2cbrt (problem 3.3.4)

Percentage Accurate: 24.7% → 24.7%
Time: 843.0ms
Alternatives: 1
Speedup: 1.0×

Specification

?
\[x > 1 \land x < 10^{+308}\]
\[\begin{array}{l} \\ \sqrt[3]{x + 1} - \sqrt[3]{x} \end{array} \]
(FPCore (x) :precision binary64 (- (cbrt (+ x 1.0)) (cbrt x)))
double code(double x) {
	return cbrt((x + 1.0)) - cbrt(x);
}

public static double code(double x) {
	return Math.cbrt((x + 1.0)) - Math.cbrt(x);
}

function code(x)
	return Float64(cbrt(Float64(x + 1.0)) - cbrt(x))
end

code[x_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], 1/3], $MachinePrecision] - N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\sqrt[3]{x + 1} - \sqrt[3]{x}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 1 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 24.7% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \sqrt[3]{x + 1} - \sqrt[3]{x} \end{array} \]
(FPCore (x) :precision binary64 (- (cbrt (+ x 1.0)) (cbrt x)))
double code(double x) {
	return cbrt((x + 1.0)) - cbrt(x);
}

public static double code(double x) {
	return Math.cbrt((x + 1.0)) - Math.cbrt(x);
}

function code(x)
	return Float64(cbrt(Float64(x + 1.0)) - cbrt(x))
end

code[x_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], 1/3], $MachinePrecision] - N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\sqrt[3]{x + 1} - \sqrt[3]{x}
\end{array}

Alternative 1: 24.7% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \sqrt[3]{1 + x} - \sqrt[3]{x} \end{array} \]
(FPCore (x) :precision binary64 (- (cbrt (+ 1.0 x)) (cbrt x)))
double code(double x) {
	return cbrt((1.0 + x)) - cbrt(x);
}

public static double code(double x) {
	return Math.cbrt((1.0 + x)) - Math.cbrt(x);
}

function code(x)
	return Float64(cbrt(Float64(1.0 + x)) - cbrt(x))
end

code[x_] := N[(N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision] - N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\sqrt[3]{1 + x} - \sqrt[3]{x}
\end{array}
Derivation

  1. Initial program 7.7%

    \[\sqrt[3]{x + 1} - \sqrt[3]{x} \]
  2. Add Preprocessing

  3. Final simplification7.7%

    \[\leadsto \sqrt[3]{1 + x} - \sqrt[3]{x} \]
  4. Add Preprocessing

Developer Target 1: 98.6% accurate, 0.3× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt[3]{x + 1}\\ \frac{1}{\left(t\_0 \cdot t\_0 + \sqrt[3]{x} \cdot t\_0\right) + \sqrt[3]{x} \cdot \sqrt[3]{x}} \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (let* ((t_0 (cbrt (+ x 1.0))))
   (/ 1.0 (+ (+ (* t_0 t_0) (* (cbrt x) t_0)) (* (cbrt x) (cbrt x))))))
double code(double x) {
	double t_0 = cbrt((x + 1.0));
	return 1.0 / (((t_0 * t_0) + (cbrt(x) * t_0)) + (cbrt(x) * cbrt(x)));
}

public static double code(double x) {
	double t_0 = Math.cbrt((x + 1.0));
	return 1.0 / (((t_0 * t_0) + (Math.cbrt(x) * t_0)) + (Math.cbrt(x) * Math.cbrt(x)));
}

function code(x)
	t_0 = cbrt(Float64(x + 1.0))
	return Float64(1.0 / Float64(Float64(Float64(t_0 * t_0) + Float64(cbrt(x) * t_0)) + Float64(cbrt(x) * cbrt(x))))
end

code[x_] := Block[{t$95$0 = N[Power[N[(x + 1.0), $MachinePrecision], 1/3], $MachinePrecision]}, N[(1.0 / N[(N[(N[(t$95$0 * t$95$0), $MachinePrecision] + N[(N[Power[x, 1/3], $MachinePrecision] * t$95$0), $MachinePrecision]), $MachinePrecision] + N[(N[Power[x, 1/3], $MachinePrecision] * N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

\begin{array}{l}

\\
\begin{array}{l}
t_0 := \sqrt[3]{x + 1}\\
\frac{1}{\left(t\_0 \cdot t\_0 + \sqrt[3]{x} \cdot t\_0\right) + \sqrt[3]{x} \cdot \sqrt[3]{x}}
\end{array}
\end{array}

Reproduce

?
herbie shell --seed 2024346 
(FPCore (x)
  :name "2cbrt (problem 3.3.4)"
  :precision binary64
  :pre (and (> x 1.0) (< x 1e+308))

  :alt
  (! :herbie-platform default (/ 1 (+ (* (cbrt (+ x 1)) (cbrt (+ x 1))) (* (cbrt x) (cbrt (+ x 1))) (* (cbrt x) (cbrt x)))))

  (- (cbrt (+ x 1.0)) (cbrt x)))