Logistic function from Lakshay Garg

Percentage Accurate: 54.4% → 99.8%

Time: 13.2s

Precision: binary64

Cost: 34244

Math

\[\frac{2}{1 + e^{-2 \cdot x}} - 1 \]

↓

\[\begin{array}{l} t_0 := 1 + e^{-2 \cdot x}\\ \mathbf{if}\;-2 \cdot x \leq -0.02:\\ \;\;\;\;\frac{1}{\frac{1 + \left(\frac{4}{{t_0}^{2}} + \frac{2}{t_0}\right)}{-1 + \frac{8}{{t_0}^{3}}}}\\ \mathbf{elif}\;-2 \cdot x \leq 0.01:\\ \;\;\;\;-0.3333333333333333 \cdot {x}^{3} + \left(x + 0.13333333333333333 \cdot {x}^{5}\right)\\ \mathbf{else}:\\ \;\;\;\;-1\\ \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (- (/ 2.0 (+ 1.0 (exp (* -2.0 x)))) 1.0))

↓

(FPCore (x y)
 :precision binary64
 (let* ((t_0 (+ 1.0 (exp (* -2.0 x)))))
   (if (<= (* -2.0 x) -0.02)
     (/
      1.0
      (/
       (+ 1.0 (+ (/ 4.0 (pow t_0 2.0)) (/ 2.0 t_0)))
       (+ -1.0 (/ 8.0 (pow t_0 3.0)))))
     (if (<= (* -2.0 x) 0.01)
       (+
        (* -0.3333333333333333 (pow x 3.0))
        (+ x (* 0.13333333333333333 (pow x 5.0))))
       -1.0))))

C

double code(double x, double y) {
	return (2.0 / (1.0 + exp((-2.0 * x)))) - 1.0;
}

↓

double code(double x, double y) {
	double t_0 = 1.0 + exp((-2.0 * x));
	double tmp;
	if ((-2.0 * x) <= -0.02) {
		tmp = 1.0 / ((1.0 + ((4.0 / pow(t_0, 2.0)) + (2.0 / t_0))) / (-1.0 + (8.0 / pow(t_0, 3.0))));
	} else if ((-2.0 * x) <= 0.01) {
		tmp = (-0.3333333333333333 * pow(x, 3.0)) + (x + (0.13333333333333333 * pow(x, 5.0)));
	} else {
		tmp = -1.0;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = (2.0d0 / (1.0d0 + exp(((-2.0d0) * x)))) - 1.0d0
end function

↓

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: t_0
    real(8) :: tmp
    t_0 = 1.0d0 + exp(((-2.0d0) * x))
    if (((-2.0d0) * x) <= (-0.02d0)) then
        tmp = 1.0d0 / ((1.0d0 + ((4.0d0 / (t_0 ** 2.0d0)) + (2.0d0 / t_0))) / ((-1.0d0) + (8.0d0 / (t_0 ** 3.0d0))))
    else if (((-2.0d0) * x) <= 0.01d0) then
        tmp = ((-0.3333333333333333d0) * (x ** 3.0d0)) + (x + (0.13333333333333333d0 * (x ** 5.0d0)))
    else
        tmp = -1.0d0
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	return (2.0 / (1.0 + Math.exp((-2.0 * x)))) - 1.0;
}

↓

public static double code(double x, double y) {
	double t_0 = 1.0 + Math.exp((-2.0 * x));
	double tmp;
	if ((-2.0 * x) <= -0.02) {
		tmp = 1.0 / ((1.0 + ((4.0 / Math.pow(t_0, 2.0)) + (2.0 / t_0))) / (-1.0 + (8.0 / Math.pow(t_0, 3.0))));
	} else if ((-2.0 * x) <= 0.01) {
		tmp = (-0.3333333333333333 * Math.pow(x, 3.0)) + (x + (0.13333333333333333 * Math.pow(x, 5.0)));
	} else {
		tmp = -1.0;
	}
	return tmp;
}

Python

def code(x, y):
	return (2.0 / (1.0 + math.exp((-2.0 * x)))) - 1.0

↓

def code(x, y):
	t_0 = 1.0 + math.exp((-2.0 * x))
	tmp = 0
	if (-2.0 * x) <= -0.02:
		tmp = 1.0 / ((1.0 + ((4.0 / math.pow(t_0, 2.0)) + (2.0 / t_0))) / (-1.0 + (8.0 / math.pow(t_0, 3.0))))
	elif (-2.0 * x) <= 0.01:
		tmp = (-0.3333333333333333 * math.pow(x, 3.0)) + (x + (0.13333333333333333 * math.pow(x, 5.0)))
	else:
		tmp = -1.0
	return tmp

Julia

function code(x, y)
	return Float64(Float64(2.0 / Float64(1.0 + exp(Float64(-2.0 * x)))) - 1.0)
end

