Falkner and Boettcher, Appendix A

Percentage Accurate: 90.3% → 98.3%

Time: 11.1s

Alternatives: 12

Speedup: 1.0×

• 23.0% of points produce a very large (infinite) output. You may want to add a precondition.

• could not determine a ground truth

  1. a = -1.9999999999999999e-154
  2. k = 0.0
  3. m = -2.00000000000000007e154

Specification

Math

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

FPCore

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

C

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

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

Java

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

Python

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

Julia

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

MATLAB

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

Initial Program: 90.3% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

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

Java

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

Python

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

Julia

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

MATLAB

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

Alternative 1: 98.3% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;\frac{t\_0}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 4 \cdot 10^{+96}:\\ \;\;\;\;\frac{{k}^{m}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{a}{\mathsf{hypot}\left(1, k\right)}\\ \mathbf{else}:\\ \;\;\;\;t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (let* ((t_0 (* a (pow k m))))
   (if (<= (/ t_0 (+ (+ 1.0 (* k 10.0)) (* k k))) 4e+96)
     (* (/ (pow k m) (hypot 1.0 k)) (/ a (hypot 1.0 k)))
     t_0)))

C

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))) <= 4e+96) {
		tmp = (pow(k, m) / hypot(1.0, k)) * (a / hypot(1.0, k));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Java

public static double code(double a, double k, double m) {
	double t_0 = a * Math.pow(k, m);
	double tmp;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 4e+96) {
		tmp = (Math.pow(k, m) / Math.hypot(1.0, k)) * (a / Math.hypot(1.0, k));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(a, k, m):
	t_0 = a * math.pow(k, m)
	tmp = 0
	if (t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 4e+96:
		tmp = (math.pow(k, m) / math.hypot(1.0, k)) * (a / math.hypot(1.0, k))
	else:
		tmp = t_0
	return tmp

Julia

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))) <= 4e+96)
		tmp = Float64(Float64((k ^ m) / hypot(1.0, k)) * Float64(a / hypot(1.0, k)));
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	t_0 = a * (k ^ m);
	tmp = 0.0;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 4e+96)
		tmp = ((k ^ m) / hypot(1.0, k)) * (a / hypot(1.0, k));
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

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], 4e+96], N[(N[(N[Power[k, m], $MachinePrecision] / N[Sqrt[1.0 ^ 2 + k ^ 2], $MachinePrecision]), $MachinePrecision] * N[(a / N[Sqrt[1.0 ^ 2 + k ^ 2], $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))) < 4.0000000000000002e96

   1. Initial program 97.7%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. *-commutative 97.7%

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

   3. Simplified 97.7%

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

   5. Taylor expanded in k around 0 97.2%

      \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1} + k \cdot k} \]
   6. Step-by-step derivation

      1. *-commutative 97.2%

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

      2. add-sqr-sqrt 97.2%

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

      3. times-frac 97.2%

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

      4. hypot-1-def 97.2%

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

      5. hypot-1-def 99.4%

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

   7. Applied egg-rr 99.4%

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

   if 4.0000000000000002e96 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k)))

   1. Initial program 56.3%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 47.9%

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

      2. sqr-neg 47.9%

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

      3. associate-+l+ 47.9%

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

      4. sqr-neg 47.9%

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

      5. distribute-rgt-out 47.9%

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

   3. Simplified 47.9%

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

   5. Taylor expanded in k around 0 99.6%

      \[\leadsto \color{blue}{a \cdot {k}^{m}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 99.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 4 \cdot 10^{+96}:\\ \;\;\;\;\frac{{k}^{m}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{a}{\mathsf{hypot}\left(1, k\right)}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 66.9% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ {\left(\frac{1}{\frac{\mathsf{hypot}\left(1, k\right)}{\sqrt{a \cdot {k}^{m}}}}\right)}^{2} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (pow (/ 1.0 (/ (hypot 1.0 k) (sqrt (* a (pow k m))))) 2.0))

C

double code(double a, double k, double m) {
	return pow((1.0 / (hypot(1.0, k) / sqrt((a * pow(k, m))))), 2.0);
}

