symmetry log of sum of exp

Percentage Accurate: 53.5% → 99.0%
Time: 19.3s
Alternatives: 11
Speedup: 2.8×

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 11 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: 99.0% accurate, 0.7× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-110}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + e^{b}\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 2e-110) (/ b (+ (exp a) 1.0)) (log (+ (exp a) (exp b)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 2e-110) {
		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) <= 2d-110) 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) <= 2e-110) {
		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) <= 2e-110:
		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) <= 2e-110)
		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) <= 2e-110)
		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], 2e-110], 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 2 \cdot 10^{-110}:\\
\;\;\;\;\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) < 2.0000000000000001e-110

    1. Initial program 6.8%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 96.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def96.6%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified96.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 96.6%

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

    if 2.0000000000000001e-110 < (exp.f64 a)

    1. Initial program 65.4%

      \[\log \left(e^{a} + e^{b}\right) \]
  3. Recombined 2 regimes into one program.
  4. Final simplification72.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-110}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + e^{b}\right)\\ \end{array} \]

Alternative 2: 97.8% accurate, 0.7× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-110}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{elif}\;e^{a} \leq 1:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{b}\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 2e-110)
   (/ 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) <= 2e-110) {
		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) <= 2e-110) {
		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) <= 2e-110:
		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) <= 2e-110)
		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], 2e-110], 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 2 \cdot 10^{-110}:\\
\;\;\;\;\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) < 2.0000000000000001e-110

    1. Initial program 6.8%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 96.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def96.6%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified96.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 96.6%

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

    if 2.0000000000000001e-110 < (exp.f64 a) < 1

    1. Initial program 64.9%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 62.5%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right)} \]
    3. Step-by-step derivation
      1. log1p-def63.0%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
    4. Simplified63.0%

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

    if 1 < (exp.f64 a)

    1. Initial program 81.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in a around 0 33.3%

      \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
    3. Step-by-step derivation
      1. log1p-def33.3%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
    4. Simplified33.3%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification69.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-110}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{elif}\;e^{a} \leq 1:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{b}\right)\\ \end{array} \]

Alternative 3: 98.3% 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 52.8%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Step-by-step derivation
    1. add-sqr-sqrt51.8%

      \[\leadsto \log \color{blue}{\left(\sqrt{e^{a} + e^{b}} \cdot \sqrt{e^{a} + e^{b}}\right)} \]
    2. log-prod52.2%

      \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}}\right) + \log \left(\sqrt{e^{a} + e^{b}}\right)} \]
  3. Applied egg-rr52.2%

    \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}}\right) + \log \left(\sqrt{e^{a} + e^{b}}\right)} \]
  4. Step-by-step derivation
    1. log-prod51.8%

      \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}} \cdot \sqrt{e^{a} + e^{b}}\right)} \]
    2. rem-square-sqrt52.8%

      \[\leadsto \log \color{blue}{\left(e^{a} + e^{b}\right)} \]
    3. log1p-expm152.8%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\log \left(e^{a} + e^{b}\right)\right)\right)} \]
    4. expm1-def52.8%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{\log \left(e^{a} + e^{b}\right)} - 1}\right) \]
    5. rem-exp-log52.8%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{\left(e^{a} + e^{b}\right)} - 1\right) \]
    6. associate--l+53.3%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{a} + \left(e^{b} - 1\right)}\right) \]
    7. expm1-def72.5%

      \[\leadsto \mathsf{log1p}\left(e^{a} + \color{blue}{\mathsf{expm1}\left(b\right)}\right) \]
  5. Simplified72.5%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right)} \]
  6. Final simplification72.5%

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

Alternative 4: 97.6% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-110}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(2 + \left(a + \left(b + 0.5 \cdot \left(a \cdot a\right)\right)\right)\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 2e-110)
   (/ b (+ (exp a) 1.0))
   (log (+ 2.0 (+ a (+ b (* 0.5 (* a a))))))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 2e-110) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log((2.0 + (a + (b + (0.5 * (a * a))))));
	}
	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) <= 2d-110) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = log((2.0d0 + (a + (b + (0.5d0 * (a * a))))))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(a) <= 2e-110) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log((2.0 + (a + (b + (0.5 * (a * a))))));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 2e-110:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log((2.0 + (a + (b + (0.5 * (a * a))))))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 2e-110)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = log(Float64(2.0 + Float64(a + Float64(b + Float64(0.5 * Float64(a * a))))));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (exp(a) <= 2e-110)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = log((2.0 + (a + (b + (0.5 * (a * a))))));
	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], 2e-110], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[Log[N[(2.0 + N[(a + N[(b + N[(0.5 * N[(a * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 2 \cdot 10^{-110}:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\log \left(2 + \left(a + \left(b + 0.5 \cdot \left(a \cdot a\right)\right)\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 a) < 2.0000000000000001e-110

