Average Error: 0.5 → 0.5
Time: 5.2s
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 r143677 = 1.0;
        double r143678 = x;
        double r143679 = exp(r143678);
        double r143680 = r143677 + r143679;
        double r143681 = log(r143680);
        double r143682 = y;
        double r143683 = r143678 * r143682;
        double r143684 = r143681 - r143683;
        return r143684;
}
↓
double f(double x, double y) {
        double r143685 = 1.0;
        double r143686 = x;
        double r143687 = exp(r143686);
        double r143688 = r143685 + r143687;
        double r143689 = log(r143688);
        double r143690 = y;
        double r143691 = r143686 * r143690;
        double r143692 = r143689 - r143691;
        return r143692;
}

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
Target0.1
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 2020065 +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)))