Average Error: 0.0 → 0.0
Time: 8.4s
Precision: 64
\[0 \le x \le 2\]
\[x \cdot \left(x \cdot x\right) + x \cdot x\]
↓
\[x \cdot x + x \cdot \left(x \cdot x\right)\]
x \cdot \left(x \cdot x\right) + x \cdot x
↓
x \cdot x + x \cdot \left(x \cdot x\right)
double f(double x) {
        double r11391992 = x;
        double r11391993 = r11391992 * r11391992;
        double r11391994 = r11391992 * r11391993;
        double r11391995 = r11391994 + r11391993;
        return r11391995;
}
↓
double f(double x) {
        double r11391996 = x;
        double r11391997 = r11391996 * r11391996;
        double r11391998 = r11391996 * r11391997;
        double r11391999 = r11391997 + r11391998;
        return r11391999;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.0
Target0.0
Herbie0.0
\[\left(\left(1.0 + x\right) \cdot x\right) \cdot x\]

Derivation

  1. Initial program 0.0

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

  2. Final simplification0.0

    \[\leadsto x \cdot x + x \cdot \left(x \cdot x\right)\]

Reproduce

herbie shell --seed 2019104 +o rules:numerics
(FPCore (x)
  :name "Expression 3, p15"
  :pre (<= 0 x 2)

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

  (+ (* x (* x x)) (* x x)))