Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.5%

Time: 11.3s

Alternatives: 9

Speedup: 1.1×

• Could not converge on a ground truth

  1. k = 8.467416492603893e+183
  2. n = -1.4843496959706205e+202

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, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 2 \cdot \left(\pi \cdot n\right)\\ \frac{\sqrt{t\_0}}{\sqrt{k} \cdot {t\_0}^{\left(k \cdot 0.5\right)}} \end{array} \end{array} \]

FPCore

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

C

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

Python

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

Julia

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

MATLAB

function tmp = code(k, n)
	t_0 = 2.0 * (pi * n);
	tmp = sqrt(t_0) / (sqrt(k) * (t_0 ^ (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[Sqrt[k], $MachinePrecision] * N[Power[t$95$0, N[(k * 0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 99.5%

   \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
2. Add Preprocessing
3. Step-by-step derivation

   1. lift-*.f64 N/A

      \[\leadsto \color{blue}{\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}} \]

   2. *-commutative N/A

      \[\leadsto \color{blue}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}} \]

   3. lift-/.f64 N/A

      \[\leadsto {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \color{blue}{\frac{1}{\sqrt{k}}} \]

   4. un-div-inv N/A

      \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]

   5. lift-pow.f64 N/A

      \[\leadsto \frac{\color{blue}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]

   6. lift-/.f64 N/A

      \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\color{blue}{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]

   7. lift--.f64 N/A

      \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{\color{blue}{1 - k}}{2}\right)}}{\sqrt{k}} \]

   8. div-sub N/A

      \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}}}{\sqrt{k}} \]

   9. metadata-eval N/A

      \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\color{blue}{\frac{1}{2}} - \frac{k}{2}\right)}}{\sqrt{k}} \]

   10. pow-sub N/A

       \[\leadsto \frac{\color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}}}{\sqrt{k}} \]

   11. associate-/l/ N/A

       \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{\sqrt{k} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \]

   12. lower-/.f64 N/A

       \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{\sqrt{k} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \]

4. Applied rewrites 99.6%

   \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{\sqrt{k} \cdot {\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}}} \]
5. Add Preprocessing

Alternative 2: 61.1% accurate, 0.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt{\pi \cdot n}\\ \mathbf{if}\;\frac{1}{\sqrt{k}} \cdot {\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1 - k}{2}\right)} \leq 2 \cdot 10^{-82}:\\ \;\;\;\;t\_0 \cdot \sqrt{2 \cdot \sqrt{\frac{1}{k} \cdot \frac{1}{k}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{t\_0}{\sqrt{k \cdot 0.5}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (sqrt (* PI n))))
   (if (<= (* (/ 1.0 (sqrt k)) (pow (* n (* 2.0 PI)) (/ (- 1.0 k) 2.0))) 2e-82)
     (* t_0 (sqrt (* 2.0 (sqrt (* (/ 1.0 k) (/ 1.0 k))))))
     (/ t_0 (sqrt (* k 0.5))))))

C

double code(double k, double n) {
	double t_0 = sqrt((((double) M_PI) * n));
	double tmp;
	if (((1.0 / sqrt(k)) * pow((n * (2.0 * ((double) M_PI))), ((1.0 - k) / 2.0))) <= 2e-82) {
		tmp = t_0 * sqrt((2.0 * sqrt(((1.0 / k) * (1.0 / k)))));
	} else {
		tmp = t_0 / sqrt((k * 0.5));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double t_0 = Math.sqrt((Math.PI * n));
	double tmp;
	if (((1.0 / Math.sqrt(k)) * Math.pow((n * (2.0 * Math.PI)), ((1.0 - k) / 2.0))) <= 2e-82) {
		tmp = t_0 * Math.sqrt((2.0 * Math.sqrt(((1.0 / k) * (1.0 / k)))));
	} else {
		tmp = t_0 / Math.sqrt((k * 0.5));
	}
	return tmp;
}

Python

def code(k, n):
	t_0 = math.sqrt((math.pi * n))
	tmp = 0
	if ((1.0 / math.sqrt(k)) * math.pow((n * (2.0 * math.pi)), ((1.0 - k) / 2.0))) <= 2e-82:
		tmp = t_0 * math.sqrt((2.0 * math.sqrt(((1.0 / k) * (1.0 / k)))))
	else:
		tmp = t_0 / math.sqrt((k * 0.5))
	return tmp

