Average Error: 0.1 → 0
Time: 3.6s
Precision: 64
\[\left(\left(d1 \cdot d1\right) \cdot d1\right) \cdot d1\]
↓
\[{d1}^{4}\]
\left(\left(d1 \cdot d1\right) \cdot d1\right) \cdot d1
↓
{d1}^{4}
double f(double d1) {
        double r20067 = d1;
        double r20068 = r20067 * r20067;
        double r20069 = r20068 * r20067;
        double r20070 = r20069 * r20067;
        return r20070;
}
↓
double f(double d1) {
        double r20071 = d1;
        double r20072 = 4.0;
        double r20073 = pow(r20071, r20072);
        return r20073;
}

Error

Bits error versus d1

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.1
Target0
Herbie0
\[{d1}^{4}\]

Derivation

  1. Initial program 0.1

    \[\left(\left(d1 \cdot d1\right) \cdot d1\right) \cdot d1\]

  2. Simplified0

    \[\leadsto \color{blue}{{d1}^{4}}\]

  3. Final simplification0

    \[\leadsto {d1}^{4}\]

Reproduce

herbie shell --seed 2019310 +o rules:numerics
(FPCore (d1)
  :name "FastMath repmul"
  :precision binary64

  :herbie-target
  (pow d1 4)

  (* (* (* d1 d1) d1) d1))