Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.5%

Time: 18.1s

Alternatives: 10

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

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 2: 99.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \left(2 \cdot n\right)\\ \mathbf{if}\;k \leq 10^{-26}:\\ \;\;\;\;\frac{\sqrt{t_0}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{t_0}^{\left(1 - k\right)}}{k}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* PI (* 2.0 n))))
   (if (<= k 1e-26) (/ (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 <= 1e-26) {
		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 <= 1e-26) {
		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 <= 1e-26:
		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 <= 1e-26)
		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 <= 1e-26)
		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, 1e-26], 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

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

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

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 4: 50.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 7 \cdot 10^{+232}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(2 \cdot \frac{n}{\frac{k}{\pi}}\right)}^{1.5}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 7e+232)
   (/ (sqrt (* PI (* 2.0 n))) (sqrt k))
   (cbrt (pow (* 2.0 (/ n (/ k PI))) 1.5))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 7e+232) {
		tmp = sqrt((((double) M_PI) * (2.0 * n))) / sqrt(k);
	} else {
		tmp = cbrt(pow((2.0 * (n / (k / ((double) M_PI)))), 1.5));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 7e+232) {
		tmp = Math.sqrt((Math.PI * (2.0 * n))) / Math.sqrt(k);
	} else {
		tmp = Math.cbrt(Math.pow((2.0 * (n / (k / Math.PI))), 1.5));
	}
	return tmp;
}

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 7e+232)
		tmp = Float64(sqrt(Float64(pi * Float64(2.0 * n))) / sqrt(k));
	else
		tmp = cbrt((Float64(2.0 * Float64(n / Float64(k / pi))) ^ 1.5));
	end
	return tmp
end

Wolfram

code[k_, n_] := If[LessEqual[k, 7e+232], N[(N[Sqrt[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision], N[Power[N[Power[N[(2.0 * N[(n / N[(k / Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1.5], $MachinePrecision], 1/3], $MachinePrecision]]

Derivation

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 5: 49.2% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 6: 49.2% 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

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 7: 38.0% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 8: 38.1% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 9: 37.5% 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

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 10: 37.5% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Reproduce

herbie shell --seed 2023340 
(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))))