Falkner and Boettcher, Appendix A

Percentage Accurate: 89.1% → 98.6%

Time: 18.1s

Precision: binary64

Cost: 20804

• 24.4% of points produce a very large (infinite) output. You may want to add a precondition.

• could not determine a ground truth

  1. a = -5.152228619319171e-14
  2. k = 2.499466027733833e+40
  3. m = -1.370950242682727e+210

Math

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

↓

\[\begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;\frac{t_0}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+219}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, k + 10, 1\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

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))))
   (if (<= (/ t_0 (+ (+ 1.0 (* k 10.0)) (* k k))) 5e+219)
     (* a (/ (pow k m) (fma k (+ k 10.0) 1.0)))
     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 tmp;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+219) {
		tmp = a * (pow(k, m) / fma(k, (k + 10.0), 1.0));
	} 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))
	tmp = 0.0
	if (Float64(t_0 / Float64(Float64(1.0 + Float64(k * 10.0)) + Float64(k * k))) <= 5e+219)
		tmp = Float64(a * Float64((k ^ m) / fma(k, Float64(k + 10.0), 1.0)));
	else
		tmp = t_0;
	end
	return 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]}, If[LessEqual[N[(t$95$0 / N[(N[(1.0 + N[(k * 10.0), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 5e+219], N[(a * N[(N[Power[k, m], $MachinePrecision] / N[(k * N[(k + 10.0), $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 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))) < 5e219

   1. Initial program 98.5%

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

   2. Simplified 98.5%

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

      [Start] 98.5%	\[ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
      associate-*r/ [<=] 98.5%	\[ \color{blue}{a \cdot \frac{{k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}} \]
      associate-+l+ [=>] 98.5%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{1 + \left(10 \cdot k + k \cdot k\right)}} \]
      +-commutative [=>] 98.5%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\left(10 \cdot k + k \cdot k\right) + 1}} \]
      distribute-rgt-out [=>] 98.5%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{k \cdot \left(10 + k\right)} + 1} \]
      fma-def [=>] 98.5%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\mathsf{fma}\left(k, 10 + k, 1\right)}} \]
      +-commutative [=>] 98.5%	\[ a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, \color{blue}{k + 10}, 1\right)} \]

   if 5e219 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k)))

   1. Initial program 73.8%

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

   2. Simplified 73.8%

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

      [Start] 73.8%	\[ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
      associate-*r/ [<=] 73.8%	\[ \color{blue}{a \cdot \frac{{k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}} \]
      associate-+l+ [=>] 73.8%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{1 + \left(10 \cdot k + k \cdot k\right)}} \]
      +-commutative [=>] 73.8%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\left(10 \cdot k + k \cdot k\right) + 1}} \]
      distribute-rgt-out [=>] 73.8%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{k \cdot \left(10 + k\right)} + 1} \]
      fma-def [=>] 73.8%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\mathsf{fma}\left(k, 10 + k, 1\right)}} \]
      +-commutative [=>] 73.8%	\[ a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, \color{blue}{k + 10}, 1\right)} \]

   3. Taylor expanded in k around 0 50.8%

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

   4. Simplified 100.0%

      \[\leadsto a \cdot \color{blue}{{k}^{m}} \]
      Step-by-step derivation

      [Start] 50.8%	\[ a \cdot e^{\log k \cdot m} \]
      exp-to-pow [=>] 100.0%	\[ a \cdot \color{blue}{{k}^{m}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 98.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+219}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, k + 10, 1\right)}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	98.6%
Cost	20804

\[\begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;\frac{t_0}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+219}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, k + 10, 1\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 2
Accuracy	98.6%
Cost	14660

\[\begin{array}{l} t_0 := a \cdot {k}^{m}\\ t_1 := \frac{t_0}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{if}\;t_1 \leq 5 \cdot 10^{+219}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 3
Accuracy	98.1%
Cost	6921

\[\begin{array}{l} \mathbf{if}\;m \leq -8.8 \cdot 10^{-17} \lor \neg \left(m \leq 3 \cdot 10^{-8}\right):\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;a \cdot \frac{1}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \]

Alternative 4
Accuracy	97.7%
Cost	6916

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

Alternative 5
Accuracy	60.0%
Cost	1096

\[\begin{array}{l} \mathbf{if}\;m \leq -0.032:\\ \;\;\;\;a \cdot \frac{--1}{k \cdot k}\\ \mathbf{elif}\;m \leq 1.75:\\ \;\;\;\;a \cdot \frac{1}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;a + a \cdot \left(k \cdot \left(k \cdot 100\right) + k \cdot -10\right)\\ \end{array} \]

Alternative 6
Accuracy	55.0%
Cost	968

\[\begin{array}{l} \mathbf{if}\;m \leq -0.075:\\ \;\;\;\;a \cdot \frac{--1}{k \cdot k}\\ \mathbf{elif}\;m \leq 0.64:\\ \;\;\;\;a \cdot \frac{1}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 7
Accuracy	45.1%
Cost	844

\[\begin{array}{l} t_0 := -10 \cdot \left(a \cdot k\right)\\ \mathbf{if}\;m \leq -0.011:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;m \leq 5.1 \cdot 10^{-118}:\\ \;\;\;\;a + t_0\\ \mathbf{elif}\;m \leq 1.15:\\ \;\;\;\;\frac{a}{k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 8
Accuracy	55.0%
Cost	840

\[\begin{array}{l} \mathbf{if}\;m \leq -0.08:\\ \;\;\;\;a \cdot \frac{--1}{k \cdot k}\\ \mathbf{elif}\;m \leq 0.92:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 9
Accuracy	45.0%
Cost	716

\[\begin{array}{l} t_0 := \frac{a}{k \cdot k}\\ \mathbf{if}\;m \leq -0.065:\\ \;\;\;\;t_0\\ \mathbf{elif}\;m \leq 8.8 \cdot 10^{-117}:\\ \;\;\;\;a\\ \mathbf{elif}\;m \leq 0.42:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 10
Accuracy	45.0%
Cost	716

\[\begin{array}{l} t_0 := \frac{a}{k \cdot k}\\ t_1 := -10 \cdot \left(a \cdot k\right)\\ \mathbf{if}\;m \leq -0.032:\\ \;\;\;\;t_0\\ \mathbf{elif}\;m \leq 6.7 \cdot 10^{-112}:\\ \;\;\;\;a + t_1\\ \mathbf{elif}\;m \leq 0.51:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]

Alternative 11
Accuracy	48.1%
Cost	712

\[\begin{array}{l} \mathbf{if}\;m \leq -0.06:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;m \leq 1.08:\\ \;\;\;\;\frac{a}{1 + k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 12
Accuracy	53.9%
Cost	712

\[\begin{array}{l} \mathbf{if}\;m \leq -0.075:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;m \leq 0.8:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 13
Accuracy	54.1%
Cost	712

\[\begin{array}{l} \mathbf{if}\;m \leq -0.011:\\ \;\;\;\;a \cdot \frac{--1}{k \cdot k}\\ \mathbf{elif}\;m \leq 0.58:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 14
Accuracy	32.9%
Cost	584

\[\begin{array}{l} \mathbf{if}\;m \leq -0.032:\\ \;\;\;\;\frac{a}{k \cdot 10}\\ \mathbf{elif}\;m \leq 0.52:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 15
Accuracy	28.6%
Cost	452

\[\begin{array}{l} \mathbf{if}\;m \leq 0.375:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 16
Accuracy	22.1%
Cost	64

\[a \]

Reproduce

herbie shell --seed 2023271 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))