expq3 (problem 3.4.2)

Average Error: 60.3 → 0.3

Time: 46.4s

Precision: 64

• could not determine a ground truth for program body

  1. a = 1.4639452118847904e+51
  2. b = -1.805670951386701e+285
  3. eps = -1.4405055651571276e-214

\[-1 \lt \varepsilon \land \varepsilon \lt 1\]

Math

\[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]

↓

\[\begin{array}{l} \mathbf{if}\;\frac{\left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right) \cdot \varepsilon}{\left(e^{\varepsilon \cdot b} - 1\right) \cdot \left(e^{\varepsilon \cdot a} - 1\right)} = -\infty:\\ \;\;\;\;\frac{1}{a} + \frac{1}{b}\\ \mathbf{elif}\;\frac{\left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right) \cdot \varepsilon}{\left(e^{\varepsilon \cdot b} - 1\right) \cdot \left(e^{\varepsilon \cdot a} - 1\right)} \le 3.502190010791832831227506580315349260513 \cdot 10^{-9}:\\ \;\;\;\;\frac{\left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right) \cdot \varepsilon}{\left(e^{\varepsilon \cdot b} - 1\right) \cdot \left(e^{\varepsilon \cdot a} - 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{a} + \frac{1}{b}\\ \end{array}\]

C

double f(double a, double b, double eps) {
        double r5696747 = eps;
        double r5696748 = a;
        double r5696749 = b;
        double r5696750 = r5696748 + r5696749;
        double r5696751 = r5696750 * r5696747;
        double r5696752 = exp(r5696751);
        double r5696753 = 1.0;
        double r5696754 = r5696752 - r5696753;
        double r5696755 = r5696747 * r5696754;
        double r5696756 = r5696748 * r5696747;
        double r5696757 = exp(r5696756);
        double r5696758 = r5696757 - r5696753;
        double r5696759 = r5696749 * r5696747;
        double r5696760 = exp(r5696759);
        double r5696761 = r5696760 - r5696753;
        double r5696762 = r5696758 * r5696761;
        double r5696763 = r5696755 / r5696762;
        return r5696763;
}

↓

double f(double a, double b, double eps) {
        double r5696764 = a;
        double r5696765 = b;
        double r5696766 = r5696764 + r5696765;
        double r5696767 = eps;
        double r5696768 = r5696766 * r5696767;
        double r5696769 = exp(r5696768);
        double r5696770 = 1.0;
        double r5696771 = r5696769 - r5696770;
        double r5696772 = r5696771 * r5696767;
        double r5696773 = r5696767 * r5696765;
        double r5696774 = exp(r5696773);
        double r5696775 = r5696774 - r5696770;
        double r5696776 = r5696767 * r5696764;
        double r5696777 = exp(r5696776);
        double r5696778 = r5696777 - r5696770;
        double r5696779 = r5696775 * r5696778;
        double r5696780 = r5696772 / r5696779;
        double r5696781 = -inf.0;
        bool r5696782 = r5696780 <= r5696781;
        double r5696783 = 1.0;
        double r5696784 = r5696783 / r5696764;
        double r5696785 = r5696783 / r5696765;
        double r5696786 = r5696784 + r5696785;
        double r5696787 = 3.502190010791833e-09;
        bool r5696788 = r5696780 <= r5696787;
        double r5696789 = r5696788 ? r5696780 : r5696786;
        double r5696790 = r5696782 ? r5696786 : r5696789;
        return r5696790;
}

Error

Bits error versus a

Bits error versus b

Bits error versus eps

Target

Original	60.3
Target	14.8
Herbie	0.3

\[\frac{a + b}{a \cdot b}\]

Derivation

1. Split input into 2 regimes

2. if (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0))) < -inf.0 or 3.502190010791833e-09 < (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0)))

   1. Initial program 63.9

      \[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]

   2. Taylor expanded around 0 0.1

      \[\leadsto \color{blue}{\frac{1}{a} + \frac{1}{b}}\]

   if -inf.0 < (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0))) < 3.502190010791833e-09

   1. Initial program 3.6

      \[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
3. Recombined 2 regimes into one program.

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right) \cdot \varepsilon}{\left(e^{\varepsilon \cdot b} - 1\right) \cdot \left(e^{\varepsilon \cdot a} - 1\right)} = -\infty:\\ \;\;\;\;\frac{1}{a} + \frac{1}{b}\\ \mathbf{elif}\;\frac{\left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right) \cdot \varepsilon}{\left(e^{\varepsilon \cdot b} - 1\right) \cdot \left(e^{\varepsilon \cdot a} - 1\right)} \le 3.502190010791832831227506580315349260513 \cdot 10^{-9}:\\ \;\;\;\;\frac{\left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right) \cdot \varepsilon}{\left(e^{\varepsilon \cdot b} - 1\right) \cdot \left(e^{\varepsilon \cdot a} - 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{a} + \frac{1}{b}\\ \end{array}\]

Reproduce

herbie shell --seed 2019170 
(FPCore (a b eps)
  :name "expq3 (problem 3.4.2)"
  :pre (and (< -1.0 eps) (< eps 1.0))

  :herbie-target
  (/ (+ a b) (* a b))

  (/ (* eps (- (exp (* (+ a b) eps)) 1.0)) (* (- (exp (* a eps)) 1.0) (- (exp (* b eps)) 1.0))))