Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.5%

Time: 13.6s

Alternatives: 15

Speedup: 1.1×

• Could not converge on a ground truth

  1. k = 3.5516130763723187e+233
  2. n = -111487480.66932766

Specification

Math

\[\begin{array}{l} \\ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))

C

double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}

Java

public static double code(double k, double n) {
	return (1.0 / Math.sqrt(k)) * Math.pow(((2.0 * Math.PI) * n), ((1.0 - k) / 2.0));
}

Python

def code(k, n):
	return (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))

Julia

function code(k, n)
	return Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0)))
end

MATLAB

function tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0));
end

Wolfram

code[k_, n_] := N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Initial Program: 99.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))

C

double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}

Java

public static double code(double k, double n) {
	return (1.0 / Math.sqrt(k)) * Math.pow(((2.0 * Math.PI) * n), ((1.0 - k) / 2.0));
}

Python

def code(k, n):
	return (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))

Julia

function code(k, n)
	return Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0)))
end

MATLAB

function tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0));
end

Wolfram

code[k_, n_] := N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.5% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 3 \cdot 10^{-41}:\\ \;\;\;\;\frac{n \cdot \sqrt{2 \cdot \frac{\pi}{n}}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 3e-41)
   (/ (* n (sqrt (* 2.0 (/ PI n)))) (sqrt k))
   (sqrt (/ (pow (* (+ PI PI) n) (- 1.0 k)) k))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 3e-41) {
		tmp = (n * sqrt((2.0 * (((double) M_PI) / n)))) / sqrt(k);
	} else {
		tmp = sqrt((pow(((((double) M_PI) + ((double) M_PI)) * n), (1.0 - k)) / k));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 3e-41) {
		tmp = (n * Math.sqrt((2.0 * (Math.PI / n)))) / Math.sqrt(k);
	} else {
		tmp = Math.sqrt((Math.pow(((Math.PI + Math.PI) * n), (1.0 - k)) / k));
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 3e-41:
		tmp = (n * math.sqrt((2.0 * (math.pi / n)))) / math.sqrt(k)
	else:
		tmp = math.sqrt((math.pow(((math.pi + math.pi) * n), (1.0 - k)) / k))
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 3e-41)
		tmp = Float64(Float64(n * sqrt(Float64(2.0 * Float64(pi / n)))) / sqrt(k));
	else
		tmp = sqrt(Float64((Float64(Float64(pi + pi) * n) ^ Float64(1.0 - k)) / k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 3e-41)
		tmp = (n * sqrt((2.0 * (pi / n)))) / sqrt(k);
	else
		tmp = sqrt(((((pi + pi) * n) ^ (1.0 - k)) / k));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 3e-41], N[(N[(n * N[Sqrt[N[(2.0 * N[(Pi / n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[Power[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision] / k), $MachinePrecision]], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 2.99999999999999989e-41

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   5. Taylor expanded in n around inf

      \[\leadsto \frac{n \cdot \sqrt{2 \cdot \frac{\mathsf{PI}\left(\right)}{n}}}{\sqrt{\color{blue}{k}}} \]

   7. Applied rewrites 50.2%

      \[\leadsto \frac{n \cdot \sqrt{2 \cdot \frac{\pi}{n}}}{\sqrt{\color{blue}{k}}} \]

   if 2.99999999999999989e-41 < k

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   3. Applied rewrites 99.4%

      \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   5. Applied rewrites 99.3%

      \[\leadsto \frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot \color{blue}{\left(\sqrt{\left(\pi + \pi\right) \cdot n} \cdot e^{\log \left(\sqrt{\left(\pi + \pi\right) \cdot n}\right) \cdot \left(-k\right)}\right)} \]

   7. Applied rewrites 87.5%

      \[\leadsto \color{blue}{\sqrt{\frac{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\left(1 - k\right)}}{k}}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 2: 99.4% accurate, 0.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt{\left(\pi + \pi\right) \cdot n}\\ \frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot \left(t\_0 \cdot e^{\log t\_0 \cdot \left(-k\right)}\right) \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (sqrt (* (+ PI PI) n))))
   (*
    (/ (fma 1.0 (sqrt k) (* (sqrt k) 0.0)) k)
    (* t_0 (exp (* (log t_0) (- k)))))))

C

double code(double k, double n) {
	double t_0 = sqrt(((((double) M_PI) + ((double) M_PI)) * n));
	return (fma(1.0, sqrt(k), (sqrt(k) * 0.0)) / k) * (t_0 * exp((log(t_0) * -k)));
}

