2nthrt (problem 3.4.6)

Average Accuracy: 38.6% → 77.0%

Time: 23.7s

Precision: binary64

Cost: 13576

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-10}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-18}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \end{array} \]

FPCore

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

↓

(FPCore (x n)
 :precision binary64
 (if (<= (/ 1.0 n) -2e-10)
   (/ (pow x (/ 1.0 n)) (* n x))
   (if (<= (/ 1.0 n) 4e-18)
     (/ (log1p (/ 1.0 x)) n)
     (- (expm1 (/ (log 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 tmp;
	if ((1.0 / n) <= -2e-10) {
		tmp = pow(x, (1.0 / n)) / (n * x);
	} else if ((1.0 / n) <= 4e-18) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = -expm1((log(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 tmp;
	if ((1.0 / n) <= -2e-10) {
		tmp = Math.pow(x, (1.0 / n)) / (n * x);
	} else if ((1.0 / n) <= 4e-18) {
		tmp = Math.log1p((1.0 / x)) / n;
	} else {
		tmp = -Math.expm1((Math.log(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):
	tmp = 0
	if (1.0 / n) <= -2e-10:
		tmp = math.pow(x, (1.0 / n)) / (n * x)
	elif (1.0 / n) <= 4e-18:
		tmp = math.log1p((1.0 / x)) / n
	else:
		tmp = -math.expm1((math.log(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)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-10)
		tmp = Float64((x ^ Float64(1.0 / n)) / Float64(n * x));
	elseif (Float64(1.0 / n) <= 4e-18)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = Float64(-expm1(Float64(log(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_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-10], N[(N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision] / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 4e-18], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], (-N[(Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]] - 1), $MachinePrecision])]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -2.00000000000000007e-10

   1. Initial program 52.9%

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

   2. Taylor expanded in x around inf 54.1%

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

   3. Simplified 54.1%

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

      [Start] 54.1	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      log-rec [=>] 54.1	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      mul-1-neg [<=] 54.1	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      mul-1-neg [=>] 54.1	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      distribute-frac-neg [=>] 54.1	\[ \frac{e^{-1 \cdot \color{blue}{\left(-\frac{\log x}{n}\right)}}}{n \cdot x} \]
      neg-mul-1 [<=] 54.1	\[ \frac{e^{\color{blue}{-\left(-\frac{\log x}{n}\right)}}}{n \cdot x} \]
      remove-double-neg [=>] 54.1	\[ \frac{e^{\color{blue}{\frac{\log x}{n}}}}{n \cdot x} \]
      *-rgt-identity [<=] 54.1	\[ \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
      associate-*r/ [<=] 54.1	\[ \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
      unpow-1 [<=] 54.1	\[ \frac{e^{\log x \cdot \color{blue}{{n}^{-1}}}}{n \cdot x} \]
      exp-to-pow [=>] 54.1	\[ \frac{\color{blue}{{x}^{\left({n}^{-1}\right)}}}{n \cdot x} \]
      unpow-1 [=>] 54.1	\[ \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      *-commutative [=>] 54.1	\[ \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   if -2.00000000000000007e-10 < (/.f64 1 n) < 4.0000000000000003e-18

   1. Initial program 25.1%

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

   2. Taylor expanded in n around inf 70.7%

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

   3. Simplified 70.7%

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

      [Start] 70.7	\[ \frac{\log \left(1 + x\right) - \log x}{n} \]
      log1p-def [=>] 70.7	\[ \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   4. Applied egg-rr 70.9%

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

      [Start] 70.7	\[ \frac{\mathsf{log1p}\left(x\right) - \log x}{n} \]
      log1p-udef [=>] 70.7	\[ \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]
      diff-log [=>] 70.9	\[ \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
      +-commutative [=>] 70.9	\[ \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   5. Taylor expanded in n around 0 70.9%

