Average Error: 0.5 → 0.5
Time: 15.3s
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 r96637 = 1.0;
        double r96638 = x;
        double r96639 = exp(r96638);
        double r96640 = r96637 + r96639;
        double r96641 = log(r96640);
        double r96642 = y;
        double r96643 = r96638 * r96642;
        double r96644 = r96641 - r96643;
        return r96644;
}
↓
double f(double x, double y) {
        double r96645 = 1.0;
        double r96646 = x;
        double r96647 = exp(r96646);
        double r96648 = r96645 + r96647;
        double r96649 = log(r96648);
        double r96650 = y;
        double r96651 = r96646 * r96650;
        double r96652 = r96649 - r96651;
        return r96652;
}

Error

Bits error versus x

Bits error versus y

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.5
Target0.0
Herbie0.5
\[\begin{array}{l} \mathbf{if}\;x \le 0.0:\\ \;\;\;\;\log \left(1 + e^{x}\right) - x \cdot y\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + e^{-x}\right) - \left(-x\right) \cdot \left(1 - y\right)\\ \end{array}\]

Derivation

  1. Initial program 0.5

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

  2. Final simplification0.5

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

Reproduce

herbie shell --seed 2019306 +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)))