AI.Clustering.Hierarchical.Internal:average from clustering-0.2.1, A

Percentage Accurate: 100.0% → 100.0%
Time: 2.0s
Alternatives: 3
Speedup: 1.0×

Specification

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

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

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

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

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

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

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

\begin{array}{l}

\\
\frac{x}{x + y}
\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?

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

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

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

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

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

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

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

\begin{array}{l}

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

Alternative 1: 100.0% accurate, 1.0× speedup?

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

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

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

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

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

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

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

\begin{array}{l}

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

  1. Initial program 100.0%

    \[\frac{x}{x + y} \]

  2. Final simplification100.0%

    \[\leadsto \frac{x}{x + y} \]

Alternative 2: 72.2% accurate, 0.4× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq -2.7 \cdot 10^{-120} \lor \neg \left(y \leq 6.5 \cdot 10^{+38}\right) \land \left(y \leq 3.4 \cdot 10^{+101} \lor \neg \left(y \leq 2.6 \cdot 10^{+142}\right)\right):\\ \;\;\;\;\frac{x}{y}\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \end{array} \]
(FPCore (x y)
 :precision binary64
 (if (or (<= y -2.7e-120)
         (and (not (<= y 6.5e+38)) (or (<= y 3.4e+101) (not (<= y 2.6e+142)))))
   (/ x y)
   1.0))
double code(double x, double y) {
	double tmp;
	if ((y <= -2.7e-120) || (!(y <= 6.5e+38) && ((y <= 3.4e+101) || !(y <= 2.6e+142)))) {
		tmp = x / y;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if ((y <= (-2.7d-120)) .or. (.not. (y <= 6.5d+38)) .and. (y <= 3.4d+101) .or. (.not. (y <= 2.6d+142))) then
        tmp = x / y
    else
        tmp = 1.0d0
    end if
    code = tmp
end function

public static double code(double x, double y) {
	double tmp;
	if ((y <= -2.7e-120) || (!(y <= 6.5e+38) && ((y <= 3.4e+101) || !(y <= 2.6e+142)))) {
		tmp = x / y;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

def code(x, y):
	tmp = 0
	if (y <= -2.7e-120) or (not (y <= 6.5e+38) and ((y <= 3.4e+101) or not (y <= 2.6e+142))):
		tmp = x / y
	else:
		tmp = 1.0
	return tmp

function code(x, y)
	tmp = 0.0
	if ((y <= -2.7e-120) || (!(y <= 6.5e+38) && ((y <= 3.4e+101) || !(y <= 2.6e+142))))
		tmp = Float64(x / y);
	else
		tmp = 1.0;
	end
	return tmp
end

function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y <= -2.7e-120) || (~((y <= 6.5e+38)) && ((y <= 3.4e+101) || ~((y <= 2.6e+142)))))
		tmp = x / y;
	else
		tmp = 1.0;
	end
	tmp_2 = tmp;
end

code[x_, y_] := If[Or[LessEqual[y, -2.7e-120], And[N[Not[LessEqual[y, 6.5e+38]], $MachinePrecision], Or[LessEqual[y, 3.4e+101], N[Not[LessEqual[y, 2.6e+142]], $MachinePrecision]]]], N[(x / y), $MachinePrecision], 1.0]

\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;y \leq -2.7 \cdot 10^{-120} \lor \neg \left(y \leq 6.5 \cdot 10^{+38}\right) \land \left(y \leq 3.4 \cdot 10^{+101} \lor \neg \left(y \leq 2.6 \cdot 10^{+142}\right)\right):\\
\;\;\;\;\frac{x}{y}\\

\mathbf{else}:\\
\;\;\;\;1\\


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

  2. if y < -2.6999999999999999e-120 or 6.5e38 < y < 3.40000000000000017e101 or 2.60000000000000021e142 < y

    1. Initial program 100.0%

      \[\frac{x}{x + y} \]

    2. Taylor expanded in x around 0 79.5%

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

    if -2.6999999999999999e-120 < y < 6.5e38 or 3.40000000000000017e101 < y < 2.60000000000000021e142

    1. Initial program 100.0%

      \[\frac{x}{x + y} \]

    2. Taylor expanded in x around inf 76.6%

      \[\leadsto \color{blue}{1} \]
  3. Recombined 2 regimes into one program.

  4. Final simplification78.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq -2.7 \cdot 10^{-120} \lor \neg \left(y \leq 6.5 \cdot 10^{+38}\right) \land \left(y \leq 3.4 \cdot 10^{+101} \lor \neg \left(y \leq 2.6 \cdot 10^{+142}\right)\right):\\ \;\;\;\;\frac{x}{y}\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 3: 50.7% accurate, 5.0× speedup?

\[\begin{array}{l} \\ 1 \end{array} \]
(FPCore (x y) :precision binary64 1.0)
double code(double x, double y) {
	return 1.0;
}

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = 1.0d0
end function

public static double code(double x, double y) {
	return 1.0;
}

def code(x, y):
	return 1.0

function code(x, y)
	return 1.0
end

function tmp = code(x, y)
	tmp = 1.0;
end

code[x_, y_] := 1.0

\begin{array}{l}

\\
1
\end{array}
Derivation

  1. Initial program 100.0%

    \[\frac{x}{x + y} \]

  2. Taylor expanded in x around inf 47.3%

    \[\leadsto \color{blue}{1} \]

  3. Final simplification47.3%

    \[\leadsto 1 \]

Reproduce

?
herbie shell --seed 2023310 
(FPCore (x y)
  :name "AI.Clustering.Hierarchical.Internal:average from clustering-0.2.1, A"
  :precision binary64
  (/ x (+ x y)))