2nthrt (problem 3.4.6)

Average Accuracy: 53.6% → 91.6%

Time: 31.8s

Precision: binary64

Cost: 13188

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;x \leq 0.62:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{{x}^{\left(\frac{1}{n}\right)}}{n}}{x}\\ \end{array} \]

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 0.62) (- (expm1 (/ (log x) n))) (/ (/ (pow x (/ 1.0 n)) 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 <= 0.62) {
		tmp = -expm1((log(x) / n));
	} else {
		tmp = (pow(x, (1.0 / n)) / 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 <= 0.62) {
		tmp = -Math.expm1((Math.log(x) / n));
	} else {
		tmp = (Math.pow(x, (1.0 / n)) / 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 <= 0.62:
		tmp = -math.expm1((math.log(x) / n))
	else:
		tmp = (math.pow(x, (1.0 / n)) / 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 <= 0.62)
		tmp = Float64(-expm1(Float64(log(x) / n)));
	else
		tmp = Float64(Float64((x ^ Float64(1.0 / n)) / 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, 0.62], (-N[(Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]] - 1), $MachinePrecision]), N[(N[(N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision] / x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 0.619999999999999996

   1. Initial program 48.5%

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

   2. Taylor expanded in x around 0 47.9%

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

   3. Taylor expanded in x around 0 47.9%

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

   4. Simplified 90.8%

      \[\leadsto \color{blue}{-\mathsf{expm1}\left(\frac{\log x}{n}\right)} \]
      Step-by-step derivation

      [Start] 47.9	\[ 1 - e^{\frac{\log x}{n}} \]
      sub-neg [=>] 47.9	\[ \color{blue}{1 + \left(-e^{\frac{\log x}{n}}\right)} \]
      +-commutative [=>] 47.9	\[ \color{blue}{\left(-e^{\frac{\log x}{n}}\right) + 1} \]
      neg-sub0 [=>] 47.9	\[ \color{blue}{\left(0 - e^{\frac{\log x}{n}}\right)} + 1 \]
      metadata-eval [<=] 47.9	\[ \left(\color{blue}{\log 1} - e^{\frac{\log x}{n}}\right) + 1 \]
      associate-+l- [=>] 47.9	\[ \color{blue}{\log 1 - \left(e^{\frac{\log x}{n}} - 1\right)} \]
      metadata-eval [=>] 47.9	\[ \color{blue}{0} - \left(e^{\frac{\log x}{n}} - 1\right) \]
      sub0-neg [=>] 47.9	\[ \color{blue}{-\left(e^{\frac{\log x}{n}} - 1\right)} \]
      expm1-def [=>] 90.8	\[ -\color{blue}{\mathsf{expm1}\left(\frac{\log x}{n}\right)} \]

   if 0.619999999999999996 < x

   1. Initial program 66.6%

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

   2. Taylor expanded in x around inf 97.8%

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

   3. Simplified 97.8%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
      Step-by-step derivation

      [Start] 97.8	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      log-rec [=>] 97.8	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      mul-1-neg [<=] 97.8	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      associate-*r/ [=>] 97.8	\[ \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
      neg-mul-1 [<=] 97.8	\[ \frac{e^{\frac{\color{blue}{--1 \cdot \log x}}{n}}}{n \cdot x} \]
      mul-1-neg [=>] 97.8	\[ \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]
      remove-double-neg [=>] 97.8	\[ \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]
      *-commutative [=>] 97.8	\[ \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Applied egg-rr 97.8%

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

      [Start] 97.8	\[ \frac{e^{\frac{\log x}{n}}}{x \cdot n} \]
      div-inv [=>] 97.8	\[ \color{blue}{e^{\frac{\log x}{n}} \cdot \frac{1}{x \cdot n}} \]
      div-inv [=>] 97.8	\[ e^{\color{blue}{\log x \cdot \frac{1}{n}}} \cdot \frac{1}{x \cdot n} \]
      exp-to-pow [=>] 97.8	\[ \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \cdot \frac{1}{x \cdot n} \]

   5. Applied egg-rr 98.5%

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

      [Start] 97.8	\[ {x}^{\left(\frac{1}{n}\right)} \cdot \frac{1}{x \cdot n} \]
      un-div-inv [=>] 97.8	\[ \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]
      *-commutative [=>] 97.8	\[ \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{n \cdot x}} \]
      associate-/r* [=>] 98.5	\[ \color{blue}{\frac{\frac{{x}^{\left(\frac{1}{n}\right)}}{n}}{x}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 94.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 0.62:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{{x}^{\left(\frac{1}{n}\right)}}{n}}{x}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	76.7%
Cost	8848

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-114}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-259}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-150}:\\ \;\;\;\;\frac{\left(\frac{1}{x} + \frac{0.3333333333333333}{{x}^{3}}\right) - \frac{0.5}{x \cdot x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-16}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n} + \left(1 + \left(\left(x \cdot x\right) \cdot \frac{\frac{0.5}{n} - 0.5}{n} - t_0\right)\right)\\ \end{array} \]

