x / (x^2 + 1)

Percentage Accurate: 76.4% → 100.0%

Time: 3.2s

Precision: binary64

Cost: 712

Math

\[\frac{x}{x \cdot x + 1} \]

↓

\[\begin{array}{l} \mathbf{if}\;x \leq -5 \cdot 10^{+21}:\\ \;\;\;\;\frac{1}{x}\\ \mathbf{elif}\;x \leq 50000000:\\ \;\;\;\;\frac{x}{1 + x \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x}\\ \end{array} \]

FPCore

(FPCore (x) :precision binary64 (/ x (+ (* x x) 1.0)))

↓

(FPCore (x)
 :precision binary64
 (if (<= x -5e+21)
   (/ 1.0 x)
   (if (<= x 50000000.0) (/ x (+ 1.0 (* x x))) (/ 1.0 x))))

C

double code(double x) {
	return x / ((x * x) + 1.0);
}

↓

double code(double x) {
	double tmp;
	if (x <= -5e+21) {
		tmp = 1.0 / x;
	} else if (x <= 50000000.0) {
		tmp = x / (1.0 + (x * x));
	} else {
		tmp = 1.0 / x;
	}
	return tmp;
}

Fortran

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

↓

real(8) function code(x)
    real(8), intent (in) :: x
    real(8) :: tmp
    if (x <= (-5d+21)) then
        tmp = 1.0d0 / x
    else if (x <= 50000000.0d0) then
        tmp = x / (1.0d0 + (x * x))
    else
        tmp = 1.0d0 / x
    end if
    code = tmp
end function

Java

public static double code(double x) {
	return x / ((x * x) + 1.0);
}

↓

public static double code(double x) {
	double tmp;
	if (x <= -5e+21) {
		tmp = 1.0 / x;
	} else if (x <= 50000000.0) {
		tmp = x / (1.0 + (x * x));
	} else {
		tmp = 1.0 / x;
	}
	return tmp;
}

Python

def code(x):
	return x / ((x * x) + 1.0)

↓

def code(x):
	tmp = 0
	if x <= -5e+21:
		tmp = 1.0 / x
	elif x <= 50000000.0:
		tmp = x / (1.0 + (x * x))
	else:
		tmp = 1.0 / x
	return tmp

Julia

function code(x)
	return Float64(x / Float64(Float64(x * x) + 1.0))
end

↓

function code(x)
	tmp = 0.0
	if (x <= -5e+21)
		tmp = Float64(1.0 / x);
	elseif (x <= 50000000.0)
		tmp = Float64(x / Float64(1.0 + Float64(x * x)));
	else
		tmp = Float64(1.0 / x);
	end
	return tmp
end

MATLAB

function tmp = code(x)
	tmp = x / ((x * x) + 1.0);
end

↓

function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= -5e+21)
		tmp = 1.0 / x;
	elseif (x <= 50000000.0)
		tmp = x / (1.0 + (x * x));
	else
		tmp = 1.0 / x;
	end
	tmp_2 = tmp;
end

Wolfram

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

↓

code[x_] := If[LessEqual[x, -5e+21], N[(1.0 / x), $MachinePrecision], If[LessEqual[x, 50000000.0], N[(x / N[(1.0 + N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(1.0 / x), $MachinePrecision]]]

Target

Original	76.4%
Target	99.8%
Herbie	100.0%

\[\frac{1}{x + \frac{1}{x}} \]

Derivation

1. Split input into 2 regimes

2. if x < -5e21 or 5e7 < x

   1. Initial program 50.0%

      \[\frac{x}{x \cdot x + 1} \]

   2. Taylor expanded in x around inf 100.0%

      \[\leadsto \color{blue}{\frac{1}{x}} \]

   if -5e21 < x < 5e7

   1. Initial program 100.0%

      \[\frac{x}{x \cdot x + 1} \]
3. Recombined 2 regimes into one program.

4. Final simplification 100.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -5 \cdot 10^{+21}:\\ \;\;\;\;\frac{1}{x}\\ \mathbf{elif}\;x \leq 50000000:\\ \;\;\;\;\frac{x}{1 + x \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	100.0%
Cost	712

\[\begin{array}{l} \mathbf{if}\;x \leq -5 \cdot 10^{+21}:\\ \;\;\;\;\frac{1}{x}\\ \mathbf{elif}\;x \leq 50000000:\\ \;\;\;\;\frac{x}{1 + x \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x}\\ \end{array} \]

Alternative 2
Accuracy	98.8%
Cost	456

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

Alternative 3
Accuracy	50.6%
Cost	64

\[x \]

Reproduce

herbie shell --seed 2023263 
(FPCore (x)
  :name "x / (x^2 + 1)"
  :precision binary64

  :herbie-target
  (/ 1.0 (+ x (/ 1.0 x)))

  (/ x (+ (* x x) 1.0)))