Migdal et al, Equation (51)

Average Accuracy: 99.4% → 99.5%

Time: 14.5s

Precision: binary64

Cost: 20032

• Unsound rule application detected in e-graph. Results may not be sound.

• could not determine a ground truth

  1. k = 2.00000000000000007e154
  2. n = 0.0

Math

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

↓

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

FPCore

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

↓

(FPCore (k n)
 :precision binary64
 (* (sqrt (/ 1.0 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));
}

↓

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

↓

public static double code(double k, double n) {
	return Math.sqrt((1.0 / 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))

↓

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

↓

function code(k, n)
	return Float64(sqrt(Float64(1.0 / 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

↓

function tmp = code(k, n)
	tmp = sqrt((1.0 / 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]

↓

code[k_, n_] := N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $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. Applied egg-rr 99.6%

   \[\leadsto \color{blue}{\sqrt{\frac{1}{k}}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   Step-by-step derivation

   [Start] 99.5	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   add-sqr-sqrt [=>] 99.3	\[ \color{blue}{\left(\sqrt{\frac{1}{\sqrt{k}}} \cdot \sqrt{\frac{1}{\sqrt{k}}}\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   sqrt-unprod [=>] 99.5	\[ \color{blue}{\sqrt{\frac{1}{\sqrt{k}} \cdot \frac{1}{\sqrt{k}}}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   frac-times [=>] 99.4	\[ \sqrt{\color{blue}{\frac{1 \cdot 1}{\sqrt{k} \cdot \sqrt{k}}}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   metadata-eval [=>] 99.4	\[ \sqrt{\frac{\color{blue}{1}}{\sqrt{k} \cdot \sqrt{k}}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   add-sqr-sqrt [<=] 99.6	\[ \sqrt{\frac{1}{\color{blue}{k}}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

3. Final simplification 99.6%

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

Alternatives

Alternative 1
Accuracy	99.4%
Cost	20036

\[\begin{array}{l} \mathbf{if}\;k \leq 5.3 \cdot 10^{-16}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{1}{k} \cdot {\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}\\ \end{array} \]

Alternative 2
Accuracy	99.4%
Cost	20036

\[\begin{array}{l} \mathbf{if}\;k \leq 6.8 \cdot 10^{-17}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\sqrt{\frac{k}{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}}}\\ \end{array} \]

Alternative 3
Accuracy	99.5%
Cost	19968

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

Alternative 4
Accuracy	99.4%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 1.02 \cdot 10^{-16}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 5
Accuracy	99.5%
Cost	19904

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

Alternative 6
Accuracy	62.8%
Cost	19716

\[\begin{array}{l} \mathbf{if}\;k \leq 3.35:\\ \;\;\;\;\frac{1}{\sqrt{\frac{k}{\pi \cdot \left(2 \cdot n\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot 0\right)}^{\left(-k\right)}}{k}}\\ \end{array} \]

Alternative 7
Accuracy	73.8%
Cost	19716

\[\begin{array}{l} \mathbf{if}\;k \leq 3.35:\\ \;\;\;\;\sqrt{\pi \cdot \frac{2}{k}} \cdot \sqrt{n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot 0\right)}^{\left(-k\right)}}{k}}\\ \end{array} \]

Alternative 8
Accuracy	73.7%
Cost	19716

\[\begin{array}{l} \mathbf{if}\;k \leq 2.95:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot 0\right)}^{\left(-k\right)}}{k}}\\ \end{array} \]

Alternative 9
Accuracy	38.3%
Cost	13312

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

Alternative 10
Accuracy	37.8%
Cost	13184

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

Alternative 11
Accuracy	37.8%
Cost	13184

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

Alternative 12
Accuracy	37.8%
Cost	13184

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

Reproduce

herbie shell --seed 2023159 
(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))))