Java

public static double code(double a, double k, double m) {
	return Math.pow((1.0 / (Math.hypot(1.0, k) / Math.sqrt((a * Math.pow(k, m))))), 2.0);
}

Python

def code(a, k, m):
	return math.pow((1.0 / (math.hypot(1.0, k) / math.sqrt((a * math.pow(k, m))))), 2.0)

Julia

function code(a, k, m)
	return Float64(1.0 / Float64(hypot(1.0, k) / sqrt(Float64(a * (k ^ m))))) ^ 2.0
end

MATLAB

function tmp = code(a, k, m)
	tmp = (1.0 / (hypot(1.0, k) / sqrt((a * (k ^ m))))) ^ 2.0;
end

Wolfram

code[a_, k_, m_] := N[Power[N[(1.0 / N[(N[Sqrt[1.0 ^ 2 + k ^ 2], $MachinePrecision] / N[Sqrt[N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]

Derivation

1. Initial program 89.9%

   \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
2. Step-by-step derivation

   1. *-commutative 89.9%

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

3. Simplified 89.9%

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

5. Taylor expanded in k around 0 89.4%

   \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1} + k \cdot k} \]
6. Step-by-step derivation

   1. add-sqr-sqrt 58.3%

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

   2. add-sqr-sqrt 58.3%

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

   3. times-frac 58.4%

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

   4. hypot-1-def 58.4%

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

   5. hypot-1-def 65.7%

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

7. Applied egg-rr 65.7%

   \[\leadsto \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}} \]
8. Step-by-step derivation

   1. unpow2 65.7%

      \[\leadsto \color{blue}{{\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2}} \]

9. Simplified 65.7%

   \[\leadsto \color{blue}{{\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2}} \]
10. Step-by-step derivation

    1. clear-num 65.7%

       \[\leadsto {\color{blue}{\left(\frac{1}{\frac{\mathsf{hypot}\left(1, k\right)}{\sqrt{a \cdot {k}^{m}}}}\right)}}^{2} \]

    2. inv-pow 65.7%

       \[\leadsto {\color{blue}{\left({\left(\frac{\mathsf{hypot}\left(1, k\right)}{\sqrt{a \cdot {k}^{m}}}\right)}^{-1}\right)}}^{2} \]

11. Applied egg-rr 65.7%

    \[\leadsto {\color{blue}{\left({\left(\frac{\mathsf{hypot}\left(1, k\right)}{\sqrt{a \cdot {k}^{m}}}\right)}^{-1}\right)}}^{2} \]
12. Step-by-step derivation

    1. unpow-1 65.7%

       \[\leadsto {\color{blue}{\left(\frac{1}{\frac{\mathsf{hypot}\left(1, k\right)}{\sqrt{a \cdot {k}^{m}}}}\right)}}^{2} \]

13. Simplified 65.7%

    \[\leadsto {\color{blue}{\left(\frac{1}{\frac{\mathsf{hypot}\left(1, k\right)}{\sqrt{a \cdot {k}^{m}}}}\right)}}^{2} \]

14. Final simplification 65.7%

    \[\leadsto {\left(\frac{1}{\frac{\mathsf{hypot}\left(1, k\right)}{\sqrt{a \cdot {k}^{m}}}}\right)}^{2} \]
15. Add Preprocessing

Alternative 3: 66.9% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ {\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (pow (/ (sqrt (* a (pow k m))) (hypot 1.0 k)) 2.0))

C

double code(double a, double k, double m) {
	return pow((sqrt((a * pow(k, m))) / hypot(1.0, k)), 2.0);
}

Java

public static double code(double a, double k, double m) {
	return Math.pow((Math.sqrt((a * Math.pow(k, m))) / Math.hypot(1.0, k)), 2.0);
}

Python

def code(a, k, m):
	return math.pow((math.sqrt((a * math.pow(k, m))) / math.hypot(1.0, k)), 2.0)

Julia

function code(a, k, m)
	return Float64(sqrt(Float64(a * (k ^ m))) / hypot(1.0, k)) ^ 2.0
end

MATLAB

function tmp = code(a, k, m)
	tmp = (sqrt((a * (k ^ m))) / hypot(1.0, k)) ^ 2.0;
end

