2nthrt (problem 3.4.6)

Average Accuracy: 48.5% → 97.8%

Time: 27.9s

Precision: binary64

Cost: 33288

Math

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

↓

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-11}:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{\left({\left(e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\right)}^{-0.5}\right)}^{2}}\\ \end{array} \]

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 (<= (/ 1.0 n) -5e-11)
     (- (exp (/ x n)) t_0)
     (if (<= (/ 1.0 n) 5e-8)
       (/ (log1p (/ 1.0 x)) n)
       (/ 1.0 (pow (pow (- (exp (/ (log1p x) n)) t_0) -0.5) 2.0))))))

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 t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -5e-11) {
		tmp = exp((x / n)) - t_0;
	} else if ((1.0 / n) <= 5e-8) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = 1.0 / pow(pow((exp((log1p(x) / n)) - t_0), -0.5), 2.0);
	}
	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 t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -5e-11) {
		tmp = Math.exp((x / n)) - t_0;
	} else if ((1.0 / n) <= 5e-8) {
		tmp = Math.log1p((1.0 / x)) / n;
	} else {
		tmp = 1.0 / Math.pow(Math.pow((Math.exp((Math.log1p(x) / n)) - t_0), -0.5), 2.0);
	}
	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):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -5e-11:
		tmp = math.exp((x / n)) - t_0
	elif (1.0 / n) <= 5e-8:
		tmp = math.log1p((1.0 / x)) / n
	else:
		tmp = 1.0 / math.pow(math.pow((math.exp((math.log1p(x) / n)) - t_0), -0.5), 2.0)
	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)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -5e-11)
		tmp = Float64(exp(Float64(x / n)) - t_0);
	elseif (Float64(1.0 / n) <= 5e-8)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = Float64(1.0 / ((Float64(exp(Float64(log1p(x) / n)) - t_0) ^ -0.5) ^ 2.0));
	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_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e-11], N[(N[Exp[N[(x / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-8], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], N[(1.0 / N[Power[N[Power[N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision], -0.5], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -5.00000000000000018e-11

   1. Initial program 95.6%

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

   2. Taylor expanded in n around 0 95.5%

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

   3. Simplified 95.5%

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

      [Start] 95.5	\[ e^{\frac{\log \left(1 + x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      log1p-def [=>] 95.5	\[ e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]

   4. Taylor expanded in x around 0 95.3%

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

   if -5.00000000000000018e-11 < (/.f64 1 n) < 4.9999999999999998e-8

   1. Initial program 29.6%

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

   2. Taylor expanded in n around inf 77.3%

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

   3. Simplified 77.3%

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

      [Start] 77.3	\[ \frac{\log \left(1 + x\right) - \log x}{n} \]
      log1p-def [=>] 77.3	\[ \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   4. Applied egg-rr 77.5%

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

   5. Applied egg-rr 77.5%

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

   6. Simplified 98.9%

      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x} + 0\right)}}{n} \]
      Proof

      [Start] 77.5	\[ \frac{\mathsf{log1p}\left(\frac{x + 1}{x} - 1\right)}{n} \]
      *-lft-identity [<=] 77.5	\[ \frac{\mathsf{log1p}\left(\frac{\color{blue}{1 \cdot \left(x + 1\right)}}{x} - 1\right)}{n} \]
      associate-*l/ [<=] 74.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} \cdot \left(x + 1\right)} - 1\right)}{n} \]
      distribute-rgt-in [=>] 74.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(x \cdot \frac{1}{x} + 1 \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      +-commutative [=>] 74.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(1 \cdot \frac{1}{x} + x \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      *-lft-identity [=>] 74.2	\[ \frac{\mathsf{log1p}\left(\left(\color{blue}{\frac{1}{x}} + x \cdot \frac{1}{x}\right) - 1\right)}{n} \]
      rgt-mult-inverse [=>] 77.5	\[ \frac{\mathsf{log1p}\left(\left(\frac{1}{x} + \color{blue}{1}\right) - 1\right)}{n} \]
      associate--l+ [=>] 98.9	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} + \left(1 - 1\right)}\right)}{n} \]
      metadata-eval [=>] 98.9	\[ \frac{\mathsf{log1p}\left(\frac{1}{x} + \color{blue}{0}\right)}{n} \]

   if 4.9999999999999998e-8 < (/.f64 1 n)

