Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.0%

Time: 14.5s

Alternatives: 11

Speedup: 1.0×

• could not determine a ground truth

  1. k = 2.00000000000000007e154
  2. n = 0.0

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: 99.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 1.7 \cdot 10^{-56}:\\ \;\;\;\;\sqrt{\frac{1}{k}} \cdot \sqrt{\left(2 \cdot \pi\right) \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 1.7e-56)
   (* (sqrt (/ 1.0 k)) (sqrt (* (* 2.0 PI) n)))
   (sqrt (/ (pow (* PI (* 2.0 n)) (- 1.0 k)) k))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 1.7e-56) {
		tmp = sqrt((1.0 / k)) * sqrt(((2.0 * ((double) M_PI)) * n));
	} else {
		tmp = sqrt((pow((((double) M_PI) * (2.0 * n)), (1.0 - k)) / k));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 1.7e-56) {
		tmp = Math.sqrt((1.0 / k)) * Math.sqrt(((2.0 * Math.PI) * n));
	} else {
		tmp = Math.sqrt((Math.pow((Math.PI * (2.0 * n)), (1.0 - k)) / k));
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 1.7e-56:
		tmp = math.sqrt((1.0 / k)) * math.sqrt(((2.0 * math.pi) * n))
	else:
		tmp = math.sqrt((math.pow((math.pi * (2.0 * n)), (1.0 - k)) / k))
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 1.7e-56)
		tmp = Float64(sqrt(Float64(1.0 / k)) * sqrt(Float64(Float64(2.0 * pi) * n)));
	else
		tmp = sqrt(Float64((Float64(pi * Float64(2.0 * n)) ^ Float64(1.0 - k)) / k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 1.7e-56)
		tmp = sqrt((1.0 / k)) * sqrt(((2.0 * pi) * n));
	else
		tmp = sqrt((((pi * (2.0 * n)) ^ (1.0 - k)) / k));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 1.7e-56], N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[Power[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision] / k), $MachinePrecision]], $MachinePrecision]]

Derivation

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Alternative 2: 99.4% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 3: 99.5% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(0.5 + k \cdot -0.5\right)}}{\sqrt{k}} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (/ (pow (* PI (* 2.0 n)) (+ 0.5 (* k -0.5))) (sqrt k)))

C

double code(double k, double n) {
	return pow((((double) M_PI) * (2.0 * n)), (0.5 + (k * -0.5))) / sqrt(k);
}

Java

public static double code(double k, double n) {
	return Math.pow((Math.PI * (2.0 * n)), (0.5 + (k * -0.5))) / Math.sqrt(k);
}

Python

def code(k, n):
	return math.pow((math.pi * (2.0 * n)), (0.5 + (k * -0.5))) / math.sqrt(k)

Julia

function code(k, n)
	return Float64((Float64(pi * Float64(2.0 * n)) ^ Float64(0.5 + Float64(k * -0.5))) / sqrt(k))
end

MATLAB

function tmp = code(k, n)
	tmp = ((pi * (2.0 * n)) ^ (0.5 + (k * -0.5))) / sqrt(k);
end

Wolfram

code[k_, n_] := N[(N[Power[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision], N[(0.5 + N[(k * -0.5), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

Derivation

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Alternative 4: 51.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 1.16 \cdot 10^{+228}:\\ \;\;\;\;\sqrt{\frac{1}{k}} \cdot \sqrt{\left(2 \cdot \pi\right) \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(2 \cdot \frac{\pi \cdot n}{k}\right)}^{1.5}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 1.16e+228)
   (* (sqrt (/ 1.0 k)) (sqrt (* (* 2.0 PI) n)))
   (cbrt (pow (* 2.0 (/ (* PI n) k)) 1.5))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 1.16e+228) {
		tmp = sqrt((1.0 / k)) * sqrt(((2.0 * ((double) M_PI)) * n));
	} else {
		tmp = cbrt(pow((2.0 * ((((double) M_PI) * n) / k)), 1.5));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 1.16e+228) {
		tmp = Math.sqrt((1.0 / k)) * Math.sqrt(((2.0 * Math.PI) * n));
	} else {
		tmp = Math.cbrt(Math.pow((2.0 * ((Math.PI * n) / k)), 1.5));
	}
	return tmp;
}

