symmetry log of sum of exp

Percentage Accurate: 89.6% → 97.3%
Time: 16.8s
Alternatives: 12
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 12 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: 89.6% 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: 97.3% accurate, 0.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} t_0 := \mathsf{log1p}\left(e^{a}\right)\\ t_1 := e^{a} + 1\\ t_2 := \frac{b}{t_1}\\ \mathbf{if}\;b \leq -4 \cdot 10^{-67}:\\ \;\;\;\;t_0 + t_2\\ \mathbf{elif}\;b \leq 3.5 \cdot 10^{-84}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\log t_1 + \left(t_2 + 0.5 \cdot \left({b}^{2} \cdot \left(\frac{1}{t_1} + \frac{-1}{{t_1}^{2}}\right)\right)\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (let* ((t_0 (log1p (exp a))) (t_1 (+ (exp a) 1.0)) (t_2 (/ b t_1)))
   (if (<= b -4e-67)
     (+ t_0 t_2)
     (if (<= b 3.5e-84)
       t_0
       (+
        (log t_1)
        (+
         t_2
         (* 0.5 (* (pow b 2.0) (+ (/ 1.0 t_1) (/ -1.0 (pow t_1 2.0)))))))))))
assert(a < b);
double code(double a, double b) {
	double t_0 = log1p(exp(a));
	double t_1 = exp(a) + 1.0;
	double t_2 = b / t_1;
	double tmp;
	if (b <= -4e-67) {
		tmp = t_0 + t_2;
	} else if (b <= 3.5e-84) {
		tmp = t_0;
	} else {
		tmp = log(t_1) + (t_2 + (0.5 * (pow(b, 2.0) * ((1.0 / t_1) + (-1.0 / pow(t_1, 2.0))))));
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double t_0 = Math.log1p(Math.exp(a));
	double t_1 = Math.exp(a) + 1.0;
	double t_2 = b / t_1;
	double tmp;
	if (b <= -4e-67) {
		tmp = t_0 + t_2;
	} else if (b <= 3.5e-84) {
		tmp = t_0;
	} else {
		tmp = Math.log(t_1) + (t_2 + (0.5 * (Math.pow(b, 2.0) * ((1.0 / t_1) + (-1.0 / Math.pow(t_1, 2.0))))));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	t_0 = math.log1p(math.exp(a))
	t_1 = math.exp(a) + 1.0
	t_2 = b / t_1
	tmp = 0
	if b <= -4e-67:
		tmp = t_0 + t_2
	elif b <= 3.5e-84:
		tmp = t_0
	else:
		tmp = math.log(t_1) + (t_2 + (0.5 * (math.pow(b, 2.0) * ((1.0 / t_1) + (-1.0 / math.pow(t_1, 2.0))))))
	return tmp
a, b = sort([a, b])
function code(a, b)
	t_0 = log1p(exp(a))
	t_1 = Float64(exp(a) + 1.0)
	t_2 = Float64(b / t_1)
	tmp = 0.0
	if (b <= -4e-67)
		tmp = Float64(t_0 + t_2);
	elseif (b <= 3.5e-84)
		tmp = t_0;
	else
		tmp = Float64(log(t_1) + Float64(t_2 + Float64(0.5 * Float64((b ^ 2.0) * Float64(Float64(1.0 / t_1) + Float64(-1.0 / (t_1 ^ 2.0)))))));
	end
	return tmp
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := Block[{t$95$0 = N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]}, Block[{t$95$2 = N[(b / t$95$1), $MachinePrecision]}, If[LessEqual[b, -4e-67], N[(t$95$0 + t$95$2), $MachinePrecision], If[LessEqual[b, 3.5e-84], t$95$0, N[(N[Log[t$95$1], $MachinePrecision] + N[(t$95$2 + N[(0.5 * N[(N[Power[b, 2.0], $MachinePrecision] * N[(N[(1.0 / t$95$1), $MachinePrecision] + N[(-1.0 / N[Power[t$95$1, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
t_0 := \mathsf{log1p}\left(e^{a}\right)\\
t_1 := e^{a} + 1\\
t_2 := \frac{b}{t_1}\\
\mathbf{if}\;b \leq -4 \cdot 10^{-67}:\\
\;\;\;\;t_0 + t_2\\

