symmetry log of sum of exp

Percentage Accurate: 53.5% → 98.9%
Time: 15.1s
Alternatives: 16
Speedup: 1.5×

Specification

?
\[\begin{array}{l} \\ \log \left(e^{a} + e^{b}\right) \end{array} \]
(FPCore (a b) :precision binary64 (log (+ (exp a) (exp b))))
double code(double a, double b) {
	return log((exp(a) + exp(b)));
}
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log((exp(a) + exp(b)))
end function
public static double code(double a, double b) {
	return Math.log((Math.exp(a) + Math.exp(b)));
}
def code(a, b):
	return math.log((math.exp(a) + math.exp(b)))
function code(a, b)
	return log(Float64(exp(a) + exp(b)))
end
function tmp = code(a, b)
	tmp = log((exp(a) + exp(b)));
end
code[a_, b_] := N[Log[N[(N[Exp[a], $MachinePrecision] + N[Exp[b], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}

\\
\log \left(e^{a} + e^{b}\right)
\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 16 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: 53.5% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \log \left(e^{a} + e^{b}\right) \end{array} \]
(FPCore (a b) :precision binary64 (log (+ (exp a) (exp b))))
double code(double a, double b) {
	return log((exp(a) + exp(b)));
}
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log((exp(a) + exp(b)))
end function
public static double code(double a, double b) {
	return Math.log((Math.exp(a) + Math.exp(b)));
}
def code(a, b):
	return math.log((math.exp(a) + math.exp(b)))
function code(a, b)
	return log(Float64(exp(a) + exp(b)))
end
function tmp = code(a, b)
	tmp = log((exp(a) + exp(b)));
end
code[a_, b_] := N[Log[N[(N[Exp[a], $MachinePrecision] + N[Exp[b], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}

\\
\log \left(e^{a} + e^{b}\right)
\end{array}

Alternative 1: 98.9% accurate, 0.7× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + e^{b}\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 0.0) (/ b (+ (exp a) 1.0)) (log (+ (exp a) (exp b)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.0) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log((exp(a) + exp(b)));
	}
	return tmp;
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8) :: tmp
    if (exp(a) <= 0.0d0) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = log((exp(a) + exp(b)))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(a) <= 0.0) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log((Math.exp(a) + Math.exp(b)));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 0.0:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log((math.exp(a) + math.exp(b)))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 0.0)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = log(Float64(exp(a) + exp(b)));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (exp(a) <= 0.0)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = log((exp(a) + exp(b)));
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[N[Exp[a], $MachinePrecision], 0.0], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[Log[N[(N[Exp[a], $MachinePrecision] + N[Exp[b], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\log \left(e^{a} + e^{b}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 a) < 0.0

    1. Initial program 8.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 100.0%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def100.0%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 100.0%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]

    if 0.0 < (exp.f64 a)

    1. Initial program 72.3%

      \[\log \left(e^{a} + e^{b}\right) \]
  3. Recombined 2 regimes into one program.
  4. Final simplification78.3%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + e^{b}\right)\\ \end{array} \]

Alternative 2: 98.4% accurate, 0.6× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{a} + 1\right)\right)} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (+ (log1p (exp a)) (/ b (expm1 (log1p (+ (exp a) 1.0))))))
assert(a < b);
double code(double a, double b) {
	return log1p(exp(a)) + (b / expm1(log1p((exp(a) + 1.0))));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p(Math.exp(a)) + (b / Math.expm1(Math.log1p((Math.exp(a) + 1.0))));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p(math.exp(a)) + (b / math.expm1(math.log1p((math.exp(a) + 1.0))))
a, b = sort([a, b])
function code(a, b)
	return Float64(log1p(exp(a)) + Float64(b / expm1(log1p(Float64(exp(a) + 1.0)))))
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision] + N[(b / N[(Exp[N[Log[1 + N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(e^{a}\right) + \frac{b}{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{a} + 1\right)\right)}
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 76.0%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  3. Step-by-step derivation
    1. log1p-def76.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
  4. Simplified76.0%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  5. Step-by-step derivation
    1. expm1-log1p-u76.0%

      \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(1 + e^{a}\right)\right)}} \]
  6. Applied egg-rr76.0%

    \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(1 + e^{a}\right)\right)}} \]
  7. Final simplification76.0%

    \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{a} + 1\right)\right)} \]