Alternative 2
Accuracy	71.2%
Cost	8344

\[\begin{array}{l} t_0 := 1 - {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ t_2 := \frac{0.3333333333333333}{n \cdot {x}^{3}}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+83}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;\frac{1}{n} \leq -50000000000:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-259}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-211}:\\ \;\;\;\;\frac{\frac{1 - \frac{0.5}{x}}{x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-16}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{+125}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;t_2\\ \end{array} \]

Alternative 3
Accuracy	79.4%
Cost	8340

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-114}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-259}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-211}:\\ \;\;\;\;\frac{\frac{1 - \frac{0.5}{x}}{x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-16}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+164}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333}{n \cdot {x}^{3}}\\ \end{array} \]

Alternative 4
Accuracy	77.7%
Cost	8340

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-114}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-259}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-150}:\\ \;\;\;\;\frac{\left(\frac{1}{x} + \frac{0.3333333333333333}{{x}^{3}}\right) - \frac{0.5}{x \cdot x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-16}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+164}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333}{n \cdot {x}^{3}}\\ \end{array} \]

Alternative 5
Accuracy	79.1%
Cost	8084

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-114}:\\ \;\;\;\;\frac{\frac{t_0}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-259}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-211}:\\ \;\;\;\;\frac{\frac{1 - \frac{0.5}{x}}{x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-16}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{+125}:\\ \;\;\;\;1 - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333}{n \cdot {x}^{3}}\\ \end{array} \]

Alternative 6
Accuracy	78.9%
Cost	8084

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-114}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-259}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-211}:\\ \;\;\;\;\frac{\frac{1 - \frac{0.5}{x}}{x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-16}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{+125}:\\ \;\;\;\;1 - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333}{n \cdot {x}^{3}}\\ \end{array} \]

Alternative 7
Accuracy	60.0%
Cost	7512

\[\begin{array}{l} t_0 := \frac{-\log x}{n}\\ t_1 := 1 - {x}^{\left(\frac{1}{n}\right)}\\ t_2 := n \cdot \left(x \cdot x\right)\\ \mathbf{if}\;x \leq 1.9 \cdot 10^{-251}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;x \leq 6.2 \cdot 10^{-193}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 3.2 \cdot 10^{-181}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;x \leq 1.6 \cdot 10^{-149}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 2.15 \cdot 10^{-128}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;x \leq 1:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 5.2 \cdot 10^{+157}:\\ \;\;\;\;\frac{1}{x \cdot n} - \frac{0.5}{t_2}\\ \mathbf{else}:\\ \;\;\;\;\frac{-0.5}{t_2}\\ \end{array} \]

Alternative 8
Accuracy	58.8%
Cost	7444

\[\begin{array}{l} t_0 := 1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;n \leq -5.7 \cdot 10^{+113}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;n \leq -1.1 \cdot 10^{-7}:\\ \;\;\;\;\frac{1}{x \cdot n} - \frac{0.5}{n \cdot \left(x \cdot x\right)}\\ \mathbf{elif}\;n \leq -6.9 \cdot 10^{-78}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 4.4 \cdot 10^{-153}:\\ \;\;\;\;\frac{0.3333333333333333}{n \cdot {x}^{3}}\\ \mathbf{elif}\;n \leq 16200000000:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \]

Alternative 9
Accuracy	59.9%
Cost	6984

\[\begin{array}{l} t_0 := \frac{1}{x \cdot n}\\ t_1 := n \cdot \left(x \cdot x\right)\\ \mathbf{if}\;x \leq 4.6 \cdot 10^{-291}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 0.96:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 10^{+159}:\\ \;\;\;\;t_0 - \frac{0.5}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{-0.5}{t_1}\\ \end{array} \]

Alternative 10
Accuracy	59.8%
Cost	6920

\[\begin{array}{l} t_0 := \frac{1}{x \cdot n}\\ t_1 := n \cdot \left(x \cdot x\right)\\ \mathbf{if}\;x \leq 3.2 \cdot 10^{-291}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 0.67:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2.7 \cdot 10^{+157}:\\ \;\;\;\;t_0 - \frac{0.5}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{-0.5}{t_1}\\ \end{array} \]

Alternative 11
Accuracy	47.8%
Cost	1868

\[\begin{array}{l} t_0 := \frac{1}{x \cdot n}\\ \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\left(1 + t_0\right) + -1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-7}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+194}:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 12
Accuracy	47.0%
Cost	836

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\left(1 + \frac{1}{x \cdot n}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \]

Alternative 13
Accuracy	45.0%
Cost	580

\[\begin{array}{l} \mathbf{if}\;x \leq 9 \cdot 10^{+156}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{-0.5}{n \cdot \left(x \cdot x\right)}\\ \end{array} \]

Alternative 14
Accuracy	40.5%
Cost	320

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

Alternative 15
Accuracy	41.1%
Cost	320

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

Alternative 16
Accuracy	4.5%
Cost	192

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

Reproduce

herbie shell --seed 2023160 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))