Julia

function code(k, n)
	t_0 = sqrt(Float64(Float64(pi + pi) * n))
	return Float64(Float64(fma(1.0, sqrt(k), Float64(sqrt(k) * 0.0)) / k) * Float64(t_0 * exp(Float64(log(t_0) * Float64(-k)))))
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision]}, N[(N[(N[(1.0 * N[Sqrt[k], $MachinePrecision] + N[(N[Sqrt[k], $MachinePrecision] * 0.0), $MachinePrecision]), $MachinePrecision] / k), $MachinePrecision] * N[(t$95$0 * N[Exp[N[(N[Log[t$95$0], $MachinePrecision] * (-k)), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 99.4%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

3. Applied rewrites 99.4%

   \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

5. Applied rewrites 99.3%

   \[\leadsto \frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot \color{blue}{\left(\sqrt{\left(\pi + \pi\right) \cdot n} \cdot e^{\log \left(\sqrt{\left(\pi + \pi\right) \cdot n}\right) \cdot \left(-k\right)}\right)} \]
6. Add Preprocessing

Alternative 3: 99.4% accurate, 0.8× speedup

Math

\[\begin{array}{l} \\ \frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (*
  (/ (fma 1.0 (sqrt k) (* (sqrt k) 0.0)) k)
  (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))

C

double code(double k, double n) {
	return (fma(1.0, sqrt(k), (sqrt(k) * 0.0)) / k) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}

Julia

function code(k, n)
	return Float64(Float64(fma(1.0, sqrt(k), Float64(sqrt(k) * 0.0)) / k) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0)))
end

Wolfram

code[k_, n_] := N[(N[(N[(1.0 * N[Sqrt[k], $MachinePrecision] + N[(N[Sqrt[k], $MachinePrecision] * 0.0), $MachinePrecision]), $MachinePrecision] / k), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

3. Applied rewrites 99.4%

   \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
4. Add Preprocessing

Alternative 4: 99.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\sqrt{{\left(\sqrt{\left(\pi + \pi\right) \cdot n}\right)}^{\left(\mathsf{fma}\left(k, -2, 2\right)\right)}}}{\sqrt{k}} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (/ (sqrt (pow (sqrt (* (+ PI PI) n)) (fma k -2.0 2.0))) (sqrt k)))

C

double code(double k, double n) {
	return sqrt(pow(sqrt(((((double) M_PI) + ((double) M_PI)) * n)), fma(k, -2.0, 2.0))) / sqrt(k);
}

Julia

function code(k, n)
	return Float64(sqrt((sqrt(Float64(Float64(pi + pi) * n)) ^ fma(k, -2.0, 2.0))) / sqrt(k))
end

Wolfram

code[k_, n_] := N[(N[Sqrt[N[Power[N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision], N[(k * -2.0 + 2.0), $MachinePrecision]], $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

3. Applied rewrites 99.4%

   \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

5. Applied rewrites 99.4%

   \[\leadsto \color{blue}{\frac{\sqrt{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\left(1 - k\right)}}}{\sqrt{k}}} \]

7. Applied rewrites 99.3%

   \[\leadsto \frac{\sqrt{\color{blue}{{\left(\sqrt{\left(\pi + \pi\right) \cdot n}\right)}^{\left(2 - \left(k + k\right)\right)}}}}{\sqrt{k}} \]

9. Applied rewrites 99.3%

   \[\leadsto \frac{\sqrt{\color{blue}{{\left(\sqrt{\left(\pi + \pi\right) \cdot n}\right)}^{\left(\mathsf{fma}\left(k, -2, 2\right)\right)}}}}{\sqrt{k}} \]
10. Add Preprocessing

Alternative 5: 99.3% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \frac{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (/ (pow (* (+ PI PI) n) (fma k -0.5 0.5)) (sqrt k)))

C

double code(double k, double n) {
	return pow(((((double) M_PI) + ((double) M_PI)) * n), fma(k, -0.5, 0.5)) / sqrt(k);
}

Julia

function code(k, n)
	return Float64((Float64(Float64(pi + pi) * n) ^ fma(k, -0.5, 0.5)) / sqrt(k))
end

Wolfram

code[k_, n_] := N[(N[Power[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision], N[(k * -0.5 + 0.5), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

3. Applied rewrites 99.5%

   \[\leadsto \color{blue}{\frac{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}}} \]
4. Add Preprocessing

Alternative 6: 99.2% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \frac{\sqrt{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\left(1 - k\right)}}}{\sqrt{k}} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (/ (sqrt (pow (* (+ PI PI) n) (- 1.0 k))) (sqrt k)))

