2nthrt (problem 3.4.6)

Percentage Accurate: 53.8% → 85.6%

Time: 27.1s

Alternatives: 16

Speedup: 1.9×

• could not determine a ground truth

  1. x = -2.4264156776130412e+291
  2. n = -6.162196612384609e+291

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: 53.8% 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: 85.6% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-86}:\\ \;\;\;\;\frac{t\_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-11}:\\ \;\;\;\;\frac{\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n}, \mathsf{log1p}\left(x\right)\right) - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -1e-86)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 5e-11)
       (/
        (-
         (fma 0.5 (/ (- (pow (log1p x) 2.0) (pow (log x) 2.0)) n) (log1p x))
         (log x))
        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-86) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 5e-11) {
		tmp = (fma(0.5, ((pow(log1p(x), 2.0) - pow(log(x), 2.0)) / n), log1p(x)) - log(x)) / n;
	} else {
		tmp = exp((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-86)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 5e-11)
		tmp = Float64(Float64(fma(0.5, Float64(Float64((log1p(x) ^ 2.0) - (log(x) ^ 2.0)) / n), log1p(x)) - log(x)) / 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-86], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-11], N[(N[(N[(0.5 * N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision] + N[Log[1 + x], $MachinePrecision]), $MachinePrecision] - N[Log[x], $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) < -1.00000000000000008e-86

   1. Initial program 86.8%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   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 93.0

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

   5. Applied rewrites 93.0%

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

   if -1.00000000000000008e-86 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000018e-11

   1. Initial program 32.3%

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

   3. Taylor expanded in n around inf

      \[\leadsto \color{blue}{\frac{\left(\log \left(1 + x\right) + \frac{1}{2} \cdot \frac{{\log \left(1 + x\right)}^{2}}{n}\right) - \left(\log x + \frac{1}{2} \cdot \frac{{\log x}^{2}}{n}\right)}{n}} \]
   4. Step-by-step derivation

      1. lower-/.f64 N/A

         \[\leadsto \color{blue}{\frac{\left(\log \left(1 + x\right) + \frac{1}{2} \cdot \frac{{\log \left(1 + x\right)}^{2}}{n}\right) - \left(\log x + \frac{1}{2} \cdot \frac{{\log x}^{2}}{n}\right)}{n}} \]

   5. Applied rewrites 86.6%

      \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n}, \mathsf{log1p}\left(x\right)\right) - \log x}{n}} \]

   if 5.00000000000000018e-11 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 62.6%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   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 97.6

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

   4. Applied rewrites 97.6%

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

4. Final simplification 90.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-86}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-11}:\\ \;\;\;\;\frac{\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n}, \mathsf{log1p}\left(x\right)\right) - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 77.9% accurate, 0.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t\_0\\ t_2 := 1 - t\_0\\ \mathbf{if}\;t\_1 \leq -0.02:\\ \;\;\;\;t\_2\\ \mathbf{elif}\;t\_1 \leq 5 \cdot 10^{-9}:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;t\_2\\ \end{array} \end{array} \]

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))
        (t_2 (- 1.0 t_0)))
   (if (<= t_1 -0.02) t_2 (if (<= t_1 5e-9) (/ (log (/ (+ x 1.0) x)) n) t_2))))

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 t_2 = 1.0 - t_0;
	double tmp;
	if (t_1 <= -0.02) {
		tmp = t_2;
	} else if (t_1 <= 5e-9) {
		tmp = log(((x + 1.0) / x)) / n;
	} else {
		tmp = t_2;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: t_2
    real(8) :: tmp
    t_0 = x ** (1.0d0 / n)
    t_1 = ((x + 1.0d0) ** (1.0d0 / n)) - t_0
    t_2 = 1.0d0 - t_0
    if (t_1 <= (-0.02d0)) then
        tmp = t_2
    else if (t_1 <= 5d-9) then
        tmp = log(((x + 1.0d0) / x)) / n
    else
        tmp = t_2
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double t_1 = Math.pow((x + 1.0), (1.0 / n)) - t_0;
	double t_2 = 1.0 - t_0;
	double tmp;
	if (t_1 <= -0.02) {
		tmp = t_2;
	} else if (t_1 <= 5e-9) {
		tmp = Math.log(((x + 1.0) / x)) / n;
	} else {
		tmp = t_2;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	t_1 = math.pow((x + 1.0), (1.0 / n)) - t_0
	t_2 = 1.0 - t_0
	tmp = 0
	if t_1 <= -0.02:
		tmp = t_2
	elif t_1 <= 5e-9:
		tmp = math.log(((x + 1.0) / x)) / n
	else:
		tmp = t_2
	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)
	t_2 = Float64(1.0 - t_0)
	tmp = 0.0
	if (t_1 <= -0.02)
		tmp = t_2;
	elseif (t_1 <= 5e-9)
		tmp = Float64(log(Float64(Float64(x + 1.0) / x)) / n);
	else
		tmp = t_2;
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	t_1 = ((x + 1.0) ^ (1.0 / n)) - t_0;
	t_2 = 1.0 - t_0;
	tmp = 0.0;
	if (t_1 <= -0.02)
		tmp = t_2;
	elseif (t_1 <= 5e-9)
		tmp = log(((x + 1.0) / x)) / n;
	else
		tmp = t_2;
	end
	tmp_2 = 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]}, Block[{t$95$2 = N[(1.0 - t$95$0), $MachinePrecision]}, If[LessEqual[t$95$1, -0.02], t$95$2, If[LessEqual[t$95$1, 5e-9], N[(N[Log[N[(N[(x + 1.0), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], t$95$2]]]]]

Derivation

1. Split input into 2 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))) < -0.0200000000000000004 or 5.0000000000000001e-9 < (-.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 76.7%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   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 74.0%

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

   if -0.0200000000000000004 < (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n))) < 5.0000000000000001e-9

   1. Initial program 44.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   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 79.5

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

   5. Applied rewrites 79.5%

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

   7. Applied rewrites 79.6%

      \[\leadsto \color{blue}{\frac{\log \left(\frac{x + 1}{x}\right)}{n}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Reproduce

herbie shell --seed 2024226 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))