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 1.16e+228)
		tmp = Float64(sqrt(Float64(1.0 / k)) * sqrt(Float64(Float64(2.0 * pi) * n)));
	else
		tmp = cbrt((Float64(2.0 * Float64(Float64(pi * n) / k)) ^ 1.5));
	end
	return tmp
end

Wolfram

code[k_, n_] := If[LessEqual[k, 1.16e+228], N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Power[N[Power[N[(2.0 * N[(N[(Pi * n), $MachinePrecision] / k), $MachinePrecision]), $MachinePrecision], 1.5], $MachinePrecision], 1/3], $MachinePrecision]]

Derivation

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Alternative 5: 51.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 1.16 \cdot 10^{+228}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \pi\right) \cdot n}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(2 \cdot \frac{\pi \cdot n}{k}\right)}^{1.5}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 1.16e+228)
   (/ (sqrt (* (* 2.0 PI) n)) (sqrt k))
   (cbrt (pow (* 2.0 (/ (* PI n) k)) 1.5))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 1.16e+228) {
		tmp = sqrt(((2.0 * ((double) M_PI)) * n)) / sqrt(k);
	} else {
		tmp = cbrt(pow((2.0 * ((((double) M_PI) * n) / k)), 1.5));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 1.16e+228) {
		tmp = Math.sqrt(((2.0 * Math.PI) * n)) / Math.sqrt(k);
	} else {
		tmp = Math.cbrt(Math.pow((2.0 * ((Math.PI * n) / k)), 1.5));
	}
	return tmp;
}

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 1.16e+228)
		tmp = Float64(sqrt(Float64(Float64(2.0 * pi) * n)) / sqrt(k));
	else
		tmp = cbrt((Float64(2.0 * Float64(Float64(pi * n) / k)) ^ 1.5));
	end
	return tmp
end

Wolfram

code[k_, n_] := If[LessEqual[k, 1.16e+228], N[(N[Sqrt[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision], N[Power[N[Power[N[(2.0 * N[(N[(Pi * n), $MachinePrecision] / k), $MachinePrecision]), $MachinePrecision], 1.5], $MachinePrecision], 1/3], $MachinePrecision]]

Derivation

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Alternative 6: 50.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\sqrt{n}}{\sqrt{0.5 \cdot \frac{k}{\pi}}} \end{array} \]

FPCore

(FPCore (k n) :precision binary64 (/ (sqrt n) (sqrt (* 0.5 (/ k PI)))))

C

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

Java

public static double code(double k, double n) {
	return Math.sqrt(n) / Math.sqrt((0.5 * (k / Math.PI)));
}

Python

def code(k, n):
	return math.sqrt(n) / math.sqrt((0.5 * (k / math.pi)))

Julia

function code(k, n)
	return Float64(sqrt(n) / sqrt(Float64(0.5 * Float64(k / pi))))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt(n) / sqrt((0.5 * (k / pi)));
end

Wolfram

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

Derivation

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Alternative 7: 50.7% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 8: 39.6% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ \frac{1}{\sqrt{0.5 \cdot \frac{k}{\pi \cdot n}}} \end{array} \]

FPCore

(FPCore (k n) :precision binary64 (/ 1.0 (sqrt (* 0.5 (/ k (* PI n))))))

C

double code(double k, double n) {
	return 1.0 / sqrt((0.5 * (k / (((double) M_PI) * n))));
}

Java

public static double code(double k, double n) {
	return 1.0 / Math.sqrt((0.5 * (k / (Math.PI * n))));
}

Python

def code(k, n):
	return 1.0 / math.sqrt((0.5 * (k / (math.pi * n))))

Julia

function code(k, n)
	return Float64(1.0 / sqrt(Float64(0.5 * Float64(k / Float64(pi * n)))))
end

MATLAB

function tmp = code(k, n)
	tmp = 1.0 / sqrt((0.5 * (k / (pi * n))));
end

Wolfram

code[k_, n_] := N[(1.0 / N[Sqrt[N[(0.5 * N[(k / N[(Pi * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

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Alternative 9: 39.6% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ \frac{1}{\sqrt{\frac{\frac{k}{n}}{2 \cdot \pi}}} \end{array} \]

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 10: 38.9% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 11: 38.9% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Reproduce

herbie shell --seed 2024003 
(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))))