Alternative 3: 97.7% accurate, 0.7× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{elif}\;e^{a} \leq 1:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{b}\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 0.0)
   (/ b (+ (exp a) 1.0))
   (if (<= (exp a) 1.0) (log1p (exp a)) (log1p (exp b)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.0) {
		tmp = b / (exp(a) + 1.0);
	} else if (exp(a) <= 1.0) {
		tmp = log1p(exp(a));
	} else {
		tmp = log1p(exp(b));
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(a) <= 0.0) {
		tmp = b / (Math.exp(a) + 1.0);
	} else if (Math.exp(a) <= 1.0) {
		tmp = Math.log1p(Math.exp(a));
	} else {
		tmp = Math.log1p(Math.exp(b));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 0.0:
		tmp = b / (math.exp(a) + 1.0)
	elif math.exp(a) <= 1.0:
		tmp = math.log1p(math.exp(a))
	else:
		tmp = math.log1p(math.exp(b))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 0.0)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	elseif (exp(a) <= 1.0)
		tmp = log1p(exp(a));
	else
		tmp = log1p(exp(b));
	end
	return tmp
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[N[Exp[a], $MachinePrecision], 0.0], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Exp[a], $MachinePrecision], 1.0], N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision], N[Log[1 + N[Exp[b], $MachinePrecision]], $MachinePrecision]]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{elif}\;e^{a} \leq 1:\\
\;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\

\mathbf{else}:\\
\;\;\;\;\mathsf{log1p}\left(e^{b}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (exp.f64 a) < 0.0

    1. Initial program 8.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 100.0%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def100.0%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 100.0%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]

    if 0.0 < (exp.f64 a) < 1

    1. Initial program 72.7%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 69.5%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right)} \]
    3. Step-by-step derivation
      1. log1p-def69.5%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
    4. Simplified69.5%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]

    if 1 < (exp.f64 a)

    1. Initial program 63.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in a around 0 38.7%

      \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
    3. Step-by-step derivation
      1. log1p-def38.7%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
    4. Simplified38.7%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification75.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{elif}\;e^{a} \leq 1:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{b}\right)\\ \end{array} \]

Alternative 4: 98.4% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (+ (log1p (exp a)) (/ b (+ (exp a) 1.0))))
assert(a < b);
double code(double a, double b) {
	return log1p(exp(a)) + (b / (exp(a) + 1.0));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p(Math.exp(a)) + (b / (Math.exp(a) + 1.0));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p(math.exp(a)) + (b / (math.exp(a) + 1.0))
a, b = sort([a, b])
function code(a, b)
	return Float64(log1p(exp(a)) + Float64(b / Float64(exp(a) + 1.0)))
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision] + N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1}
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 76.0%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  3. Step-by-step derivation
    1. log1p-def76.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
  4. Simplified76.0%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  5. Final simplification76.0%

    \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1} \]

Alternative 5: 97.7% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 0.0) (/ b (+ (exp a) 1.0)) (log1p (exp a))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.0) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log1p(exp(a));
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(a) <= 0.0) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log1p(Math.exp(a));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 0.0:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log1p(math.exp(a))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 0.0)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = log1p(exp(a));
	end
	return tmp
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[N[Exp[a], $MachinePrecision], 0.0], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 a) < 0.0

    1. Initial program 8.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 100.0%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def100.0%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 100.0%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]

    if 0.0 < (exp.f64 a)

    1. Initial program 72.3%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 68.8%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right)} \]
    3. Step-by-step derivation
      1. log1p-def68.9%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
    4. Simplified68.9%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification75.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \end{array} \]

Alternative 6: 98.2% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log1p (+ (exp a) (expm1 b))))
assert(a < b);
double code(double a, double b) {
	return log1p((exp(a) + expm1(b)));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p((Math.exp(a) + Math.expm1(b)));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p((math.exp(a) + math.expm1(b)))
a, b = sort([a, b])
function code(a, b)
	return log1p(Float64(exp(a) + expm1(b)))
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[1 + N[(N[Exp[a], $MachinePrecision] + N[(Exp[b] - 1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right)
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Step-by-step derivation
    1. add-sqr-sqrt57.0%

      \[\leadsto \log \color{blue}{\left(\sqrt{e^{a} + e^{b}} \cdot \sqrt{e^{a} + e^{b}}\right)} \]
    2. log-prod57.4%

      \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}}\right) + \log \left(\sqrt{e^{a} + e^{b}}\right)} \]
  3. Applied egg-rr57.4%

    \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}}\right) + \log \left(\sqrt{e^{a} + e^{b}}\right)} \]
  4. Step-by-step derivation
    1. log-prod57.0%

