Falkner and Boettcher, Appendix A

Average Accuracy: 97.1% → 99.9%

Time: 13.9s

Precision: binary64

Cost: 13572.00

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{+23}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, k + 10, 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{\frac{\mathsf{hypot}\left(1, k\right)}{{k}^{m}}}\\ \end{array} \]

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 6e+23)
   (* a (/ (pow k m) (fma k (+ k 10.0) 1.0)))
   (/ (/ a k) (/ (hypot 1.0 k) (pow k m)))))

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 <= 6e+23) {
		tmp = a * (pow(k, m) / fma(k, (k + 10.0), 1.0));
	} else {
		tmp = (a / k) / (hypot(1.0, k) / pow(k, m));
	}
	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 <= 6e+23)
		tmp = Float64(a * Float64((k ^ m) / fma(k, Float64(k + 10.0), 1.0)));
	else
		tmp = Float64(Float64(a / k) / Float64(hypot(1.0, k) / (k ^ m)));
	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_] := If[LessEqual[k, 6e+23], N[(a * N[(N[Power[k, m], $MachinePrecision] / N[(k * N[(k + 10.0), $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(a / k), $MachinePrecision] / N[(N[Sqrt[1.0 ^ 2 + k ^ 2], $MachinePrecision] / N[Power[k, m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 6.0000000000000002e23

   1. Initial program 99.9%

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

   2. Simplified 100.0%

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

      [Start] 99.9	\[ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
      associate-*r/ [<=] 99.9	\[ \color{blue}{a \cdot \frac{{k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}} \]
      associate-+l+ [=>] 99.9	\[ a \cdot \frac{{k}^{m}}{\color{blue}{1 + \left(10 \cdot k + k \cdot k\right)}} \]
      +-commutative [=>] 99.9	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\left(10 \cdot k + k \cdot k\right) + 1}} \]
      distribute-rgt-out [=>] 100.0	\[ a \cdot \frac{{k}^{m}}{\color{blue}{k \cdot \left(10 + k\right)} + 1} \]
      fma-def [=>] 100.0	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\mathsf{fma}\left(k, 10 + k, 1\right)}} \]
      +-commutative [=>] 100.0	\[ a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, \color{blue}{k + 10}, 1\right)} \]

   if 6.0000000000000002e23 < k

   1. Initial program 91.8%

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

   2. Simplified 91.8%

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

      [Start] 91.8	\[ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
      associate-/l* [=>] 91.8	\[ \color{blue}{\frac{a}{\frac{\left(1 + 10 \cdot k\right) + k \cdot k}{{k}^{m}}}} \]
      associate-+l+ [=>] 91.8	\[ \frac{a}{\frac{\color{blue}{1 + \left(10 \cdot k + k \cdot k\right)}}{{k}^{m}}} \]
      *-commutative [=>] 91.8	\[ \frac{a}{\frac{1 + \left(\color{blue}{k \cdot 10} + k \cdot k\right)}{{k}^{m}}} \]

   3. Taylor expanded in k around inf 91.8%

      \[\leadsto \frac{a}{\frac{1 + \color{blue}{{k}^{2}}}{{k}^{m}}} \]

   4. Simplified 91.8%

      \[\leadsto \frac{a}{\frac{1 + \color{blue}{k \cdot k}}{{k}^{m}}} \]
      Proof

      [Start] 91.8	\[ \frac{a}{\frac{1 + {k}^{2}}{{k}^{m}}} \]
      unpow2 [=>] 91.8	\[ \frac{a}{\frac{1 + \color{blue}{k \cdot k}}{{k}^{m}}} \]

   5. Applied egg-rr 99.8%

      \[\leadsto \color{blue}{\frac{\frac{a}{\mathsf{hypot}\left(1, k\right)} \cdot \sqrt{{k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \sqrt{{k}^{m}}} \]

   6. Applied egg-rr 75.1%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\frac{a}{\mathsf{hypot}\left(1, k\right)} \cdot {k}^{m}}{\mathsf{hypot}\left(1, k\right)}\right)} - 1} \]

