Average Error: 0.6 → 0.6
Time: 5.5s
Precision: 64
\[\log \left(1 + e^{x}\right) - x \cdot y\]
↓
\[\log \left(1 + e^{x}\right) - x \cdot y\]
\log \left(1 + e^{x}\right) - x \cdot y
↓
\log \left(1 + e^{x}\right) - x \cdot y
double f(double x, double y) {
        double r165383 = 1.0;
        double r165384 = x;
        double r165385 = exp(r165384);
        double r165386 = r165383 + r165385;
        double r165387 = log(r165386);
        double r165388 = y;
        double r165389 = r165384 * r165388;
        double r165390 = r165387 - r165389;
        return r165390;
}
↓
double f(double x, double y) {
        double r165391 = 1.0;
        double r165392 = x;
        double r165393 = exp(r165392);
        double r165394 = r165391 + r165393;
        double r165395 = log(r165394);
        double r165396 = y;
        double r165397 = r165392 * r165396;
        double r165398 = r165395 - r165397;
        return r165398;
}

Error

Bits error versus x

Bits error versus y

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.6
Target0.1
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;x \le 0.0:\\ \;\;\;\;\log \left(1 + e^{x}\right) - x \cdot y\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + e^{-x}\right) - \left(-x\right) \cdot \left(1 - y\right)\\ \end{array}\]

Derivation

  1. Initial program 0.6

    \[\log \left(1 + e^{x}\right) - x \cdot y\]

  2. Final simplification0.6

    \[\leadsto \log \left(1 + e^{x}\right) - x \cdot y\]

Reproduce

herbie shell --seed 2020046 +o rules:numerics
(FPCore (x y)
  :name "Logistic regression 2"
  :precision binary64

  :herbie-target
  (if (<= x 0.0) (- (log (+ 1 (exp x))) (* x y)) (- (log (+ 1 (exp (- x)))) (* (- x) (- 1 y))))

  (- (log (+ 1 (exp x))) (* x y)))