell --seed 2023339 
(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))))