Wolfram

code[a_, k_, m_] := N[Power[N[(N[Sqrt[N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Sqrt[1.0 ^ 2 + k ^ 2], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]

Derivation

1. Initial program 89.9%

   \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
2. Step-by-step derivation

   1. *-commutative 89.9%

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

3. Simplified 89.9%

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

5. Taylor expanded in k around 0 89.4%

   \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1} + k \cdot k} \]
6. Step-by-step derivation

   1. add-sqr-sqrt 58.3%

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

   2. add-sqr-sqrt 58.3%

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

   3. times-frac 58.4%

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

   4. hypot-1-def 58.4%

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

   5. hypot-1-def 65.7%

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

7. Applied egg-rr 65.7%

   \[\leadsto \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}} \]
8. Step-by-step derivation

   1. unpow2 65.7%

      \[\leadsto \color{blue}{{\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2}} \]

9. Simplified 65.7%

   \[\leadsto \color{blue}{{\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2}} \]

10. Final simplification 65.7%

    \[\leadsto {\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2} \]
11. Add Preprocessing

Alternative 4: 96.9% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;\frac{t\_0}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+300}:\\ \;\;\;\;{k}^{m} \cdot \frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (let* ((t_0 (* a (pow k m))))
   (if (<= (/ t_0 (+ (+ 1.0 (* k 10.0)) (* k k))) 5e+300)
     (* (pow k m) (/ a (+ 1.0 (* k (+ k 10.0)))))
     t_0)))

C

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+300) {
		tmp = pow(k, m) * (a / (1.0 + (k * (k + 10.0))));
	} 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
    real(8) :: t_0
    real(8) :: tmp
    t_0 = a * (k ** m)
    if ((t_0 / ((1.0d0 + (k * 10.0d0)) + (k * k))) <= 5d+300) then
        tmp = (k ** m) * (a / (1.0d0 + (k * (k + 10.0d0))))
    else
        tmp = t_0
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double t_0 = a * Math.pow(k, m);
	double tmp;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+300) {
		tmp = Math.pow(k, m) * (a / (1.0 + (k * (k + 10.0))));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(a, k, m):
	t_0 = a * math.pow(k, m)
	tmp = 0
	if (t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+300:
		tmp = math.pow(k, m) * (a / (1.0 + (k * (k + 10.0))))
	else:
		tmp = t_0
	return tmp

Julia

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+300)
		tmp = Float64((k ^ m) * Float64(a / Float64(1.0 + Float64(k * Float64(k + 10.0)))));
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	t_0 = a * (k ^ m);
	tmp = 0.0;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+300)
		tmp = (k ^ m) * (a / (1.0 + (k * (k + 10.0))));
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

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+300], N[(N[Power[k, m], $MachinePrecision] * N[(a / N[(1.0 + N[(k * N[(k + 10.0), $MachinePrecision]), $MachinePrecision]), $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))) < 5.00000000000000026e300

   1. Initial program 97.7%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 96.8%

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

      2. sqr-neg 96.8%

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

      3. associate-+l+ 96.8%

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

      4. sqr-neg 96.8%

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

      5. distribute-rgt-out 96.8%

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

   3. Simplified 96.8%

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

   if 5.00000000000000026e300 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k)))

   1. Initial program 50.0%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 40.5%

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

      2. sqr-neg 40.5%

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

      3. associate-+l+ 40.5%

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

      4. sqr-neg 40.5%

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

      5. distribute-rgt-out 40.5%

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

   3. Simplified 40.5%

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

   5. Taylor expanded in k around 0 100.0%

      \[\leadsto \color{blue}{a \cdot {k}^{m}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 97.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+300}:\\ \;\;\;\;{k}^{m} \cdot \frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 96.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;m \leq 2.2:\\ \;\;\;\;\frac{t\_0}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (let* ((t_0 (* a (pow k m)))) (if (<= m 2.2) (/ t_0 (+ 1.0 (* k k))) t_0)))

