2nthrt (problem 3.4.6)

Percentage Accurate: 52.5% → 84.7%

Time: 25.7s

Alternatives: 17

Speedup: 5.2×

• could not determine a ground truth

  1. x = -3.998858881728447e+205
  2. n = 1.2492783728651992e-8

Specification

Math

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

FPCore

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

C

double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}

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

Java

public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}

Python

def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))

Julia

function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end

MATLAB

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

Initial Program: 52.5% accurate, 1.0× speedup

Math

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

FPCore

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

C

double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}

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

Java

public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}

Python

def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))

Julia

function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end

MATLAB

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

Alternative 1: 84.7% accurate, 0.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-104}:\\ \;\;\;\;\frac{t\_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -1e-104)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 4.0)
       (/ (log (/ x (+ x 1.0))) (- n))
       (- (exp (/ (log1p x) n)) t_0)))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -1e-104) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 4.0) {
		tmp = log((x / (x + 1.0))) / -n;
	} else {
		tmp = exp((log1p(x) / n)) - t_0;
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -1e-104) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 4.0) {
		tmp = Math.log((x / (x + 1.0))) / -n;
	} else {
		tmp = Math.exp((Math.log1p(x) / n)) - t_0;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -1e-104:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 4.0:
		tmp = math.log((x / (x + 1.0))) / -n
	else:
		tmp = math.exp((math.log1p(x) / n)) - t_0
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -1e-104)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 4.0)
		tmp = Float64(log(Float64(x / Float64(x + 1.0))) / Float64(-n));
	else
		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
	end
	return tmp
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -1e-104], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 4.0], N[(N[Log[N[(x / N[(x + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-n)), $MachinePrecision], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -9.99999999999999927e-105

   1. Initial program 76.0%

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

   3. Taylor expanded in x around inf

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

      1. lower-/.f64 N/A

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

      2. log-rec N/A

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

      3. mul-1-neg N/A

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

      4. associate-*r/ N/A

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

      5. associate-*r* N/A

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

      6. metadata-eval N/A

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

      7. *-commutative N/A

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

      8. associate-/l* N/A

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

      9. exp-to-pow N/A

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

      10. lower-pow.f64 N/A

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

      11. lower-/.f64 N/A

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

      12. *-commutative N/A

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

      13. lower-*.f64 88.2

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

   5. Applied rewrites 88.2%

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

   if -9.99999999999999927e-105 < (/.f64 #s(literal 1 binary64) n) < 4

   1. Initial program 32.5%

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

   3. Taylor expanded in n around inf

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

      1. lower-/.f64 N/A

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

      2. lower--.f64 N/A

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

      3. lower-log1p.f64 N/A

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

      4. lower-log.f64 86.6

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

   5. Applied rewrites 86.6%

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

   7. Applied rewrites 86.7%

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

   if 4 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 56.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. lift-pow.f64 N/A

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

      2. pow-to-exp N/A

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

      3. lower-exp.f64 N/A

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

      4. lift-/.f64 N/A

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

      5. un-div-inv N/A

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

      6. lower-/.f64 N/A

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

      7. lift-+.f64 N/A

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

      8. +-commutative N/A

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

      9. lower-log1p.f64 100.0

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

   4. Applied rewrites 100.0%

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

4. Final simplification 89.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-104}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 77.0% accurate, 0.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t\_0\\ \mathbf{if}\;t\_1 \leq -5 \cdot 10^{-7}:\\ \;\;\;\;1 - t\_0\\ \mathbf{elif}\;t\_1 \leq 2 \cdot 10^{-16}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left(x, n \cdot -0.5, n \cdot 0.3333333333333333\right)}{x \cdot \left(x \cdot \left(n \cdot n\right)\right)} + \frac{1}{n}}{x}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))) (t_1 (- (pow (+ x 1.0) (/ 1.0 n)) t_0)))
   (if (<= t_1 -5e-7)
     (- 1.0 t_0)
     (if (<= t_1 2e-16)
       (/ (log (/ x (+ x 1.0))) (- n))
       (/
        (+
         (/ (fma x (* n -0.5) (* n 0.3333333333333333)) (* x (* x (* n n))))
         (/ 1.0 n))
        x)))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double t_1 = pow((x + 1.0), (1.0 / n)) - t_0;
	double tmp;
	if (t_1 <= -5e-7) {
		tmp = 1.0 - t_0;
	} else if (t_1 <= 2e-16) {
		tmp = log((x / (x + 1.0))) / -n;
	} else {
		tmp = ((fma(x, (n * -0.5), (n * 0.3333333333333333)) / (x * (x * (n * n)))) + (1.0 / n)) / x;
	}
	return tmp;
}