↓

function code(x, y)
	t_0 = Float64(1.0 + exp(Float64(-2.0 * x)))
	tmp = 0.0
	if (Float64(-2.0 * x) <= -0.02)
		tmp = Float64(1.0 / Float64(Float64(1.0 + Float64(Float64(4.0 / (t_0 ^ 2.0)) + Float64(2.0 / t_0))) / Float64(-1.0 + Float64(8.0 / (t_0 ^ 3.0)))));
	elseif (Float64(-2.0 * x) <= 0.01)
		tmp = Float64(Float64(-0.3333333333333333 * (x ^ 3.0)) + Float64(x + Float64(0.13333333333333333 * (x ^ 5.0))));
	else
		tmp = -1.0;
	end
	return tmp
end

MATLAB

function tmp = code(x, y)
	tmp = (2.0 / (1.0 + exp((-2.0 * x)))) - 1.0;
end

↓

function tmp_2 = code(x, y)
	t_0 = 1.0 + exp((-2.0 * x));
	tmp = 0.0;
	if ((-2.0 * x) <= -0.02)
		tmp = 1.0 / ((1.0 + ((4.0 / (t_0 ^ 2.0)) + (2.0 / t_0))) / (-1.0 + (8.0 / (t_0 ^ 3.0))));
	elseif ((-2.0 * x) <= 0.01)
		tmp = (-0.3333333333333333 * (x ^ 3.0)) + (x + (0.13333333333333333 * (x ^ 5.0)));
	else
		tmp = -1.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := N[(N[(2.0 / N[(1.0 + N[Exp[N[(-2.0 * x), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - 1.0), $MachinePrecision]

↓

code[x_, y_] := Block[{t$95$0 = N[(1.0 + N[Exp[N[(-2.0 * x), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(-2.0 * x), $MachinePrecision], -0.02], N[(1.0 / N[(N[(1.0 + N[(N[(4.0 / N[Power[t$95$0, 2.0], $MachinePrecision]), $MachinePrecision] + N[(2.0 / t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(-1.0 + N[(8.0 / N[Power[t$95$0, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(-2.0 * x), $MachinePrecision], 0.01], N[(N[(-0.3333333333333333 * N[Power[x, 3.0], $MachinePrecision]), $MachinePrecision] + N[(x + N[(0.13333333333333333 * N[Power[x, 5.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], -1.0]]]

Derivation

1. Split input into 3 regimes

2. if (*.f64 -2 x) < -0.0200000000000000004

   1. Initial program 99.9%

      \[\frac{2}{1 + e^{-2 \cdot x}} - 1 \]

   2. Applied egg-rr 100.0%

      \[\leadsto \color{blue}{\frac{1}{\frac{\mathsf{fma}\left(4, {\left(1 + {\left(e^{-2}\right)}^{x}\right)}^{-2}, 1\right) + \frac{2}{1 + {\left(e^{-2}\right)}^{x}}}{\mathsf{fma}\left(8, {\left(1 + {\left(e^{-2}\right)}^{x}\right)}^{-3}, -1\right)}}} \]
      Step-by-step derivation

      [Start] 99.9	\[ \frac{2}{1 + e^{-2 \cdot x}} - 1 \]
      flip3-- [=>] 99.9	\[ \color{blue}{\frac{{\left(\frac{2}{1 + e^{-2 \cdot x}}\right)}^{3} - {1}^{3}}{\frac{2}{1 + e^{-2 \cdot x}} \cdot \frac{2}{1 + e^{-2 \cdot x}} + \left(1 \cdot 1 + \frac{2}{1 + e^{-2 \cdot x}} \cdot 1\right)}} \]
      clear-num [=>] 99.9	\[ \color{blue}{\frac{1}{\frac{\frac{2}{1 + e^{-2 \cdot x}} \cdot \frac{2}{1 + e^{-2 \cdot x}} + \left(1 \cdot 1 + \frac{2}{1 + e^{-2 \cdot x}} \cdot 1\right)}{{\left(\frac{2}{1 + e^{-2 \cdot x}}\right)}^{3} - {1}^{3}}}} \]

   3. Taylor expanded in x around inf 100.0%

      \[\leadsto \frac{1}{\color{blue}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}}} \]

   4. Applied egg-rr 100.0%

      \[\leadsto \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\color{blue}{\frac{8}{{\left(1 + {\left(e^{-2}\right)}^{x}\right)}^{3}} + -1}}} \]
      Step-by-step derivation