\mathbf{elif}\;b \leq 3.5 \cdot 10^{-84}:\\
\;\;\;\;t_0\\

\mathbf{else}:\\
\;\;\;\;\log t_1 + \left(t_2 + 0.5 \cdot \left({b}^{2} \cdot \left(\frac{1}{t_1} + \frac{-1}{{t_1}^{2}}\right)\right)\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if b < -3.99999999999999977e-67

    1. Initial program 85.2%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 29.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def30.1%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified30.1%

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

    if -3.99999999999999977e-67 < b < 3.5000000000000001e-84

    1. Initial program 100.0%

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

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

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
    5. Simplified100.0%

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

    if 3.5000000000000001e-84 < b

    1. Initial program 70.9%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 93.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \left(0.5 \cdot \left({b}^{2} \cdot \left(\frac{1}{1 + e^{a}} - \frac{1}{{\left(1 + e^{a}\right)}^{2}}\right)\right) + \frac{b}{1 + e^{a}}\right)} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification76.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq -4 \cdot 10^{-67}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1}\\ \mathbf{elif}\;b \leq 3.5 \cdot 10^{-84}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + 1\right) + \left(\frac{b}{e^{a} + 1} + 0.5 \cdot \left({b}^{2} \cdot \left(\frac{1}{e^{a} + 1} + \frac{-1}{{\left(e^{a} + 1\right)}^{2}}\right)\right)\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 97.1% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} t_0 := \mathsf{log1p}\left(e^{a}\right)\\ \mathbf{if}\;b \leq -4 \cdot 10^{-67} \lor \neg \left(b \leq 3.5 \cdot 10^{-84}\right):\\ \;\;\;\;t_0 + \frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (let* ((t_0 (log1p (exp a))))
   (if (or (<= b -4e-67) (not (<= b 3.5e-84)))
     (+ t_0 (/ b (+ (exp a) 1.0)))
     t_0)))
assert(a < b);
double code(double a, double b) {
	double t_0 = log1p(exp(a));
	double tmp;
	if ((b <= -4e-67) || !(b <= 3.5e-84)) {
		tmp = t_0 + (b / (exp(a) + 1.0));
	} else {
		tmp = t_0;
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double t_0 = Math.log1p(Math.exp(a));
	double tmp;
	if ((b <= -4e-67) || !(b <= 3.5e-84)) {
		tmp = t_0 + (b / (Math.exp(a) + 1.0));
	} else {
		tmp = t_0;
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	t_0 = math.log1p(math.exp(a))
	tmp = 0
	if (b <= -4e-67) or not (b <= 3.5e-84):
		tmp = t_0 + (b / (math.exp(a) + 1.0))
	else:
		tmp = t_0
	return tmp
a, b = sort([a, b])
function code(a, b)
	t_0 = log1p(exp(a))
	tmp = 0.0
	if ((b <= -4e-67) || !(b <= 3.5e-84))
		tmp = Float64(t_0 + Float64(b / Float64(exp(a) + 1.0)));
	else
		tmp = t_0;
	end
	return tmp
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := Block[{t$95$0 = N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision]}, If[Or[LessEqual[b, -4e-67], N[Not[LessEqual[b, 3.5e-84]], $MachinePrecision]], N[(t$95$0 + N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], t$95$0]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
t_0 := \mathsf{log1p}\left(e^{a}\right)\\
\mathbf{if}\;b \leq -4 \cdot 10^{-67} \lor \neg \left(b \leq 3.5 \cdot 10^{-84}\right):\\
\;\;\;\;t_0 + \frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;t_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if b < -3.99999999999999977e-67 or 3.5000000000000001e-84 < b