      \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}} \cdot \sqrt{e^{a} + e^{b}}\right)} \]
    2. rem-square-sqrt58.2%

      \[\leadsto \log \color{blue}{\left(e^{a} + e^{b}\right)} \]
    3. log1p-expm158.2%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\log \left(e^{a} + e^{b}\right)\right)\right)} \]
    4. expm1-def58.2%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{\log \left(e^{a} + e^{b}\right)} - 1}\right) \]
    5. rem-exp-log58.2%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{\left(e^{a} + e^{b}\right)} - 1\right) \]
    6. associate--l+58.3%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{a} + \left(e^{b} - 1\right)}\right) \]
    7. expm1-def78.0%

      \[\leadsto \mathsf{log1p}\left(e^{a} + \color{blue}{\mathsf{expm1}\left(b\right)}\right) \]
  5. Simplified78.0%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right)} \]
  6. Final simplification78.0%

    \[\leadsto \mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right) \]

Alternative 7: 98.3% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;a \leq -38:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + \left(e^{a} + \left(b + 0.5 \cdot \left(b \cdot b\right)\right)\right)\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= a -38.0)
   (/ b (+ (exp a) 1.0))
   (log (+ 1.0 (+ (exp a) (+ b (* 0.5 (* b b))))))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (a <= -38.0) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log((1.0 + (exp(a) + (b + (0.5 * (b * b))))));
	}
	return tmp;
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8) :: tmp
    if (a <= (-38.0d0)) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = log((1.0d0 + (exp(a) + (b + (0.5d0 * (b * b))))))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (a <= -38.0) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log((1.0 + (Math.exp(a) + (b + (0.5 * (b * b))))));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if a <= -38.0:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log((1.0 + (math.exp(a) + (b + (0.5 * (b * b))))))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (a <= -38.0)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = log(Float64(1.0 + Float64(exp(a) + Float64(b + Float64(0.5 * Float64(b * b))))));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (a <= -38.0)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = log((1.0 + (exp(a) + (b + (0.5 * (b * b))))));
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[a, -38.0], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[Log[N[(1.0 + N[(N[Exp[a], $MachinePrecision] + N[(b + N[(0.5 * N[(b * b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;a \leq -38:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\log \left(1 + \left(e^{a} + \left(b + 0.5 \cdot \left(b \cdot b\right)\right)\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if a < -38

    1. Initial program 8.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 100.0%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def100.0%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 100.0%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]

    if -38 < a

    1. Initial program 72.3%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 69.4%

      \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + \left(b + 0.5 \cdot {b}^{2}\right)\right)\right)} \]
    3. Step-by-step derivation
      1. unpow269.4%

        \[\leadsto \log \left(1 + \left(e^{a} + \left(b + 0.5 \cdot \color{blue}{\left(b \cdot b\right)}\right)\right)\right) \]
    4. Simplified69.4%

      \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + \left(b + 0.5 \cdot \left(b \cdot b\right)\right)\right)\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification76.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq -38:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + \left(e^{a} + \left(b + 0.5 \cdot \left(b \cdot b\right)\right)\right)\right)\\ \end{array} \]

Alternative 8: 97.4% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.005:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{b + 2} + \log \left(b + 2\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 0.005)
   (/ b (+ (exp a) 1.0))
   (+ (/ a (+ b 2.0)) (log (+ b 2.0)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.005) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = (a / (b + 2.0)) + log((b + 2.0));
	}
	return tmp;
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8) :: tmp
    if (exp(a) <= 0.005d0) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = (a / (b + 2.0d0)) + log((b + 2.0d0))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(a) <= 0.005) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = (a / (b + 2.0)) + Math.log((b + 2.0));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 0.005:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = (a / (b + 2.0)) + math.log((b + 2.0))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 0.005)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = Float64(Float64(a / Float64(b + 2.0)) + log(Float64(b + 2.0)));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (exp(a) <= 0.005)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = (a / (b + 2.0)) + log((b + 2.0));
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[N[Exp[a], $MachinePrecision], 0.005], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[(N[(a / N[(b + 2.0), $MachinePrecision]), $MachinePrecision] + N[Log[N[(b + 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0.005:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\frac{a}{b + 2} + \log \left(b + 2\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 a) < 0.0050000000000000001

    1. Initial program 9.6%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 99.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def100.0%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 98.3%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]

    if 0.0050000000000000001 < (exp.f64 a)