      [Start] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      sub-neg [=>] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\color{blue}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} + \left(-1\right)}}} \]
      un-div-inv [=>] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\color{blue}{\frac{8}{{\left(1 + e^{-2 \cdot x}\right)}^{3}}} + \left(-1\right)}} \]
      exp-prod [=>] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\frac{8}{{\left(1 + \color{blue}{{\left(e^{-2}\right)}^{x}}\right)}^{3}} + \left(-1\right)}} \]
      metadata-eval [=>] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\frac{8}{{\left(1 + {\left(e^{-2}\right)}^{x}\right)}^{3}} + \color{blue}{-1}}} \]

   5. Simplified 100.0%

      \[\leadsto \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\color{blue}{\frac{8}{{\left(1 + e^{x \cdot -2}\right)}^{3}} + -1}}} \]
      Step-by-step derivation

      [Start] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\frac{8}{{\left(1 + {\left(e^{-2}\right)}^{x}\right)}^{3}} + -1}} \]
      exp-prod [<=] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\frac{8}{{\left(1 + \color{blue}{e^{-2 \cdot x}}\right)}^{3}} + -1}} \]
      *-commutative [=>] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{\frac{8}{{\left(1 + e^{\color{blue}{x \cdot -2}}\right)}^{3}} + -1}} \]

   6. Taylor expanded in x around inf 100.0%

      \[\leadsto \frac{1}{\color{blue}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}}} \]

   7. Simplified 100.0%

      \[\leadsto \frac{1}{\color{blue}{\frac{1 + \left(\frac{4}{{\left(1 + e^{x \cdot -2}\right)}^{2}} + \frac{2}{1 + e^{x \cdot -2}}\right)}{-1 + \frac{8}{{\left(1 + e^{x \cdot -2}\right)}^{3}}}}} \]
      Step-by-step derivation

      [Start] 100.0	\[ \frac{1}{\frac{1 + \left(2 \cdot \frac{1}{1 + e^{-2 \cdot x}} + 4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      +-commutative [=>] 100.0	\[ \frac{1}{\frac{1 + \color{blue}{\left(4 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}} + 2 \cdot \frac{1}{1 + e^{-2 \cdot x}}\right)}}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      associate-*r/ [=>] 100.0	\[ \frac{1}{\frac{1 + \left(\color{blue}{\frac{4 \cdot 1}{{\left(1 + e^{-2 \cdot x}\right)}^{2}}} + 2 \cdot \frac{1}{1 + e^{-2 \cdot x}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      metadata-eval [=>] 100.0	\[ \frac{1}{\frac{1 + \left(\frac{\color{blue}{4}}{{\left(1 + e^{-2 \cdot x}\right)}^{2}} + 2 \cdot \frac{1}{1 + e^{-2 \cdot x}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      *-commutative [=>] 100.0	\[ \frac{1}{\frac{1 + \left(\frac{4}{{\left(1 + e^{\color{blue}{x \cdot -2}}\right)}^{2}} + 2 \cdot \frac{1}{1 + e^{-2 \cdot x}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      associate-*r/ [=>] 100.0	\[ \frac{1}{\frac{1 + \left(\frac{4}{{\left(1 + e^{x \cdot -2}\right)}^{2}} + \color{blue}{\frac{2 \cdot 1}{1 + e^{-2 \cdot x}}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      metadata-eval [=>] 100.0	\[ \frac{1}{\frac{1 + \left(\frac{4}{{\left(1 + e^{x \cdot -2}\right)}^{2}} + \frac{\color{blue}{2}}{1 + e^{-2 \cdot x}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      *-commutative [=>] 100.0	\[ \frac{1}{\frac{1 + \left(\frac{4}{{\left(1 + e^{x \cdot -2}\right)}^{2}} + \frac{2}{1 + e^{\color{blue}{x \cdot -2}}}\right)}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} - 1}} \]
      sub-neg [=>] 100.0	\[ \frac{1}{\frac{1 + \left(\frac{4}{{\left(1 + e^{x \cdot -2}\right)}^{2}} + \frac{2}{1 + e^{x \cdot -2}}\right)}{\color{blue}{8 \cdot \frac{1}{{\left(1 + e^{-2 \cdot x}\right)}^{3}} + \left(-1\right)}}} \]

   if -0.0200000000000000004 < (*.f64 -2 x) < 0.0100000000000000002

   1. Initial program 7.3%

      \[\frac{2}{1 + e^{-2 \cdot x}} - 1 \]

   2. Taylor expanded in x around 0 100.0%

      \[\leadsto \color{blue}{-0.3333333333333333 \cdot {x}^{3} + \left(0.13333333333333333 \cdot {x}^{5} + x\right)} \]

   if 0.0100000000000000002 < (*.f64 -2 x)

   1. Initial program 100.0%

      \[\frac{2}{1 + e^{-2 \cdot x}} - 1 \]

   2. Taylor expanded in x around 0 97.1%

      \[\leadsto \frac{2}{\color{blue}{2 + -2 \cdot x}} - 1 \]

   3. Simplified 97.1%

      \[\leadsto \frac{2}{\color{blue}{2 + x \cdot -2}} - 1 \]
      Step-by-step derivation

      [Start] 97.1	\[ \frac{2}{2 + -2 \cdot x} - 1 \]
      *-commutative [=>] 97.1	\[ \frac{2}{2 + \color{blue}{x \cdot -2}} - 1 \]

   4. Taylor expanded in x around inf 100.0%

      \[\leadsto \color{blue}{-1} \]
3. Recombined 3 regimes into one program.