    1. Initial program 81.3%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 46.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def46.8%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified46.8%

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

    if -3.99999999999999977e-67 < b < 3.5000000000000001e-84

    1. Initial program 100.0%

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

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

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
    5. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification76.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq -4 \cdot 10^{-67} \lor \neg \left(b \leq 3.5 \cdot 10^{-84}\right):\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 93.8% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq -2.3 \cdot 10^{-66}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + e^{b}\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b -2.3e-66) (/ b (+ (exp a) 1.0)) (log (+ (exp a) (exp b)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= -2.3e-66) {
		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 (b <= (-2.3d-66)) 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 (b <= -2.3e-66) {
		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 b <= -2.3e-66:
		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 (b <= -2.3e-66)
		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 (b <= -2.3e-66)
		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[b, -2.3e-66], 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}\;b \leq -2.3 \cdot 10^{-66}:\\
\;\;\;\;\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 b < -2.29999999999999992e-66

    1. Initial program 85.2%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 29.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def30.1%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified30.1%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 10.4%

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

    if -2.29999999999999992e-66 < b

    1. Initial program 95.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
  3. Recombined 2 regimes into one program.
  4. Final simplification67.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq -2.3 \cdot 10^{-66}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + e^{b}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 4: 79.0% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-27}:\\ \;\;\;\;\frac{b}{2 + \left(a + 0.5 \cdot {a}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + 0.5 \cdot \left(b + a\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 2e-27)
   (/ b (+ 2.0 (+ a (* 0.5 (pow a 2.0)))))
   (+ (log 2.0) (* 0.5 (+ b a)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 2e-27) {
		tmp = b / (2.0 + (a + (0.5 * pow(a, 2.0))));
	} else {
		tmp = log(2.0) + (0.5 * (b + 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-27) then
        tmp = b / (2.0d0 + (a + (0.5d0 * (a ** 2.0d0))))
    else
        tmp = log(2.0d0) + (0.5d0 * (b + 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-27) {
		tmp = b / (2.0 + (a + (0.5 * Math.pow(a, 2.0))));
	} else {
		tmp = Math.log(2.0) + (0.5 * (b + a));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 2e-27:
		tmp = b / (2.0 + (a + (0.5 * math.pow(a, 2.0))))
	else:
		tmp = math.log(2.0) + (0.5 * (b + a))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 2e-27)
		tmp = Float64(b / Float64(2.0 + Float64(a + Float64(0.5 * (a ^ 2.0)))));
	else
		tmp = Float64(log(2.0) + Float64(0.5 * Float64(b + a)));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (exp(a) <= 2e-27)
		tmp = b / (2.0 + (a + (0.5 * (a ^ 2.0))));
	else
		tmp = log(2.0) + (0.5 * (b + 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-27], N[(b / N[(2.0 + N[(a + N[(0.5 * N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Log[2.0], $MachinePrecision] + N[(0.5 * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 2 \cdot 10^{-27}:\\
\;\;\;\;\frac{b}{2 + \left(a + 0.5 \cdot {a}^{2}\right)}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 a) < 2.0000000000000001e-27

    1. Initial program 77.9%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 29.1%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def29.4%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified29.4%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 29.1%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
    7. Taylor expanded in a around 0 65.2%

      \[\leadsto \frac{b}{\color{blue}{2 + \left(a + 0.5 \cdot {a}^{2}\right)}} \]

    if 2.0000000000000001e-27 < (exp.f64 a)

    1. Initial program 95.7%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 70.2%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def70.2%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified70.2%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in a around 0 70.2%