C

double code(double k, double n) {
	return sqrt(pow(((((double) M_PI) + ((double) M_PI)) * n), (1.0 - k))) / sqrt(k);
}

Java

public static double code(double k, double n) {
	return Math.sqrt(Math.pow(((Math.PI + Math.PI) * n), (1.0 - k))) / Math.sqrt(k);
}

Python

def code(k, n):
	return math.sqrt(math.pow(((math.pi + math.pi) * n), (1.0 - k))) / math.sqrt(k)

Julia

function code(k, n)
	return Float64(sqrt((Float64(Float64(pi + pi) * n) ^ Float64(1.0 - k))) / sqrt(k))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt((((pi + pi) * n) ^ (1.0 - k))) / sqrt(k);
end

Wolfram

code[k_, n_] := N[(N[Sqrt[N[Power[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

3. Applied rewrites 99.4%

   \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

5. Applied rewrites 99.4%

   \[\leadsto \color{blue}{\frac{\sqrt{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\left(1 - k\right)}}}{\sqrt{k}}} \]
6. Add Preprocessing

Alternative 7: 79.2% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\\ \mathbf{if}\;t\_0 \leq 0:\\ \;\;\;\;\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{\sqrt{\left(k \cdot k\right) \cdot \left(k \cdot k\right)}}}}\\ \mathbf{elif}\;t\_0 \leq 4 \cdot 10^{+296}:\\ \;\;\;\;\frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot \sqrt{2 \cdot \left(n \cdot \pi\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{\left(\pi + \pi\right) \cdot n}}{k \cdot \sqrt{\sqrt{\frac{1}{k} \cdot \frac{1}{k}}}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0)))))
   (if (<= t_0 0.0)
     (sqrt (* (+ PI PI) (/ n (sqrt (sqrt (* (* k k) (* k k)))))))
     (if (<= t_0 4e+296)
       (* (/ (fma 1.0 (sqrt k) (* (sqrt k) 0.0)) k) (sqrt (* 2.0 (* n PI))))
       (/
        (sqrt (* (+ PI PI) n))
        (* k (sqrt (sqrt (* (/ 1.0 k) (/ 1.0 k))))))))))

C

double code(double k, double n) {
	double t_0 = (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
	double tmp;
	if (t_0 <= 0.0) {
		tmp = sqrt(((((double) M_PI) + ((double) M_PI)) * (n / sqrt(sqrt(((k * k) * (k * k)))))));
	} else if (t_0 <= 4e+296) {
		tmp = (fma(1.0, sqrt(k), (sqrt(k) * 0.0)) / k) * sqrt((2.0 * (n * ((double) M_PI))));
	} else {
		tmp = sqrt(((((double) M_PI) + ((double) M_PI)) * n)) / (k * sqrt(sqrt(((1.0 / k) * (1.0 / k)))));
	}
	return tmp;
}

Julia

function code(k, n)
	t_0 = Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0)))
	tmp = 0.0
	if (t_0 <= 0.0)
		tmp = sqrt(Float64(Float64(pi + pi) * Float64(n / sqrt(sqrt(Float64(Float64(k * k) * Float64(k * k)))))));
	elseif (t_0 <= 4e+296)
		tmp = Float64(Float64(fma(1.0, sqrt(k), Float64(sqrt(k) * 0.0)) / k) * sqrt(Float64(2.0 * Float64(n * pi))));
	else
		tmp = Float64(sqrt(Float64(Float64(pi + pi) * n)) / Float64(k * sqrt(sqrt(Float64(Float64(1.0 / k) * Float64(1.0 / k))))));
	end
	return tmp
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$0, 0.0], N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * N[(n / N[Sqrt[N[Sqrt[N[(N[(k * k), $MachinePrecision] * N[(k * k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], If[LessEqual[t$95$0, 4e+296], N[(N[(N[(1.0 * N[Sqrt[k], $MachinePrecision] + N[(N[Sqrt[k], $MachinePrecision] * 0.0), $MachinePrecision]), $MachinePrecision] / k), $MachinePrecision] * N[Sqrt[N[(2.0 * N[(n * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision] / N[(k * N[Sqrt[N[Sqrt[N[(N[(1.0 / k), $MachinePrecision] * N[(1.0 / k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64)))) < 0.0

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   8. Applied rewrites 35.9%

      \[\leadsto \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{k \cdot k}}} \]

   10. Applied rewrites 31.8%

       \[\leadsto \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{\sqrt{\left(k \cdot k\right) \cdot \left(k \cdot k\right)}}}} \]

   if 0.0 < (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64)))) < 3.99999999999999993e296

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   3. Applied rewrites 99.4%

