Average Error: 0.6 → 0.6
Time: 4.9s
Precision: 64
\[\log \left(1 + e^{x}\right) - x \cdot y\]
↓
\[\log \left(\frac{{1}^{3} + {\left(e^{x}\right)}^{3}}{e^{x} \cdot \left(e^{x} - 1\right) + 1 \cdot 1}\right) - x \cdot y\]
\log \left(1 + e^{x}\right) - x \cdot y
↓
\log \left(\frac{{1}^{3} + {\left(e^{x}\right)}^{3}}{e^{x} \cdot \left(e^{x} - 1\right) + 1 \cdot 1}\right) - x \cdot y
double f(double x, double y) {
        double r149274 = 1.0;
        double r149275 = x;
        double r149276 = exp(r149275);
        double r149277 = r149274 + r149276;
        double r149278 = log(r149277);
        double r149279 = y;
        double r149280 = r149275 * r149279;
        double r149281 = r149278 - r149280;
        return r149281;
}
↓
double f(double x, double y) {
        double r149282 = 1.0;
        double r149283 = 3.0;
        double r149284 = pow(r149282, r149283);
        double r149285 = x;
        double r149286 = exp(r149285);
        double r149287 = pow(r149286, r149283);
        double r149288 = r149284 + r149287;
        double r149289 = r149286 - r149282;
        double r149290 = r149286 * r149289;
        double r149291 = r149282 * r149282;
        double r149292 = r149290 + r149291;
        double r149293 = r149288 / r149292;
        double r149294 = log(r149293);
        double r149295 = y;
        double r149296 = r149285 * r149295;
        double r149297 = r149294 - r149296;
        return r149297;
}

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. Using strategy rm

  3. Applied flip3-+0.6

    \[\leadsto \log \color{blue}{\left(\frac{{1}^{3} + {\left(e^{x}\right)}^{3}}{1 \cdot 1 + \left(e^{x} \cdot e^{x} - 1 \cdot e^{x}\right)}\right)} - x \cdot y\]

  4. Simplified0.6

    \[\leadsto \log \left(\frac{{1}^{3} + {\left(e^{x}\right)}^{3}}{\color{blue}{e^{x} \cdot \left(e^{x} - 1\right) + 1 \cdot 1}}\right) - x \cdot y\]

  5. Final simplification0.6

    \[\leadsto \log \left(\frac{{1}^{3} + {\left(e^{x}\right)}^{3}}{e^{x} \cdot \left(e^{x} - 1\right) + 1 \cdot 1}\right) - x \cdot y\]

Reproduce

herbie shell --seed 2019353 
(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)))