      \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\color{blue}{2}} \]
    7. Taylor expanded in a around 0 69.8%

      \[\leadsto \color{blue}{\log 2 + \left(0.5 \cdot a + 0.5 \cdot b\right)} \]
    8. Step-by-step derivation
      1. distribute-lft-out69.8%

        \[\leadsto \log 2 + \color{blue}{0.5 \cdot \left(a + b\right)} \]
    9. Simplified69.8%

      \[\leadsto \color{blue}{\log 2 + 0.5 \cdot \left(a + b\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification68.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-27}:\\ \;\;\;\;\frac{b}{2 + \left(a + 0.5 \cdot {a}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + 0.5 \cdot \left(b + a\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 71.6% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{a + 2}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + 0.5 \cdot \left(b + a\right)\\ \end{array} \end{array} \]
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 (+ a 2.0)) (+ (log 2.0) (* 0.5 (+ b a)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.0) {
		tmp = b / (a + 2.0);
	} else {
		tmp = log(2.0) + (0.5 * (b + 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) <= 0.0d0) then
        tmp = b / (a + 2.0d0)
    else
        tmp = log(2.0d0) + (0.5d0 * (b + a))
    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 / (a + 2.0);
	} else {
		tmp = Math.log(2.0) + (0.5 * (b + a));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 0.0:
		tmp = b / (a + 2.0)
	else:
		tmp = math.log(2.0) + (0.5 * (b + a))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 0.0)
		tmp = Float64(b / Float64(a + 2.0));
	else
		tmp = Float64(log(2.0) + Float64(0.5 * Float64(b + a)));
	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 / (a + 2.0);
	else
		tmp = log(2.0) + (0.5 * (b + 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], 0.0], N[(b / N[(a + 2.0), $MachinePrecision]), $MachinePrecision], N[(N[Log[2.0], $MachinePrecision] + N[(0.5 * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0:\\
\;\;\;\;\frac{b}{a + 2}\\

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


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

    1. Initial program 78.9%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 28.2%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def28.2%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified28.2%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 28.2%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
    7. Taylor expanded in a around 0 48.0%

      \[\leadsto \frac{b}{\color{blue}{2 + a}} \]
    8. Step-by-step derivation
      1. +-commutative48.0%

        \[\leadsto \frac{b}{\color{blue}{a + 2}} \]
    9. Simplified48.0%

      \[\leadsto \frac{b}{\color{blue}{a + 2}} \]

    if 0.0 < (exp.f64 a)

    1. Initial program 95.4%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 70.3%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def70.4%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified70.4%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in a around 0 70.0%

      \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\color{blue}{2}} \]
    7. Taylor expanded in a around 0 69.5%

      \[\leadsto \color{blue}{\log 2 + \left(0.5 \cdot a + 0.5 \cdot b\right)} \]
    8. Step-by-step derivation
      1. distribute-lft-out69.5%

        \[\leadsto \log 2 + \color{blue}{0.5 \cdot \left(a + b\right)} \]
    9. Simplified69.5%

      \[\leadsto \color{blue}{\log 2 + 0.5 \cdot \left(a + b\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification64.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{a + 2}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + 0.5 \cdot \left(b + a\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 6: 71.5% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-27}:\\ \;\;\;\;\frac{b}{a + 2}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + \left(a + 2\right)\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 2e-27) (/ b (+ a 2.0)) (log (+ b (+ a 2.0)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 2e-27) {
		tmp = b / (a + 2.0);
	} 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) <= 2d-27) then
        tmp = b / (a + 2.0d0)
    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) <= 2e-27) {
		tmp = b / (a + 2.0);
	} 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) <= 2e-27:
		tmp = b / (a + 2.0)
	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) <= 2e-27)
		tmp = Float64(b / Float64(a + 2.0));
	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) <= 2e-27)
		tmp = b / (a + 2.0);
	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], 2e-27], N[(b / N[(a + 2.0), $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 2 \cdot 10^{-27}:\\
\;\;\;\;\frac{b}{a + 2}\\

