2nthrt (problem 3.4.6)

Average Accuracy: 49.0% → 97.7%

Time: 21.4s

Precision: binary64

Cost: 13188

Math

\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]

↓

\[\begin{array}{l} \mathbf{if}\;x \leq 0.55:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{{x}^{\left(\frac{1}{n}\right)}}{x}\\ \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))

↓

(FPCore (x n)
 :precision binary64
 (if (<= x 0.55)
   (- (expm1 (/ (log x) n)))
   (* (/ 1.0 n) (/ (pow x (/ 1.0 n)) x))))

C

double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}

↓

double code(double x, double n) {
	double tmp;
	if (x <= 0.55) {
		tmp = -expm1((log(x) / n));
	} else {
		tmp = (1.0 / n) * (pow(x, (1.0 / n)) / x);
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}

↓

public static double code(double x, double n) {
	double tmp;
	if (x <= 0.55) {
		tmp = -Math.expm1((Math.log(x) / n));
	} else {
		tmp = (1.0 / n) * (Math.pow(x, (1.0 / n)) / x);
	}
	return tmp;
}

Python

def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))

↓

def code(x, n):
	tmp = 0
	if x <= 0.55:
		tmp = -math.expm1((math.log(x) / n))
	else:
		tmp = (1.0 / n) * (math.pow(x, (1.0 / n)) / x)
	return tmp

Julia

function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end

↓

function code(x, n)
	tmp = 0.0
	if (x <= 0.55)
		tmp = Float64(-expm1(Float64(log(x) / n)));
	else
		tmp = Float64(Float64(1.0 / n) * Float64((x ^ Float64(1.0 / n)) / x));
	end
	return tmp
end

Wolfram

code[x_, n_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

↓

code[x_, n_] := If[LessEqual[x, 0.55], (-N[(Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]] - 1), $MachinePrecision]), N[(N[(1.0 / n), $MachinePrecision] * N[(N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 0.55000000000000004

   1. Initial program 26.8%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]

   2. Taylor expanded in x around 0 26.7%

      \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]

   3. Taylor expanded in x around 0 26.7%

      \[\leadsto \color{blue}{1 - e^{\frac{\log x}{n}}} \]

   4. Simplified 97.0%

      \[\leadsto \color{blue}{-\mathsf{expm1}\left(\frac{\log x}{n}\right)} \]
      Proof

      [Start] 26.7	\[ 1 - e^{\frac{\log x}{n}} \]
      sub-neg [=>] 26.7	\[ \color{blue}{1 + \left(-e^{\frac{\log x}{n}}\right)} \]
      +-commutative [=>] 26.7	\[ \color{blue}{\left(-e^{\frac{\log x}{n}}\right) + 1} \]
      neg-sub0 [=>] 26.7	\[ \color{blue}{\left(0 - e^{\frac{\log x}{n}}\right)} + 1 \]
      metadata-eval [<=] 26.7	\[ \left(\color{blue}{\log 1} - e^{\frac{\log x}{n}}\right) + 1 \]
      associate-+l- [=>] 26.7	\[ \color{blue}{\log 1 - \left(e^{\frac{\log x}{n}} - 1\right)} \]
      metadata-eval [=>] 26.7	\[ \color{blue}{0} - \left(e^{\frac{\log x}{n}} - 1\right) \]
      sub0-neg [=>] 26.7	\[ \color{blue}{-\left(e^{\frac{\log x}{n}} - 1\right)} \]
      expm1-def [=>] 97.0	\[ -\color{blue}{\mathsf{expm1}\left(\frac{\log x}{n}\right)} \]

   if 0.55000000000000004 < x

   1. Initial program 67.5%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]

   2. Taylor expanded in x around inf 97.2%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]

   3. Simplified 97.2%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
      Proof

      [Start] 97.2	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      log-rec [=>] 97.2	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      mul-1-neg [<=] 97.2	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      associate-*r/ [=>] 97.2	\[ \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
      neg-mul-1 [<=] 97.2	\[ \frac{e^{\frac{\color{blue}{--1 \cdot \log x}}{n}}}{n \cdot x} \]
      mul-1-neg [=>] 97.2	\[ \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]
      remove-double-neg [=>] 97.2	\[ \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]
      *-commutative [=>] 97.2	\[ \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Applied egg-rr 98.3%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x} \cdot \frac{1}{n}} \]
      Proof

