Falkner and Boettcher, Appendix A

Percentage Accurate: 89.0% → 98.5%

Time: 18.7s

Precision: binary64

Cost: 14660

• 24.1% of points produce a very large (infinite) output. You may want to add a precondition.

• could not determine a ground truth

  1. a = -1.5613951578394885e+295
  2. k = 1.8263930114072074e+65
  3. m = -1.9310125239979484e-272

Math

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

↓

\[\begin{array}{l} t_0 := a \cdot {k}^{m}\\ t_1 := \frac{t_0}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{if}\;t_1 \leq 2 \cdot 10^{+184}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

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+184) 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+184) {
		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+184) 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+184) {
		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+184:
		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+184)
		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+184)
		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+184], 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))) < 2.00000000000000003e184

   1. Initial program 99.5%

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

   if 2.00000000000000003e184 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k)))

   1. Initial program 69.7%

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

   2. Simplified 69.7%

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

      [Start] 69.7%	\[ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
      associate-*r/ [<=] 69.7%	\[ \color{blue}{a \cdot \frac{{k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}} \]
      associate-+l+ [=>] 69.7%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{1 + \left(10 \cdot k + k \cdot k\right)}} \]
      +-commutative [=>] 69.7%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\left(10 \cdot k + k \cdot k\right) + 1}} \]
      distribute-rgt-out [=>] 69.7%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{k \cdot \left(10 + k\right)} + 1} \]
      fma-def [=>] 69.7%	\[ a \cdot \frac{{k}^{m}}{\color{blue}{\mathsf{fma}\left(k, 10 + k, 1\right)}} \]
      +-commutative [=>] 69.7%	\[ a \cdot \frac{{k}^{m}}{\mathsf{fma}\left(k, \color{blue}{k + 10}, 1\right)} \]

   3. Taylor expanded in k around 0 57.6%

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

   4. Simplified 100.0%

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

      [Start] 57.6%	\[ a \cdot e^{\log k \cdot m} \]
      exp-to-pow [=>] 100.0%	\[ a \cdot \color{blue}{{k}^{m}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 99.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 2 \cdot 10^{+184}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	98.5%
Cost	14660

\[\begin{array}{l} t_0 := a \cdot {k}^{m}\\ t_1 := \frac{t_0}{\left(1 + k \cdot 10\right) + k \cdot k}\\ \mathbf{if}\;t_1 \leq 2 \cdot 10^{+184}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 2
Accuracy	98.4%
Cost	6921

\[\begin{array}{l} \mathbf{if}\;m \leq -9 \cdot 10^{-18} \lor \neg \left(m \leq 3 \cdot 10^{-6}\right):\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + \left(k \cdot k + k \cdot 10\right)}\\ \end{array} \]

Alternative 3
Accuracy	58.3%
Cost	1745

\[\begin{array}{l} \mathbf{if}\;m \leq -0.095:\\ \;\;\;\;a \cdot \frac{1}{k \cdot k}\\ \mathbf{elif}\;m \leq 11000000000000:\\ \;\;\;\;\frac{a}{1 + \left(k \cdot k + k \cdot 10\right)}\\ \mathbf{elif}\;m \leq 4 \cdot 10^{+110} \lor \neg \left(m \leq 1.5 \cdot 10^{+215}\right):\\ \;\;\;\;\frac{a \cdot \left(a - \left(k \cdot k\right) \cdot \left(a \cdot 100\right)\right)}{a + a \cdot \left(k \cdot 10\right)}\\ \mathbf{else}:\\ \;\;\;\;a + a \cdot \left(k \cdot \left(k \cdot 100\right) + k \cdot -10\right)\\ \end{array} \]

Alternative 4
Accuracy	60.9%
Cost	1096

\[\begin{array}{l} \mathbf{if}\;m \leq -0.122:\\ \;\;\;\;a \cdot \frac{1}{k \cdot k}\\ \mathbf{elif}\;m \leq 2.2:\\ \;\;\;\;\frac{a}{1 + \left(k \cdot k + k \cdot 10\right)}\\ \mathbf{else}:\\ \;\;\;\;a + a \cdot \left(k \cdot \left(k \cdot 100\right) + k \cdot -10\right)\\ \end{array} \]

Alternative 5
Accuracy	47.9%
Cost	976

\[\begin{array}{l} t_0 := \frac{a}{1 + k \cdot 10}\\ \mathbf{if}\;m \leq -0.072:\\ \;\;\;\;a \cdot \frac{1}{k \cdot k}\\ \mathbf{elif}\;m \leq 9.2 \cdot 10^{-269}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;m \leq 3.15 \cdot 10^{-122}:\\ \;\;\;\;\frac{1}{k} \cdot \frac{a}{k}\\ \mathbf{elif}\;m \leq 5900000000:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 6
Accuracy	55.8%
Cost	968

\[\begin{array}{l} \mathbf{if}\;m \leq -0.16:\\ \;\;\;\;a \cdot \frac{1}{k \cdot k}\\ \mathbf{elif}\;m \leq 5900000000:\\ \;\;\;\;\frac{a}{1 + \left(k \cdot k + k \cdot 10\right)}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 7
Accuracy	55.8%
Cost	840

\[\begin{array}{l} \mathbf{if}\;m \leq -0.075:\\ \;\;\;\;a \cdot \frac{1}{k \cdot k}\\ \mathbf{elif}\;m \leq 5900000000:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 8
Accuracy	46.4%
Cost	716

\[\begin{array}{l} \mathbf{if}\;m \leq -0.08:\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{elif}\;m \leq 9.2 \cdot 10^{-269}:\\ \;\;\;\;a\\ \mathbf{elif}\;m \leq 5900000000:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 9
Accuracy	46.7%
Cost	716

\[\begin{array}{l} \mathbf{if}\;m \leq -0.07:\\ \;\;\;\;a \cdot \frac{1}{k \cdot k}\\ \mathbf{elif}\;m \leq 9.2 \cdot 10^{-269}:\\ \;\;\;\;a\\ \mathbf{elif}\;m \leq 5900000000:\\ \;\;\;\;\frac{\frac{a}{k}}{k}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 10
Accuracy	54.8%
Cost	712

\[\begin{array}{l} \mathbf{if}\;m \leq -0.27:\\ \;\;\;\;a \cdot \frac{1}{k \cdot k}\\ \mathbf{elif}\;m \leq 5900000000:\\ \;\;\;\;\frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 11
Accuracy	40.3%
Cost	585

\[\begin{array}{l} \mathbf{if}\;k \leq 6 \cdot 10^{-273} \lor \neg \left(k \leq 16\right):\\ \;\;\;\;\frac{a}{k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a\\ \end{array} \]

Alternative 12
Accuracy	29.1%
Cost	452

\[\begin{array}{l} \mathbf{if}\;m \leq 6200000000:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]

Alternative 13
Accuracy	22.5%
Cost	64

\[a \]

Reproduce

herbie shell --seed 2023272 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))