2nthrt (problem 3.4.6)

Average Accuracy: 39.0% → 77.4%

Time: 23.5s

Precision: binary64

Cost: 20040

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-7}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left({\left(\frac{n}{\log x}\right)}^{-1}\right)\\ \end{array} \]

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-7)
   (/ (pow x (/ 1.0 n)) (* n x))
   (if (<= (/ 1.0 n) 2e-12)
     (/ (log1p (/ 1.0 x)) n)
     (- (expm1 (pow (/ n (log x)) -1.0))))))

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-7) {
		tmp = pow(x, (1.0 / n)) / (n * x);
	} else if ((1.0 / n) <= 2e-12) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = -expm1(pow((n / log(x)), -1.0));
	}
	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-7) {
		tmp = Math.pow(x, (1.0 / n)) / (n * x);
	} else if ((1.0 / n) <= 2e-12) {
		tmp = Math.log1p((1.0 / x)) / n;
	} else {
		tmp = -Math.expm1(Math.pow((n / Math.log(x)), -1.0));
	}
	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-7:
		tmp = math.pow(x, (1.0 / n)) / (n * x)
	elif (1.0 / n) <= 2e-12:
		tmp = math.log1p((1.0 / x)) / n
	else:
		tmp = -math.expm1(math.pow((n / math.log(x)), -1.0))
	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-7)
		tmp = Float64((x ^ Float64(1.0 / n)) / Float64(n * x));
	elseif (Float64(1.0 / n) <= 2e-12)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = Float64(-expm1((Float64(n / log(x)) ^ -1.0)));
	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-7], N[(N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision] / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-12], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], (-N[(Exp[N[Power[N[(n / N[Log[x], $MachinePrecision]), $MachinePrecision], -1.0], $MachinePrecision]] - 1), $MachinePrecision])]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -1.9999999999999999e-7

   1. Initial program 49.8%

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

   2. Taylor expanded in x around inf 49.7%

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

   3. Simplified 49.7%

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

      [Start] 49.7	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      log-rec [=>] 49.7	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      mul-1-neg [<=] 49.7	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      mul-1-neg [=>] 49.7	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      distribute-frac-neg [=>] 49.7	\[ \frac{e^{-1 \cdot \color{blue}{\left(-\frac{\log x}{n}\right)}}}{n \cdot x} \]
      neg-mul-1 [<=] 49.7	\[ \frac{e^{\color{blue}{-\left(-\frac{\log x}{n}\right)}}}{n \cdot x} \]
      remove-double-neg [=>] 49.7	\[ \frac{e^{\color{blue}{\frac{\log x}{n}}}}{n \cdot x} \]
      *-rgt-identity [<=] 49.7	\[ \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
      associate-*r/ [<=] 49.7	\[ \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
      unpow-1 [<=] 49.7	\[ \frac{e^{\log x \cdot \color{blue}{{n}^{-1}}}}{n \cdot x} \]
      exp-to-pow [=>] 49.7	\[ \frac{\color{blue}{{x}^{\left({n}^{-1}\right)}}}{n \cdot x} \]
      unpow-1 [=>] 49.7	\[ \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      *-commutative [=>] 49.7	\[ \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   if -1.9999999999999999e-7 < (/.f64 1 n) < 1.99999999999999996e-12

   1. Initial program 30.2%

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

   2. Taylor expanded in n around inf 76.6%

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

   3. Simplified 76.6%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
      Proof

      [Start] 76.6	\[ \frac{\log \left(1 + x\right) - \log x}{n} \]
      log1p-def [=>] 76.6	\[ \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   4. Applied egg-rr 76.7%

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

      [Start] 76.6	\[ \frac{\mathsf{log1p}\left(x\right) - \log x}{n} \]
      log1p-udef [=>] 76.6	\[ \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]
      diff-log [=>] 76.7	\[ \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
      +-commutative [=>] 76.7	\[ \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   5. Taylor expanded in n around 0 76.7%

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

   6. Simplified 98.9%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}} \]
      Proof

      [Start] 76.7	\[ \frac{\log \left(\frac{1 + x}{x}\right)}{n} \]
      log-div [=>] 76.6	\[ \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
      log1p-def [=>] 76.6	\[ \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
      log1p-def [<=] 76.6	\[ \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]
      log-div [<=] 76.7	\[ \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
      +-commutative [<=] 76.7	\[ \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]
      *-lft-identity [<=] 76.7	\[ \frac{\log \left(\frac{\color{blue}{1 \cdot \left(x + 1\right)}}{x}\right)}{n} \]
      associate-*l/ [<=] 72.9	\[ \frac{\log \color{blue}{\left(\frac{1}{x} \cdot \left(x + 1\right)\right)}}{n} \]
      distribute-rgt-in [=>] 72.9	\[ \frac{\log \color{blue}{\left(x \cdot \frac{1}{x} + 1 \cdot \frac{1}{x}\right)}}{n} \]
      *-lft-identity [=>] 72.9	\[ \frac{\log \left(x \cdot \frac{1}{x} + \color{blue}{\frac{1}{x}}\right)}{n} \]
      rgt-mult-inverse [=>] 76.7	\[ \frac{\log \left(\color{blue}{1} + \frac{1}{x}\right)}{n} \]
      log1p-def [=>] 98.9	\[ \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x}\right)}}{n} \]

   if 1.99999999999999996e-12 < (/.f64 1 n)