      [Start] 97.2	\[ \frac{e^{\frac{\log x}{n}}}{x \cdot n} \]
      associate-/r* [=>] 98.3	\[ \color{blue}{\frac{\frac{e^{\frac{\log x}{n}}}{x}}{n}} \]
      div-inv [=>] 98.3	\[ \color{blue}{\frac{e^{\frac{\log x}{n}}}{x} \cdot \frac{1}{n}} \]
      div-inv [=>] 98.3	\[ \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{x} \cdot \frac{1}{n} \]
      exp-to-pow [=>] 98.3	\[ \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{x} \cdot \frac{1}{n} \]

3. Recombined 2 regimes into one program.

4. Final simplification 97.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 0.55:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{{x}^{\left(\frac{1}{n}\right)}}{x}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	80.4%
Cost	8668

\[\begin{array}{l} t_0 := \frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -50:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{-35}:\\ \;\;\;\;\frac{1}{-\frac{x}{\frac{-1}{n}}}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{-65}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq -1 \cdot 10^{-93}:\\ \;\;\;\;\frac{\frac{1}{x} - \frac{0.5}{x \cdot x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-61}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-35}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.05:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 2
Accuracy	63.3%
Cost	8344

\[\begin{array}{l} t_0 := \frac{\frac{1}{n}}{x}\\ \mathbf{if}\;\frac{1}{n} \leq -50:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{-35}:\\ \;\;\;\;\frac{1}{-\frac{x}{\frac{-1}{n}}}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{-51}:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-285}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-212}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.05:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 3
Accuracy	89.3%
Cost	7172

\[\begin{array}{l} \mathbf{if}\;x \leq 1700:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x} \cdot \frac{{x}^{\left(\frac{1}{n}\right)}}{n}\\ \end{array} \]

Alternative 4
Accuracy	89.3%
Cost	7172

\[\begin{array}{l} \mathbf{if}\;x \leq 22500:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{{x}^{\left(\frac{1}{n}\right)}}{x}\\ \end{array} \]

Alternative 5
Accuracy	88.7%
Cost	7044

\[\begin{array}{l} \mathbf{if}\;x \leq 13500:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \end{array} \]

Alternative 6
Accuracy	75.5%
Cost	6852

\[\begin{array}{l} \mathbf{if}\;x \leq 0.95:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 1.15 \cdot 10^{+163}:\\ \;\;\;\;\frac{\frac{1}{n}}{x} - \frac{0.5}{n \cdot \left(x \cdot x\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 7
Accuracy	72.0%
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.68:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2.8 \cdot 10^{+253}:\\ \;\;\;\;\frac{\frac{1}{n}}{x} - \frac{0.5}{n \cdot \left(x \cdot x\right)}\\ \mathbf{else}:\\ \;\;\;\;-1 + \left(1 + \frac{1}{x \cdot n}\right)\\ \end{array} \]

Alternative 8
Accuracy	75.1%
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.68:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 1.12 \cdot 10^{+163}:\\ \;\;\;\;\frac{\frac{1}{n}}{x} - \frac{0.5}{n \cdot \left(x \cdot x\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 9
Accuracy	44.9%
Cost	836

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -50:\\ \;\;\;\;-1 + \left(1 + \frac{1}{x \cdot n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \]

Alternative 10
Accuracy	35.9%
Cost	320

\[\frac{1}{x \cdot n} \]

Alternative 11
Accuracy	36.5%
Cost	320

\[\frac{\frac{1}{n}}{x} \]

Reproduce

herbie shell --seed 2023151 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))