Average Error: 0.5 → 0.5
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 r197142 = 1.0;
        double r197143 = x;
        double r197144 = exp(r197143);
        double r197145 = r197142 + r197144;
        double r197146 = log(r197145);
        double r197147 = y;
        double r197148 = r197143 * r197147;
        double r197149 = r197146 - r197148;
        return r197149;
}
↓
double f(double x, double y) {
        double r197150 = 1.0;
        double r197151 = x;
        double r197152 = exp(r197151);
        double r197153 = r197150 + r197152;
        double r197154 = log(r197153);
        double r197155 = y;
        double r197156 = r197151 * r197155;
        double r197157 = r197154 - r197156;
        return r197157;
}

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 2020057 +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)))