Falkner and Boettcher, Appendix A

Average Accuracy: 90.3% → 97.5%

Time: 13.5s

Precision: binary64

Cost: 14660

• 22.9% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]

↓

\[\begin{array}{l} t_0 := a \cdot {k}^{m}\\ t_1 := \frac{t_0}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{if}\;t_1 \leq 2 \cdot 10^{+214}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))

↓

(FPCore (a k m)
 :precision binary64
 (let* ((t_0 (* a (pow k m))) (t_1 (/ t_0 (+ (+ 1.0 (* k 10.0)) (* k k)))))
   (if (<= t_1 2e+214) t_1 t_0)))

C

double code(double a, double k, double m) {
	return (a * pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k));
}

↓

double code(double a, double k, double m) {
	double t_0 = a * pow(k, m);
	double t_1 = t_0 / ((1.0 + (k * 10.0)) + (k * k));
	double tmp;
	if (t_1 <= 2e+214) {
		tmp = t_1;
	} else {
		tmp = t_0;
	}
	return tmp;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    code = (a * (k ** m)) / ((1.0d0 + (10.0d0 * k)) + (k * k))
end function

↓

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = a * (k ** m)
    t_1 = t_0 / ((1.0d0 + (k * 10.0d0)) + (k * k))
    if (t_1 <= 2d+214) then
        tmp = t_1
    else
        tmp = t_0
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	return (a * Math.pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k));
}

↓

public static double code(double a, double k, double m) {
	double t_0 = a * Math.pow(k, m);
	double t_1 = t_0 / ((1.0 + (k * 10.0)) + (k * k));
	double tmp;
	if (t_1 <= 2e+214) {
		tmp = t_1;
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(a, k, m):
	return (a * math.pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k))

↓

def code(a, k, m):
	t_0 = a * math.pow(k, m)
	t_1 = t_0 / ((1.0 + (k * 10.0)) + (k * k))
	tmp = 0
	if t_1 <= 2e+214:
		tmp = t_1
	else:
		tmp = t_0
	return tmp

Julia

function code(a, k, m)
	return Float64(Float64(a * (k ^ m)) / Float64(Float64(1.0 + Float64(10.0 * k)) + Float64(k * k)))
end

↓

function code(a, k, m)
	t_0 = Float64(a * (k ^ m))
	t_1 = Float64(t_0 / Float64(Float64(1.0 + Float64(k * 10.0)) + Float64(k * k)))
	tmp = 0.0
	if (t_1 <= 2e+214)
		tmp = t_1;
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp = code(a, k, m)
	tmp = (a * (k ^ m)) / ((1.0 + (10.0 * k)) + (k * k));
end

↓

function tmp_2 = code(a, k, m)
	t_0 = a * (k ^ m);
	t_1 = t_0 / ((1.0 + (k * 10.0)) + (k * k));
	tmp = 0.0;
	if (t_1 <= 2e+214)
		tmp = t_1;
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := N[(N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision] / N[(N[(1.0 + N[(10.0 * k), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

code[a_, k_, m_] := Block[{t$95$0 = N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(N[(1.0 + N[(k * 10.0), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 2e+214], t$95$1, t$95$0]]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k))) < 1.9999999999999999e214

   1. Initial program 96.4%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]

   if 1.9999999999999999e214 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k)))

   1. Initial program 55.3%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]

   2. Simplified 55.3%

      \[\leadsto \color{blue}{a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, k + 10, 1\right)}} \]
      Proof

      [Start] 55.3	\[ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
      associate-*r/ [<=] 55.3	\[ \color{blue}{a \cdot \frac{{k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}} \]
      associate-+l+ [=>] 55.3	\[ a \cdot \frac{{k}^{m}}{\color{blue}{1 + \left(10 \cdot k + k \cdot k\right)}} \]
      +-commutative [=>] 55.3	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\left(10 \cdot k + k \cdot k\right) + 1}} \]
      distribute-rgt-out [=>] 55.3	\[ a \cdot \frac{{k}^{m}}{\color{blue}{k \cdot \left(10 + k\right)} + 1} \]
      fma-def [=>] 55.3	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\mathsf{fma}\left(k, 10 + k, 1\right)}} \]
      +-commutative [=>] 55.3	\[ a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, \color{blue}{k + 10}, 1\right)} \]

   3. Taylor expanded in k around 0 66.0%

      \[\leadsto a \cdot \color{blue}{e^{\log k \cdot m}} \]