\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) < 2.0000000000000001e-27

    1. Initial program 77.9%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 29.1%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def29.4%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified29.4%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 29.1%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
    7. Taylor expanded in a around 0 47.1%

      \[\leadsto \frac{b}{\color{blue}{2 + a}} \]
    8. Step-by-step derivation
      1. +-commutative47.1%

        \[\leadsto \frac{b}{\color{blue}{a + 2}} \]
    9. Simplified47.1%

      \[\leadsto \frac{b}{\color{blue}{a + 2}} \]

    if 2.0000000000000001e-27 < (exp.f64 a)

    1. Initial program 95.7%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in a around 0 94.4%

      \[\leadsto \log \color{blue}{\left(1 + \left(a + e^{b}\right)\right)} \]
    4. Step-by-step derivation
      1. +-commutative94.4%

        \[\leadsto \log \left(1 + \color{blue}{\left(e^{b} + a\right)}\right) \]
    5. Simplified94.4%

      \[\leadsto \log \color{blue}{\left(1 + \left(e^{b} + a\right)\right)} \]
    6. Taylor expanded in b around 0 69.0%

      \[\leadsto \log \color{blue}{\left(2 + \left(a + b\right)\right)} \]
    7. Step-by-step derivation
      1. +-commutative69.0%

        \[\leadsto \log \color{blue}{\left(\left(a + b\right) + 2\right)} \]
      2. +-commutative69.0%

        \[\leadsto \log \left(\color{blue}{\left(b + a\right)} + 2\right) \]
      3. associate-+l+69.0%

        \[\leadsto \log \color{blue}{\left(b + \left(a + 2\right)\right)} \]
    8. Simplified69.0%

      \[\leadsto \log \color{blue}{\left(b + \left(a + 2\right)\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification64.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 2 \cdot 10^{-27}:\\ \;\;\;\;\frac{b}{a + 2}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + \left(a + 2\right)\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 7: 92.5% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq -2.7 \cdot 10^{-66}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + \left(b + 1\right)\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b -2.7e-66) (/ b (+ (exp a) 1.0)) (log (+ (exp a) (+ b 1.0)))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= -2.7e-66) {
		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 (b <= (-2.7d-66)) 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 (b <= -2.7e-66) {
		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 b <= -2.7e-66:
		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 (b <= -2.7e-66)
		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 (b <= -2.7e-66)
		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[b, -2.7e-66], 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}\;b \leq -2.7 \cdot 10^{-66}:\\
\;\;\;\;\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 b < -2.69999999999999996e-66

    1. Initial program 85.2%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 29.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def30.1%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified30.1%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 10.4%

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

    if -2.69999999999999996e-66 < b

    1. Initial program 95.0%

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

      \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
    4. Step-by-step derivation
      1. associate-+r+94.9%

        \[\leadsto \log \color{blue}{\left(\left(1 + b\right) + e^{a}\right)} \]
      2. +-commutative94.9%

        \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
    5. Simplified94.9%

      \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification67.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq -2.7 \cdot 10^{-66}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + \left(b + 1\right)\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 8: 91.4% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;b \leq -2.7 \cdot 10^{-66}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= b -2.7e-66) (/ b (+ (exp a) 1.0)) (log1p (exp a))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (b <= -2.7e-66) {
		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 (b <= -2.7e-66) {
		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 b <= -2.7e-66:
		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 (b <= -2.7e-66)
		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[b, -2.7e-66], 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}\;b \leq -2.7 \cdot 10^{-66}:\\
\;\;\;\;\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 b < -2.69999999999999996e-66

    1. Initial program 85.2%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 29.9%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def30.1%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified30.1%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 10.4%

