Migdal et al, Equation (64)

Percentage Accurate: 99.6% → 99.6%

Time: 7.2s

Alternatives: 8

Speedup: 2.0×

• 44.1% 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 99.7%

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

   1. distribute-lft-out 99.7%

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

   2. cos-neg 99.7%

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

   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* 99.7%

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

   5. cos-neg 99.7%

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

   6. +-commutative 99.7%

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

   7. fma-define 99.7%

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

3. Simplified 99.7%

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

Alternative 2: 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 99.7%

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

   1. distribute-lft-out 99.7%

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

3. Simplified 99.7%

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

Alternative 3: 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 99.7%

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

   1. distribute-lft-out 99.7%

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

3. Simplified 99.7%

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

   1. clear-num 99.7%

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

   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 99.6%

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

   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 4: 57.8% accurate, 2.0× speedup

Math

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

FPCore

NOTE: a1, a2, and th should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (* (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(cos(th) * Float64(a2 * Float64(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[Cos[th], $MachinePrecision] * N[(a2 * N[(a2 / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.7%

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

   1. distribute-lft-out 99.7%

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

   2. cos-neg 99.7%

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

   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* 99.7%

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

   5. cos-neg 99.7%

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

   6. +-commutative 99.7%

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

   7. fma-define 99.7%

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

3. Simplified 99.7%

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

5. Taylor expanded in a2 around inf 61.5%

   \[\leadsto \cos th \cdot \color{blue}{\frac{{a2}^{2}}{\sqrt{2}}} \]
6. Step-by-step derivation

   1. pow2 61.5%

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

   2. associate-/l* 61.6%

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

7. Applied egg-rr 61.6%

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

Alternative 5: 66.0% accurate, 3.7× speedup

Math

\[\begin{array}{l} [a1, a2, th] = \mathsf{sort}([a1, a2, th])\\ \\ \left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \frac{1}{\sqrt{2}} \end{array} \]

FPCore

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

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return ((a1 * a1) + (a2 * a2)) * (1.0 / 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 = ((a1 * a1) + (a2 * a2)) * (1.0d0 / sqrt(2.0d0))
end function

Java

assert a1 < a2 && a2 < th;
public static double code(double a1, double a2, double th) {
	return ((a1 * a1) + (a2 * a2)) * (1.0 / Math.sqrt(2.0));
}

Python

[a1, a2, th] = sort([a1, a2, th])
def code(a1, a2, th):
	return ((a1 * a1) + (a2 * a2)) * (1.0 / math.sqrt(2.0))

Julia

a1, a2, th = sort([a1, a2, th])
function code(a1, a2, th)
	return Float64(Float64(Float64(a1 * a1) + Float64(a2 * a2)) * Float64(1.0 / sqrt(2.0)))
end

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = ((a1 * a1) + (a2 * a2)) * (1.0 / 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[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision] * N[(1.0 / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.7%

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

   1. distribute-lft-out 99.7%

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

3. Simplified 99.7%

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

5. Taylor expanded in th around 0 67.3%

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

6. Final simplification 67.3%

   \[\leadsto \left(a1 \cdot a1 + a2 \cdot a2\right) \cdot \frac{1}{\sqrt{2}} \]
7. Add Preprocessing

Alternative 6: 66.1% accurate, 3.8× speedup

Math

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

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)) (sqrt 0.5)))

C

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

Python

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

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = ((a1 * a1) + (a2 * a2)) * 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[Sqrt[0.5], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.7%

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

   1. distribute-lft-out 99.7%

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

3. Simplified 99.7%

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

   1. clear-num 99.7%

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

   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 99.6%

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

   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 67.3%

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

8. Final simplification 67.3%

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

Alternative 7: 40.1% accurate, 4.0× speedup

Math

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

FPCore

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

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return (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 = (a2 * a2) / sqrt(2.0d0)
end function

Java

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

Python

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

Julia

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

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = (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[(a2 * a2), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.7%

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

   1. distribute-lft-out 99.7%

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

3. Simplified 99.7%

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

5. Taylor expanded in th around 0 67.3%

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

6. Taylor expanded in a1 around 0 43.3%

   \[\leadsto \color{blue}{\frac{{a2}^{2}}{\sqrt{2}}} \]
7. Step-by-step derivation

   1. pow2 61.5%

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

   2. associate-/l* 61.6%

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

8. Applied egg-rr 43.3%

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

   1. *-commutative 43.3%

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

   2. frac-2neg 43.3%

      \[\leadsto \color{blue}{\frac{-a2}{-\sqrt{2}}} \cdot a2 \]

   3. associate-*l/ 43.3%

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

10. Applied egg-rr 43.3%

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

11. Final simplification 43.3%

    \[\leadsto \frac{a2 \cdot a2}{\sqrt{2}} \]
12. Add Preprocessing

Alternative 8: 40.1% accurate, 4.0× speedup

Math

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

FPCore

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

C

assert(a1 < a2 && a2 < th);
double code(double a1, double a2, double th) {
	return 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 = a2 * (a2 / sqrt(2.0d0))
end function

Java

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

Python

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

Julia

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

MATLAB

a1, a2, th = num2cell(sort([a1, a2, th])){:}
function tmp = code(a1, a2, th)
	tmp = 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[(a2 * N[(a2 / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.7%

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

   1. distribute-lft-out 99.7%

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

3. Simplified 99.7%

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

5. Taylor expanded in th around 0 67.3%

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

6. Taylor expanded in a1 around 0 43.3%

   \[\leadsto \color{blue}{\frac{{a2}^{2}}{\sqrt{2}}} \]
7. Step-by-step derivation

   1. pow2 61.5%

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

   2. associate-/l* 61.6%

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

8. Applied egg-rr 43.3%

   \[\leadsto \color{blue}{a2 \cdot \frac{a2}{\sqrt{2}}} \]
9. Add Preprocessing

Reproduce

herbie shell --seed 2024107 
(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))))