C

double code(double a, double k, double m) {
	double t_0 = a * pow(k, m);
	double tmp;
	if (m <= 2.2) {
		tmp = t_0 / (1.0 + (k * k));
	} 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
    real(8) :: t_0
    real(8) :: tmp
    t_0 = a * (k ** m)
    if (m <= 2.2d0) then
        tmp = t_0 / (1.0d0 + (k * k))
    else
        tmp = t_0
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double t_0 = a * Math.pow(k, m);
	double tmp;
	if (m <= 2.2) {
		tmp = t_0 / (1.0 + (k * k));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(a, k, m):
	t_0 = a * math.pow(k, m)
	tmp = 0
	if m <= 2.2:
		tmp = t_0 / (1.0 + (k * k))
	else:
		tmp = t_0
	return tmp

Julia

function code(a, k, m)
	t_0 = Float64(a * (k ^ m))
	tmp = 0.0
	if (m <= 2.2)
		tmp = Float64(t_0 / Float64(1.0 + Float64(k * k)));
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	t_0 = a * (k ^ m);
	tmp = 0.0;
	if (m <= 2.2)
		tmp = t_0 / (1.0 + (k * k));
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := Block[{t$95$0 = N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[m, 2.2], N[(t$95$0 / N[(1.0 + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], t$95$0]]

Derivation

1. Split input into 2 regimes

2. if m < 2.2000000000000002

   1. Initial program 97.3%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. *-commutative 97.3%

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

   3. Simplified 97.3%

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

   5. Taylor expanded in k around 0 96.6%

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

   if 2.2000000000000002 < m

   1. Initial program 73.1%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 65.4%

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

      2. sqr-neg 65.4%

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

      3. associate-+l+ 65.4%

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

      4. sqr-neg 65.4%

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

      5. distribute-rgt-out 65.4%

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

   3. Simplified 65.4%

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

   5. Taylor expanded in k around 0 100.0%

      \[\leadsto \color{blue}{a \cdot {k}^{m}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 97.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq 2.2:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 96.9% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;m \leq -3.7 \lor \neg \left(m \leq 0.042\right):\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (if (or (<= m -3.7) (not (<= m 0.042)))
   (* a (pow k m))
   (/ a (+ 1.0 (* k (+ k 10.0))))))

C

double code(double a, double k, double m) {
	double tmp;
	if ((m <= -3.7) || !(m <= 0.042)) {
		tmp = a * pow(k, m);
	} else {
		tmp = a / (1.0 + (k * (k + 10.0)));
	}
	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
    real(8) :: tmp
    if ((m <= (-3.7d0)) .or. (.not. (m <= 0.042d0))) then
        tmp = a * (k ** m)
    else
        tmp = a / (1.0d0 + (k * (k + 10.0d0)))
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if ((m <= -3.7) || !(m <= 0.042)) {
		tmp = a * Math.pow(k, m);
	} else {
		tmp = a / (1.0 + (k * (k + 10.0)));
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if (m <= -3.7) or not (m <= 0.042):
		tmp = a * math.pow(k, m)
	else:
		tmp = a / (1.0 + (k * (k + 10.0)))
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if ((m <= -3.7) || !(m <= 0.042))
		tmp = Float64(a * (k ^ m));
	else
		tmp = Float64(a / Float64(1.0 + Float64(k * Float64(k + 10.0))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if ((m <= -3.7) || ~((m <= 0.042)))
		tmp = a * (k ^ m);
	else
		tmp = a / (1.0 + (k * (k + 10.0)));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[Or[LessEqual[m, -3.7], N[Not[LessEqual[m, 0.042]], $MachinePrecision]], N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision], N[(a / N[(1.0 + N[(k * N[(k + 10.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if m < -3.7000000000000002 or 0.0420000000000000026 < m

   1. Initial program 87.3%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 83.6%

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

      2. sqr-neg 83.6%

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

      3. associate-+l+ 83.6%

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

      4. sqr-neg 83.6%

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

      5. distribute-rgt-out 83.6%

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

   3. Simplified 83.6%

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

   5. Taylor expanded in k around 0 100.0%

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

   if -3.7000000000000002 < m < 0.0420000000000000026

   1. Initial program 94.7%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 94.7%

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

      2. sqr-neg 94.7%

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

      3. associate-+l+ 94.7%

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

      4. sqr-neg 94.7%

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

      5. distribute-rgt-out 94.7%

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

   3. Simplified 94.7%

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

   5. Taylor expanded in m around 0 94.0%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 97.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq -3.7 \lor \neg \left(m \leq 0.042\right):\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 32.7% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;m \leq -3.7:\\ \;\;\;\;\frac{0.1}{\frac{k}{a}}\\ \mathbf{elif}\;m \leq 5 \cdot 10^{+50}:\\ \;\;\;\;\frac{a}{1 + k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (if (<= m -3.7)
   (/ 0.1 (/ k a))
   (if (<= m 5e+50) (/ a (+ 1.0 (* k 10.0))) (+ a (* -10.0 (* k a))))))

