Data.Colour.CIE:cieLAB from colour-2.3.3, C

Percentage Accurate: 100.0% → 100.0%
Time: 3.5s
Alternatives: 3
Speedup: 1.0×

Specification

?
\[\begin{array}{l} \\ x + \frac{y}{500} \end{array} \]
(FPCore (x y) :precision binary64 (+ x (/ y 500.0)))
double code(double x, double y) {
	return x + (y / 500.0);
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x + (y / 500.0d0)
end function

public static double code(double x, double y) {
	return x + (y / 500.0);
}

def code(x, y):
	return x + (y / 500.0)

function code(x, y)
	return Float64(x + Float64(y / 500.0))
end

function tmp = code(x, y)
	tmp = x + (y / 500.0);
end

code[x_, y_] := N[(x + N[(y / 500.0), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
x + \frac{y}{500}
\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 3 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: 100.0% accurate, 1.0× speedup?

\[\begin{array}{l} \\ x + \frac{y}{500} \end{array} \]
(FPCore (x y) :precision binary64 (+ x (/ y 500.0)))
double code(double x, double y) {
	return x + (y / 500.0);
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x + (y / 500.0d0)
end function

public static double code(double x, double y) {
	return x + (y / 500.0);
}

def code(x, y):
	return x + (y / 500.0)

function code(x, y)
	return Float64(x + Float64(y / 500.0))
end

function tmp = code(x, y)
	tmp = x + (y / 500.0);
end

code[x_, y_] := N[(x + N[(y / 500.0), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
x + \frac{y}{500}
\end{array}

Alternative 1: 100.0% accurate, 1.0× speedup?

\[\begin{array}{l} \\ x + \frac{y}{500} \end{array} \]
(FPCore (x y) :precision binary64 (+ x (/ y 500.0)))
double code(double x, double y) {
	return x + (y / 500.0);
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x + (y / 500.0d0)
end function

public static double code(double x, double y) {
	return x + (y / 500.0);
}

def code(x, y):
	return x + (y / 500.0)

function code(x, y)
	return Float64(x + Float64(y / 500.0))
end

function tmp = code(x, y)
	tmp = x + (y / 500.0);
end

code[x_, y_] := N[(x + N[(y / 500.0), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
x + \frac{y}{500}
\end{array}
Derivation

  1. Initial program 100.0%

    \[x + \frac{y}{500} \]
  2. Add Preprocessing

  3. Final simplification100.0%

    \[\leadsto x + \frac{y}{500} \]
  4. Add Preprocessing

Alternative 2: 74.9% accurate, 0.2× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq -9.8 \cdot 10^{+113} \lor \neg \left(y \leq -1.76 \cdot 10^{+68}\right) \land \left(y \leq -74000000000 \lor \neg \left(y \leq 2500\right)\right):\\ \;\;\;\;y \cdot 0.002\\ \mathbf{else}:\\ \;\;\;\;x\\ \end{array} \end{array} \]
(FPCore (x y)
 :precision binary64
 (if (or (<= y -9.8e+113)
         (and (not (<= y -1.76e+68))
              (or (<= y -74000000000.0) (not (<= y 2500.0)))))
   (* y 0.002)
   x))
double code(double x, double y) {
	double tmp;
	if ((y <= -9.8e+113) || (!(y <= -1.76e+68) && ((y <= -74000000000.0) || !(y <= 2500.0)))) {
		tmp = y * 0.002;
	} else {
		tmp = x;
	}
	return tmp;
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if ((y <= (-9.8d+113)) .or. (.not. (y <= (-1.76d+68))) .and. (y <= (-74000000000.0d0)) .or. (.not. (y <= 2500.0d0))) then
        tmp = y * 0.002d0
    else
        tmp = x
    end if
    code = tmp
end function

public static double code(double x, double y) {
	double tmp;
	if ((y <= -9.8e+113) || (!(y <= -1.76e+68) && ((y <= -74000000000.0) || !(y <= 2500.0)))) {
		tmp = y * 0.002;
	} else {
		tmp = x;
	}
	return tmp;
}

def code(x, y):
	tmp = 0
	if (y <= -9.8e+113) or (not (y <= -1.76e+68) and ((y <= -74000000000.0) or not (y <= 2500.0))):
		tmp = y * 0.002
	else:
		tmp = x
	return tmp

function code(x, y)
	tmp = 0.0
	if ((y <= -9.8e+113) || (!(y <= -1.76e+68) && ((y <= -74000000000.0) || !(y <= 2500.0))))
		tmp = Float64(y * 0.002);
	else
		tmp = x;
	end
	return tmp
end

function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y <= -9.8e+113) || (~((y <= -1.76e+68)) && ((y <= -74000000000.0) || ~((y <= 2500.0)))))
		tmp = y * 0.002;
	else
		tmp = x;
	end
	tmp_2 = tmp;
end

code[x_, y_] := If[Or[LessEqual[y, -9.8e+113], And[N[Not[LessEqual[y, -1.76e+68]], $MachinePrecision], Or[LessEqual[y, -74000000000.0], N[Not[LessEqual[y, 2500.0]], $MachinePrecision]]]], N[(y * 0.002), $MachinePrecision], x]

\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;y \leq -9.8 \cdot 10^{+113} \lor \neg \left(y \leq -1.76 \cdot 10^{+68}\right) \land \left(y \leq -74000000000 \lor \neg \left(y \leq 2500\right)\right):\\
\;\;\;\;y \cdot 0.002\\

\mathbf{else}:\\
\;\;\;\;x\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes

  2. if y < -9.80000000000000043e113 or -1.76e68 < y < -7.4e10 or 2500 < y

    1. Initial program 100.0%

      \[x + \frac{y}{500} \]
    2. Add Preprocessing

    3. Taylor expanded in x around 0 82.6%

      \[\leadsto \color{blue}{0.002 \cdot y} \]

    if -9.80000000000000043e113 < y < -1.76e68 or -7.4e10 < y < 2500

    1. Initial program 100.0%

      \[x + \frac{y}{500} \]
    2. Add Preprocessing

    3. Taylor expanded in x around inf 79.2%

      \[\leadsto \color{blue}{x} \]
  3. Recombined 2 regimes into one program.

  4. Final simplification80.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq -9.8 \cdot 10^{+113} \lor \neg \left(y \leq -1.76 \cdot 10^{+68}\right) \land \left(y \leq -74000000000 \lor \neg \left(y \leq 2500\right)\right):\\ \;\;\;\;y \cdot 0.002\\ \mathbf{else}:\\ \;\;\;\;x\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 51.2% accurate, 5.0× speedup?

\[\begin{array}{l} \\ x \end{array} \]
(FPCore (x y) :precision binary64 x)
double code(double x, double y) {
	return x;
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x
end function

public static double code(double x, double y) {
	return x;
}

def code(x, y):
	return x

function code(x, y)
	return x
end

function tmp = code(x, y)
	tmp = x;
end

code[x_, y_] := x

\begin{array}{l}

\\
x
\end{array}
Derivation

  1. Initial program 100.0%

    \[x + \frac{y}{500} \]
  2. Add Preprocessing

  3. Taylor expanded in x around inf 53.8%

    \[\leadsto \color{blue}{x} \]

  4. Final simplification53.8%

    \[\leadsto x \]
  5. Add Preprocessing

Reproduce

?
herbie shell --seed 2024026 
(FPCore (x y)
  :name "Data.Colour.CIE:cieLAB from colour-2.3.3, C"
  :precision binary64
  (+ x (/ y 500.0)))