ENA, Section 1.4, Mentioned, B

Percentage Accurate: 87.8% → 99.6%
Time: 11.5s
Alternatives: 5
Speedup: 1.1×

Specification

?
\[0.999 \leq x \land x \leq 1.001\]
\[\begin{array}{l} \\ \frac{10}{1 - x \cdot x} \end{array} \]
(FPCore (x) :precision binary64 (/ 10.0 (- 1.0 (* x x))))
double code(double x) {
	return 10.0 / (1.0 - (x * x));
}
real(8) function code(x)
    real(8), intent (in) :: x
    code = 10.0d0 / (1.0d0 - (x * x))
end function
public static double code(double x) {
	return 10.0 / (1.0 - (x * x));
}
def code(x):
	return 10.0 / (1.0 - (x * x))
function code(x)
	return Float64(10.0 / Float64(1.0 - Float64(x * x)))
end
function tmp = code(x)
	tmp = 10.0 / (1.0 - (x * x));
end
code[x_] := N[(10.0 / N[(1.0 - N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\frac{10}{1 - x \cdot 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 5 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: 87.8% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \frac{10}{1 - x \cdot x} \end{array} \]
(FPCore (x) :precision binary64 (/ 10.0 (- 1.0 (* x x))))
double code(double x) {
	return 10.0 / (1.0 - (x * x));
}
real(8) function code(x)
    real(8), intent (in) :: x
    code = 10.0d0 / (1.0d0 - (x * x))
end function
public static double code(double x) {
	return 10.0 / (1.0 - (x * x));
}
def code(x):
	return 10.0 / (1.0 - (x * x))
function code(x)
	return Float64(10.0 / Float64(1.0 - Float64(x * x)))
end
function tmp = code(x)
	tmp = 10.0 / (1.0 - (x * x));
end
code[x_] := N[(10.0 / N[(1.0 - N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\frac{10}{1 - x \cdot x}
\end{array}

Alternative 1: 99.6% accurate, 1.1× speedup?

\[\begin{array}{l} \\ \frac{-10}{\mathsf{fma}\left(x, x, -1\right)} \end{array} \]
(FPCore (x) :precision binary64 (/ -10.0 (fma x x -1.0)))
double code(double x) {
	return -10.0 / fma(x, x, -1.0);
}
function code(x)
	return Float64(-10.0 / fma(x, x, -1.0))
end
code[x_] := N[(-10.0 / N[(x * x + -1.0), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\frac{-10}{\mathsf{fma}\left(x, x, -1\right)}
\end{array}
Derivation
  1. Initial program 88.0%

    \[\frac{10}{1 - x \cdot x} \]
  2. Add Preprocessing
  3. Applied rewrites99.6%

    \[\leadsto \color{blue}{\frac{-10}{\mathsf{fma}\left(x, x, -1\right)}} \]
  4. Add Preprocessing

Alternative 2: 13.6% accurate, 0.8× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;1 - x \cdot x \leq -5 \cdot 10^{-5}:\\ \;\;\;\;-10\\ \mathbf{else}:\\ \;\;\;\;10 \cdot \mathsf{fma}\left(x, x, 1\right)\\ \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (if (<= (- 1.0 (* x x)) -5e-5) -10.0 (* 10.0 (fma x x 1.0))))
double code(double x) {
	double tmp;
	if ((1.0 - (x * x)) <= -5e-5) {
		tmp = -10.0;
	} else {
		tmp = 10.0 * fma(x, x, 1.0);
	}
	return tmp;
}
function code(x)
	tmp = 0.0
	if (Float64(1.0 - Float64(x * x)) <= -5e-5)
		tmp = -10.0;
	else
		tmp = Float64(10.0 * fma(x, x, 1.0));
	end
	return tmp
end
code[x_] := If[LessEqual[N[(1.0 - N[(x * x), $MachinePrecision]), $MachinePrecision], -5e-5], -10.0, N[(10.0 * N[(x * x + 1.0), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;1 - x \cdot x \leq -5 \cdot 10^{-5}:\\
\;\;\;\;-10\\