Julia

function code(k, n)
	t_0 = sqrt(Float64(pi * n))
	tmp = 0.0
	if (Float64(Float64(1.0 / sqrt(k)) * (Float64(n * Float64(2.0 * pi)) ^ Float64(Float64(1.0 - k) / 2.0))) <= 2e-82)
		tmp = Float64(t_0 * sqrt(Float64(2.0 * sqrt(Float64(Float64(1.0 / k) * Float64(1.0 / k))))));
	else
		tmp = Float64(t_0 / sqrt(Float64(k * 0.5)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	t_0 = sqrt((pi * n));
	tmp = 0.0;
	if (((1.0 / sqrt(k)) * ((n * (2.0 * pi)) ^ ((1.0 - k) / 2.0))) <= 2e-82)
		tmp = t_0 * sqrt((2.0 * sqrt(((1.0 / k) * (1.0 / k)))));
	else
		tmp = t_0 / sqrt((k * 0.5));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[Sqrt[N[(Pi * n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(n * N[(2.0 * Pi), $MachinePrecision]), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2e-82], N[(t$95$0 * N[Sqrt[N[(2.0 * N[Sqrt[N[(N[(1.0 / k), $MachinePrecision] * N[(1.0 / k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(t$95$0 / N[Sqrt[N[(k * 0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64)))) < 2e-82

   1. Initial program 99.8%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
   4. Step-by-step derivation

      1. lower-*.f64 N/A

         \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]

      2. lower-sqrt.f64 N/A

         \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]

      3. lower-/.f64 N/A

         \[\leadsto \sqrt{\color{blue}{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]

      4. lower-*.f64 N/A

         \[\leadsto \sqrt{\frac{\color{blue}{n \cdot \mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]

      5. lower-PI.f64 N/A

         \[\leadsto \sqrt{\frac{n \cdot \color{blue}{\mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]

      6. lower-sqrt.f64 11.8

         \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]

   5. Applied rewrites 11.8%

      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
   6. Step-by-step derivation

   7. Applied rewrites 11.8%

      \[\leadsto \sqrt{\pi \cdot n} \cdot \color{blue}{\sqrt{\frac{1}{k} \cdot 2}} \]
   8. Step-by-step derivation

   9. Applied rewrites 52.0%

      \[\leadsto \sqrt{\pi \cdot n} \cdot \sqrt{\sqrt{\frac{1}{k} \cdot \frac{1}{k}} \cdot 2} \]

   if 2e-82 < (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64))))

   1. Initial program 99.3%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
   4. Step-by-step derivation

      1. lower-*.f64 N/A

         \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]

      2. lower-sqrt.f64 N/A

         \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]

      3. lower-/.f64 N/A

         \[\leadsto \sqrt{\color{blue}{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]

      4. lower-*.f64 N/A

         \[\leadsto \sqrt{\frac{\color{blue}{n \cdot \mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]

      5. lower-PI.f64 N/A

         \[\leadsto \sqrt{\frac{n \cdot \color{blue}{\mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]

      6. lower-sqrt.f64 48.5

         \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]

   5. Applied rewrites 48.5%

      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
   6. Step-by-step derivation

   7. Applied rewrites 48.6%

      \[\leadsto \color{blue}{\sqrt{\frac{2 \cdot \left(\pi \cdot n\right)}{k}}} \]
   8. Step-by-step derivation

   9. Applied rewrites 48.5%

      \[\leadsto \sqrt{n \cdot \left(\pi \cdot \frac{2}{k}\right)} \]
   10. Step-by-step derivation

   11. Applied rewrites 64.5%

       \[\leadsto \frac{\sqrt{\pi \cdot n}}{\color{blue}{\sqrt{k \cdot 0.5}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 61.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{\sqrt{k}} \cdot {\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1 - k}{2}\right)} \leq 2 \cdot 10^{-82}:\\ \;\;\;\;\sqrt{\pi \cdot n} \cdot \sqrt{2 \cdot \sqrt{\frac{1}{k} \cdot \frac{1}{k}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot n}}{\sqrt{k \cdot 0.5}}\\ \end{array} \]
5. Add Preprocessing

Reproduce

herbie shell --seed 2024227 
(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))))