C

double code(double a, double k, double m) {
	double tmp;
	if (m <= -3.7) {
		tmp = 0.1 / (k / a);
	} else if (m <= 5e+50) {
		tmp = a / (1.0 + (k * 10.0));
	} else {
		tmp = a + (-10.0 * (k * a));
	}
	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
    real(8) :: tmp
    if (m <= (-3.7d0)) then
        tmp = 0.1d0 / (k / a)
    else if (m <= 5d+50) then
        tmp = a / (1.0d0 + (k * 10.0d0))
    else
        tmp = a + ((-10.0d0) * (k * a))
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (m <= -3.7) {
		tmp = 0.1 / (k / a);
	} else if (m <= 5e+50) {
		tmp = a / (1.0 + (k * 10.0));
	} else {
		tmp = a + (-10.0 * (k * a));
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if m <= -3.7:
		tmp = 0.1 / (k / a)
	elif m <= 5e+50:
		tmp = a / (1.0 + (k * 10.0))
	else:
		tmp = a + (-10.0 * (k * a))
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (m <= -3.7)
		tmp = Float64(0.1 / Float64(k / a));
	elseif (m <= 5e+50)
		tmp = Float64(a / Float64(1.0 + Float64(k * 10.0)));
	else
		tmp = Float64(a + Float64(-10.0 * Float64(k * a)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (m <= -3.7)
		tmp = 0.1 / (k / a);
	elseif (m <= 5e+50)
		tmp = a / (1.0 + (k * 10.0));
	else
		tmp = a + (-10.0 * (k * a));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[m, -3.7], N[(0.1 / N[(k / a), $MachinePrecision]), $MachinePrecision], If[LessEqual[m, 5e+50], N[(a / N[(1.0 + N[(k * 10.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(a + N[(-10.0 * N[(k * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if m < -3.7000000000000002

   1. Initial program 100.0%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 100.0%

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

      2. sqr-neg 100.0%

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

      3. associate-+l+ 100.0%

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

      4. sqr-neg 100.0%

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

      5. distribute-rgt-out 100.0%

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

   3. Simplified 100.0%

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

   5. Taylor expanded in m around 0 38.1%

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

   6. Taylor expanded in k around 0 18.1%

      \[\leadsto \frac{a}{1 + \color{blue}{10 \cdot k}} \]
   7. Step-by-step derivation

      1. *-commutative 18.1%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   8. Simplified 18.1%

      \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   9. Taylor expanded in k around inf 27.6%

      \[\leadsto \color{blue}{0.1 \cdot \frac{a}{k}} \]
   10. Step-by-step derivation

       1. clear-num 27.9%

          \[\leadsto 0.1 \cdot \color{blue}{\frac{1}{\frac{k}{a}}} \]

       2. un-div-inv 27.9%

          \[\leadsto \color{blue}{\frac{0.1}{\frac{k}{a}}} \]

   11. Applied egg-rr 27.9%

       \[\leadsto \color{blue}{\frac{0.1}{\frac{k}{a}}} \]

   if -3.7000000000000002 < m < 5e50

   1. Initial program 92.5%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 90.5%

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

      2. sqr-neg 90.5%

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

      3. associate-+l+ 90.5%

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

      4. sqr-neg 90.5%

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

      5. distribute-rgt-out 90.5%

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

   3. Simplified 90.5%

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

   5. Taylor expanded in m around 0 82.7%

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

   6. Taylor expanded in k around 0 57.5%

      \[\leadsto \frac{a}{1 + \color{blue}{10 \cdot k}} \]
   7. Step-by-step derivation

      1. *-commutative 57.5%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   8. Simplified 57.5%

