Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.4%

Time: 15.6s

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: 99.4% 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: 99.4% accurate, 0.6× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 2: 99.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \left(n \cdot 2\right)\\ \mathbf{if}\;k \leq 2.4 \cdot 10^{-53}:\\ \;\;\;\;\sqrt{t_0} \cdot {k}^{-0.5}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\sqrt{\frac{k}{{t_0}^{\left(1 - k\right)}}}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* PI (* n 2.0))))
   (if (<= k 2.4e-53)
     (* (sqrt t_0) (pow k -0.5))
     (/ 1.0 (sqrt (/ k (pow t_0 (- 1.0 k))))))))

C

double code(double k, double n) {
	double t_0 = ((double) M_PI) * (n * 2.0);
	double tmp;
	if (k <= 2.4e-53) {
		tmp = sqrt(t_0) * pow(k, -0.5);
	} else {
		tmp = 1.0 / sqrt((k / pow(t_0, (1.0 - k))));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double t_0 = Math.PI * (n * 2.0);
	double tmp;
	if (k <= 2.4e-53) {
		tmp = Math.sqrt(t_0) * Math.pow(k, -0.5);
	} else {
		tmp = 1.0 / Math.sqrt((k / Math.pow(t_0, (1.0 - k))));
	}
	return tmp;
}

Python

def code(k, n):
	t_0 = math.pi * (n * 2.0)
	tmp = 0
	if k <= 2.4e-53:
		tmp = math.sqrt(t_0) * math.pow(k, -0.5)
	else:
		tmp = 1.0 / math.sqrt((k / math.pow(t_0, (1.0 - k))))
	return tmp

Julia

function code(k, n)
	t_0 = Float64(pi * Float64(n * 2.0))
	tmp = 0.0
	if (k <= 2.4e-53)
		tmp = Float64(sqrt(t_0) * (k ^ -0.5));
	else
		tmp = Float64(1.0 / sqrt(Float64(k / (t_0 ^ Float64(1.0 - k)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	t_0 = pi * (n * 2.0);
	tmp = 0.0;
	if (k <= 2.4e-53)
		tmp = sqrt(t_0) * (k ^ -0.5);
	else
		tmp = 1.0 / sqrt((k / (t_0 ^ (1.0 - k))));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[(Pi * N[(n * 2.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[k, 2.4e-53], N[(N[Sqrt[t$95$0], $MachinePrecision] * N[Power[k, -0.5], $MachinePrecision]), $MachinePrecision], N[(1.0 / N[Sqrt[N[(k / N[Power[t$95$0, N[(1.0 - k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]

Derivation

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Alternative 3: 99.5% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ {k}^{-0.5} \cdot {\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(\frac{1 - k}{2}\right)} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (* (pow k -0.5) (pow (* n (* PI 2.0)) (/ (- 1.0 k) 2.0))))

C

double code(double k, double n) {
	return pow(k, -0.5) * pow((n * (((double) M_PI) * 2.0)), ((1.0 - k) / 2.0));
}

Java

public static double code(double k, double n) {
	return Math.pow(k, -0.5) * Math.pow((n * (Math.PI * 2.0)), ((1.0 - k) / 2.0));
}

Python

def code(k, n):
	return math.pow(k, -0.5) * math.pow((n * (math.pi * 2.0)), ((1.0 - k) / 2.0))

Julia

function code(k, n)
	return Float64((k ^ -0.5) * (Float64(n * Float64(pi * 2.0)) ^ Float64(Float64(1.0 - k) / 2.0)))
end

MATLAB

function tmp = code(k, n)
	tmp = (k ^ -0.5) * ((n * (pi * 2.0)) ^ ((1.0 - k) / 2.0));
end

Wolfram

code[k_, n_] := N[(N[Power[k, -0.5], $MachinePrecision] * N[Power[N[(n * N[(Pi * 2.0), $MachinePrecision]), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

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Alternative 4: 99.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \left(n \cdot 2\right)\\ \mathbf{if}\;k \leq 3.6 \cdot 10^{-53}:\\ \;\;\;\;\sqrt{t_0} \cdot {k}^{-0.5}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{t_0}^{\left(1 - k\right)}}{k}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* PI (* n 2.0))))
   (if (<= k 3.6e-53)
     (* (sqrt t_0) (pow k -0.5))
     (sqrt (/ (pow t_0 (- 1.0 k)) k)))))

C

double code(double k, double n) {
	double t_0 = ((double) M_PI) * (n * 2.0);
	double tmp;
	if (k <= 3.6e-53) {
		tmp = sqrt(t_0) * pow(k, -0.5);
	} else {
		tmp = sqrt((pow(t_0, (1.0 - k)) / k));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double t_0 = Math.PI * (n * 2.0);
	double tmp;
	if (k <= 3.6e-53) {
		tmp = Math.sqrt(t_0) * Math.pow(k, -0.5);
	} else {
		tmp = Math.sqrt((Math.pow(t_0, (1.0 - k)) / k));
	}
	return tmp;
}

Python

def code(k, n):
	t_0 = math.pi * (n * 2.0)
	tmp = 0
	if k <= 3.6e-53:
		tmp = math.sqrt(t_0) * math.pow(k, -0.5)
	else:
		tmp = math.sqrt((math.pow(t_0, (1.0 - k)) / k))
	return tmp

Julia

function code(k, n)
	t_0 = Float64(pi * Float64(n * 2.0))
	tmp = 0.0
	if (k <= 3.6e-53)
		tmp = Float64(sqrt(t_0) * (k ^ -0.5));
	else
		tmp = sqrt(Float64((t_0 ^ Float64(1.0 - k)) / k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	t_0 = pi * (n * 2.0);
	tmp = 0.0;
	if (k <= 3.6e-53)
		tmp = sqrt(t_0) * (k ^ -0.5);
	else
		tmp = sqrt(((t_0 ^ (1.0 - k)) / k));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[(Pi * N[(n * 2.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[k, 3.6e-53], N[(N[Sqrt[t$95$0], $MachinePrecision] * N[Power[k, -0.5], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[Power[t$95$0, N[(1.0 - k), $MachinePrecision]], $MachinePrecision] / k), $MachinePrecision]], $MachinePrecision]]]

Derivation

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Alternative 5: 99.5% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 6: 49.9% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 7: 49.9% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 8: 50.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 9: 38.4% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 10: 38.4% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 11: 37.7% 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 12: 37.7% accurate, 1.5× speedup

Math

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

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

Derivation

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Reproduce

herbie shell --seed 2024008 
(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))))