Numeric.Log:$cexpm1 from log-domain-0.10.2.1, B

Percentage Accurate: 100.0% → 100.0%
Time: 1.6s
Alternatives: 5
Speedup: 1.0×

Specification

?
\[\begin{array}{l} \\ \left(x \cdot y + x\right) + y \end{array} \]
(FPCore (x y) :precision binary64 (+ (+ (* x y) x) y))
double code(double x, double y) {
	return ((x * y) + x) + y;
}
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = ((x * y) + x) + y
end function
public static double code(double x, double y) {
	return ((x * y) + x) + y;
}
def code(x, y):
	return ((x * y) + x) + y
function code(x, y)
	return Float64(Float64(Float64(x * y) + x) + y)
end
function tmp = code(x, y)
	tmp = ((x * y) + x) + y;
end
code[x_, y_] := N[(N[(N[(x * y), $MachinePrecision] + x), $MachinePrecision] + y), $MachinePrecision]
\begin{array}{l}

\\
\left(x \cdot y + x\right) + 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 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: 100.0% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \left(x \cdot y + x\right) + y \end{array} \]
(FPCore (x y) :precision binary64 (+ (+ (* x y) x) y))
double code(double x, double y) {
	return ((x * y) + x) + y;
}
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = ((x * y) + x) + y
end function
public static double code(double x, double y) {
	return ((x * y) + x) + y;
}
def code(x, y):
	return ((x * y) + x) + y
function code(x, y)
	return Float64(Float64(Float64(x * y) + x) + y)
end
function tmp = code(x, y)
	tmp = ((x * y) + x) + y;
end
code[x_, y_] := N[(N[(N[(x * y), $MachinePrecision] + x), $MachinePrecision] + y), $MachinePrecision]
\begin{array}{l}

\\
\left(x \cdot y + x\right) + y
\end{array}

Alternative 1: 100.0% accurate, 1.0× speedup?

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ y + \left(y + 1\right) \cdot x \end{array} \]
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y) :precision binary64 (+ y (* (+ y 1.0) x)))
assert(x < y);
double code(double x, double y) {
	return y + ((y + 1.0) * x);
}
NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = y + ((y + 1.0d0) * x)
end function
assert x < y;
public static double code(double x, double y) {
	return y + ((y + 1.0) * x);
}
[x, y] = sort([x, y])
def code(x, y):
	return y + ((y + 1.0) * x)
x, y = sort([x, y])
function code(x, y)
	return Float64(y + Float64(Float64(y + 1.0) * x))
end
x, y = num2cell(sort([x, y])){:}
function tmp = code(x, y)
	tmp = y + ((y + 1.0) * x);
end
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(y + N[(N[(y + 1.0), $MachinePrecision] * x), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[x, y] = \mathsf{sort}([x, y])\\
\\
y + \left(y + 1\right) \cdot x
\end{array}
Derivation
  1. Initial program 100.0%

    \[\left(x \cdot y + x\right) + y \]
  2. Step-by-step derivation
    1. *-commutative100.0%

      \[\leadsto \left(\color{blue}{y \cdot x} + x\right) + y \]
    2. distribute-lft1-in100.0%

      \[\leadsto \color{blue}{\left(y + 1\right) \cdot x} + y \]
  3. Applied egg-rr100.0%

    \[\leadsto \color{blue}{\left(y + 1\right) \cdot x} + y \]
  4. Final simplification100.0%

    \[\leadsto y + \left(y + 1\right) \cdot x \]

Alternative 2: 79.9% accurate, 0.5× speedup?