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

    if -2.69999999999999996e-66 < b

    1. Initial program 95.0%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 94.3%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right)} \]
    4. Step-by-step derivation
      1. log1p-def94.3%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
    5. Simplified94.3%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification67.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq -2.7 \cdot 10^{-66}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 9: 71.0% accurate, 2.8× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;a \leq -1700000000:\\ \;\;\;\;\frac{b}{a + 2}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + 2\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= a -1700000000.0) (/ b (+ a 2.0)) (log (+ b 2.0))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (a <= -1700000000.0) {
		tmp = b / (a + 2.0);
	} else {
		tmp = 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 (a <= (-1700000000.0d0)) then
        tmp = b / (a + 2.0d0)
    else
        tmp = log((b + 2.0d0))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (a <= -1700000000.0) {
		tmp = b / (a + 2.0);
	} else {
		tmp = Math.log((b + 2.0));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if a <= -1700000000.0:
		tmp = b / (a + 2.0)
	else:
		tmp = math.log((b + 2.0))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (a <= -1700000000.0)
		tmp = Float64(b / Float64(a + 2.0));
	else
		tmp = 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 (a <= -1700000000.0)
		tmp = b / (a + 2.0);
	else
		tmp = 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[a, -1700000000.0], N[(b / N[(a + 2.0), $MachinePrecision]), $MachinePrecision], N[Log[N[(b + 2.0), $MachinePrecision]], $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;a \leq -1700000000:\\
\;\;\;\;\frac{b}{a + 2}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if a < -1.7e9

    1. Initial program 78.5%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 28.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def28.6%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified28.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 28.6%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
    7. Taylor expanded in a around 0 48.8%

      \[\leadsto \frac{b}{\color{blue}{2 + a}} \]
    8. Step-by-step derivation
      1. +-commutative48.8%

        \[\leadsto \frac{b}{\color{blue}{a + 2}} \]
    9. Simplified48.8%

      \[\leadsto \frac{b}{\color{blue}{a + 2}} \]

    if -1.7e9 < a

    1. Initial program 95.4%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in a around 0 93.4%

      \[\leadsto \log \color{blue}{\left(1 + \left(a + e^{b}\right)\right)} \]
    4. Step-by-step derivation
      1. +-commutative93.4%

        \[\leadsto \log \left(1 + \color{blue}{\left(e^{b} + a\right)}\right) \]
    5. Simplified93.4%

      \[\leadsto \log \color{blue}{\left(1 + \left(e^{b} + a\right)\right)} \]
    6. Taylor expanded in b around 0 68.3%

      \[\leadsto \log \color{blue}{\left(2 + \left(a + b\right)\right)} \]
    7. Step-by-step derivation
      1. +-commutative68.3%

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

        \[\leadsto \log \left(\color{blue}{\left(b + a\right)} + 2\right) \]
      3. associate-+l+68.3%

        \[\leadsto \log \color{blue}{\left(b + \left(a + 2\right)\right)} \]
    8. Simplified68.3%

      \[\leadsto \log \color{blue}{\left(b + \left(a + 2\right)\right)} \]
    9. Taylor expanded in a around 0 68.0%

      \[\leadsto \color{blue}{\log \left(2 + b\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification63.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq -1700000000:\\ \;\;\;\;\frac{b}{a + 2}\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + 2\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 10: 70.5% accurate, 2.9× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;a \leq -4200000000:\\ \;\;\;\;\frac{b}{a + 2}\\ \mathbf{else}:\\ \;\;\;\;\log 2\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= a -4200000000.0) (/ b (+ a 2.0)) (log 2.0)))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (a <= -4200000000.0) {
		tmp = b / (a + 2.0);
	} else {
		tmp = log(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 <= (-4200000000.0d0)) then
        tmp = b / (a + 2.0d0)
    else
        tmp = log(2.0d0)
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (a <= -4200000000.0) {
		tmp = b / (a + 2.0);
	} else {
		tmp = Math.log(2.0);
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if a <= -4200000000.0:
		tmp = b / (a + 2.0)
	else:
		tmp = math.log(2.0)
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (a <= -4200000000.0)
		tmp = Float64(b / Float64(a + 2.0));
	else
		tmp = log(2.0);
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (a <= -4200000000.0)
		tmp = b / (a + 2.0);
	else
		tmp = log(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, -4200000000.0], N[(b / N[(a + 2.0), $MachinePrecision]), $MachinePrecision], N[Log[2.0], $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;a \leq -4200000000:\\
\;\;\;\;\frac{b}{a + 2}\\