      \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   5. Applied rewrites 99.3%

      \[\leadsto \frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot \color{blue}{\left(\sqrt{\left(\pi + \pi\right) \cdot n} \cdot e^{\log \left(\sqrt{\left(\pi + \pi\right) \cdot n}\right) \cdot \left(-k\right)}\right)} \]

   6. Taylor expanded in k around 0

      \[\leadsto \frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot \color{blue}{\sqrt{2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)}} \]

   8. Applied rewrites 50.1%

      \[\leadsto \frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot \color{blue}{\sqrt{2 \cdot \left(n \cdot \pi\right)}} \]

   if 3.99999999999999993e296 < (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64))))

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   5. Taylor expanded in k around inf

      \[\leadsto \frac{\sqrt{2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)}}{\color{blue}{k \cdot \sqrt{\frac{1}{k}}}} \]

   7. Applied rewrites 50.1%

      \[\leadsto \frac{\sqrt{2 \cdot \left(n \cdot \pi\right)}}{\color{blue}{k \cdot \sqrt{\frac{1}{k}}}} \]

   9. Applied rewrites 50.1%

      \[\leadsto \frac{\sqrt{\left(\pi + \pi\right) \cdot n}}{k \cdot \color{blue}{\sqrt{\frac{1}{k}}}} \]

   11. Applied rewrites 36.6%

       \[\leadsto \frac{\sqrt{\left(\pi + \pi\right) \cdot n}}{k \cdot \sqrt{\sqrt{\frac{1}{k} \cdot \frac{1}{k}}}} \]
3. Recombined 3 regimes into one program.
4. Add Preprocessing

Alternative 8: 68.3% accurate, 0.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \leq 5 \cdot 10^{-124}:\\ \;\;\;\;\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{\sqrt{\left(k \cdot k\right) \cdot \left(k \cdot k\right)}}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{n \cdot \sqrt{2 \cdot \frac{\pi}{n}}}{\sqrt{k}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))) 5e-124)
   (sqrt (* (+ PI PI) (/ n (sqrt (sqrt (* (* k k) (* k k)))))))
   (/ (* n (sqrt (* 2.0 (/ PI n)))) (sqrt k))))

C

double code(double k, double n) {
	double tmp;
	if (((1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0))) <= 5e-124) {
		tmp = sqrt(((((double) M_PI) + ((double) M_PI)) * (n / sqrt(sqrt(((k * k) * (k * k)))))));
	} else {
		tmp = (n * sqrt((2.0 * (((double) M_PI) / n)))) / sqrt(k);
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (((1.0 / Math.sqrt(k)) * Math.pow(((2.0 * Math.PI) * n), ((1.0 - k) / 2.0))) <= 5e-124) {
		tmp = Math.sqrt(((Math.PI + Math.PI) * (n / Math.sqrt(Math.sqrt(((k * k) * (k * k)))))));
	} else {
		tmp = (n * Math.sqrt((2.0 * (Math.PI / n)))) / Math.sqrt(k);
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if ((1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))) <= 5e-124:
		tmp = math.sqrt(((math.pi + math.pi) * (n / math.sqrt(math.sqrt(((k * k) * (k * k)))))))
	else:
		tmp = (n * math.sqrt((2.0 * (math.pi / n)))) / math.sqrt(k)
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0))) <= 5e-124)
		tmp = sqrt(Float64(Float64(pi + pi) * Float64(n / sqrt(sqrt(Float64(Float64(k * k) * Float64(k * k)))))));
	else
		tmp = Float64(Float64(n * sqrt(Float64(2.0 * Float64(pi / n)))) / sqrt(k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (((1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0))) <= 5e-124)
		tmp = sqrt(((pi + pi) * (n / sqrt(sqrt(((k * k) * (k * k)))))));
	else
		tmp = (n * sqrt((2.0 * (pi / n)))) / sqrt(k);
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 5e-124], N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * N[(n / N[Sqrt[N[Sqrt[N[(N[(k * k), $MachinePrecision] * N[(k * k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], N[(N[(n * N[Sqrt[N[(2.0 * N[(Pi / n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64)))) < 5.0000000000000003e-124

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   8. Applied rewrites 35.9%

      \[\leadsto \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{k \cdot k}}} \]

   10. Applied rewrites 31.8%

       \[\leadsto \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{\sqrt{\left(k \cdot k\right) \cdot \left(k \cdot k\right)}}}} \]

   if 5.0000000000000003e-124 < (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64))))