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ \begin{array}{l} \mathbf{if}\;y \leq -4400000000:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq 1.05 \cdot 10^{+175}:\\ \;\;\;\;y + x\\ \mathbf{elif}\;y \leq 1.3 \cdot 10^{+210}:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq 2 \cdot 10^{+225}:\\ \;\;\;\;y\\ \mathbf{elif}\;y \leq 6.5 \cdot 10^{+247}:\\ \;\;\;\;y \cdot x\\ \mathbf{else}:\\ \;\;\;\;y + x\\ \end{array} \end{array} \]
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (if (<= y -4400000000.0)
   (* y x)
   (if (<= y 1.05e+175)
     (+ y x)
     (if (<= y 1.3e+210)
       (* y x)
       (if (<= y 2e+225) y (if (<= y 6.5e+247) (* y x) (+ y x)))))))
assert(x < y);
double code(double x, double y) {
	double tmp;
	if (y <= -4400000000.0) {
		tmp = y * x;
	} else if (y <= 1.05e+175) {
		tmp = y + x;
	} else if (y <= 1.3e+210) {
		tmp = y * x;
	} else if (y <= 2e+225) {
		tmp = y;
	} else if (y <= 6.5e+247) {
		tmp = y * x;
	} else {
		tmp = y + x;
	}
	return tmp;
}
NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= (-4400000000.0d0)) then
        tmp = y * x
    else if (y <= 1.05d+175) then
        tmp = y + x
    else if (y <= 1.3d+210) then
        tmp = y * x
    else if (y <= 2d+225) then
        tmp = y
    else if (y <= 6.5d+247) then
        tmp = y * x
    else
        tmp = y + x
    end if
    code = tmp
end function
assert x < y;
public static double code(double x, double y) {
	double tmp;
	if (y <= -4400000000.0) {
		tmp = y * x;
	} else if (y <= 1.05e+175) {
		tmp = y + x;
	} else if (y <= 1.3e+210) {
		tmp = y * x;
	} else if (y <= 2e+225) {
		tmp = y;
	} else if (y <= 6.5e+247) {
		tmp = y * x;
	} else {
		tmp = y + x;
	}
	return tmp;
}
[x, y] = sort([x, y])
def code(x, y):
	tmp = 0
	if y <= -4400000000.0:
		tmp = y * x
	elif y <= 1.05e+175:
		tmp = y + x
	elif y <= 1.3e+210:
		tmp = y * x
	elif y <= 2e+225:
		tmp = y
	elif y <= 6.5e+247:
		tmp = y * x
	else:
		tmp = y + x
	return tmp
x, y = sort([x, y])
function code(x, y)
	tmp = 0.0
	if (y <= -4400000000.0)
		tmp = Float64(y * x);
	elseif (y <= 1.05e+175)
		tmp = Float64(y + x);
	elseif (y <= 1.3e+210)
		tmp = Float64(y * x);
	elseif (y <= 2e+225)
		tmp = y;
	elseif (y <= 6.5e+247)
		tmp = Float64(y * x);
	else
		tmp = Float64(y + x);
	end
	return tmp
end
x, y = num2cell(sort([x, y])){:}
function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= -4400000000.0)
		tmp = y * x;
	elseif (y <= 1.05e+175)
		tmp = y + x;
	elseif (y <= 1.3e+210)
		tmp = y * x;
	elseif (y <= 2e+225)
		tmp = y;
	elseif (y <= 6.5e+247)
		tmp = y * x;
	else
		tmp = y + x;
	end
	tmp_2 = tmp;
end
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := If[LessEqual[y, -4400000000.0], N[(y * x), $MachinePrecision], If[LessEqual[y, 1.05e+175], N[(y + x), $MachinePrecision], If[LessEqual[y, 1.3e+210], N[(y * x), $MachinePrecision], If[LessEqual[y, 2e+225], y, If[LessEqual[y, 6.5e+247], N[(y * x), $MachinePrecision], N[(y + x), $MachinePrecision]]]]]]
\begin{array}{l}
[x, y] = \mathsf{sort}([x, y])\\
\\
\begin{array}{l}
\mathbf{if}\;y \leq -4400000000:\\
\;\;\;\;y \cdot x\\

\mathbf{elif}\;y \leq 1.05 \cdot 10^{+175}:\\
\;\;\;\;y + x\\

\mathbf{elif}\;y \leq 1.3 \cdot 10^{+210}:\\
\;\;\;\;y \cdot x\\

\mathbf{elif}\;y \leq 2 \cdot 10^{+225}:\\
\;\;\;\;y\\

\mathbf{elif}\;y \leq 6.5 \cdot 10^{+247}:\\
\;\;\;\;y \cdot x\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if y < -4.4e9 or 1.05e175 < y < 1.29999999999999995e210 or 1.99999999999999986e225 < y < 6.50000000000000023e247

    1. Initial program 100.0%

      \[\left(x \cdot y + x\right) + y \]
    2. Taylor expanded in y around inf 99.5%

      \[\leadsto \color{blue}{y \cdot x} + y \]
    3. Taylor expanded in x around inf 52.5%

