?

Average Accuracy: 99.1% → 99.1%
Time: 5.2s
Precision: binary64
Cost: 19520

?

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

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original99.1%
Target100.0%
Herbie99.1%
\[\frac{1}{1 + e^{b - a}} \]

Derivation?

  1. Initial program 99.2%

    \[\frac{e^{a}}{e^{a} + e^{b}} \]
  2. Final simplification99.2%

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

Alternatives

Alternative 1
Accuracy98.6%
Cost26185
\[\begin{array}{l} \mathbf{if}\;e^{b} \leq 0.5 \lor \neg \left(e^{b} \leq 2\right):\\ \;\;\;\;\frac{1}{e^{b} + 1}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{a}}{e^{a} + 1}\\ \end{array} \]
Alternative 2
Accuracy98.6%
Cost26184
\[\begin{array}{l} \mathbf{if}\;e^{b} \leq 0.5:\\ \;\;\;\;\frac{\mathsf{expm1}\left(b\right)}{\mathsf{expm1}\left(b \cdot 2\right)}\\ \mathbf{elif}\;e^{b} \leq 2:\\ \;\;\;\;\frac{e^{a}}{e^{a} + 1}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{e^{b} + 1}\\ \end{array} \]
Alternative 3
Accuracy98.6%
Cost13252
\[\begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{e^{a}}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{e^{b} + 1}\\ \end{array} \]
Alternative 4
Accuracy98.3%
Cost13248
\[\frac{e^{a}}{1 + \left(a + e^{b}\right)} \]
Alternative 5
Accuracy62.5%
Cost6920
\[\begin{array}{l} \mathbf{if}\;a \leq -9 \cdot 10^{-203}:\\ \;\;\;\;\frac{e^{a}}{2}\\ \mathbf{elif}\;a \leq -3.1 \cdot 10^{-261}:\\ \;\;\;\;-0.020833333333333332 \cdot {a}^{3}\\ \mathbf{else}:\\ \;\;\;\;0.5 + a \cdot 0.25\\ \end{array} \]
Alternative 6
Accuracy64.4%
Cost6592
\[\frac{e^{a}}{2} \]
Alternative 7
Accuracy38.8%
Cost320
\[0.5 + a \cdot 0.25 \]
Alternative 8
Accuracy38.6%
Cost64
\[0.5 \]

Error

Reproduce?

herbie shell --seed 2023157 -o generate:proofs
(FPCore (a b)
  :name "Quotient of sum of exps"
  :precision binary64

  :herbie-target
  (/ 1.0 (+ 1.0 (exp (- b a))))

  (/ (exp a) (+ (exp a) (exp b))))