Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.5%

Time: 11.5s

Alternatives: 7

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: 99.5% accurate, 0.6× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 2: 99.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 2.45 \cdot 10^{-32}:\\ \;\;\;\;\sqrt{2 \cdot \left(\pi \cdot n\right)} \cdot \sqrt{\frac{1}{k}}\\ \mathbf{else}:\\ \;\;\;\;{\left(\frac{k}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}\right)}^{-0.5}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 2.45e-32)
   (* (sqrt (* 2.0 (* PI n))) (sqrt (/ 1.0 k)))
   (pow (/ k (pow (* PI (* 2.0 n)) (- 1.0 k))) -0.5)))

C

double code(double k, double n) {
	double tmp;
	if (k <= 2.45e-32) {
		tmp = sqrt((2.0 * (((double) M_PI) * n))) * sqrt((1.0 / k));
	} else {
		tmp = pow((k / pow((((double) M_PI) * (2.0 * n)), (1.0 - k))), -0.5);
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 2.45e-32) {
		tmp = Math.sqrt((2.0 * (Math.PI * n))) * Math.sqrt((1.0 / k));
	} else {
		tmp = Math.pow((k / Math.pow((Math.PI * (2.0 * n)), (1.0 - k))), -0.5);
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 2.45e-32:
		tmp = math.sqrt((2.0 * (math.pi * n))) * math.sqrt((1.0 / k))
	else:
		tmp = math.pow((k / math.pow((math.pi * (2.0 * n)), (1.0 - k))), -0.5)
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 2.45e-32)
		tmp = Float64(sqrt(Float64(2.0 * Float64(pi * n))) * sqrt(Float64(1.0 / k)));
	else
		tmp = Float64(k / (Float64(pi * Float64(2.0 * n)) ^ Float64(1.0 - k))) ^ -0.5;
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 2.45e-32)
		tmp = sqrt((2.0 * (pi * n))) * sqrt((1.0 / k));
	else
		tmp = (k / ((pi * (2.0 * n)) ^ (1.0 - k))) ^ -0.5;
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 2.45e-32], N[(N[Sqrt[N[(2.0 * N[(Pi * n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Power[N[(k / N[Power[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], -0.5], $MachinePrecision]]

Derivation

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Alternative 3: 99.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \left(2 \cdot n\right)\\ \mathbf{if}\;k \leq 7.6 \cdot 10^{-41}:\\ \;\;\;\;\frac{\sqrt{t_0}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{t_0}^{\left(1 - k\right)}}{k}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* PI (* 2.0 n))))
   (if (<= k 7.6e-41)
     (/ (sqrt t_0) (sqrt k))
     (sqrt (/ (pow t_0 (- 1.0 k)) k)))))

C

double code(double k, double n) {
	double t_0 = ((double) M_PI) * (2.0 * n);
	double tmp;
	if (k <= 7.6e-41) {
		tmp = sqrt(t_0) / sqrt(k);
	} 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 * (2.0 * n);
	double tmp;
	if (k <= 7.6e-41) {
		tmp = Math.sqrt(t_0) / Math.sqrt(k);
	} else {
		tmp = Math.sqrt((Math.pow(t_0, (1.0 - k)) / k));
	}
	return tmp;
}

Python

def code(k, n):
	t_0 = math.pi * (2.0 * n)
	tmp = 0
	if k <= 7.6e-41:
		tmp = math.sqrt(t_0) / math.sqrt(k)
	else:
		tmp = math.sqrt((math.pow(t_0, (1.0 - k)) / k))
	return tmp

Julia

function code(k, n)
	t_0 = Float64(pi * Float64(2.0 * n))
	tmp = 0.0
	if (k <= 7.6e-41)
		tmp = Float64(sqrt(t_0) / sqrt(k));
	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 * (2.0 * n);
	tmp = 0.0;
	if (k <= 7.6e-41)
		tmp = sqrt(t_0) / sqrt(k);
	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[(2.0 * n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[k, 7.6e-41], N[(N[Sqrt[t$95$0], $MachinePrecision] / N[Sqrt[k], $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 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: 49.4% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 6: 49.3% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 7: 37.9% 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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1. Add Preprocessing

Reproduce

herbie shell --seed 2023350 
(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))))