      \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   if 5e50 < m

   1. Initial program 72.3%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 66.2%

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

      2. sqr-neg 66.2%

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

      3. associate-+l+ 66.2%

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

      4. sqr-neg 66.2%

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

      5. distribute-rgt-out 66.2%

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

   3. Simplified 66.2%

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

   5. Taylor expanded in k around 0 73.8%

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

   6. Taylor expanded in m around 0 12.4%

      \[\leadsto \color{blue}{a + -10 \cdot \left(a \cdot k\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 36.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq -3.7:\\ \;\;\;\;\frac{0.1}{\frac{k}{a}}\\ \mathbf{elif}\;m \leq 5 \cdot 10^{+50}:\\ \;\;\;\;\frac{a}{1 + k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 47.0% accurate, 8.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;m \leq 1.05 \cdot 10^{+53}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (if (<= m 1.05e+53) (/ a (+ 1.0 (* k (+ k 10.0)))) (+ a (* -10.0 (* k a)))))

C

double code(double a, double k, double m) {
	double tmp;
	if (m <= 1.05e+53) {
		tmp = a / (1.0 + (k * (k + 10.0)));
	} else {
		tmp = a + (-10.0 * (k * a));
	}
	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
    real(8) :: tmp
    if (m <= 1.05d+53) then
        tmp = a / (1.0d0 + (k * (k + 10.0d0)))
    else
        tmp = a + ((-10.0d0) * (k * a))
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (m <= 1.05e+53) {
		tmp = a / (1.0 + (k * (k + 10.0)));
	} else {
		tmp = a + (-10.0 * (k * a));
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if m <= 1.05e+53:
		tmp = a / (1.0 + (k * (k + 10.0)))
	else:
		tmp = a + (-10.0 * (k * a))
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (m <= 1.05e+53)
		tmp = Float64(a / Float64(1.0 + Float64(k * Float64(k + 10.0))));
	else
		tmp = Float64(a + Float64(-10.0 * Float64(k * a)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (m <= 1.05e+53)
		tmp = a / (1.0 + (k * (k + 10.0)));
	else
		tmp = a + (-10.0 * (k * a));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[m, 1.05e+53], N[(a / N[(1.0 + N[(k * N[(k + 10.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(a + N[(-10.0 * N[(k * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if m < 1.0500000000000001e53

   1. Initial program 95.9%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 94.9%

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

      2. sqr-neg 94.9%

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

      3. associate-+l+ 94.9%

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

      4. sqr-neg 94.9%

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

      5. distribute-rgt-out 94.9%

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

   3. Simplified 94.9%

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

   5. Taylor expanded in m around 0 62.3%

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

   if 1.0500000000000001e53 < m

   1. Initial program 72.3%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 66.2%

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

      2. sqr-neg 66.2%

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

      3. associate-+l+ 66.2%

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

      4. sqr-neg 66.2%

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

      5. distribute-rgt-out 66.2%

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

   3. Simplified 66.2%

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

   5. Taylor expanded in k around 0 73.8%

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

   6. Taylor expanded in m around 0 12.4%

      \[\leadsto \color{blue}{a + -10 \cdot \left(a \cdot k\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 49.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq 1.05 \cdot 10^{+53}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 28.7% accurate, 9.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 0.075:\\ \;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.1}{\frac{k}{a}}\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (if (<= k 0.075) (+ a (* -10.0 (* k a))) (/ 0.1 (/ k a))))

C

double code(double a, double k, double m) {
	double tmp;
	if (k <= 0.075) {
		tmp = a + (-10.0 * (k * a));
	} else {
		tmp = 0.1 / (k / a);
	}
	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
    real(8) :: tmp
    if (k <= 0.075d0) then
        tmp = a + ((-10.0d0) * (k * a))
    else
        tmp = 0.1d0 / (k / a)
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (k <= 0.075) {
		tmp = a + (-10.0 * (k * a));
	} else {
		tmp = 0.1 / (k / a);
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if k <= 0.075:
		tmp = a + (-10.0 * (k * a))
	else:
		tmp = 0.1 / (k / a)
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (k <= 0.075)
		tmp = Float64(a + Float64(-10.0 * Float64(k * a)));
	else
		tmp = Float64(0.1 / Float64(k / a));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (k <= 0.075)
		tmp = a + (-10.0 * (k * a));
	else
		tmp = 0.1 / (k / a);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[k, 0.075], N[(a + N[(-10.0 * N[(k * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(0.1 / N[(k / a), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 0.0749999999999999972

