Average Error: 0.3 → 0.3
Time: 1.3s
Precision: 64
\[\left(x \cdot 27\right) \cdot y\]
↓
\[\left(x \cdot 27\right) \cdot y\]
\left(x \cdot 27\right) \cdot y
↓
\left(x \cdot 27\right) \cdot y
double f(double x, double y) {
        double r204768 = x;
        double r204769 = 27.0;
        double r204770 = r204768 * r204769;
        double r204771 = y;
        double r204772 = r204770 * r204771;
        return r204772;
}
↓
double f(double x, double y) {
        double r204773 = x;
        double r204774 = 27.0;
        double r204775 = r204773 * r204774;
        double r204776 = y;
        double r204777 = r204775 * r204776;
        return r204777;
}

Error

Bits error versus x

Bits error versus y

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 0.3

    \[\left(x \cdot 27\right) \cdot y\]

  2. Final simplification0.3

    \[\leadsto \left(x \cdot 27\right) \cdot y\]

Reproduce

herbie shell --seed 2020062 +o rules:numerics
(FPCore (x y)
  :name "Diagrams.Solve.Polynomial:cubForm  from diagrams-solve-0.1, F"
  :precision binary64
  (* (* x 27) y))