   1. Initial program 91.3%

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

   2. Applied egg-rr 94.8%

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

   3. Applied egg-rr 94.8%

      \[\leadsto \frac{1}{\color{blue}{{\left({\left(e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\right)}^{-0.5}\right)}^{2}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 97.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-11}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{\left({\left(e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\right)}^{-0.5}\right)}^{2}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	97.8%
Cost	26824

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-11}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\frac{1}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left({n}^{-1}\right)}}}\\ \end{array} \]

Alternative 2
Accuracy	97.8%
Cost	20232

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-11}:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \end{array} \]

Alternative 3
Accuracy	97.8%
Cost	13833

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-11} \lor \neg \left(\frac{1}{n} \leq 5 \cdot 10^{-8}\right):\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \end{array} \]

Alternative 4
Accuracy	97.3%
Cost	13512

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -290:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -1 \cdot 10^{-302}:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{n \cdot x}\\ \mathbf{elif}\;n \leq 600000:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 5
Accuracy	63.4%
Cost	9644

\[\begin{array}{l} t_0 := \frac{\frac{1}{x}}{n}\\ t_1 := \frac{\frac{1}{n}}{x}\\ t_2 := \frac{x - \log x}{n}\\ t_3 := 1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -0.004:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -1 \cdot 10^{-11}:\\ \;\;\;\;t_3\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{-75}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-200}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-256}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-242}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-191}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-159}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-140}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-88}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;t_3\\ \end{array} \]

Alternative 6
Accuracy	63.2%
Cost	8996

\[\begin{array}{l} t_0 := \frac{\frac{1}{n}}{x}\\ t_1 := \frac{-\log x}{n}\\ t_2 := \frac{\frac{1}{x}}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -200:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{-75}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;\frac{1}{n} \leq -1 \cdot 10^{-92}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-256}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-242}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-191}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-159}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-140}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-88}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 0.005:\\ \;\;\;\;t_2\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 7
Accuracy	80.4%
Cost	7628

\[\begin{array}{l} t_0 := \frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-23}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{-75}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-8}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 8
Accuracy	97.2%
Cost	7436

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -5.4:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -1 \cdot 10^{-302}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;n \leq 10000000:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 9
Accuracy	97.1%
Cost	7180

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -7:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -1 \cdot 10^{-302}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;n \leq 1400000:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 10
Accuracy	74.9%
Cost	6852

\[\begin{array}{l} \mathbf{if}\;x \leq 0.95:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 1.52 \cdot 10^{+94}:\\ \;\;\;\;\frac{\frac{1}{x} + \frac{-0.5}{x \cdot x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 11
Accuracy	57.7%
Cost	848

\[\begin{array}{l} \mathbf{if}\;n \leq -7.5:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;n \leq -1.8 \cdot 10^{-138}:\\ \;\;\;\;-1 + \left(1 + \frac{1}{n \cdot x}\right)\\ \mathbf{elif}\;n \leq -1 \cdot 10^{-302}:\\ \;\;\;\;1 + \left(-1 + n \cdot x\right)\\ \mathbf{elif}\;n \leq 106:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \]

Alternative 12
Accuracy	53.0%
Cost	716

\[\begin{array}{l} \mathbf{if}\;n \leq -1.36 \cdot 10^{-71}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;n \leq -1 \cdot 10^{-302}:\\ \;\;\;\;1 + \left(-1 + n \cdot x\right)\\ \mathbf{elif}\;n \leq 230:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \]

Alternative 13
Accuracy	62.7%
Cost	716

\[\begin{array}{l} \mathbf{if}\;n \leq -9.5:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;n \leq -1 \cdot 10^{-302}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;n \leq 175:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \]

Alternative 14
Accuracy	45.0%
Cost	585

\[\begin{array}{l} \mathbf{if}\;n \leq -3 \cdot 10^{-138} \lor \neg \left(n \leq 220\right):\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 15
Accuracy	45.7%
Cost	585

\[\begin{array}{l} \mathbf{if}\;n \leq -6.2 \cdot 10^{-139} \lor \neg \left(n \leq 100\right):\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 16
Accuracy	45.7%
Cost	584

\[\begin{array}{l} \mathbf{if}\;n \leq -7.5 \cdot 10^{-139}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;n \leq 120:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \]

Alternative 17
Accuracy	11.9%
Cost	64

\[1 \]

Reproduce

herbie shell --seed 2023122 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))