   7. Simplified 99.8%

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

      [Start] 75.1	\[ e^{\mathsf{log1p}\left(\frac{\frac{a}{\mathsf{hypot}\left(1, k\right)} \cdot {k}^{m}}{\mathsf{hypot}\left(1, k\right)}\right)} - 1 \]
      expm1-def [=>] 96.4	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\frac{a}{\mathsf{hypot}\left(1, k\right)} \cdot {k}^{m}}{\mathsf{hypot}\left(1, k\right)}\right)\right)} \]
      expm1-log1p [=>] 99.8	\[ \color{blue}{\frac{\frac{a}{\mathsf{hypot}\left(1, k\right)} \cdot {k}^{m}}{\mathsf{hypot}\left(1, k\right)}} \]
      associate-/l* [=>] 99.8	\[ \color{blue}{\frac{\frac{a}{\mathsf{hypot}\left(1, k\right)}}{\frac{\mathsf{hypot}\left(1, k\right)}{{k}^{m}}}} \]

   8. Taylor expanded in k around inf 99.8%

      \[\leadsto \frac{\color{blue}{\frac{a}{k}}}{\frac{\mathsf{hypot}\left(1, k\right)}{{k}^{m}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 99.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{+23}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, k + 10, 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{\frac{\mathsf{hypot}\left(1, k\right)}{{k}^{m}}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	99.9%
Cost	13508.00

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{+35}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k \cdot 10 + k \cdot k\right)}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{\frac{\mathsf{hypot}\left(1, k\right)}{{k}^{m}}}\\ \end{array} \]

Alternative 2
Accuracy	99.9%
Cost	7428.00

\[\begin{array}{l} \mathbf{if}\;k \leq 5 \cdot 10^{+49}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k \cdot 10 + k \cdot k\right)}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{\frac{1}{\frac{1}{{k}^{m}}}}{k}\\ \end{array} \]

Alternative 3
Accuracy	95.9%
Cost	7304.00

\[\begin{array}{l} \mathbf{if}\;k \leq 4.8 \cdot 10^{-22}:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{elif}\;k \leq 35000000000:\\ \;\;\;\;a \cdot \frac{1}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{\frac{1 + k \cdot k}{{k}^{m}}}\\ \end{array} \]

Alternative 4
Accuracy	99.1%
Cost	7300.00

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

Alternative 5
Accuracy	96.1%
Cost	6788.00

\[\begin{array}{l} \mathbf{if}\;m \leq -1.15 \cdot 10^{-12}:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{elif}\;m \leq 0.52:\\ \;\;\;\;\frac{a}{1 + \frac{k}{10 - k} \cdot \left(100 - k \cdot k\right)}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 6
Accuracy	96.1%
Cost	1224.00

\[\begin{array}{l} \mathbf{if}\;m \leq -0.68:\\ \;\;\;\;0\\ \mathbf{elif}\;m \leq 0.14:\\ \;\;\;\;\frac{a}{1 + \frac{k}{10 - k} \cdot \left(100 - k \cdot k\right)}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 7
Accuracy	96.2%
Cost	840.00

\[\begin{array}{l} \mathbf{if}\;m \leq -2.35:\\ \;\;\;\;0\\ \mathbf{elif}\;m \leq 1.05:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 8
Accuracy	79.9%
Cost	712.00

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{-301}:\\ \;\;\;\;0\\ \mathbf{elif}\;k \leq 10.2:\\ \;\;\;\;\frac{a}{1 + k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 9
Accuracy	95.1%
Cost	712.00

\[\begin{array}{l} \mathbf{if}\;m \leq -2.4:\\ \;\;\;\;0\\ \mathbf{elif}\;m \leq 0.28:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 10
Accuracy	78.5%
Cost	584.00

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{-301}:\\ \;\;\;\;0\\ \mathbf{elif}\;k \leq 1:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \end{array} \]

Alternative 11
Accuracy	79.6%
Cost	584.00

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{-301}:\\ \;\;\;\;0\\ \mathbf{elif}\;k \leq 1:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \end{array} \]

Alternative 12
Accuracy	74.7%
Cost	328.00

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{-301}:\\ \;\;\;\;0\\ \mathbf{elif}\;k \leq 210:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 13
Accuracy	26.8%
Cost	64.00

\[a \]

Reproduce

herbie shell --seed 2023096 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))