symmetry log of sum of exp

Percentage Accurate: 96.5% → 95.5%
Time: 19.8s
Alternatives: 9
Speedup: 1.0×

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 9 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: 96.5% 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: 95.5% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{b} \leq 1.000005:\\ \;\;\;\;\log \left(e^{a} + \left(b + 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp b) 1.000005) (log (+ (exp a) (+ b 1.0))) (/ b (+ (exp a) 1.0))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(b) <= 1.000005) {
		tmp = log((exp(a) + (b + 1.0)));
	} else {
		tmp = b / (exp(a) + 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 (exp(b) <= 1.000005d0) then
        tmp = log((exp(a) + (b + 1.0d0)))
    else
        tmp = b / (exp(a) + 1.0d0)
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(b) <= 1.000005) {
		tmp = Math.log((Math.exp(a) + (b + 1.0)));
	} else {
		tmp = b / (Math.exp(a) + 1.0);
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(b) <= 1.000005:
		tmp = math.log((math.exp(a) + (b + 1.0)))
	else:
		tmp = b / (math.exp(a) + 1.0)
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(b) <= 1.000005)
		tmp = log(Float64(exp(a) + Float64(b + 1.0)));
	else
		tmp = Float64(b / Float64(exp(a) + 1.0));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (exp(b) <= 1.000005)
		tmp = log((exp(a) + (b + 1.0)));
	else
		tmp = b / (exp(a) + 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[N[Exp[b], $MachinePrecision], 1.000005], N[Log[N[(N[Exp[a], $MachinePrecision] + N[(b + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{b} \leq 1.000005:\\
\;\;\;\;\log \left(e^{a} + \left(b + 1\right)\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 b) < 1.00000500000000003

    1. Initial program 97.1%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 69.9%

      \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
    4. Step-by-step derivation
      1. associate-+r+69.9%

        \[\leadsto \log \color{blue}{\left(\left(1 + b\right) + e^{a}\right)} \]
      2. +-commutative69.9%

        \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
    5. Simplified69.9%

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

    if 1.00000500000000003 < (exp.f64 b)

    1. Initial program 52.3%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 50.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-define50.6%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified50.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 50.1%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification69.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{b} \leq 1.000005:\\ \;\;\;\;\log \left(e^{a} + \left(b + 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 94.2% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{b} \leq 1.000005:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp b) 1.000005) (log1p (exp a)) (/ b (+ (exp a) 1.0))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(b) <= 1.000005) {
		tmp = log1p(exp(a));
	} else {
		tmp = b / (exp(a) + 1.0);
	}
	return tmp;
}
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(b) <= 1.000005) {
		tmp = Math.log1p(Math.exp(a));
	} else {
		tmp = b / (Math.exp(a) + 1.0);
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(b) <= 1.000005:
		tmp = math.log1p(math.exp(a))
	else:
		tmp = b / (math.exp(a) + 1.0)
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(b) <= 1.000005)
		tmp = log1p(exp(a));
	else
		tmp = Float64(b / Float64(exp(a) + 1.0));
	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[b], $MachinePrecision], 1.000005], N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{b} \leq 1.000005:\\
\;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 b) < 1.00000500000000003

    1. Initial program 97.1%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 70.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right)} \]
    4. Step-by-step derivation
      1. log1p-define70.6%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
    5. Simplified70.6%

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

    if 1.00000500000000003 < (exp.f64 b)

    1. Initial program 52.3%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 50.6%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-define50.6%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified50.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 50.1%

      \[\leadsto \color{blue}{\frac{b}{1 + e^{a}}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification70.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{b} \leq 1.000005:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 96.5% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \log \left(e^{a} + e^{b}\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log (+ (exp a) (exp b))))
assert(a < b);
double code(double a, double b) {
	return log((exp(a) + exp(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 = log((exp(a) + exp(b)))
end function
assert a < b;
public static double code(double a, double b) {
	return Math.log((Math.exp(a) + Math.exp(b)));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log((math.exp(a) + math.exp(b)))
a, b = sort([a, b])
function code(a, b)
	return log(Float64(exp(a) + exp(b)))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = log((exp(a) + exp(b)));
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[N[(N[Exp[a], $MachinePrecision] + N[Exp[b], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\log \left(e^{a} + e^{b}\right)
\end{array}
Derivation
  1. Initial program 96.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Add Preprocessing
  3. Final simplification96.2%

