?

Average Accuracy: 91.2% → 99.9%
Time: 9.4s
Precision: binary64
Cost: 713

?

\[x \cdot \left(1 + y \cdot y\right) \]
\[\begin{array}{l} \mathbf{if}\;y \leq -5.4 \cdot 10^{+135} \lor \neg \left(y \leq 5000000\right):\\ \;\;\;\;y \cdot \left(x \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;x + x \cdot \left(y \cdot y\right)\\ \end{array} \]
(FPCore (x y) :precision binary64 (* x (+ 1.0 (* y y))))
(FPCore (x y)
 :precision binary64
 (if (or (<= y -5.4e+135) (not (<= y 5000000.0)))
   (* y (* x y))
   (+ x (* x (* y y)))))
double code(double x, double y) {
	return x * (1.0 + (y * y));
}
double code(double x, double y) {
	double tmp;
	if ((y <= -5.4e+135) || !(y <= 5000000.0)) {
		tmp = y * (x * y);
	} else {
		tmp = x + (x * (y * y));
	}
	return tmp;
}
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x * (1.0d0 + (y * y))
end function
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if ((y <= (-5.4d+135)) .or. (.not. (y <= 5000000.0d0))) then
        tmp = y * (x * y)
    else
        tmp = x + (x * (y * y))
    end if
    code = tmp
end function
public static double code(double x, double y) {
	return x * (1.0 + (y * y));
}
public static double code(double x, double y) {
	double tmp;
	if ((y <= -5.4e+135) || !(y <= 5000000.0)) {
		tmp = y * (x * y);
	} else {
		tmp = x + (x * (y * y));
	}
	return tmp;
}
def code(x, y):
	return x * (1.0 + (y * y))
def code(x, y):
	tmp = 0
	if (y <= -5.4e+135) or not (y <= 5000000.0):
		tmp = y * (x * y)
	else:
		tmp = x + (x * (y * y))
	return tmp
function code(x, y)
	return Float64(x * Float64(1.0 + Float64(y * y)))
end
function code(x, y)
	tmp = 0.0
	if ((y <= -5.4e+135) || !(y <= 5000000.0))
		tmp = Float64(y * Float64(x * y));
	else
		tmp = Float64(x + Float64(x * Float64(y * y)));
	end
	return tmp
end
function tmp = code(x, y)
	tmp = x * (1.0 + (y * y));
end
function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y <= -5.4e+135) || ~((y <= 5000000.0)))
		tmp = y * (x * y);
	else
		tmp = x + (x * (y * y));
	end
	tmp_2 = tmp;
end
code[x_, y_] := N[(x * N[(1.0 + N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
code[x_, y_] := If[Or[LessEqual[y, -5.4e+135], N[Not[LessEqual[y, 5000000.0]], $MachinePrecision]], N[(y * N[(x * y), $MachinePrecision]), $MachinePrecision], N[(x + N[(x * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
x \cdot \left(1 + y \cdot y\right)
\begin{array}{l}
\mathbf{if}\;y \leq -5.4 \cdot 10^{+135} \lor \neg \left(y \leq 5000000\right):\\
\;\;\;\;y \cdot \left(x \cdot y\right)\\

\mathbf{else}:\\
\;\;\;\;x + x \cdot \left(y \cdot y\right)\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original91.2%
Target99.9%
Herbie99.9%
\[x + \left(x \cdot y\right) \cdot y \]

Derivation?

  1. Split input into 2 regimes
  2. if y < -5.3999999999999997e135 or 5e6 < y

    1. Initial program 60.2%

      \[x \cdot \left(1 + y \cdot y\right) \]
    2. Taylor expanded in y around inf 60.2%

      \[\leadsto \color{blue}{{y}^{2} \cdot x} \]
    3. Simplified99.6%

      \[\leadsto \color{blue}{y \cdot \left(y \cdot x\right)} \]
      Proof

      [Start]60.2

      \[ {y}^{2} \cdot x \]

      unpow2 [=>]60.2

      \[ \color{blue}{\left(y \cdot y\right)} \cdot x \]

      associate-*l* [=>]99.6

      \[ \color{blue}{y \cdot \left(y \cdot x\right)} \]

    if -5.3999999999999997e135 < y < 5e6

    1. Initial program 99.9%

      \[x \cdot \left(1 + y \cdot y\right) \]
    2. Applied egg-rr99.9%

      \[\leadsto \color{blue}{x \cdot \left(y \cdot y\right) + x} \]
      Proof

      [Start]99.9

      \[ x \cdot \left(1 + y \cdot y\right) \]

      distribute-rgt-in [=>]99.9

      \[ \color{blue}{1 \cdot x + \left(y \cdot y\right) \cdot x} \]

      *-un-lft-identity [<=]99.9

      \[ \color{blue}{x} + \left(y \cdot y\right) \cdot x \]

      +-commutative [=>]99.9

      \[ \color{blue}{\left(y \cdot y\right) \cdot x + x} \]

      *-commutative [=>]99.9

      \[ \color{blue}{x \cdot \left(y \cdot y\right)} + x \]
  3. Recombined 2 regimes into one program.
  4. Final simplification99.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq -5.4 \cdot 10^{+135} \lor \neg \left(y \leq 5000000\right):\\ \;\;\;\;y \cdot \left(x \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;x + x \cdot \left(y \cdot y\right)\\ \end{array} \]

Alternatives

Alternative 1
Accuracy99.8%
Cost13376
\[\mathsf{hypot}\left(1, y\right) \cdot \frac{x}{\frac{1}{\mathsf{hypot}\left(1, y\right)}} \]
Alternative 2
Accuracy99.9%
Cost713
\[\begin{array}{l} \mathbf{if}\;y \leq -5.4 \cdot 10^{+135} \lor \neg \left(y \leq 4200000000000\right):\\ \;\;\;\;y \cdot \left(x \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(1 + y \cdot y\right)\\ \end{array} \]
Alternative 3
Accuracy89.9%
Cost580
\[\begin{array}{l} \mathbf{if}\;y \cdot y \leq 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(y \cdot y\right)\\ \end{array} \]
Alternative 4
Accuracy98.6%
Cost580
\[\begin{array}{l} \mathbf{if}\;y \cdot y \leq 0.1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(x \cdot y\right)\\ \end{array} \]
Alternative 5
Accuracy67.4%
Cost64
\[x \]

Error

Reproduce?

herbie shell --seed 2023143 
(FPCore (x y)
  :name "Numeric.Integration.TanhSinh:everywhere from integration-0.2.1"
  :precision binary64

  :herbie-target
  (+ x (* (* x y) y))

  (* x (+ 1.0 (* y y))))