2nthrt (problem 3.4.6)

Percentage Accurate: 52.3% → 98.6%

Time: 1.2min

Precision: binary64

Cost: 19844

• could not determine a ground truth

  1. x = 1.5004564603928934e-240
  2. n = -1.593288660425601e+149

Math

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

↓

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;x \leq 6.4 \cdot 10^{-6}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \end{array} \]

FPCore

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

↓

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= x 6.4e-6) (- (exp (/ (log1p x) n)) t_0) (/ t_0 (* x n)))))

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 t_0 = pow(x, (1.0 / n));
	double tmp;
	if (x <= 6.4e-6) {
		tmp = exp((log1p(x) / n)) - t_0;
	} else {
		tmp = t_0 / (x * n);
	}
	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 t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if (x <= 6.4e-6) {
		tmp = Math.exp((Math.log1p(x) / n)) - t_0;
	} else {
		tmp = t_0 / (x * n);
	}
	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):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if x <= 6.4e-6:
		tmp = math.exp((math.log1p(x) / n)) - t_0
	else:
		tmp = t_0 / (x * n)
	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)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (x <= 6.4e-6)
		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
	else
		tmp = Float64(t_0 / Float64(x * n));
	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_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[x, 6.4e-6], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if x < 6.3999999999999997e-6

   1. Initial program 77.0%

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

   2. Taylor expanded in n around 0 76.9%

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

   3. Simplified 99.1%

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}} \]
      Step-by-step derivation

      [Start] 76.9%	\[ e^{\frac{\log \left(1 + x\right)}{n}} - e^{\frac{\log x}{n}} \]
      log1p-def [=>] 99.1%	\[ e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - e^{\frac{\log x}{n}} \]
      *-rgt-identity [<=] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]
      associate-*r/ [<=] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]
      unpow-1 [<=] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\log x \cdot \color{blue}{{n}^{-1}}} \]
      exp-to-pow [=>] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - \color{blue}{{x}^{\left({n}^{-1}\right)}} \]
      /-rgt-identity [<=] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{{n}^{-1}}{1}\right)}} \]
      metadata-eval [<=] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{{n}^{-1}}{\color{blue}{\frac{2}{2}}}\right)} \]
      associate-/l* [<=] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{{n}^{-1} \cdot 2}{2}\right)}} \]
      *-commutative [<=] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{\color{blue}{2 \cdot {n}^{-1}}}{2}\right)} \]
      *-commutative [=>] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{\color{blue}{{n}^{-1} \cdot 2}}{2}\right)} \]
      associate-/l* [=>] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{{n}^{-1}}{\frac{2}{2}}\right)}} \]
      metadata-eval [=>] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{{n}^{-1}}{\color{blue}{1}}\right)} \]
      /-rgt-identity [=>] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left({n}^{-1}\right)}} \]
      unpow-1 [=>] 99.1%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{1}{n}\right)}} \]

   if 6.3999999999999997e-6 < x

   1. Initial program 8.9%

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

   2. Taylor expanded in x around inf 100.0%

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

   3. Simplified 100.0%

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

      [Start] 100.0%	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      log-rec [=>] 100.0%	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      mul-1-neg [<=] 100.0%	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      mul-1-neg [=>] 100.0%	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      distribute-frac-neg [=>] 100.0%	\[ \frac{e^{-1 \cdot \color{blue}{\left(-\frac{\log x}{n}\right)}}}{n \cdot x} \]
      neg-mul-1 [<=] 100.0%	\[ \frac{e^{\color{blue}{-\left(-\frac{\log x}{n}\right)}}}{n \cdot x} \]
      remove-double-neg [=>] 100.0%	\[ \frac{e^{\color{blue}{\frac{\log x}{n}}}}{n \cdot x} \]
      *-rgt-identity [<=] 100.0%	\[ \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
      associate-*r/ [<=] 100.0%	\[ \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
      unpow-1 [<=] 100.0%	\[ \frac{e^{\log x \cdot \color{blue}{{n}^{-1}}}}{n \cdot x} \]
      exp-to-pow [=>] 100.0%	\[ \frac{\color{blue}{{x}^{\left({n}^{-1}\right)}}}{n \cdot x} \]
      unpow-1 [=>] 100.0%	\[ \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      *-commutative [=>] 100.0%	\[ \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 99.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 6.4 \cdot 10^{-6}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	98.6%
Cost	19844

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;x \leq 6.4 \cdot 10^{-6}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \end{array} \]

Alternative 2
Accuracy	88.7%
Cost	13648

\[\begin{array}{l} t_0 := \mathsf{log1p}\left(\mathsf{expm1}\left(\frac{1}{x \cdot n}\right)\right)\\ t_1 := {x}^{\left(\frac{1}{n}\right)}\\ t_2 := \left(1 + \frac{x}{n}\right) - t_1\\ \mathbf{if}\;x \leq 5.3 \cdot 10^{-192}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;x \leq 4.6 \cdot 10^{-177}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 2.15 \cdot 10^{-116}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;x \leq 6.4 \cdot 10^{-6}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_1}{x \cdot n}\\ \end{array} \]

Alternative 3
Accuracy	87.2%
Cost	13520

\[\begin{array}{l} t_0 := \log \left(e^{\frac{x}{n}}\right)\\ t_1 := {x}^{\left(\frac{1}{n}\right)}\\ t_2 := \left(1 + \frac{x}{n}\right) - t_1\\ \mathbf{if}\;x \leq 6.8 \cdot 10^{-115}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;x \leq 5 \cdot 10^{-102}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 4.1 \cdot 10^{-92}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;x \leq 6.4 \cdot 10^{-6}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_1}{x \cdot n}\\ \end{array} \]

Alternative 4
Accuracy	82.1%
Cost	8200

\[\begin{array}{l} t_0 := 1 + \frac{x}{n}\\ t_1 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;x \leq 2.85 \cdot 10^{-114}:\\ \;\;\;\;t_0 - t_1\\ \mathbf{elif}\;x \leq 5.1 \cdot 10^{-74}:\\ \;\;\;\;\left(t_0 + \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right) \cdot \left(x \cdot x\right)\right) - t_1\\ \mathbf{else}:\\ \;\;\;\;\frac{t_1}{x \cdot n}\\ \end{array} \]

Alternative 5
Accuracy	83.9%
Cost	7172

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;x \leq 6.4 \cdot 10^{-6}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \end{array} \]

Alternative 6
Accuracy	83.4%
Cost	7044

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;x \leq 6.4 \cdot 10^{-6}:\\ \;\;\;\;1 - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \end{array} \]

Alternative 7
Accuracy	65.7%
Cost	6916

\[\begin{array}{l} \mathbf{if}\;x \leq 75000:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n}\\ \end{array} \]

Alternative 8
Accuracy	42.0%
Cost	452

\[\begin{array}{l} \mathbf{if}\;x \leq 1.4 \cdot 10^{+23}:\\ \;\;\;\;\frac{1}{x \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n}\\ \end{array} \]

Alternative 9
Accuracy	19.5%
Cost	192

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

Reproduce

herbie shell --seed 2023277 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))