Average Error: 5.7 → 0
Time: 2.8s
Precision: 64
\[e^{\log a + \log b}\]
\[a \cdot b\]
e^{\log a + \log b}
a \cdot b
double f(double a, double b) {
        double r159403 = a;
        double r159404 = log(r159403);
        double r159405 = b;
        double r159406 = log(r159405);
        double r159407 = r159404 + r159406;
        double r159408 = exp(r159407);
        return r159408;
}

double f(double a, double b) {
        double r159409 = a;
        double r159410 = b;
        double r159411 = r159409 * r159410;
        return r159411;
}

Error

Bits error versus a

Bits error versus b

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original5.7
Target0
Herbie0
\[a \cdot b\]

Derivation

  1. Initial program 5.7

    \[e^{\log a + \log b}\]
  2. Using strategy rm
  3. Applied exp-sum5.4

    \[\leadsto \color{blue}{e^{\log a} \cdot e^{\log b}}\]
  4. Simplified4.7

    \[\leadsto \color{blue}{a} \cdot e^{\log b}\]
  5. Simplified0

    \[\leadsto a \cdot \color{blue}{b}\]
  6. Final simplification0

    \[\leadsto a \cdot b\]

Reproduce

herbie shell --seed 2020033 +o rules:numerics
(FPCore (a b)
  :name "Exp of sum of logs"
  :precision binary64

  :herbie-target
  (* a b)

  (exp (+ (log a) (log b))))