2nthrt (problem 3.4.6)

Average Accuracy: 49.6% → 88.6%

Time: 22.0s

Precision: binary64

Cost: 13380

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;x \leq 330000000:\\ \;\;\;\;\frac{-\log \left(\frac{x}{x + 1}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{x \cdot n}\\ \end{array} \]

FPCore

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

↓

(FPCore (x n)
 :precision binary64
 (if (<= x 330000000.0)
   (/ (- (log (/ x (+ x 1.0)))) n)
   (/ (exp (/ (log x) n)) (* 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 (x <= 330000000.0) {
		tmp = -log((x / (x + 1.0))) / n;
	} else {
		tmp = exp((log(x) / n)) / (x * n);
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = ((x + 1.0d0) ** (1.0d0 / n)) - (x ** (1.0d0 / n))
end function

↓

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: tmp
    if (x <= 330000000.0d0) then
        tmp = -log((x / (x + 1.0d0))) / n
    else
        tmp = exp((log(x) / n)) / (x * n)
    end if
    code = tmp
end function

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 <= 330000000.0) {
		tmp = -Math.log((x / (x + 1.0))) / n;
	} else {
		tmp = Math.exp((Math.log(x) / n)) / (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 x <= 330000000.0:
		tmp = -math.log((x / (x + 1.0))) / n
	else:
		tmp = math.exp((math.log(x) / n)) / (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 (x <= 330000000.0)
		tmp = Float64(Float64(-log(Float64(x / Float64(x + 1.0)))) / n);
	else
		tmp = Float64(exp(Float64(log(x) / n)) / Float64(x * n));
	end
	return tmp
end

MATLAB

function tmp = code(x, n)
	tmp = ((x + 1.0) ^ (1.0 / n)) - (x ^ (1.0 / n));
end

↓

function tmp_2 = code(x, n)
	tmp = 0.0;
	if (x <= 330000000.0)
		tmp = -log((x / (x + 1.0))) / n;
	else
		tmp = exp((log(x) / n)) / (x * n);
	end
	tmp_2 = 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, 330000000.0], N[((-N[Log[N[(x / N[(x + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision], N[(N[Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] / N[(x * n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 3.3e8

   1. Initial program 26.9%

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

   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 77.4%

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

      [Start] 77.3	\[ \frac{\mathsf{log1p}\left(x\right) - \log x}{n} \]
      add-log-exp [=>] 77.3	\[ \frac{\color{blue}{\log \left(e^{\mathsf{log1p}\left(x\right) - \log x}\right)}}{n} \]
      exp-diff [=>] 77.3	\[ \frac{\log \color{blue}{\left(\frac{e^{\mathsf{log1p}\left(x\right)}}{e^{\log x}}\right)}}{n} \]
      log1p-udef [=>] 77.3	\[ \frac{\log \left(\frac{e^{\color{blue}{\log \left(1 + x\right)}}}{e^{\log x}}\right)}{n} \]
      add-exp-log [<=] 77.3	\[ \frac{\log \left(\frac{\color{blue}{1 + x}}{e^{\log x}}\right)}{n} \]
      +-commutative [=>] 77.3	\[ \frac{\log \left(\frac{\color{blue}{x + 1}}{e^{\log x}}\right)}{n} \]
      add-exp-log [<=] 77.4	\[ \frac{\log \left(\frac{x + 1}{\color{blue}{x}}\right)}{n} \]

   5. Applied egg-rr 77.4%

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

      [Start] 77.4	\[ \frac{\log \left(\frac{x + 1}{x}\right)}{n} \]
      clear-num [=>] 77.4	\[ \frac{\log \color{blue}{\left(\frac{1}{\frac{x}{x + 1}}\right)}}{n} \]
      log-rec [=>] 77.4	\[ \frac{\color{blue}{-\log \left(\frac{x}{x + 1}\right)}}{n} \]

   if 3.3e8 < x

   1. Initial program 69.2%

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

   2. Taylor expanded in x around inf 98.2%

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

   3. Simplified 98.2%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
      Proof

      [Start] 98.2	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      mul-1-neg [=>] 98.2	\[ \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]
      log-rec [=>] 98.2	\[ \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      mul-1-neg [<=] 98.2	\[ \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      distribute-neg-frac [=>] 98.2	\[ \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]
      mul-1-neg [=>] 98.2	\[ \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]
      remove-double-neg [=>] 98.2	\[ \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]
      *-commutative [=>] 98.2	\[ \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 88.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 330000000:\\ \;\;\;\;\frac{-\log \left(\frac{x}{x + 1}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{x \cdot n}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	80.3%
Cost	7820

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -200000:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -1 \cdot 10^{-115}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{1}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-12}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{x + 1}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 2
Accuracy	80.1%
Cost	7692

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -200000:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -1 \cdot 10^{-115}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{1}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{x + 1}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 3
Accuracy	79.3%
Cost	7377

\[\begin{array}{l} t_0 := \frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -2.3 \cdot 10^{+126}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -3:\\ \;\;\;\;\frac{1}{n} \cdot \frac{1}{x}\\ \mathbf{elif}\;n \leq -1 \cdot 10^{-293} \lor \neg \left(n \leq 2900000000\right):\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 4
Accuracy	70.9%
Cost	6980

\[\begin{array}{l} \mathbf{if}\;x \leq 0.98:\\ \;\;\;\;\frac{x}{n} - \frac{\log x}{n}\\ \mathbf{elif}\;x \leq 2.75 \cdot 10^{+118}:\\ \;\;\;\;\frac{\frac{1}{x} - \frac{\frac{0.5}{x}}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{1}{x \cdot n}\right) + -1\\ \end{array} \]

Alternative 5
Accuracy	70.9%
Cost	6852

\[\begin{array}{l} \mathbf{if}\;x \leq 0.98:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 2.75 \cdot 10^{+118}:\\ \;\;\;\;\frac{\frac{1}{x} - \frac{\frac{0.5}{x}}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{1}{x \cdot n}\right) + -1\\ \end{array} \]

Alternative 6
Accuracy	70.6%
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.68:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2.75 \cdot 10^{+118}:\\ \;\;\;\;\frac{\frac{1}{x} - \frac{\frac{0.5}{x}}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{1}{x \cdot n}\right) + -1\\ \end{array} \]

Alternative 7
Accuracy	44.7%
Cost	840

\[\begin{array}{l} \mathbf{if}\;n \leq -3:\\ \;\;\;\;\frac{1}{n} \cdot \frac{1}{x}\\ \mathbf{elif}\;n \leq -1.22 \cdot 10^{-225}:\\ \;\;\;\;\left(1 + \frac{1}{x \cdot n}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]

Alternative 8
Accuracy	37.3%
Cost	448

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

Alternative 9
Accuracy	36.6%
Cost	320

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

Alternative 10
Accuracy	37.3%
Cost	320

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

Alternative 11
Accuracy	4.6%
Cost	192

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

Reproduce

herbie shell --seed 2023131 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))