\mathbf{else}:\\
\;\;\;\;\log 2\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if a < -4.2e9

    1. Initial program 78.5%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 28.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-def28.6%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified28.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 28.6%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
    7. Taylor expanded in a around 0 48.8%

      \[\leadsto \frac{b}{\color{blue}{2 + a}} \]
    8. Step-by-step derivation
      1. +-commutative48.8%

        \[\leadsto \frac{b}{\color{blue}{a + 2}} \]
    9. Simplified48.8%

      \[\leadsto \frac{b}{\color{blue}{a + 2}} \]

    if -4.2e9 < a

    1. Initial program 95.4%

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

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

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

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
    6. Taylor expanded in a around 0 68.2%

      \[\leadsto \color{blue}{\log 2} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification64.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq -4200000000:\\ \;\;\;\;\frac{b}{a + 2}\\ \mathbf{else}:\\ \;\;\;\;\log 2\\ \end{array} \]
  5. Add Preprocessing

Alternative 11: 26.3% accurate, 60.6× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \frac{b}{a + 2} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (/ b (+ a 2.0)))
assert(a < b);
double code(double a, double b) {
	return b / (a + 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 / (a + 2.0d0)
end function
assert a < b;
public static double code(double a, double b) {
	return b / (a + 2.0);
}
[a, b] = sort([a, b])
def code(a, b):
	return b / (a + 2.0)
a, b = sort([a, b])
function code(a, b)
	return Float64(b / Float64(a + 2.0))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = b / (a + 2.0);
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(b / N[(a + 2.0), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\frac{b}{a + 2}
\end{array}
Derivation
  1. Initial program 91.8%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Add Preprocessing
  3. Taylor expanded in b around 0 61.2%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  4. Step-by-step derivation
    1. log1p-def61.3%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
  5. Simplified61.3%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  6. Taylor expanded in b around inf 9.2%

    \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
  7. Taylor expanded in a around 0 13.1%

    \[\leadsto \frac{b}{\color{blue}{2 + a}} \]
  8. Step-by-step derivation
    1. +-commutative13.1%

      \[\leadsto \frac{b}{\color{blue}{a + 2}} \]
  9. Simplified13.1%

    \[\leadsto \frac{b}{\color{blue}{a + 2}} \]
  10. Final simplification13.1%

    \[\leadsto \frac{b}{a + 2} \]
  11. Add Preprocessing

Alternative 12: 7.1% accurate, 101.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \frac{b}{2} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (/ b 2.0))
assert(a < b);
double code(double a, double b) {
	return 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 = b / 2.0d0
end function
assert a < b;
public static double code(double a, double b) {
	return b / 2.0;
}
[a, b] = sort([a, b])
def code(a, b):
	return b / 2.0
a, b = sort([a, b])
function code(a, b)
	return Float64(b / 2.0)
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = b / 2.0;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(b / 2.0), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\frac{b}{2}
\end{array}
Derivation
  1. Initial program 91.8%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Add Preprocessing
  3. Taylor expanded in b around 0 61.2%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  4. Step-by-step derivation
    1. log1p-def61.3%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
  5. Simplified61.3%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  6. Taylor expanded in b around inf 9.2%

    \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
  7. Taylor expanded in a around 0 4.8%

    \[\leadsto \frac{b}{\color{blue}{2}} \]
  8. Final simplification4.8%

    \[\leadsto \frac{b}{2} \]
  9. Add Preprocessing

Reproduce

?
herbie shell --seed 2024023 
(FPCore (a b)
  :name "symmetry log of sum of exp"
  :precision binary64
  (log (+ (exp a) (exp b))))