   4. Simplified 100.0%

      \[\leadsto a \cdot \color{blue}{{k}^{m}} \]
      Proof

      [Start] 66.0	\[ a \cdot e^{\log k \cdot m} \]
      exp-to-pow [=>] 100.0	\[ a \cdot \color{blue}{{k}^{m}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 97.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 2 \cdot 10^{+214}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	96.4%
Cost	7048

\[\begin{array}{l} \mathbf{if}\;m \leq -4.5 \cdot 10^{-17}:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{elif}\;m \leq 1.9 \cdot 10^{-34}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{{k}^{\left(2 - m\right)}}\\ \end{array} \]

Alternative 2
Accuracy	96.4%
Cost	7048

\[\begin{array}{l} \mathbf{if}\;m \leq -4.5 \cdot 10^{-17}:\\ \;\;\;\;\frac{1}{\frac{{k}^{\left(-m\right)}}{a}}\\ \mathbf{elif}\;m \leq 1.9 \cdot 10^{-34}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{{k}^{\left(2 - m\right)}}\\ \end{array} \]

Alternative 3
Accuracy	96.9%
Cost	6921

\[\begin{array}{l} \mathbf{if}\;m \leq -4.5 \cdot 10^{-17} \lor \neg \left(m \leq 0.0038\right):\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \]

Alternative 4
Accuracy	66.5%
Cost	1096

\[\begin{array}{l} \mathbf{if}\;m \leq -82000:\\ \;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\ \mathbf{elif}\;m \leq 1.82:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(\left(1 + k \cdot -10\right) - \left(k \cdot k\right) \cdot -99\right)\\ \end{array} \]

Alternative 5
Accuracy	55.5%
Cost	840

\[\begin{array}{l} \mathbf{if}\;m \leq -82000:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;m \leq 5.2 \cdot 10^{+30}:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + a \cdot \frac{0.1}{k}\right) + -1\\ \end{array} \]

Alternative 6
Accuracy	59.5%
Cost	840

\[\begin{array}{l} \mathbf{if}\;m \leq -82000:\\ \;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\ \mathbf{elif}\;m \leq 7.6 \cdot 10^{+29}:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + a \cdot \frac{0.1}{k}\right) + -1\\ \end{array} \]

Alternative 7
Accuracy	60.3%
Cost	840

\[\begin{array}{l} \mathbf{if}\;m \leq -82000:\\ \;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\ \mathbf{elif}\;m \leq 5.8 \cdot 10^{+29}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + a \cdot \frac{0.1}{k}\right) + -1\\ \end{array} \]

Alternative 8
Accuracy	48.0%
Cost	712

\[\begin{array}{l} \mathbf{if}\;k \leq -1.06 \cdot 10^{-303}:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 6.6 \cdot 10^{-6}:\\ \;\;\;\;a \cdot \left(1 + k \cdot -10\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 9
Accuracy	48.0%
Cost	712

\[\begin{array}{l} \mathbf{if}\;k \leq -1.06 \cdot 10^{-303}:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 6.6 \cdot 10^{-6}:\\ \;\;\;\;a \cdot \left(1 + k \cdot -10\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{1}{k}\\ \end{array} \]

Alternative 10
Accuracy	48.1%
Cost	712

\[\begin{array}{l} \mathbf{if}\;k \leq -1.06 \cdot 10^{-303}:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 10.2:\\ \;\;\;\;\frac{a}{1 + k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{1}{k}\\ \end{array} \]

Alternative 11
Accuracy	54.1%
Cost	712

\[\begin{array}{l} \mathbf{if}\;m \leq -82000:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;m \leq 1.5 \cdot 10^{+18}:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(1 + k \cdot -10\right)\\ \end{array} \]

Alternative 12
Accuracy	28.9%
Cost	585

\[\begin{array}{l} \mathbf{if}\;k \leq -1.5 \cdot 10^{+90} \lor \neg \left(k \leq 6.6 \cdot 10^{-6}\right):\\ \;\;\;\;\frac{a}{k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;a\\ \end{array} \]

Alternative 13
Accuracy	46.7%
Cost	585

\[\begin{array}{l} \mathbf{if}\;k \leq -1.06 \cdot 10^{-303} \lor \neg \left(k \leq 6.6 \cdot 10^{-6}\right):\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a\\ \end{array} \]

Alternative 14
Accuracy	47.8%
Cost	584

\[\begin{array}{l} \mathbf{if}\;k \leq -1.06 \cdot 10^{-303}:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 6.6 \cdot 10^{-6}:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 15
Accuracy	20.5%
Cost	64

\[a \]

Reproduce

herbie shell --seed 2023160 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))