    1. Initial program 72.2%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 68.3%

      \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + b\right)\right)} \]
    3. Step-by-step derivation
      1. associate-+r+68.3%

        \[\leadsto \log \color{blue}{\left(\left(1 + e^{a}\right) + b\right)} \]
      2. +-commutative68.3%

        \[\leadsto \log \left(\color{blue}{\left(e^{a} + 1\right)} + b\right) \]
      3. associate-+l+68.3%

        \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
    4. Simplified68.3%

      \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
    5. Taylor expanded in a around 0 67.2%

      \[\leadsto \color{blue}{\frac{a}{2 + b} + \log \left(2 + b\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification74.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.005:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{b + 2} + \log \left(b + 2\right)\\ \end{array} \]

Alternative 9: 97.4% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.005:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + \left(a + 2\right)\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 0.005) (/ b (+ (exp a) 1.0)) (log (+ b (+ a 2.0)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.005) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log((b + (a + 2.0)));
	}
	return tmp;
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8) :: tmp
    if (exp(a) <= 0.005d0) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = log((b + (a + 2.0d0)))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(a) <= 0.005) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log((b + (a + 2.0)));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 0.005:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log((b + (a + 2.0)))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 0.005)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = log(Float64(b + Float64(a + 2.0)));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (exp(a) <= 0.005)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = log((b + (a + 2.0)));
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[N[Exp[a], $MachinePrecision], 0.005], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[Log[N[(b + N[(a + 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0.005:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\log \left(b + \left(a + 2\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 a) < 0.0050000000000000001

    1. Initial program 9.6%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 99.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def100.0%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 98.3%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]

    if 0.0050000000000000001 < (exp.f64 a)

    1. Initial program 72.2%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 68.3%

      \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + b\right)\right)} \]
    3. Step-by-step derivation
      1. associate-+r+68.3%

        \[\leadsto \log \color{blue}{\left(\left(1 + e^{a}\right) + b\right)} \]
      2. +-commutative68.3%

        \[\leadsto \log \left(\color{blue}{\left(e^{a} + 1\right)} + b\right) \]
      3. associate-+l+68.3%

        \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
    4. Simplified68.3%

      \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
    5. Taylor expanded in a around 0 67.1%

      \[\leadsto \log \color{blue}{\left(2 + \left(a + b\right)\right)} \]
    6. Step-by-step derivation
      1. associate-+r+67.1%

        \[\leadsto \log \color{blue}{\left(\left(2 + a\right) + b\right)} \]
      2. +-commutative67.1%

        \[\leadsto \log \left(\color{blue}{\left(a + 2\right)} + b\right) \]
    7. Simplified67.1%

      \[\leadsto \log \color{blue}{\left(\left(a + 2\right) + b\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification74.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.005:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + \left(a + 2\right)\right)\\ \end{array} \]

Alternative 10: 98.1% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;a \leq -38:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + \left(b + 1\right)\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= a -38.0) (/ b (+ (exp a) 1.0)) (log (+ (exp a) (+ b 1.0)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (a <= -38.0) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log((exp(a) + (b + 1.0)));
	}
	return tmp;
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8) :: tmp
    if (a <= (-38.0d0)) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = log((exp(a) + (b + 1.0d0)))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (a <= -38.0) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log((Math.exp(a) + (b + 1.0)));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if a <= -38.0:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log((math.exp(a) + (b + 1.0)))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (a <= -38.0)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = log(Float64(exp(a) + Float64(b + 1.0)));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (a <= -38.0)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = log((exp(a) + (b + 1.0)));
	end
	tmp_2 = tmp;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := If[LessEqual[a, -38.0], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[Log[N[(N[Exp[a], $MachinePrecision] + N[(b + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;a \leq -38:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\log \left(e^{a} + \left(b + 1\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if a < -38

    1. Initial program 8.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 100.0%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def100.0%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 100.0%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]

    if -38 < a

    1. Initial program 72.3%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 68.5%

      \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + b\right)\right)} \]
    3. Step-by-step derivation
      1. associate-+r+68.5%

        \[\leadsto \log \color{blue}{\left(\left(1 + e^{a}\right) + b\right)} \]
      2. +-commutative68.5%

        \[\leadsto \log \left(\color{blue}{\left(e^{a} + 1\right)} + b\right) \]
      3. associate-+l+68.5%

        \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
    4. Simplified68.5%

      \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification75.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq -38:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + \left(b + 1\right)\right)\\ \end{array} \]