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

    if -4.4e9 < y < 1.05e175 or 6.50000000000000023e247 < y

    1. Initial program 100.0%

      \[\left(x \cdot y + x\right) + y \]
    2. Taylor expanded in y around 0 90.1%

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

    if 1.29999999999999995e210 < y < 1.99999999999999986e225

    1. Initial program 100.0%

      \[\left(x \cdot y + x\right) + y \]
    2. Taylor expanded in y around inf 100.0%

      \[\leadsto \color{blue}{y \cdot x} + y \]
    3. Taylor expanded in x around 0 100.0%

      \[\leadsto \color{blue}{y} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification80.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq -4400000000:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq 1.05 \cdot 10^{+175}:\\ \;\;\;\;y + x\\ \mathbf{elif}\;y \leq 1.3 \cdot 10^{+210}:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq 2 \cdot 10^{+225}:\\ \;\;\;\;y\\ \mathbf{elif}\;y \leq 6.5 \cdot 10^{+247}:\\ \;\;\;\;y \cdot x\\ \mathbf{else}:\\ \;\;\;\;y + x\\ \end{array} \]

Alternative 3: 97.6% accurate, 0.8× speedup?

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ \begin{array}{l} \mathbf{if}\;y \leq -4400000000:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq 4.6 \cdot 10^{-32}:\\ \;\;\;\;y + x\\ \mathbf{else}:\\ \;\;\;\;y + y \cdot x\\ \end{array} \end{array} \]
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (if (<= y -4400000000.0) (* y x) (if (<= y 4.6e-32) (+ y x) (+ y (* y x)))))
assert(x < y);
double code(double x, double y) {
	double tmp;
	if (y <= -4400000000.0) {
		tmp = y * x;
	} else if (y <= 4.6e-32) {
		tmp = y + x;
	} else {
		tmp = y + (y * x);
	}
	return tmp;
}
NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= (-4400000000.0d0)) then
        tmp = y * x
    else if (y <= 4.6d-32) then
        tmp = y + x
    else
        tmp = y + (y * x)
    end if
    code = tmp
end function
assert x < y;
public static double code(double x, double y) {
	double tmp;
	if (y <= -4400000000.0) {
		tmp = y * x;
	} else if (y <= 4.6e-32) {
		tmp = y + x;
	} else {
		tmp = y + (y * x);
	}
	return tmp;
}
[x, y] = sort([x, y])
def code(x, y):
	tmp = 0
	if y <= -4400000000.0:
		tmp = y * x
	elif y <= 4.6e-32:
		tmp = y + x
	else:
		tmp = y + (y * x)
	return tmp
x, y = sort([x, y])
function code(x, y)
	tmp = 0.0
	if (y <= -4400000000.0)
		tmp = Float64(y * x);
	elseif (y <= 4.6e-32)
		tmp = Float64(y + x);
	else
		tmp = Float64(y + Float64(y * x));
	end
	return tmp
end
x, y = num2cell(sort([x, y])){:}
function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= -4400000000.0)
		tmp = y * x;
	elseif (y <= 4.6e-32)
		tmp = y + x;
	else
		tmp = y + (y * x);
	end
	tmp_2 = tmp;
end
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := If[LessEqual[y, -4400000000.0], N[(y * x), $MachinePrecision], If[LessEqual[y, 4.6e-32], N[(y + x), $MachinePrecision], N[(y + N[(y * x), $MachinePrecision]), $MachinePrecision]]]
\begin{array}{l}
[x, y] = \mathsf{sort}([x, y])\\
\\
\begin{array}{l}
\mathbf{if}\;y \leq -4400000000:\\
\;\;\;\;y \cdot x\\

\mathbf{elif}\;y \leq 4.6 \cdot 10^{-32}:\\
\;\;\;\;y + x\\

\mathbf{else}:\\
\;\;\;\;y + y \cdot x\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if y < -4.4e9

    1. Initial program 100.0%

      \[\left(x \cdot y + x\right) + y \]
    2. Taylor expanded in y around inf 99.4%

      \[\leadsto \color{blue}{y \cdot x} + y \]
    3. Taylor expanded in x around inf 45.2%

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

    if -4.4e9 < y < 4.6000000000000001e-32

    1. Initial program 100.0%

      \[\left(x \cdot y + x\right) + y \]
    2. Taylor expanded in y around 0 100.0%

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

    if 4.6000000000000001e-32 < y

    1. Initial program 100.0%

      \[\left(x \cdot y + x\right) + y \]
    2. Taylor expanded in y around inf 93.5%

      \[\leadsto \color{blue}{y \cdot x} + y \]
  3. Recombined 3 regimes into one program.
  4. Final simplification85.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq -4400000000:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq 4.6 \cdot 10^{-32}:\\ \;\;\;\;y + x\\ \mathbf{else}:\\ \;\;\;\;y + y \cdot x\\ \end{array} \]

Alternative 4: 61.5% accurate, 1.0× speedup?