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

   6. Simplified 99.5%

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

      [Start] 70.9	\[ \frac{\log \left(\frac{1 + x}{x}\right)}{n} \]
      log-div [=>] 70.7	\[ \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
      log1p-def [=>] 70.7	\[ \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
      log1p-def [<=] 70.7	\[ \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]
      log-div [<=] 70.9	\[ \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
      +-commutative [<=] 70.9	\[ \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]
      *-lft-identity [<=] 70.9	\[ \frac{\log \left(\frac{\color{blue}{1 \cdot \left(x + 1\right)}}{x}\right)}{n} \]
      associate-*l/ [<=] 65.8	\[ \frac{\log \color{blue}{\left(\frac{1}{x} \cdot \left(x + 1\right)\right)}}{n} \]
      distribute-rgt-in [=>] 65.8	\[ \frac{\log \color{blue}{\left(x \cdot \frac{1}{x} + 1 \cdot \frac{1}{x}\right)}}{n} \]
      *-lft-identity [=>] 65.8	\[ \frac{\log \left(x \cdot \frac{1}{x} + \color{blue}{\frac{1}{x}}\right)}{n} \]
      rgt-mult-inverse [=>] 70.9	\[ \frac{\log \left(\color{blue}{1} + \frac{1}{x}\right)}{n} \]
      log1p-def [=>] 99.5	\[ \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x}\right)}}{n} \]

   if 4.0000000000000003e-18 < (/.f64 1 n)

   1. Initial program 54.9%

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

   2. Taylor expanded in x around 0 55.2%

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

   3. Simplified 55.2%

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

      [Start] 55.2	\[ 1 - e^{\frac{\log x}{n}} \]
      *-rgt-identity [<=] 55.2	\[ 1 - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]
      associate-*r/ [<=] 55.2	\[ 1 - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]
      unpow-1 [<=] 55.2	\[ 1 - e^{\log x \cdot \color{blue}{{n}^{-1}}} \]
      exp-to-pow [=>] 55.2	\[ 1 - \color{blue}{{x}^{\left({n}^{-1}\right)}} \]
      unpow-1 [=>] 55.2	\[ 1 - {x}^{\color{blue}{\left(\frac{1}{n}\right)}} \]

   4. Taylor expanded in x around 0 55.2%

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

   5. Simplified 58.6%

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

      [Start] 55.2	\[ 1 - e^{\frac{\log x}{n}} \]
      sub-neg [=>] 55.2	\[ \color{blue}{1 + \left(-e^{\frac{\log x}{n}}\right)} \]
      +-commutative [=>] 55.2	\[ \color{blue}{\left(-e^{\frac{\log x}{n}}\right) + 1} \]
      neg-sub0 [=>] 55.2	\[ \color{blue}{\left(0 - e^{\frac{\log x}{n}}\right)} + 1 \]
      metadata-eval [<=] 55.2	\[ \left(\color{blue}{\log 1} - e^{\frac{\log x}{n}}\right) + 1 \]
      associate-+l- [=>] 55.2	\[ \color{blue}{\log 1 - \left(e^{\frac{\log x}{n}} - 1\right)} \]
      metadata-eval [=>] 55.2	\[ \color{blue}{0} - \left(e^{\frac{\log x}{n}} - 1\right) \]
      sub0-neg [=>] 55.2	\[ \color{blue}{-\left(e^{\frac{\log x}{n}} - 1\right)} \]
      expm1-def [=>] 58.6	\[ -\color{blue}{\mathsf{expm1}\left(\frac{\log x}{n}\right)} \]

3. Recombined 3 regimes into one program.

4. Final simplification 79.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-10}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-18}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	77.3%
Cost	7560

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-10}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-14}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \end{array} \]

Alternative 2
Accuracy	77.0%
Cost	7304

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-10}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-14}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - t_0\\ \end{array} \]

Alternative 3
Accuracy	77.3%
Cost	7180

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -4.4:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -5 \cdot 10^{-310}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;n \leq 520000000:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 4
Accuracy	76.8%
Cost	7180

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -700:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 2.45 \cdot 10^{-240}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;n \leq 102000000:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 5
Accuracy	71.0%
Cost	6980

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -20:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \end{array} \]

Alternative 6
Accuracy	60.1%
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.66:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 1.05 \cdot 10^{+205}:\\ \;\;\;\;\frac{\frac{1}{x} - \frac{0.5}{x \cdot x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 7
Accuracy	36.0%
Cost	836

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -20:\\ \;\;\;\;\left(1 + \frac{1}{n \cdot x}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]

Alternative 8
Accuracy	43.7%
Cost	584

\[\begin{array}{l} \mathbf{if}\;n \leq -16:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{elif}\;n \leq 1.4 \cdot 10^{-112}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]

Alternative 9
Accuracy	29.6%
Cost	320

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

Alternative 10
Accuracy	30.0%
Cost	320

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

Alternative 11
Accuracy	30.0%
Cost	320

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

Reproduce

herbie shell --seed 2023157 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))