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   5. Taylor expanded in n around inf

      \[\leadsto \frac{n \cdot \sqrt{2 \cdot \frac{\mathsf{PI}\left(\right)}{n}}}{\sqrt{\color{blue}{k}}} \]

   7. Applied rewrites 50.2%

      \[\leadsto \frac{n \cdot \sqrt{2 \cdot \frac{\pi}{n}}}{\sqrt{\color{blue}{k}}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 9: 64.6% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 2.8 \cdot 10^{-31}:\\ \;\;\;\;\frac{n \cdot \sqrt{2 \cdot \frac{\pi}{n}}}{\sqrt{k}}\\ \mathbf{elif}\;k \leq 1.9 \cdot 10^{+164}:\\ \;\;\;\;n \cdot \sqrt{2 \cdot \frac{\pi}{k \cdot n}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{k \cdot k}}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 2.8e-31)
   (/ (* n (sqrt (* 2.0 (/ PI n)))) (sqrt k))
   (if (<= k 1.9e+164)
     (* n (sqrt (* 2.0 (/ PI (* k n)))))
     (sqrt (* (+ PI PI) (/ n (sqrt (* k k))))))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 2.8e-31) {
		tmp = (n * sqrt((2.0 * (((double) M_PI) / n)))) / sqrt(k);
	} else if (k <= 1.9e+164) {
		tmp = n * sqrt((2.0 * (((double) M_PI) / (k * n))));
	} else {
		tmp = sqrt(((((double) M_PI) + ((double) M_PI)) * (n / sqrt((k * k)))));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 2.8e-31) {
		tmp = (n * Math.sqrt((2.0 * (Math.PI / n)))) / Math.sqrt(k);
	} else if (k <= 1.9e+164) {
		tmp = n * Math.sqrt((2.0 * (Math.PI / (k * n))));
	} else {
		tmp = Math.sqrt(((Math.PI + Math.PI) * (n / Math.sqrt((k * k)))));
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 2.8e-31:
		tmp = (n * math.sqrt((2.0 * (math.pi / n)))) / math.sqrt(k)
	elif k <= 1.9e+164:
		tmp = n * math.sqrt((2.0 * (math.pi / (k * n))))
	else:
		tmp = math.sqrt(((math.pi + math.pi) * (n / math.sqrt((k * k)))))
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 2.8e-31)
		tmp = Float64(Float64(n * sqrt(Float64(2.0 * Float64(pi / n)))) / sqrt(k));
	elseif (k <= 1.9e+164)
		tmp = Float64(n * sqrt(Float64(2.0 * Float64(pi / Float64(k * n)))));
	else
		tmp = sqrt(Float64(Float64(pi + pi) * Float64(n / sqrt(Float64(k * k)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 2.8e-31)
		tmp = (n * sqrt((2.0 * (pi / n)))) / sqrt(k);
	elseif (k <= 1.9e+164)
		tmp = n * sqrt((2.0 * (pi / (k * n))));
	else
		tmp = sqrt(((pi + pi) * (n / sqrt((k * k)))));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 2.8e-31], N[(N[(n * N[Sqrt[N[(2.0 * N[(Pi / n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 1.9e+164], N[(n * N[Sqrt[N[(2.0 * N[(Pi / N[(k * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * N[(n / N[Sqrt[N[(k * k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if k < 2.7999999999999999e-31

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   5. Taylor expanded in n around inf

      \[\leadsto \frac{n \cdot \sqrt{2 \cdot \frac{\mathsf{PI}\left(\right)}{n}}}{\sqrt{\color{blue}{k}}} \]

   7. Applied rewrites 50.2%

      \[\leadsto \frac{n \cdot \sqrt{2 \cdot \frac{\pi}{n}}}{\sqrt{\color{blue}{k}}} \]

   if 2.7999999999999999e-31 < k < 1.90000000000000011e164

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   7. Taylor expanded in n around inf

      \[\leadsto n \cdot \color{blue}{\sqrt{2 \cdot \frac{\mathsf{PI}\left(\right)}{k \cdot n}}} \]

   9. Applied rewrites 50.4%

      \[\leadsto n \cdot \color{blue}{\sqrt{2 \cdot \frac{\pi}{k \cdot n}}} \]

   if 1.90000000000000011e164 < k

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   8. Applied rewrites 35.9%

      \[\leadsto \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{k \cdot k}}} \]
3. Recombined 3 regimes into one program.
4. Add Preprocessing