\mathbf{else}:\\
\;\;\;\;10 \cdot \mathsf{fma}\left(x, x, 1\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (-.f64 #s(literal 1 binary64) (*.f64 x x)) < -5.00000000000000024e-5

    1. Initial program 86.9%

      \[\frac{10}{1 - x \cdot x} \]
    2. Add Preprocessing
    3. Taylor expanded in x around 0

      \[\leadsto \color{blue}{10 + 10 \cdot {x}^{2}} \]
    4. Step-by-step derivation
      1. +-commutativeN/A

        \[\leadsto \color{blue}{10 \cdot {x}^{2} + 10} \]
      2. lower-fma.f64N/A

        \[\leadsto \color{blue}{\mathsf{fma}\left(10, {x}^{2}, 10\right)} \]
      3. unpow2N/A

        \[\leadsto \mathsf{fma}\left(10, \color{blue}{x \cdot x}, 10\right) \]
      4. lower-*.f641.5

        \[\leadsto \mathsf{fma}\left(10, \color{blue}{x \cdot x}, 10\right) \]
    5. Applied rewrites1.5%

      \[\leadsto \color{blue}{\mathsf{fma}\left(10, x \cdot x, 10\right)} \]
    6. Applied rewrites1.6%

      \[\leadsto \mathsf{fma}\left(x, x, -1\right) \cdot \color{blue}{10} \]
    7. Taylor expanded in x around 0

      \[\leadsto -10 \]
    8. Step-by-step derivation
      1. Applied rewrites13.5%

        \[\leadsto -10 \]

      if -5.00000000000000024e-5 < (-.f64 #s(literal 1 binary64) (*.f64 x x))

      1. Initial program 88.2%

        \[\frac{10}{1 - x \cdot x} \]
      2. Add Preprocessing
      3. Taylor expanded in x around 0

        \[\leadsto \color{blue}{10 + 10 \cdot {x}^{2}} \]
      4. Step-by-step derivation
        1. +-commutativeN/A

          \[\leadsto \color{blue}{10 \cdot {x}^{2} + 10} \]
        2. lower-fma.f64N/A

          \[\leadsto \color{blue}{\mathsf{fma}\left(10, {x}^{2}, 10\right)} \]
        3. unpow2N/A

          \[\leadsto \mathsf{fma}\left(10, \color{blue}{x \cdot x}, 10\right) \]
        4. lower-*.f6413.6

          \[\leadsto \mathsf{fma}\left(10, \color{blue}{x \cdot x}, 10\right) \]
      5. Applied rewrites13.6%

        \[\leadsto \color{blue}{\mathsf{fma}\left(10, x \cdot x, 10\right)} \]
      6. Step-by-step derivation
        1. Applied rewrites13.6%

          \[\leadsto 10 \cdot \left(x \cdot x\right) + \color{blue}{10} \]
        2. Step-by-step derivation
          1. Applied rewrites13.6%

            \[\leadsto \mathsf{fma}\left(x, x, 1\right) \cdot \color{blue}{10} \]
        3. Recombined 2 regimes into one program.
        4. Final simplification13.6%

          \[\leadsto \begin{array}{l} \mathbf{if}\;1 - x \cdot x \leq -5 \cdot 10^{-5}:\\ \;\;\;\;-10\\ \mathbf{else}:\\ \;\;\;\;10 \cdot \mathsf{fma}\left(x, x, 1\right)\\ \end{array} \]
        5. Add Preprocessing

        Reproduce

        ?
        herbie shell --seed 2024228 
        (FPCore (x)
          :name "ENA, Section 1.4, Mentioned, B"
          :precision binary64
          :pre (and (<= 0.999 x) (<= x 1.001))
          (/ 10.0 (- 1.0 (* x x))))