Falkner and Boettcher, Appendix A

Average Accuracy: 96.7% → 99.8%

Time: 14.9s

Precision: binary64

Cost: 7428

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;k \leq 1000000000:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{{k}^{m} \cdot \frac{-a}{k}}{-k}\\ \end{array} \]

FPCore

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

↓

(FPCore (a k m)
 :precision binary64
 (if (<= k 1000000000.0)
   (/ (* a (pow k m)) (+ (+ 1.0 (* k 10.0)) (* k k)))
   (/ (* (pow k m) (/ (- a) k)) (- k))))

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 tmp;
	if (k <= 1000000000.0) {
		tmp = (a * pow(k, m)) / ((1.0 + (k * 10.0)) + (k * k));
	} else {
		tmp = (pow(k, m) * (-a / k)) / -k;
	}
	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) :: tmp
    if (k <= 1000000000.0d0) then
        tmp = (a * (k ** m)) / ((1.0d0 + (k * 10.0d0)) + (k * k))
    else
        tmp = ((k ** m) * (-a / k)) / -k
    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 tmp;
	if (k <= 1000000000.0) {
		tmp = (a * Math.pow(k, m)) / ((1.0 + (k * 10.0)) + (k * k));
	} else {
		tmp = (Math.pow(k, m) * (-a / k)) / -k;
	}
	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):
	tmp = 0
	if k <= 1000000000.0:
		tmp = (a * math.pow(k, m)) / ((1.0 + (k * 10.0)) + (k * k))
	else:
		tmp = (math.pow(k, m) * (-a / k)) / -k
	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)
	tmp = 0.0
	if (k <= 1000000000.0)
		tmp = Float64(Float64(a * (k ^ m)) / Float64(Float64(1.0 + Float64(k * 10.0)) + Float64(k * k)));
	else
		tmp = Float64(Float64((k ^ m) * Float64(Float64(-a) / k)) / Float64(-k));
	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)
	tmp = 0.0;
	if (k <= 1000000000.0)
		tmp = (a * (k ^ m)) / ((1.0 + (k * 10.0)) + (k * k));
	else
		tmp = ((k ^ m) * (-a / k)) / -k;
	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_] := If[LessEqual[k, 1000000000.0], N[(N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision] / N[(N[(1.0 + N[(k * 10.0), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Power[k, m], $MachinePrecision] * N[((-a) / k), $MachinePrecision]), $MachinePrecision] / (-k)), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 1e9

   1. Initial program 99.9%

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

   if 1e9 < k

   1. Initial program 91.2%

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

   2. Taylor expanded in k around inf 91.0%

      \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{{k}^{2}}} \]

   3. Simplified 91.0%

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

      [Start] 91.0	\[ \frac{a \cdot {k}^{m}}{{k}^{2}} \]
      unpow2 [=>] 91.0	\[ \frac{a \cdot {k}^{m}}{\color{blue}{k \cdot k}} \]

   4. Applied egg-rr 99.6%

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

      [Start] 91.0	\[ \frac{a \cdot {k}^{m}}{k \cdot k} \]
      associate-/r* [=>] 99.5	\[ \color{blue}{\frac{\frac{a \cdot {k}^{m}}{k}}{k}} \]
      frac-2neg [=>] 99.5	\[ \color{blue}{\frac{-\frac{a \cdot {k}^{m}}{k}}{-k}} \]
      distribute-frac-neg [=>] 99.5	\[ \color{blue}{-\frac{\frac{a \cdot {k}^{m}}{k}}{-k}} \]
      associate-*l/ [<=] 99.6	\[ -\frac{\color{blue}{\frac{a}{k} \cdot {k}^{m}}}{-k} \]
      *-commutative [=>] 99.6	\[ -\frac{\color{blue}{{k}^{m} \cdot \frac{a}{k}}}{-k} \]

3. Recombined 2 regimes into one program.

4. Final simplification 99.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1000000000:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{{k}^{m} \cdot \frac{-a}{k}}{-k}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	98.8%
Cost	7304

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{-11}:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{elif}\;k \leq 370000000000:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{{k}^{m} \cdot \frac{-a}{k}}{-k}\\ \end{array} \]

Alternative 2
Accuracy	96.9%
Cost	7180

\[\begin{array}{l} \mathbf{if}\;k \leq 3.9 \cdot 10^{-11}:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{elif}\;k \leq 1000000000:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{elif}\;k \leq 5.4 \cdot 10^{+147}:\\ \;\;\;\;\frac{a}{{k}^{\left(2 - m\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 3
Accuracy	98.7%
Cost	7176

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{-12}:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{elif}\;k \leq 33000000000:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{{k}^{m}}{k}\\ \end{array} \]

Alternative 4
Accuracy	98.9%
Cost	7172

\[\begin{array}{l} \mathbf{if}\;k \leq 0.34:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{1 + k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;\frac{{k}^{m} \cdot \frac{-a}{k}}{-k}\\ \end{array} \]

Alternative 5
Accuracy	95.9%
Cost	6921

\[\begin{array}{l} \mathbf{if}\;m \leq -2.85 \cdot 10^{-9} \lor \neg \left(m \leq 1.85 \cdot 10^{-5}\right):\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \]

Alternative 6
Accuracy	69.3%
Cost	841

\[\begin{array}{l} \mathbf{if}\;m \leq -3.1 \lor \neg \left(m \leq 8 \cdot 10^{+58}\right):\\ \;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \end{array} \]

Alternative 7
Accuracy	70.6%
Cost	841

\[\begin{array}{l} \mathbf{if}\;m \leq -1.55 \lor \neg \left(m \leq 8 \cdot 10^{+58}\right):\\ \;\;\;\;\left(1 + \frac{a}{k \cdot k}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \]

Alternative 8
Accuracy	63.9%
Cost	712

\[\begin{array}{l} \mathbf{if}\;k \leq -0.43:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 0.35:\\ \;\;\;\;a \cdot \left(1 - k \cdot 10\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 9
Accuracy	63.9%
Cost	712

\[\begin{array}{l} \mathbf{if}\;k \leq -0.43:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 0.35:\\ \;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 10
Accuracy	63.9%
Cost	712

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

Alternative 11
Accuracy	62.1%
Cost	585

\[\begin{array}{l} \mathbf{if}\;k \leq -1 \lor \neg \left(k \leq 1\right):\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a\\ \end{array} \]

Alternative 12
Accuracy	63.6%
Cost	584

\[\begin{array}{l} \mathbf{if}\;k \leq -1:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;k \leq 1:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 13
Accuracy	63.6%
Cost	580

\[\begin{array}{l} \mathbf{if}\;k \leq 1000000000:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 14
Accuracy	36.0%
Cost	452

\[\begin{array}{l} \mathbf{if}\;k \leq 0.1:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k \cdot 10}\\ \end{array} \]

Alternative 15
Accuracy	26.7%
Cost	64

\[a \]

Reproduce

herbie shell --seed 2023138 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))