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	t_1 = Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - t_0)
	tmp = 0.0
	if (t_1 <= -5e-7)
		tmp = Float64(1.0 - t_0);
	elseif (t_1 <= 2e-16)
		tmp = Float64(log(Float64(x / Float64(x + 1.0))) / Float64(-n));
	else
		tmp = Float64(Float64(Float64(fma(x, Float64(n * -0.5), Float64(n * 0.3333333333333333)) / Float64(x * Float64(x * Float64(n * n)))) + Float64(1.0 / n)) / x);
	end
	return tmp
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(N[Power[N[(x + 1.0), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]}, If[LessEqual[t$95$1, -5e-7], N[(1.0 - t$95$0), $MachinePrecision], If[LessEqual[t$95$1, 2e-16], N[(N[Log[N[(x / N[(x + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-n)), $MachinePrecision], N[(N[(N[(N[(x * N[(n * -0.5), $MachinePrecision] + N[(n * 0.3333333333333333), $MachinePrecision]), $MachinePrecision] / N[(x * N[(x * N[(n * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n))) < -4.99999999999999977e-7

   1. Initial program 99.6%

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

   3. Taylor expanded in x around 0

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

   5. Applied rewrites 99.5%

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

   if -4.99999999999999977e-7 < (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n))) < 2e-16

   1. Initial program 42.5%

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

   3. Taylor expanded in n around inf

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

      1. lower-/.f64 N/A

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

      2. lower--.f64 N/A

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

      3. lower-log1p.f64 N/A

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

      4. lower-log.f64 79.9

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

   5. Applied rewrites 79.9%

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

   7. Applied rewrites 80.1%

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

   if 2e-16 < (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n)))

   1. Initial program 54.5%

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

   3. Taylor expanded in n around inf

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

      1. lower-/.f64 N/A

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

      2. lower--.f64 N/A

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

      3. lower-log1p.f64 N/A

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

      4. lower-log.f64 9.7

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

   5. Applied rewrites 9.7%

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

   6. Taylor expanded in x around -inf

      \[\leadsto -1 \cdot \color{blue}{\frac{-1 \cdot \frac{\frac{1}{3} \cdot \frac{1}{n \cdot x} - \frac{1}{2} \cdot \frac{1}{n}}{x} - \frac{1}{n}}{x}} \]
   7. Step-by-step derivation

   8. Applied rewrites 37.4%

      \[\leadsto \frac{\frac{\frac{0.3333333333333333}{x \cdot n} + \frac{-0.5}{n}}{-x} + \frac{-1}{n}}{\color{blue}{-x}} \]
   9. Step-by-step derivation

   10. Applied rewrites 41.7%

       \[\leadsto \frac{\frac{\mathsf{fma}\left(x, n \cdot -0.5, n \cdot 0.3333333333333333\right)}{\left(-x\right) \cdot \left(x \cdot \left(n \cdot n\right)\right)} + \frac{-1}{n}}{-x} \]
3. Recombined 3 regimes into one program.

4. Final simplification 77.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq -5 \cdot 10^{-7}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq 2 \cdot 10^{-16}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left(x, n \cdot -0.5, n \cdot 0.3333333333333333\right)}{x \cdot \left(x \cdot \left(n \cdot n\right)\right)} + \frac{1}{n}}{x}\\ \end{array} \]
5. Add Preprocessing

Reproduce

herbie shell --seed 2024230 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))