Migdal et al, Equation (51)

Percentage Accurate: 99.5% → 99.5%

Time: 14.4s

Alternatives: 12

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: 99.5% 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, 0.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 2 \cdot \left(\pi \cdot n\right)\\ \sqrt{t_0} \cdot \left({t_0}^{\left(k \cdot -0.5\right)} \cdot {k}^{-0.5}\right) \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* 2.0 (* PI n))))
   (* (sqrt t_0) (* (pow t_0 (* k -0.5)) (pow k -0.5)))))

C

double code(double k, double n) {
	double t_0 = 2.0 * (((double) M_PI) * n);
	return sqrt(t_0) * (pow(t_0, (k * -0.5)) * pow(k, -0.5));
}

Java

public static double code(double k, double n) {
	double t_0 = 2.0 * (Math.PI * n);
	return Math.sqrt(t_0) * (Math.pow(t_0, (k * -0.5)) * Math.pow(k, -0.5));
}

Python

def code(k, n):
	t_0 = 2.0 * (math.pi * n)
	return math.sqrt(t_0) * (math.pow(t_0, (k * -0.5)) * math.pow(k, -0.5))

Julia

function code(k, n)
	t_0 = Float64(2.0 * Float64(pi * n))
	return Float64(sqrt(t_0) * Float64((t_0 ^ Float64(k * -0.5)) * (k ^ -0.5)))
end

MATLAB

function tmp = code(k, n)
	t_0 = 2.0 * (pi * n);
	tmp = sqrt(t_0) * ((t_0 ^ (k * -0.5)) * (k ^ -0.5));
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[(2.0 * N[(Pi * n), $MachinePrecision]), $MachinePrecision]}, N[(N[Sqrt[t$95$0], $MachinePrecision] * N[(N[Power[t$95$0, N[(k * -0.5), $MachinePrecision]], $MachinePrecision] * N[Power[k, -0.5], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

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1. Add Preprocessing

Alternative 2: 99.5% accurate, 0.6× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 3: 98.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 4.3 \cdot 10^{-65}:\\ \;\;\;\;\sqrt{\frac{\pi}{k}} \cdot \sqrt{2 \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 4.3e-65)
   (* (sqrt (/ PI k)) (sqrt (* 2.0 n)))
   (sqrt (/ (pow (* PI (* 2.0 n)) (- 1.0 k)) k))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 4.3e-65) {
		tmp = sqrt((((double) M_PI) / k)) * sqrt((2.0 * 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 <= 4.3e-65) {
		tmp = Math.sqrt((Math.PI / k)) * Math.sqrt((2.0 * 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 <= 4.3e-65:
		tmp = math.sqrt((math.pi / k)) * math.sqrt((2.0 * 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 <= 4.3e-65)
		tmp = Float64(sqrt(Float64(pi / k)) * sqrt(Float64(2.0 * 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 <= 4.3e-65)
		tmp = sqrt((pi / k)) * sqrt((2.0 * n));
	else
		tmp = sqrt((((pi * (2.0 * n)) ^ (1.0 - k)) / k));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 4.3e-65], N[(N[Sqrt[N[(Pi / k), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(2.0 * 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 4: 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 5: 49.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 6: 49.1% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 7: 49.1% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 8: 38.4% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 9: 38.4% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 10: 37.7% accurate, 1.5× speedup

Math

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

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(2.0 * Float64(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[(2.0 * N[(Pi * N[(n / k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]

Derivation

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Alternative 11: 37.7% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 12: 37.7% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Reproduce

herbie shell --seed 2023348 
(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))))