4. Final simplification 100.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;-2 \cdot x \leq -0.02:\\ \;\;\;\;\frac{1}{\frac{1 + \left(\frac{4}{{\left(1 + e^{-2 \cdot x}\right)}^{2}} + \frac{2}{1 + e^{-2 \cdot x}}\right)}{-1 + \frac{8}{{\left(1 + e^{-2 \cdot x}\right)}^{3}}}}\\ \mathbf{elif}\;-2 \cdot x \leq 0.01:\\ \;\;\;\;-0.3333333333333333 \cdot {x}^{3} + \left(x + 0.13333333333333333 \cdot {x}^{5}\right)\\ \mathbf{else}:\\ \;\;\;\;-1\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	99.8%
Cost	34116

\[\begin{array}{l} t_0 := 1 + e^{-2 \cdot x}\\ \mathbf{if}\;-2 \cdot x \leq -0.4:\\ \;\;\;\;\frac{-1 + \frac{8}{{t_0}^{3}}}{\frac{4}{{t_0}^{2}} + \left(1 + \frac{2}{t_0}\right)}\\ \mathbf{elif}\;-2 \cdot x \leq 0.01:\\ \;\;\;\;\left(-0.3333333333333333 \cdot {x}^{3} + \left(x + 0.13333333333333333 \cdot {x}^{5}\right)\right) + -0.05396825396825397 \cdot {x}^{7}\\ \mathbf{else}:\\ \;\;\;\;-1\\ \end{array} \]

Alternative 2
Accuracy	99.8%
Cost	20744

\[\begin{array}{l} \mathbf{if}\;-2 \cdot x \leq -0.4:\\ \;\;\;\;\frac{2}{1 + e^{-2 \cdot x}} + -1\\ \mathbf{elif}\;-2 \cdot x \leq 0.01:\\ \;\;\;\;\left(-0.3333333333333333 \cdot {x}^{3} + \left(x + 0.13333333333333333 \cdot {x}^{5}\right)\right) + -0.05396825396825397 \cdot {x}^{7}\\ \mathbf{else}:\\ \;\;\;\;-1\\ \end{array} \]

Alternative 3
Accuracy	99.8%
Cost	14024

\[\begin{array}{l} \mathbf{if}\;-2 \cdot x \leq -0.02:\\ \;\;\;\;\frac{2}{1 + e^{-2 \cdot x}} + -1\\ \mathbf{elif}\;-2 \cdot x \leq 0.01:\\ \;\;\;\;-0.3333333333333333 \cdot {x}^{3} + \left(x + 0.13333333333333333 \cdot {x}^{5}\right)\\ \mathbf{else}:\\ \;\;\;\;-1\\ \end{array} \]

Alternative 4
Accuracy	99.9%
Cost	7497

\[\begin{array}{l} \mathbf{if}\;-2 \cdot x \leq -0.02 \lor \neg \left(-2 \cdot x \leq 0.0005\right):\\ \;\;\;\;\frac{2}{1 + e^{-2 \cdot x}} + -1\\ \mathbf{else}:\\ \;\;\;\;x + -0.3333333333333333 \cdot {x}^{3}\\ \end{array} \]

Alternative 5
Accuracy	75.8%
Cost	964

\[\begin{array}{l} \mathbf{if}\;x \leq -1.65:\\ \;\;\;\;-1\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\left(-1 + \left(1 + x \cdot 0.3333333333333333\right)\right) + \frac{1}{x}}\\ \end{array} \]

Alternative 6
Accuracy	75.8%
Cost	708

\[\begin{array}{l} \mathbf{if}\;x \leq -1.65:\\ \;\;\;\;-1\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x \cdot 0.3333333333333333 + \frac{1}{x}}\\ \end{array} \]

Alternative 7
Accuracy	75.7%
Cost	196

\[\begin{array}{l} \mathbf{if}\;x \leq -1:\\ \;\;\;\;-1\\ \mathbf{else}:\\ \;\;\;\;x\\ \end{array} \]

Alternative 8
Accuracy	27.4%
Cost	64

\[-1 \]

Reproduce

herbie shell --seed 2023160 
(FPCore (x y)
  :name "Logistic function from Lakshay Garg"
  :precision binary64
  (- (/ 2.0 (+ 1.0 (exp (* -2.0 x)))) 1.0))