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ \begin{array}{l} \mathbf{if}\;x \leq -95000:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;x \leq 1:\\ \;\;\;\;y\\ \mathbf{else}:\\ \;\;\;\;y \cdot x\\ \end{array} \end{array} \]
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (if (<= x -95000.0) (* y x) (if (<= x 1.0) y (* y x))))
assert(x < y);
double code(double x, double y) {
	double tmp;
	if (x <= -95000.0) {
		tmp = y * x;
	} else if (x <= 1.0) {
		tmp = y;
	} else {
		tmp = y * x;
	}
	return tmp;
}
NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (x <= (-95000.0d0)) then
        tmp = y * x
    else if (x <= 1.0d0) then
        tmp = y
    else
        tmp = y * x
    end if
    code = tmp
end function
assert x < y;
public static double code(double x, double y) {
	double tmp;
	if (x <= -95000.0) {
		tmp = y * x;
	} else if (x <= 1.0) {
		tmp = y;
	} else {
		tmp = y * x;
	}
	return tmp;
}
[x, y] = sort([x, y])
def code(x, y):
	tmp = 0
	if x <= -95000.0:
		tmp = y * x
	elif x <= 1.0:
		tmp = y
	else:
		tmp = y * x
	return tmp
x, y = sort([x, y])
function code(x, y)
	tmp = 0.0
	if (x <= -95000.0)
		tmp = Float64(y * x);
	elseif (x <= 1.0)
		tmp = y;
	else
		tmp = Float64(y * x);
	end
	return tmp
end
x, y = num2cell(sort([x, y])){:}
function tmp_2 = code(x, y)
	tmp = 0.0;
	if (x <= -95000.0)
		tmp = y * x;
	elseif (x <= 1.0)
		tmp = y;
	else
		tmp = y * x;
	end
	tmp_2 = tmp;
end
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := If[LessEqual[x, -95000.0], N[(y * x), $MachinePrecision], If[LessEqual[x, 1.0], y, N[(y * x), $MachinePrecision]]]
\begin{array}{l}
[x, y] = \mathsf{sort}([x, y])\\
\\
\begin{array}{l}
\mathbf{if}\;x \leq -95000:\\
\;\;\;\;y \cdot x\\

\mathbf{elif}\;x \leq 1:\\
\;\;\;\;y\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if x < -95000 or 1 < x

    1. Initial program 100.0%

      \[\left(x \cdot y + x\right) + y \]
    2. Taylor expanded in y around inf 47.5%

      \[\leadsto \color{blue}{y \cdot x} + y \]
    3. Taylor expanded in x around inf 47.5%

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

    if -95000 < x < 1

    1. Initial program 100.0%

      \[\left(x \cdot y + x\right) + y \]
    2. Taylor expanded in y around inf 70.7%

      \[\leadsto \color{blue}{y \cdot x} + y \]
    3. Taylor expanded in x around 0 70.6%

      \[\leadsto \color{blue}{y} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification60.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -95000:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;x \leq 1:\\ \;\;\;\;y\\ \mathbf{else}:\\ \;\;\;\;y \cdot x\\ \end{array} \]

Alternative 5: 37.9% accurate, 7.0× speedup?

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ y \end{array} \]
NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y) :precision binary64 y)
assert(x < y);
double code(double x, double y) {
	return y;
}
NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = y
end function
assert x < y;
public static double code(double x, double y) {
	return y;
}
[x, y] = sort([x, y])
def code(x, y):
	return y
x, y = sort([x, y])
function code(x, y)
	return y
end
x, y = num2cell(sort([x, y])){:}
function tmp = code(x, y)
	tmp = y;
end
NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := y
\begin{array}{l}
[x, y] = \mathsf{sort}([x, y])\\
\\
y
\end{array}
Derivation
  1. Initial program 100.0%

    \[\left(x \cdot y + x\right) + y \]
  2. Taylor expanded in y around inf 60.5%

    \[\leadsto \color{blue}{y \cdot x} + y \]
  3. Taylor expanded in x around 0 40.9%

    \[\leadsto \color{blue}{y} \]
  4. Final simplification40.9%

    \[\leadsto y \]

Reproduce

?
herbie shell --seed 2023199 
(FPCore (x y)
  :name "Numeric.Log:$cexpm1 from log-domain-0.10.2.1, B"
  :precision binary64
  (+ (+ (* x y) x) y))