    1. Initial program 6.8%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 96.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def96.6%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified96.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 96.6%

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

    if 2.0000000000000001e-110 < (exp.f64 a)

    1. Initial program 65.4%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 61.9%

      \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
    3. Taylor expanded in a around 0 61.3%

      \[\leadsto \log \color{blue}{\left(2 + \left(a + \left(b + 0.5 \cdot {a}^{2}\right)\right)\right)} \]
    4. Step-by-step derivation
      1. unpow261.3%

        \[\leadsto \log \left(2 + \left(a + \left(b + 0.5 \cdot \color{blue}{\left(a \cdot a\right)}\right)\right)\right) \]
    5. Simplified61.3%

      \[\leadsto \log \color{blue}{\left(2 + \left(a + \left(b + 0.5 \cdot \left(a \cdot a\right)\right)\right)\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification68.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-110}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(2 + \left(a + \left(b + 0.5 \cdot \left(a \cdot a\right)\right)\right)\right)\\ \end{array} \]

Alternative 5: 96.2% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(e^{a} + b\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log1p (+ (exp a) b)))
assert(a < b);
double code(double a, double b) {
	return log1p((exp(a) + b));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p((Math.exp(a) + b));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p((math.exp(a) + b))
a, b = sort([a, b])
function code(a, b)
	return log1p(Float64(exp(a) + 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] + b), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(e^{a} + b\right)
\end{array}
Derivation
  1. Initial program 52.8%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 50.0%

    \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
  3. Step-by-step derivation
    1. log1p-def69.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(b + e^{a}\right)} \]
    2. +-commutative69.6%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{a} + b}\right) \]
  4. Applied egg-rr69.6%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a} + b\right)} \]
  5. Final simplification69.6%

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

Alternative 6: 97.5% accurate, 2.7× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;a \leq -1.35:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + 2\right) + \frac{a}{b + 2}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= a -1.35) (/ b (+ (exp a) 1.0)) (+ (log (+ b 2.0)) (/ a (+ b 2.0)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (a <= -1.35) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log((b + 2.0)) + (a / (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 (a <= (-1.35d0)) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = log((b + 2.0d0)) + (a / (b + 2.0d0))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (a <= -1.35) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log((b + 2.0)) + (a / (b + 2.0));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if a <= -1.35:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log((b + 2.0)) + (a / (b + 2.0))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (a <= -1.35)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = Float64(log(Float64(b + 2.0)) + Float64(a / Float64(b + 2.0)));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (a <= -1.35)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = log((b + 2.0)) + (a / (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[a, -1.35], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[(N[Log[N[(b + 2.0), $MachinePrecision]], $MachinePrecision] + N[(a / N[(b + 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;a \leq -1.35:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if a < -1.3500000000000001

    1. Initial program 6.8%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 94.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def96.7%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified96.7%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 94.9%

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

    if -1.3500000000000001 < a

    1. Initial program 65.7%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 62.2%

      \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
    3. Taylor expanded in a around 0 60.9%

      \[\leadsto \color{blue}{\log \left(2 + b\right) + \frac{a}{2 + b}} \]
    4. Step-by-step derivation
      1. +-commutative60.9%

        \[\leadsto \log \color{blue}{\left(b + 2\right)} + \frac{a}{2 + b} \]
      2. +-commutative60.9%

        \[\leadsto \log \left(b + 2\right) + \frac{a}{\color{blue}{b + 2}} \]
    5. Simplified60.9%

      \[\leadsto \color{blue}{\log \left(b + 2\right) + \frac{a}{b + 2}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification68.3%

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq -1.35:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + 2\right) + \frac{a}{b + 2}\\ \end{array} \]

Alternative 7: 97.5% accurate, 2.8× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;a \leq -1:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + \left(a + 2\right)\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= a -1.0) (/ b (+ (exp a) 1.0)) (log (+ b (+ a 2.0)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (a <= -1.0) {
		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 (a <= (-1.0d0)) 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 (a <= -1.0) {
		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 a <= -1.0:
		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 (a <= -1.0)
		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 (a <= -1.0)
		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[a, -1.0], 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}\;a \leq -1:\\
\;\;\;\;\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 a < -1

