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

Percentage Accurate: 100.0% → 100.0%
Time: 2.7s
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. Add Preprocessing

  3. Final simplification100.0%

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

Alternative 2: 72.1% accurate, 0.2× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq -2.5 \cdot 10^{+111}:\\ \;\;\;\;1\\ \mathbf{elif}\;x \leq -0.16 \lor \neg \left(x \leq -6.2 \cdot 10^{-108}\right) \land x \leq 1.5 \cdot 10^{+53}:\\ \;\;\;\;\frac{x}{y}\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \end{array} \]
(FPCore (x y)
 :precision binary64
 (if (<= x -2.5e+111)
   1.0
   (if (or (<= x -0.16) (and (not (<= x -6.2e-108)) (<= x 1.5e+53)))
     (/ x y)
     1.0)))
double code(double x, double y) {
	double tmp;
	if (x <= -2.5e+111) {
		tmp = 1.0;
	} else if ((x <= -0.16) || (!(x <= -6.2e-108) && (x <= 1.5e+53))) {
		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 (x <= (-2.5d+111)) then
        tmp = 1.0d0
    else if ((x <= (-0.16d0)) .or. (.not. (x <= (-6.2d-108))) .and. (x <= 1.5d+53)) 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 (x <= -2.5e+111) {
		tmp = 1.0;
	} else if ((x <= -0.16) || (!(x <= -6.2e-108) && (x <= 1.5e+53))) {
		tmp = x / y;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

def code(x, y):
	tmp = 0
	if x <= -2.5e+111:
		tmp = 1.0
	elif (x <= -0.16) or (not (x <= -6.2e-108) and (x <= 1.5e+53)):
		tmp = x / y
	else:
		tmp = 1.0
	return tmp

function code(x, y)
	tmp = 0.0
	if (x <= -2.5e+111)
		tmp = 1.0;
	elseif ((x <= -0.16) || (!(x <= -6.2e-108) && (x <= 1.5e+53)))
		tmp = Float64(x / y);
	else
		tmp = 1.0;
	end
	return tmp
end

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (x <= -2.5e+111)
		tmp = 1.0;
	elseif ((x <= -0.16) || (~((x <= -6.2e-108)) && (x <= 1.5e+53)))
		tmp = x / y;
	else
		tmp = 1.0;
	end
	tmp_2 = tmp;
end

code[x_, y_] := If[LessEqual[x, -2.5e+111], 1.0, If[Or[LessEqual[x, -0.16], And[N[Not[LessEqual[x, -6.2e-108]], $MachinePrecision], LessEqual[x, 1.5e+53]]], N[(x / y), $MachinePrecision], 1.0]]

\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;x \leq -2.5 \cdot 10^{+111}:\\
\;\;\;\;1\\

\mathbf{elif}\;x \leq -0.16 \lor \neg \left(x \leq -6.2 \cdot 10^{-108}\right) \land x \leq 1.5 \cdot 10^{+53}:\\
\;\;\;\;\frac{x}{y}\\

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


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

  2. if x < -2.4999999999999998e111 or -0.160000000000000003 < x < -6.20000000000000028e-108 or 1.49999999999999999e53 < x

    1. Initial program 100.0%

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

    3. Taylor expanded in x around inf 82.7%

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

    if -2.4999999999999998e111 < x < -0.160000000000000003 or -6.20000000000000028e-108 < x < 1.49999999999999999e53

    1. Initial program 100.0%

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

    3. Taylor expanded in x around 0 80.6%

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

  4. Final simplification81.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -2.5 \cdot 10^{+111}:\\ \;\;\;\;1\\ \mathbf{elif}\;x \leq -0.16 \lor \neg \left(x \leq -6.2 \cdot 10^{-108}\right) \land x \leq 1.5 \cdot 10^{+53}:\\ \;\;\;\;\frac{x}{y}\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]
  5. Add Preprocessing

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. Add Preprocessing

  3. Taylor expanded in x around inf 51.4%

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

  4. Final simplification51.4%

    \[\leadsto 1 \]
  5. Add Preprocessing

Reproduce

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herbie shell --seed 2024020 
(FPCore (x y)
  :name "AI.Clustering.Hierarchical.Internal:average from clustering-0.2.1, A"
  :precision binary64
  (/ x (+ x y)))