   1. Initial program 92.9%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 91.0%

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

      2. sqr-neg 91.0%

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

      3. associate-+l+ 91.0%

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

      4. sqr-neg 91.0%

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

      5. distribute-rgt-out 91.0%

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

   3. Simplified 91.0%

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

   5. Taylor expanded in k around 0 95.8%

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

   6. Taylor expanded in m around 0 31.9%

      \[\leadsto \color{blue}{a + -10 \cdot \left(a \cdot k\right)} \]

   if 0.0749999999999999972 < k

   1. Initial program 85.2%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 82.2%

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

      2. sqr-neg 82.2%

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

      3. associate-+l+ 82.2%

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

      4. sqr-neg 82.2%

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

      5. distribute-rgt-out 82.2%

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

   3. Simplified 82.2%

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

   5. Taylor expanded in m around 0 63.5%

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

   6. Taylor expanded in k around 0 30.8%

      \[\leadsto \frac{a}{1 + \color{blue}{10 \cdot k}} \]
   7. Step-by-step derivation

      1. *-commutative 30.8%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   8. Simplified 30.8%

      \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   9. Taylor expanded in k around inf 29.9%

      \[\leadsto \color{blue}{0.1 \cdot \frac{a}{k}} \]
   10. Step-by-step derivation

       1. clear-num 30.2%

          \[\leadsto 0.1 \cdot \color{blue}{\frac{1}{\frac{k}{a}}} \]

       2. un-div-inv 30.2%

          \[\leadsto \color{blue}{\frac{0.1}{\frac{k}{a}}} \]

   11. Applied egg-rr 30.2%

       \[\leadsto \color{blue}{\frac{0.1}{\frac{k}{a}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 31.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 0.075:\\ \;\;\;\;a + -10 \cdot \left(k \cdot a\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.1}{\frac{k}{a}}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 26.4% accurate, 11.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 35000000:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;0.1 \cdot \frac{a}{k}\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m) :precision binary64 (if (<= k 35000000.0) a (* 0.1 (/ a k))))

C

double code(double a, double k, double m) {
	double tmp;
	if (k <= 35000000.0) {
		tmp = a;
	} else {
		tmp = 0.1 * (a / 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
    real(8) :: tmp
    if (k <= 35000000.0d0) then
        tmp = a
    else
        tmp = 0.1d0 * (a / k)
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (k <= 35000000.0) {
		tmp = a;
	} else {
		tmp = 0.1 * (a / k);
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if k <= 35000000.0:
		tmp = a
	else:
		tmp = 0.1 * (a / k)
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (k <= 35000000.0)
		tmp = a;
	else
		tmp = Float64(0.1 * Float64(a / k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (k <= 35000000.0)
		tmp = a;
	else
		tmp = 0.1 * (a / k);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[k, 35000000.0], a, N[(0.1 * N[(a / k), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 3.5e7

   1. Initial program 93.1%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 91.2%

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

      2. sqr-neg 91.2%

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

      3. associate-+l+ 91.2%

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

      4. sqr-neg 91.2%

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

      5. distribute-rgt-out 91.2%

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

   3. Simplified 91.2%

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

   5. Taylor expanded in m around 0 36.3%

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

   6. Taylor expanded in k around 0 27.5%

      \[\leadsto \color{blue}{a} \]

   if 3.5e7 < k

   1. Initial program 84.7%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 81.6%

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

      2. sqr-neg 81.6%

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

      3. associate-+l+ 81.6%

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

      4. sqr-neg 81.6%

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

      5. distribute-rgt-out 81.6%

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

   3. Simplified 81.6%

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

   5. Taylor expanded in m around 0 65.4%

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

   6. Taylor expanded in k around 0 31.7%

      \[\leadsto \frac{a}{1 + \color{blue}{10 \cdot k}} \]
   7. Step-by-step derivation

      1. *-commutative 31.7%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   8. Simplified 31.7%