    \[\leadsto \log \left(e^{a} + e^{b}\right) \]
  4. Add Preprocessing

Alternative 4: 54.6% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + \left(a \cdot \left(0.5 - b \cdot 0.25\right) + b \cdot 0.5\right)\\ \end{array} \end{array} \]
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 2.0) (+ (* a (- 0.5 (* b 0.25))) (* b 0.5)))))
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(2.0) + ((a * (0.5 - (b * 0.25))) + (b * 0.5));
	}
	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(2.0d0) + ((a * (0.5d0 - (b * 0.25d0))) + (b * 0.5d0))
    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(2.0) + ((a * (0.5 - (b * 0.25))) + (b * 0.5));
	}
	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(2.0) + ((a * (0.5 - (b * 0.25))) + (b * 0.5))
	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 = Float64(log(2.0) + Float64(Float64(a * Float64(0.5 - Float64(b * 0.25))) + Float64(b * 0.5)));
	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(2.0) + ((a * (0.5 - (b * 0.25))) + (b * 0.5));
	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[(N[Log[2.0], $MachinePrecision] + N[(N[(a * N[(0.5 - N[(b * 0.25), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(b * 0.5), $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 2 + \left(a \cdot \left(0.5 - b \cdot 0.25\right) + b \cdot 0.5\right)\\


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

    1. Initial program 95.9%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 12.3%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-define12.3%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified12.3%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 12.3%

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

    if 0.0 < (exp.f64 a)

    1. Initial program 96.4%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 60.8%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-define60.8%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified60.8%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in a around 0 60.4%

      \[\leadsto \color{blue}{\log 2 + \left(0.5 \cdot b + a \cdot \left(0.5 - 0.25 \cdot b\right)\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification47.3%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + \left(a \cdot \left(0.5 - b \cdot 0.25\right) + b \cdot 0.5\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 54.3% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.8:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + b \cdot \left(0.5 + b \cdot 0.125\right)\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 0.8)
   (/ b (+ (exp a) 1.0))
   (+ (log 2.0) (* b (+ 0.5 (* b 0.125))))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.8) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log(2.0) + (b * (0.5 + (b * 0.125)));
	}
	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) <= 0.8d0) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = log(2.0d0) + (b * (0.5d0 + (b * 0.125d0)))
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(a) <= 0.8) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log(2.0) + (b * (0.5 + (b * 0.125)));
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 0.8:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log(2.0) + (b * (0.5 + (b * 0.125)))
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 0.8)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = Float64(log(2.0) + Float64(b * Float64(0.5 + Float64(b * 0.125))));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (exp(a) <= 0.8)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = log(2.0) + (b * (0.5 + (b * 0.125)));
	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], 0.8], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[(N[Log[2.0], $MachinePrecision] + N[(b * N[(0.5 + N[(b * 0.125), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0.8:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\log 2 + b \cdot \left(0.5 + b \cdot 0.125\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 a) < 0.80000000000000004

    1. Initial program 95.9%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 12.3%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-define12.3%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified12.3%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 12.3%

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

    if 0.80000000000000004 < (exp.f64 a)

    1. Initial program 96.4%

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

      \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
    4. Step-by-step derivation
      1. log1p-define93.9%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
    5. Simplified93.9%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
    6. Taylor expanded in b around 0 59.9%

      \[\leadsto \color{blue}{\log 2 + \left(0.125 \cdot {b}^{2} + 0.5 \cdot b\right)} \]
    7. Step-by-step derivation
      1. +-commutative59.9%

        \[\leadsto \log 2 + \color{blue}{\left(0.5 \cdot b + 0.125 \cdot {b}^{2}\right)} \]
      2. unpow259.9%

        \[\leadsto \log 2 + \left(0.5 \cdot b + 0.125 \cdot \color{blue}{\left(b \cdot b\right)}\right) \]
      3. associate-*r*59.9%

        \[\leadsto \log 2 + \left(0.5 \cdot b + \color{blue}{\left(0.125 \cdot b\right) \cdot b}\right) \]
      4. distribute-rgt-out59.9%

        \[\leadsto \log 2 + \color{blue}{b \cdot \left(0.5 + 0.125 \cdot b\right)} \]
      5. *-commutative59.9%

        \[\leadsto \log 2 + b \cdot \left(0.5 + \color{blue}{b \cdot 0.125}\right) \]
    8. Simplified59.9%

      \[\leadsto \color{blue}{\log 2 + b \cdot \left(0.5 + b \cdot 0.125\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification46.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.8:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + b \cdot \left(0.5 + b \cdot 0.125\right)\\ \end{array} \]
  5. Add Preprocessing