Alternative 10: 64.6% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 2.8 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{k} \cdot \sqrt{2 \cdot \left(n \cdot \pi\right)}}{k}\\ \mathbf{elif}\;k \leq 1.9 \cdot 10^{+164}:\\ \;\;\;\;n \cdot \sqrt{2 \cdot \frac{\pi}{k \cdot n}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{k \cdot k}}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 2.8e-31)
   (/ (* (sqrt k) (sqrt (* 2.0 (* n PI)))) k)
   (if (<= k 1.9e+164)
     (* n (sqrt (* 2.0 (/ PI (* k n)))))
     (sqrt (* (+ PI PI) (/ n (sqrt (* k k))))))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 2.8e-31) {
		tmp = (sqrt(k) * sqrt((2.0 * (n * ((double) M_PI))))) / k;
	} else if (k <= 1.9e+164) {
		tmp = n * sqrt((2.0 * (((double) M_PI) / (k * n))));
	} else {
		tmp = sqrt(((((double) M_PI) + ((double) M_PI)) * (n / sqrt((k * k)))));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 2.8e-31) {
		tmp = (Math.sqrt(k) * Math.sqrt((2.0 * (n * Math.PI)))) / k;
	} else if (k <= 1.9e+164) {
		tmp = n * Math.sqrt((2.0 * (Math.PI / (k * n))));
	} else {
		tmp = Math.sqrt(((Math.PI + Math.PI) * (n / Math.sqrt((k * k)))));
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 2.8e-31:
		tmp = (math.sqrt(k) * math.sqrt((2.0 * (n * math.pi)))) / k
	elif k <= 1.9e+164:
		tmp = n * math.sqrt((2.0 * (math.pi / (k * n))))
	else:
		tmp = math.sqrt(((math.pi + math.pi) * (n / math.sqrt((k * k)))))
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 2.8e-31)
		tmp = Float64(Float64(sqrt(k) * sqrt(Float64(2.0 * Float64(n * pi)))) / k);
	elseif (k <= 1.9e+164)
		tmp = Float64(n * sqrt(Float64(2.0 * Float64(pi / Float64(k * n)))));
	else
		tmp = sqrt(Float64(Float64(pi + pi) * Float64(n / sqrt(Float64(k * k)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 2.8e-31)
		tmp = (sqrt(k) * sqrt((2.0 * (n * pi)))) / k;
	elseif (k <= 1.9e+164)
		tmp = n * sqrt((2.0 * (pi / (k * n))));
	else
		tmp = sqrt(((pi + pi) * (n / sqrt((k * k)))));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 2.8e-31], N[(N[(N[Sqrt[k], $MachinePrecision] * N[Sqrt[N[(2.0 * N[(n * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / k), $MachinePrecision], If[LessEqual[k, 1.9e+164], N[(n * N[Sqrt[N[(2.0 * N[(Pi / N[(k * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * N[(n / N[Sqrt[N[(k * k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if k < 2.7999999999999999e-31

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   3. Applied rewrites 99.4%

      \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   5. Applied rewrites 99.3%

      \[\leadsto \frac{\mathsf{fma}\left(1, \sqrt{k}, \sqrt{k} \cdot 0\right)}{k} \cdot \color{blue}{\left(\sqrt{\left(\pi + \pi\right) \cdot n} \cdot e^{\log \left(\sqrt{\left(\pi + \pi\right) \cdot n}\right) \cdot \left(-k\right)}\right)} \]

   6. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{\sqrt{k} \cdot \sqrt{2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)}}{k}} \]

   8. Applied rewrites 50.1%

      \[\leadsto \color{blue}{\frac{\sqrt{k} \cdot \sqrt{2 \cdot \left(n \cdot \pi\right)}}{k}} \]

   if 2.7999999999999999e-31 < k < 1.90000000000000011e164

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   7. Taylor expanded in n around inf

      \[\leadsto n \cdot \color{blue}{\sqrt{2 \cdot \frac{\mathsf{PI}\left(\right)}{k \cdot n}}} \]

   9. Applied rewrites 50.4%

      \[\leadsto n \cdot \color{blue}{\sqrt{2 \cdot \frac{\pi}{k \cdot n}}} \]

   if 1.90000000000000011e164 < k

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   8. Applied rewrites 35.9%

      \[\leadsto \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{k \cdot k}}} \]
3. Recombined 3 regimes into one program.
4. Add Preprocessing