    1. Initial program 6.8%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 94.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    3. Step-by-step derivation
      1. log1p-def96.7%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    4. Simplified96.7%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    5. Taylor expanded in b around inf 94.9%

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

    if -1 < a

    1. Initial program 65.7%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Taylor expanded in b around 0 62.2%

      \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
    3. Taylor expanded in a around 0 60.8%

      \[\leadsto \log \color{blue}{\left(2 + \left(a + b\right)\right)} \]
    4. Step-by-step derivation
      1. +-commutative60.8%

        \[\leadsto \log \color{blue}{\left(\left(a + b\right) + 2\right)} \]
      2. +-commutative60.8%

        \[\leadsto \log \left(\color{blue}{\left(b + a\right)} + 2\right) \]
      3. associate-+l+60.8%

        \[\leadsto \log \color{blue}{\left(b + \left(a + 2\right)\right)} \]
    5. Simplified60.8%

      \[\leadsto \log \color{blue}{\left(b + \left(a + 2\right)\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification68.3%

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

Alternative 8: 49.2% accurate, 2.9× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \log 2 + b \cdot 0.5 \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (+ (log 2.0) (* b 0.5)))
assert(a < b);
double code(double a, double b) {
	return log(2.0) + (b * 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 = log(2.0d0) + (b * 0.5d0)
end function
assert a < b;
public static double code(double a, double b) {
	return Math.log(2.0) + (b * 0.5);
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log(2.0) + (b * 0.5)
a, b = sort([a, b])
function code(a, b)
	return Float64(log(2.0) + Float64(b * 0.5))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = log(2.0) + (b * 0.5);
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(N[Log[2.0], $MachinePrecision] + N[(b * 0.5), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\log 2 + b \cdot 0.5
\end{array}
Derivation
  1. Initial program 52.8%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in a around 0 49.6%

    \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
  3. Step-by-step derivation
    1. log1p-def49.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  4. Simplified49.6%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  5. Taylor expanded in b around 0 47.8%

    \[\leadsto \color{blue}{\log 2 + 0.5 \cdot b} \]
  6. Step-by-step derivation
    1. *-commutative47.8%

      \[\leadsto \log 2 + \color{blue}{b \cdot 0.5} \]
  7. Simplified47.8%

    \[\leadsto \color{blue}{\log 2 + b \cdot 0.5} \]
  8. Final simplification47.8%

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

Alternative 9: 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 52.8%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 50.0%

    \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
  3. Taylor expanded in a around 0 47.1%

    \[\leadsto \color{blue}{\log \left(2 + b\right)} \]
  4. Step-by-step derivation
    1. +-commutative47.1%

      \[\leadsto \log \color{blue}{\left(b + 2\right)} \]
  5. Simplified47.1%

    \[\leadsto \color{blue}{\log \left(b + 2\right)} \]
  6. Final simplification47.1%

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

Alternative 10: 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 52.8%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 50.0%

    \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
  3. Taylor expanded in a around 0 47.1%

    \[\leadsto \color{blue}{\log \left(2 + b\right)} \]
  4. Step-by-step derivation
    1. +-commutative47.1%

      \[\leadsto \log \color{blue}{\left(b + 2\right)} \]
  5. Simplified47.1%

    \[\leadsto \color{blue}{\log \left(b + 2\right)} \]
  6. Step-by-step derivation
    1. log1p-expm1-u47.1%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\log \left(b + 2\right)\right)\right)} \]
    2. expm1-udef47.1%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{\log \left(b + 2\right)} - 1}\right) \]
    3. add-exp-log47.1%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{\left(b + 2\right)} - 1\right) \]
  7. Applied egg-rr47.1%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(\left(b + 2\right) - 1\right)} \]
  8. Step-by-step derivation
    1. associate--l+47.1%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{b + \left(2 - 1\right)}\right) \]
    2. metadata-eval47.1%

      \[\leadsto \mathsf{log1p}\left(b + \color{blue}{1}\right) \]
  9. Simplified47.1%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(b + 1\right)} \]
  10. Final simplification47.1%

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

Alternative 11: 48.5% 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 52.8%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in a around 0 49.6%

    \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
  3. Step-by-step derivation
    1. log1p-def49.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  4. Simplified49.6%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  5. Taylor expanded in b around 0 47.8%

    \[\leadsto \color{blue}{\log 2} \]
  6. Final simplification47.8%

    \[\leadsto \log 2 \]

Reproduce

?
herbie shell --seed 2023287 
(FPCore (a b)
  :name "symmetry log of sum of exp"
  :precision binary64
  (log (+ (exp a) (exp b))))