Alternative 11: 49.2% accurate, 2.9× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ b \cdot 0.5 + \log 2 \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (+ (* b 0.5) (log 2.0)))
assert(a < b);
double code(double a, double b) {
	return (b * 0.5) + log(2.0);
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = (b * 0.5d0) + log(2.0d0)
end function
assert a < b;
public static double code(double a, double b) {
	return (b * 0.5) + Math.log(2.0);
}
[a, b] = sort([a, b])
def code(a, b):
	return (b * 0.5) + math.log(2.0)
a, b = sort([a, b])
function code(a, b)
	return Float64(Float64(b * 0.5) + log(2.0))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = (b * 0.5) + log(2.0);
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(N[(b * 0.5), $MachinePrecision] + N[Log[2.0], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
b \cdot 0.5 + \log 2
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 76.0%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  3. Step-by-step derivation
    1. log1p-def76.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
  4. Simplified76.0%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  5. Taylor expanded in a around 0 53.2%

    \[\leadsto \color{blue}{0.5 \cdot b + \log 2} \]
  6. Final simplification53.2%

    \[\leadsto b \cdot 0.5 + \log 2 \]

Alternative 12: 49.0% accurate, 2.9× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \log \left(b + 2\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log (+ b 2.0)))
assert(a < b);
double code(double a, double b) {
	return log((b + 2.0));
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log((b + 2.0d0))
end function
assert a < b;
public static double code(double a, double b) {
	return Math.log((b + 2.0));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log((b + 2.0))
a, b = sort([a, b])
function code(a, b)
	return log(Float64(b + 2.0))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = log((b + 2.0));
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[N[(b + 2.0), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\log \left(b + 2\right)
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.9%

    \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + b\right)\right)} \]
  3. Step-by-step derivation
    1. associate-+r+54.9%

      \[\leadsto \log \color{blue}{\left(\left(1 + e^{a}\right) + b\right)} \]
    2. +-commutative54.9%

      \[\leadsto \log \left(\color{blue}{\left(e^{a} + 1\right)} + b\right) \]
    3. associate-+l+54.9%

      \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  4. Simplified54.9%

    \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  5. Taylor expanded in a around 0 52.5%

    \[\leadsto \color{blue}{\log \left(2 + b\right)} \]
  6. Step-by-step derivation
    1. +-commutative52.5%

      \[\leadsto \log \color{blue}{\left(b + 2\right)} \]
  7. Simplified52.5%

    \[\leadsto \color{blue}{\log \left(b + 2\right)} \]
  8. Final simplification52.5%

    \[\leadsto \log \left(b + 2\right) \]

Alternative 13: 49.0% accurate, 2.9× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(b + 1\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log1p (+ b 1.0)))
assert(a < b);
double code(double a, double b) {
	return log1p((b + 1.0));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p((b + 1.0));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p((b + 1.0))
a, b = sort([a, b])
function code(a, b)
	return log1p(Float64(b + 1.0))
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[1 + N[(b + 1.0), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(b + 1\right)
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.9%

    \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + b\right)\right)} \]
  3. Step-by-step derivation
    1. associate-+r+54.9%

      \[\leadsto \log \color{blue}{\left(\left(1 + e^{a}\right) + b\right)} \]
    2. +-commutative54.9%

      \[\leadsto \log \left(\color{blue}{\left(e^{a} + 1\right)} + b\right) \]
    3. associate-+l+54.9%

      \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  4. Simplified54.9%

    \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  5. Taylor expanded in a around 0 52.5%

    \[\leadsto \color{blue}{\log \left(2 + b\right)} \]
  6. Step-by-step derivation
    1. +-commutative52.5%

      \[\leadsto \log \color{blue}{\left(b + 2\right)} \]
  7. Simplified52.5%

    \[\leadsto \color{blue}{\log \left(b + 2\right)} \]
  8. Step-by-step derivation
    1. +-commutative52.5%

      \[\leadsto \log \color{blue}{\left(2 + b\right)} \]
    2. log1p-expm1-u52.5%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\log \left(2 + b\right)\right)\right)} \]
    3. expm1-udef52.5%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{\log \left(2 + b\right)} - 1}\right) \]
    4. add-exp-log52.5%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{\left(2 + b\right)} - 1\right) \]
    5. +-commutative52.5%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{\left(b + 2\right)} - 1\right) \]
  9. Applied egg-rr52.5%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(\left(b + 2\right) - 1\right)} \]
  10. Step-by-step derivation
    1. associate--l+52.5%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{b + \left(2 - 1\right)}\right) \]
    2. metadata-eval52.5%

      \[\leadsto \mathsf{log1p}\left(b + \color{blue}{1}\right) \]
  11. Simplified52.5%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(b + 1\right)} \]
  12. Final simplification52.5%

    \[\leadsto \mathsf{log1p}\left(b + 1\right) \]