Alternative 11: 64.6% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 2.8 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{\left(\pi + \pi\right) \cdot n}}{\sqrt{k}}\\ \mathbf{elif}\;k \leq 1.9 \cdot 10^{+164}:\\ \;\;\;\;n \cdot \sqrt{2 \cdot \frac{\pi}{k \cdot n}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{k \cdot k}}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 2.8e-31)
   (/ (sqrt (* (+ PI PI) n)) (sqrt k))
   (if (<= k 1.9e+164)
     (* n (sqrt (* 2.0 (/ PI (* k n)))))
     (sqrt (* (+ PI PI) (/ n (sqrt (* k k))))))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 2.8e-31) {
		tmp = sqrt(((((double) M_PI) + ((double) M_PI)) * n)) / sqrt(k);
	} else if (k <= 1.9e+164) {
		tmp = n * sqrt((2.0 * (((double) M_PI) / (k * n))));
	} else {
		tmp = sqrt(((((double) M_PI) + ((double) M_PI)) * (n / sqrt((k * k)))));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 2.8e-31) {
		tmp = Math.sqrt(((Math.PI + Math.PI) * n)) / Math.sqrt(k);
	} else if (k <= 1.9e+164) {
		tmp = n * Math.sqrt((2.0 * (Math.PI / (k * n))));
	} else {
		tmp = Math.sqrt(((Math.PI + Math.PI) * (n / Math.sqrt((k * k)))));
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 2.8e-31:
		tmp = math.sqrt(((math.pi + math.pi) * n)) / math.sqrt(k)
	elif k <= 1.9e+164:
		tmp = n * math.sqrt((2.0 * (math.pi / (k * n))))
	else:
		tmp = math.sqrt(((math.pi + math.pi) * (n / math.sqrt((k * k)))))
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 2.8e-31)
		tmp = Float64(sqrt(Float64(Float64(pi + pi) * n)) / sqrt(k));
	elseif (k <= 1.9e+164)
		tmp = Float64(n * sqrt(Float64(2.0 * Float64(pi / Float64(k * n)))));
	else
		tmp = sqrt(Float64(Float64(pi + pi) * Float64(n / sqrt(Float64(k * k)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 2.8e-31)
		tmp = sqrt(((pi + pi) * n)) / sqrt(k);
	elseif (k <= 1.9e+164)
		tmp = n * sqrt((2.0 * (pi / (k * n))));
	else
		tmp = sqrt(((pi + pi) * (n / sqrt((k * k)))));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 2.8e-31], N[(N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 1.9e+164], N[(n * N[Sqrt[N[(2.0 * N[(Pi / N[(k * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * N[(n / N[Sqrt[N[(k * k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if k < 2.7999999999999999e-31

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 50.2%

      \[\leadsto \frac{\sqrt{\left(\pi + \pi\right) \cdot n}}{\color{blue}{\sqrt{k}}} \]

   if 2.7999999999999999e-31 < k < 1.90000000000000011e164

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   7. Taylor expanded in n around inf

      \[\leadsto n \cdot \color{blue}{\sqrt{2 \cdot \frac{\mathsf{PI}\left(\right)}{k \cdot n}}} \]

   9. Applied rewrites 50.4%

      \[\leadsto n \cdot \color{blue}{\sqrt{2 \cdot \frac{\pi}{k \cdot n}}} \]

   if 1.90000000000000011e164 < k

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   8. Applied rewrites 35.9%

      \[\leadsto \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{\sqrt{k \cdot k}}} \]
3. Recombined 3 regimes into one program.
4. Add Preprocessing

Alternative 12: 61.8% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 3.2 \cdot 10^{-57}:\\ \;\;\;\;\sqrt{\frac{n + n}{k} \cdot \pi}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \sqrt{2 \cdot \frac{\pi}{k \cdot n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= n 3.2e-57)
   (sqrt (* (/ (+ n n) k) PI))
   (* n (sqrt (* 2.0 (/ PI (* k n)))))))

C

double code(double k, double n) {
	double tmp;
	if (n <= 3.2e-57) {
		tmp = sqrt((((n + n) / k) * ((double) M_PI)));
	} else {
		tmp = n * sqrt((2.0 * (((double) M_PI) / (k * n))));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (n <= 3.2e-57) {
		tmp = Math.sqrt((((n + n) / k) * Math.PI));
	} else {
		tmp = n * Math.sqrt((2.0 * (Math.PI / (k * n))));
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if n <= 3.2e-57:
		tmp = math.sqrt((((n + n) / k) * math.pi))
	else:
		tmp = n * math.sqrt((2.0 * (math.pi / (k * n))))
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (n <= 3.2e-57)
		tmp = sqrt(Float64(Float64(Float64(n + n) / k) * pi));
	else
		tmp = Float64(n * sqrt(Float64(2.0 * Float64(pi / Float64(k * n)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (n <= 3.2e-57)
		tmp = sqrt((((n + n) / k) * pi));
	else
		tmp = n * sqrt((2.0 * (pi / (k * n))));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[n, 3.2e-57], N[Sqrt[N[(N[(N[(n + n), $MachinePrecision] / k), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision], N[(n * N[Sqrt[N[(2.0 * N[(Pi / N[(k * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 3.2000000000000001e-57

