Quotient of sum of exps

Average Accuracy: 99.0% → 99.0%

Time: 9.0s

Precision: binary64

Cost: 19784

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;b \leq -380000000:\\ \;\;\;\;\frac{1}{1 + e^{b}}\\ \mathbf{elif}\;b \leq -9:\\ \;\;\;\;\frac{e^{a}}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{a}}{e^{b} + e^{a}}\\ \end{array} \]

FPCore

(FPCore (a b) :precision binary64 (/ (exp a) (+ (exp a) (exp b))))

↓

(FPCore (a b)
 :precision binary64
 (if (<= b -380000000.0)
   (/ 1.0 (+ 1.0 (exp b)))
   (if (<= b -9.0) (/ (exp a) 2.0) (/ (exp a) (+ (exp b) (exp a))))))

C

double code(double a, double b) {
	return exp(a) / (exp(a) + exp(b));
}

↓

double code(double a, double b) {
	double tmp;
	if (b <= -380000000.0) {
		tmp = 1.0 / (1.0 + exp(b));
	} else if (b <= -9.0) {
		tmp = exp(a) / 2.0;
	} else {
		tmp = exp(a) / (exp(b) + exp(a));
	}
	return tmp;
}

Fortran

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
    real(8) :: tmp
    if (b <= (-380000000.0d0)) then
        tmp = 1.0d0 / (1.0d0 + exp(b))
    else if (b <= (-9.0d0)) then
        tmp = exp(a) / 2.0d0
    else
        tmp = exp(a) / (exp(b) + exp(a))
    end if
    code = tmp
end function

Java

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) {
	double tmp;
	if (b <= -380000000.0) {
		tmp = 1.0 / (1.0 + Math.exp(b));
	} else if (b <= -9.0) {
		tmp = Math.exp(a) / 2.0;
	} else {
		tmp = Math.exp(a) / (Math.exp(b) + Math.exp(a));
	}
	return tmp;
}

Python

def code(a, b):
	return math.exp(a) / (math.exp(a) + math.exp(b))

↓

def code(a, b):
	tmp = 0
	if b <= -380000000.0:
		tmp = 1.0 / (1.0 + math.exp(b))
	elif b <= -9.0:
		tmp = math.exp(a) / 2.0
	else:
		tmp = math.exp(a) / (math.exp(b) + math.exp(a))
	return tmp

Julia

function code(a, b)
	return Float64(exp(a) / Float64(exp(a) + exp(b)))
end

↓

function code(a, b)
	tmp = 0.0
	if (b <= -380000000.0)
		tmp = Float64(1.0 / Float64(1.0 + exp(b)));
	elseif (b <= -9.0)
		tmp = Float64(exp(a) / 2.0);
	else
		tmp = Float64(exp(a) / Float64(exp(b) + exp(a)));
	end
	return tmp
end

MATLAB

function tmp = code(a, b)
	tmp = exp(a) / (exp(a) + exp(b));
end

↓

function tmp_2 = code(a, b)
	tmp = 0.0;
	if (b <= -380000000.0)
		tmp = 1.0 / (1.0 + exp(b));
	elseif (b <= -9.0)
		tmp = exp(a) / 2.0;
	else
		tmp = exp(a) / (exp(b) + exp(a));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_] := N[(N[Exp[a], $MachinePrecision] / N[(N[Exp[a], $MachinePrecision] + N[Exp[b], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

code[a_, b_] := If[LessEqual[b, -380000000.0], N[(1.0 / N[(1.0 + N[Exp[b], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[b, -9.0], N[(N[Exp[a], $MachinePrecision] / 2.0), $MachinePrecision], N[(N[Exp[a], $MachinePrecision] / N[(N[Exp[b], $MachinePrecision] + N[Exp[a], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Target

Original	99.0%
Target	100.0%
Herbie	99.0%

\[\frac{1}{1 + e^{b - a}} \]

Derivation

1. Split input into 3 regimes

2. if b < -3.8e8

   1. Initial program 98.4%

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

   2. Taylor expanded in a around 0 100.0%

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

   if -3.8e8 < b < -9

   1. Initial program 80.0%

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

   2. Taylor expanded in b around 0 38.8%

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

   3. Taylor expanded in a around 0 38.8%

      \[\leadsto \frac{e^{a}}{\color{blue}{2}} \]

   if -9 < b

   1. Initial program 99.4%

      \[\frac{e^{a}}{e^{a} + e^{b}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 99.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq -380000000:\\ \;\;\;\;\frac{1}{1 + e^{b}}\\ \mathbf{elif}\;b \leq -9:\\ \;\;\;\;\frac{e^{a}}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{a}}{e^{b} + e^{a}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	98.7%
Cost	19652

\[\begin{array}{l} \mathbf{if}\;e^{a} \leq 0.98:\\ \;\;\;\;\frac{e^{a}}{1 + e^{a}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1 + e^{b}}\\ \end{array} \]

Alternative 2
Accuracy	98.7%
Cost	13252

\[\begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;\frac{e^{a}}{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1 + e^{b}}\\ \end{array} \]

Alternative 3
Accuracy	81.6%
Cost	6724

\[\begin{array}{l} \mathbf{if}\;b \leq 64000000:\\ \;\;\;\;\frac{e^{a}}{2}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{2}{b \cdot b}\right) + -1\\ \end{array} \]

Alternative 4
Accuracy	64.5%
Cost	708

\[\begin{array}{l} \mathbf{if}\;b \leq -3300:\\ \;\;\;\;0.5\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{1}{b + 2}\right) + -1\\ \end{array} \]

Alternative 5
Accuracy	52.4%
Cost	452

\[\begin{array}{l} \mathbf{if}\;b \leq 2:\\ \;\;\;\;0.5\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{b \cdot b}\\ \end{array} \]

Alternative 6
Accuracy	39.1%
Cost	64

\[0.5 \]

Reproduce

herbie shell --seed 2023131 
(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))))