Alternative 14: 48.4% accurate, 3.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \log 2 \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log 2.0))
assert(a < b);
double code(double a, double b) {
	return log(2.0);
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log(2.0d0)
end function
assert a < b;
public static double code(double a, double b) {
	return Math.log(2.0);
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log(2.0)
a, b = sort([a, b])
function code(a, b)
	return log(2.0)
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = log(2.0);
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[2.0], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\log 2
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in a around 0 54.9%

    \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
  3. Step-by-step derivation
    1. log1p-def54.9%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  4. Simplified54.9%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  5. Taylor expanded in b around 0 52.9%

    \[\leadsto \color{blue}{\log 2} \]
  6. Final simplification52.9%

    \[\leadsto \log 2 \]

Alternative 15: 2.6% accurate, 101.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ a \cdot 0.5 \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (* a 0.5))
assert(a < b);
double code(double a, double b) {
	return a * 0.5;
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = a * 0.5d0
end function
assert a < b;
public static double code(double a, double b) {
	return a * 0.5;
}
[a, b] = sort([a, b])
def code(a, b):
	return a * 0.5
a, b = sort([a, b])
function code(a, b)
	return Float64(a * 0.5)
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = a * 0.5;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(a * 0.5), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
a \cdot 0.5
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.9%

    \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + b\right)\right)} \]
  3. Step-by-step derivation
    1. associate-+r+54.9%

      \[\leadsto \log \color{blue}{\left(\left(1 + e^{a}\right) + b\right)} \]
    2. +-commutative54.9%

      \[\leadsto \log \left(\color{blue}{\left(e^{a} + 1\right)} + b\right) \]
    3. associate-+l+54.9%

      \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  4. Simplified54.9%

    \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  5. Taylor expanded in a around 0 52.8%

    \[\leadsto \color{blue}{\frac{a}{2 + b} + \log \left(2 + b\right)} \]
  6. Taylor expanded in b around 0 53.2%

    \[\leadsto \color{blue}{0.5 \cdot a + \log 2} \]
  7. Taylor expanded in a around inf 7.0%

    \[\leadsto \color{blue}{0.5 \cdot a} \]
  8. Step-by-step derivation
    1. *-commutative7.0%

      \[\leadsto \color{blue}{a \cdot 0.5} \]
  9. Simplified7.0%

    \[\leadsto \color{blue}{a \cdot 0.5} \]
  10. Final simplification7.0%

    \[\leadsto a \cdot 0.5 \]

Alternative 16: 2.6% accurate, 101.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \frac{a}{b} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (/ a b))
assert(a < b);
double code(double a, double b) {
	return a / b;
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = a / b
end function
assert a < b;
public static double code(double a, double b) {
	return a / b;
}
[a, b] = sort([a, b])
def code(a, b):
	return a / b
a, b = sort([a, b])
function code(a, b)
	return Float64(a / b)
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = a / b;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(a / b), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\frac{a}{b}
\end{array}
Derivation
  1. Initial program 58.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.9%

    \[\leadsto \log \color{blue}{\left(1 + \left(e^{a} + b\right)\right)} \]
  3. Step-by-step derivation
    1. associate-+r+54.9%

      \[\leadsto \log \color{blue}{\left(\left(1 + e^{a}\right) + b\right)} \]
    2. +-commutative54.9%

      \[\leadsto \log \left(\color{blue}{\left(e^{a} + 1\right)} + b\right) \]
    3. associate-+l+54.9%

      \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  4. Simplified54.9%

    \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  5. Taylor expanded in a around 0 52.8%

    \[\leadsto \color{blue}{\frac{a}{2 + b} + \log \left(2 + b\right)} \]
  6. Taylor expanded in a around inf 3.9%

    \[\leadsto \color{blue}{\frac{a}{2 + b}} \]
  7. Taylor expanded in b around inf 3.6%

    \[\leadsto \color{blue}{\frac{a}{b}} \]
  8. Final simplification3.6%

    \[\leadsto \frac{a}{b} \]

Reproduce

?
herbie shell --seed 2023181 
(FPCore (a b)
  :name "symmetry log of sum of exp"
  :precision binary64
  (log (+ (exp a) (exp b))))