   1. Initial program 50.5%

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

   2. Taylor expanded in x around 0 50.4%

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

   3. Simplified 50.4%

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

      [Start] 50.4	\[ 1 - e^{\frac{\log x}{n}} \]
      *-rgt-identity [<=] 50.4	\[ 1 - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]
      associate-*r/ [<=] 50.4	\[ 1 - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]
      unpow-1 [<=] 50.4	\[ 1 - e^{\log x \cdot \color{blue}{{n}^{-1}}} \]
      exp-to-pow [=>] 50.4	\[ 1 - \color{blue}{{x}^{\left({n}^{-1}\right)}} \]
      unpow-1 [=>] 50.4	\[ 1 - {x}^{\color{blue}{\left(\frac{1}{n}\right)}} \]

   4. Taylor expanded in x around 0 50.4%

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

   5. Simplified 50.7%

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

      [Start] 50.4	\[ 1 - e^{\frac{\log x}{n}} \]
      sub-neg [=>] 50.4	\[ \color{blue}{1 + \left(-e^{\frac{\log x}{n}}\right)} \]
      +-commutative [=>] 50.4	\[ \color{blue}{\left(-e^{\frac{\log x}{n}}\right) + 1} \]
      neg-sub0 [=>] 50.4	\[ \color{blue}{\left(0 - e^{\frac{\log x}{n}}\right)} + 1 \]
      metadata-eval [<=] 50.4	\[ \left(\color{blue}{\log 1} - e^{\frac{\log x}{n}}\right) + 1 \]
      associate-+l- [=>] 50.4	\[ \color{blue}{\log 1 - \left(e^{\frac{\log x}{n}} - 1\right)} \]
      metadata-eval [=>] 50.4	\[ \color{blue}{0} - \left(e^{\frac{\log x}{n}} - 1\right) \]
      sub0-neg [=>] 50.4	\[ \color{blue}{-\left(e^{\frac{\log x}{n}} - 1\right)} \]
      expm1-def [=>] 50.7	\[ -\color{blue}{\mathsf{expm1}\left(\frac{\log x}{n}\right)} \]

   6. Applied egg-rr 50.8%

      \[\leadsto -\mathsf{expm1}\left(\color{blue}{{\left(\frac{n}{\log x}\right)}^{-1}}\right) \]
      Proof

      [Start] 50.7	\[ -\mathsf{expm1}\left(\frac{\log x}{n}\right) \]
      clear-num [=>] 50.8	\[ -\mathsf{expm1}\left(\color{blue}{\frac{1}{\frac{n}{\log x}}}\right) \]
      inv-pow [=>] 50.8	\[ -\mathsf{expm1}\left(\color{blue}{{\left(\frac{n}{\log x}\right)}^{-1}}\right) \]

3. Recombined 3 regimes into one program.

4. Final simplification 77.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-7}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left({\left(\frac{n}{\log x}\right)}^{-1}\right)\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	77.4%
Cost	13576

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-7}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \end{array} \]

Alternative 2
Accuracy	76.7%
Cost	8076

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -12200:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 2.25 \cdot 10^{-201}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;n \leq 23000000000:\\ \;\;\;\;\left(\left(1 + \frac{x}{n}\right) + 0.5 \cdot \frac{\frac{x \cdot x}{n}}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 3
Accuracy	76.7%
Cost	7436

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -12200:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 2.25 \cdot 10^{-201}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;n \leq 23000000000:\\ \;\;\;\;1 + \left(\frac{x}{n} - {x}^{\left(\frac{1}{n}\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 4
Accuracy	77.0%
Cost	7304

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -50000000000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 5
Accuracy	76.6%
Cost	7180

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -12200:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 2.25 \cdot 10^{-201}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;n \leq 23000000000:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 6
Accuracy	70.1%
Cost	6985

\[\begin{array}{l} \mathbf{if}\;n \leq -1.95 \cdot 10^{-10} \lor \neg \left(n \leq -6 \cdot 10^{-299}\right):\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 7
Accuracy	60.3%
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.66:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 1.8 \cdot 10^{+203}:\\ \;\;\;\;\frac{\frac{1}{x} - \frac{0.5}{x \cdot x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 8
Accuracy	36.2%
Cost	836

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -50000000000:\\ \;\;\;\;-1 + \left(1 + \frac{1}{n \cdot x}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]

Alternative 9
Accuracy	43.8%
Cost	585

\[\begin{array}{l} \mathbf{if}\;n \leq -1.85 \cdot 10^{-5} \lor \neg \left(n \leq 3.8 \cdot 10^{-274}\right):\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 10
Accuracy	30.2%
Cost	320

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

Alternative 11
Accuracy	30.8%
Cost	320

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

Alternative 12
Accuracy	30.8%
Cost	320

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

Reproduce

herbie shell --seed 2023153 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))