      \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   9. Taylor expanded in k around inf 30.7%

      \[\leadsto \color{blue}{0.1 \cdot \frac{a}{k}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 28.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 35000000:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;0.1 \cdot \frac{a}{k}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 26.7% accurate, 11.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 35000000:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;\frac{0.1}{\frac{k}{a}}\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m) :precision binary64 (if (<= k 35000000.0) a (/ 0.1 (/ k a))))

C

double code(double a, double k, double m) {
	double tmp;
	if (k <= 35000000.0) {
		tmp = a;
	} else {
		tmp = 0.1 / (k / a);
	}
	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
    real(8) :: tmp
    if (k <= 35000000.0d0) then
        tmp = a
    else
        tmp = 0.1d0 / (k / a)
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (k <= 35000000.0) {
		tmp = a;
	} else {
		tmp = 0.1 / (k / a);
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if k <= 35000000.0:
		tmp = a
	else:
		tmp = 0.1 / (k / a)
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (k <= 35000000.0)
		tmp = a;
	else
		tmp = Float64(0.1 / Float64(k / a));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (k <= 35000000.0)
		tmp = a;
	else
		tmp = 0.1 / (k / a);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[k, 35000000.0], a, N[(0.1 / N[(k / a), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 3.5e7

   1. Initial program 93.1%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 91.2%

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

      2. sqr-neg 91.2%

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

      3. associate-+l+ 91.2%

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

      4. sqr-neg 91.2%

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

      5. distribute-rgt-out 91.2%

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

   3. Simplified 91.2%

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

   5. Taylor expanded in m around 0 36.3%

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

   6. Taylor expanded in k around 0 27.5%

      \[\leadsto \color{blue}{a} \]

   if 3.5e7 < k

   1. Initial program 84.7%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 81.6%

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

      2. sqr-neg 81.6%

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

      3. associate-+l+ 81.6%

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

      4. sqr-neg 81.6%

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

      5. distribute-rgt-out 81.6%

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

   3. Simplified 81.6%

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

   5. Taylor expanded in m around 0 65.4%

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

   6. Taylor expanded in k around 0 31.7%

      \[\leadsto \frac{a}{1 + \color{blue}{10 \cdot k}} \]
   7. Step-by-step derivation

      1. *-commutative 31.7%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   8. Simplified 31.7%

      \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   9. Taylor expanded in k around inf 30.7%

      \[\leadsto \color{blue}{0.1 \cdot \frac{a}{k}} \]
   10. Step-by-step derivation

       1. clear-num 31.0%

          \[\leadsto 0.1 \cdot \color{blue}{\frac{1}{\frac{k}{a}}} \]

       2. un-div-inv 31.0%

          \[\leadsto \color{blue}{\frac{0.1}{\frac{k}{a}}} \]

   11. Applied egg-rr 31.0%

       \[\leadsto \color{blue}{\frac{0.1}{\frac{k}{a}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 28.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 35000000:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;\frac{0.1}{\frac{k}{a}}\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 20.0% accurate, 114.0× speedup

Math

\[\begin{array}{l} \\ a \end{array} \]

FPCore

(FPCore (a k m) :precision binary64 a)

C

double code(double a, double k, double m) {
	return a;
}

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
end function

Java

public static double code(double a, double k, double m) {
	return a;
}

Python

def code(a, k, m):
	return a

Julia

function code(a, k, m)
	return a
end

MATLAB

function tmp = code(a, k, m)
	tmp = a;
end

Wolfram

code[a_, k_, m_] := a

Derivation

1. Initial program 89.9%

   \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
2. Step-by-step derivation

   1. associate-*l/ 87.6%

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

   2. sqr-neg 87.6%

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

   3. associate-+l+ 87.6%

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

   4. sqr-neg 87.6%

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

   5. distribute-rgt-out 87.6%

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

3. Simplified 87.6%

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

5. Taylor expanded in m around 0 47.3%

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

6. Taylor expanded in k around 0 18.6%

   \[\leadsto \color{blue}{a} \]

7. Final simplification 18.6%

   \[\leadsto a \]
8. Add Preprocessing

Reproduce

herbie shell --seed 2024027 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))