Alternative 6: 54.1% accurate, 1.4× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.8:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + b \cdot 0.5\\ \end{array} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b)
 :precision binary64
 (if (<= (exp a) 0.8) (/ b (+ (exp a) 1.0)) (+ (log 2.0) (* b 0.5))))
assert(a < b);
double code(double a, double b) {
	double tmp;
	if (exp(a) <= 0.8) {
		tmp = b / (exp(a) + 1.0);
	} else {
		tmp = log(2.0) + (b * 0.5);
	}
	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) <= 0.8d0) then
        tmp = b / (exp(a) + 1.0d0)
    else
        tmp = log(2.0d0) + (b * 0.5d0)
    end if
    code = tmp
end function
assert a < b;
public static double code(double a, double b) {
	double tmp;
	if (Math.exp(a) <= 0.8) {
		tmp = b / (Math.exp(a) + 1.0);
	} else {
		tmp = Math.log(2.0) + (b * 0.5);
	}
	return tmp;
}
[a, b] = sort([a, b])
def code(a, b):
	tmp = 0
	if math.exp(a) <= 0.8:
		tmp = b / (math.exp(a) + 1.0)
	else:
		tmp = math.log(2.0) + (b * 0.5)
	return tmp
a, b = sort([a, b])
function code(a, b)
	tmp = 0.0
	if (exp(a) <= 0.8)
		tmp = Float64(b / Float64(exp(a) + 1.0));
	else
		tmp = Float64(log(2.0) + Float64(b * 0.5));
	end
	return tmp
end
a, b = num2cell(sort([a, b])){:}
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (exp(a) <= 0.8)
		tmp = b / (exp(a) + 1.0);
	else
		tmp = log(2.0) + (b * 0.5);
	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], 0.8], N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[(N[Log[2.0], $MachinePrecision] + N[(b * 0.5), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\begin{array}{l}
\mathbf{if}\;e^{a} \leq 0.8:\\
\;\;\;\;\frac{b}{e^{a} + 1}\\

\mathbf{else}:\\
\;\;\;\;\log 2 + b \cdot 0.5\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (exp.f64 a) < 0.80000000000000004

    1. Initial program 95.9%

      \[\log \left(e^{a} + e^{b}\right) \]
    2. Add Preprocessing
    3. Taylor expanded in b around 0 12.3%

      \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    4. Step-by-step derivation
      1. log1p-define12.3%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
    5. Simplified12.3%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    6. Taylor expanded in b around inf 12.3%

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

    if 0.80000000000000004 < (exp.f64 a)

    1. Initial program 96.4%

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

      \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
    4. Step-by-step derivation
      1. log1p-define93.9%

        \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
    5. Simplified93.9%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
    6. Taylor expanded in b around 0 59.9%

      \[\leadsto \color{blue}{\log 2 + 0.5 \cdot b} \]
    7. Step-by-step derivation
      1. *-commutative59.9%

        \[\leadsto \log 2 + \color{blue}{b \cdot 0.5} \]
    8. Simplified59.9%

      \[\leadsto \color{blue}{\log 2 + b \cdot 0.5} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification46.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{a} \leq 0.8:\\ \;\;\;\;\frac{b}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\log 2 + b \cdot 0.5\\ \end{array} \]
  5. Add Preprocessing

Alternative 7: 50.2% 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 96.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Add Preprocessing
  3. Taylor expanded in a around 0 69.1%

    \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
  4. Step-by-step derivation
    1. log1p-define69.1%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  5. Simplified69.1%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  6. Taylor expanded in b around 0 44.6%

    \[\leadsto \color{blue}{\log 2 + 0.5 \cdot b} \]
  7. Step-by-step derivation
    1. *-commutative44.6%

      \[\leadsto \log 2 + \color{blue}{b \cdot 0.5} \]
  8. Simplified44.6%

    \[\leadsto \color{blue}{\log 2 + b \cdot 0.5} \]
  9. Final simplification44.6%

    \[\leadsto \log 2 + b \cdot 0.5 \]
  10. Add Preprocessing

Alternative 8: 50.0% 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 96.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Add Preprocessing
  3. Taylor expanded in b around 0 69.1%

    \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
  4. Step-by-step derivation
    1. associate-+r+69.1%

      \[\leadsto \log \color{blue}{\left(\left(1 + b\right) + e^{a}\right)} \]
    2. +-commutative69.1%

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

    \[\leadsto \log \color{blue}{\left(e^{a} + \left(1 + b\right)\right)} \]
  6. Taylor expanded in a around 0 43.7%

    \[\leadsto \color{blue}{\log \left(2 + b\right)} \]
  7. Final simplification43.7%

    \[\leadsto \log \left(b + 2\right) \]
  8. Add Preprocessing

Alternative 9: 49.4% 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 96.2%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Add Preprocessing
  3. Taylor expanded in a around 0 69.1%

    \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
  4. Step-by-step derivation
    1. log1p-define69.1%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  5. Simplified69.1%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  6. Taylor expanded in b around 0 44.6%

    \[\leadsto \color{blue}{\log 2} \]
  7. Final simplification44.6%

    \[\leadsto \log 2 \]
  8. Add Preprocessing

Reproduce

?
herbie shell --seed 2024046 
(FPCore (a b)
  :name "symmetry log of sum of exp"
  :precision binary64
  (log (+ (exp a) (exp b))))