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   8. Applied rewrites 38.2%

      \[\leadsto \sqrt{\frac{n + n}{k} \cdot \pi} \]

   if 3.2000000000000001e-57 < n

   1. Initial program 99.4%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

   4. Applied rewrites 50.2%

      \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

   6. Applied rewrites 38.2%

      \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

   7. Taylor expanded in n around inf

      \[\leadsto n \cdot \color{blue}{\sqrt{2 \cdot \frac{\mathsf{PI}\left(\right)}{k \cdot n}}} \]

   9. Applied rewrites 50.4%

      \[\leadsto n \cdot \color{blue}{\sqrt{2 \cdot \frac{\pi}{k \cdot n}}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 13: 50.2% accurate, 2.7× speedup

Math

\[\begin{array}{l} \\ \frac{\sqrt{\left(\pi + \pi\right) \cdot n}}{\sqrt{k}} \end{array} \]

FPCore

(FPCore (k n) :precision binary64 (/ (sqrt (* (+ PI PI) n)) (sqrt k)))

C

double code(double k, double n) {
	return sqrt(((((double) M_PI) + ((double) M_PI)) * n)) / sqrt(k);
}

Java

public static double code(double k, double n) {
	return Math.sqrt(((Math.PI + Math.PI) * n)) / Math.sqrt(k);
}

Python

def code(k, n):
	return math.sqrt(((math.pi + math.pi) * n)) / math.sqrt(k)

Julia

function code(k, n)
	return Float64(sqrt(Float64(Float64(pi + pi) * n)) / sqrt(k))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt(((pi + pi) * n)) / sqrt(k);
end

Wolfram

code[k_, n_] := N[(N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

4. Applied rewrites 50.2%

   \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

6. Applied rewrites 50.2%

   \[\leadsto \frac{\sqrt{\left(\pi + \pi\right) \cdot n}}{\color{blue}{\sqrt{k}}} \]
7. Add Preprocessing

Alternative 14: 38.2% accurate, 3.1× speedup

Math

\[\begin{array}{l} \\ \sqrt{\frac{n + n}{k} \cdot \pi} \end{array} \]

FPCore

(FPCore (k n) :precision binary64 (sqrt (* (/ (+ n n) k) PI)))

C

double code(double k, double n) {
	return sqrt((((n + n) / k) * ((double) M_PI)));
}

Java

public static double code(double k, double n) {
	return Math.sqrt((((n + n) / k) * Math.PI));
}

Python

def code(k, n):
	return math.sqrt((((n + n) / k) * math.pi))

Julia

function code(k, n)
	return sqrt(Float64(Float64(Float64(n + n) / k) * pi))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt((((n + n) / k) * pi));
end

Wolfram

code[k_, n_] := N[Sqrt[N[(N[(N[(n + n), $MachinePrecision] / k), $MachinePrecision] * Pi), $MachinePrecision]], $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

4. Applied rewrites 50.2%

   \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

6. Applied rewrites 38.2%

   \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]

8. Applied rewrites 38.2%

   \[\leadsto \sqrt{\frac{n + n}{k} \cdot \pi} \]
9. Add Preprocessing

Alternative 15: 38.2% accurate, 3.1× speedup

Math

\[\begin{array}{l} \\ \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}} \end{array} \]

FPCore

(FPCore (k n) :precision binary64 (sqrt (* (+ PI PI) (/ n k))))

C

double code(double k, double n) {
	return sqrt(((((double) M_PI) + ((double) M_PI)) * (n / k)));
}

Java

public static double code(double k, double n) {
	return Math.sqrt(((Math.PI + Math.PI) * (n / k)));
}

Python

def code(k, n):
	return math.sqrt(((math.pi + math.pi) * (n / k)))

Julia

function code(k, n)
	return sqrt(Float64(Float64(pi + pi) * Float64(n / k)))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt(((pi + pi) * (n / k)));
end

Wolfram

code[k_, n_] := N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * N[(n / k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right)}^{\frac{1}{2}}}{\sqrt{k}}} \]

4. Applied rewrites 50.2%

   \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{0.5}}{\sqrt{k}}} \]

6. Applied rewrites 38.2%

   \[\leadsto \color{blue}{\sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}}} \]
7. Add Preprocessing

Reproduce

herbie shell --seed 2025161 
(FPCore (k n)
  :name "Migdal et al, Equation (51)"
  :precision binary64
  (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))