= Solution
Let $S=\beta/\alpha$ be the unrepressed <steady state>, $x=X/S,y=Y/S,z=Z/S$, $h=K/S$, and $\tau=\alpha t$. The <logic approximation for gene regulation> gives the <repressilator> equations
$$
\boxed{x'=H(h-z)-x,\qquad y'=H(h-x)-y,\qquad z'=H(h-y)-z.}
$$
An active production term makes its <concentration> approach one exponentially; an inactive term makes it decay exponentially. Oscillations require an accessible threshold, $0<h<1$. Equal thresholds and rate constants alone do not specify a numerical period.
Initially $(x,y,z)=(0,1,0)$, so $x=1-e^{-\tau}$, $y=1$, $z=0$ until $x$ reaches $h$ at $\tau_1=-\log(1-h)$. Then $Y$ production stops and $y=e^{-(\tau-\tau_1)}$. It reaches $h$ after another $\log(1/h)$, allowing $Z$ production. When $z$ rises through $h$, $X$ production stops. As $x$ falls through $h$, $Y$ production resumes; its next upward crossing represses $Z$. The downward crossing of $z$ then restarts $X$. This ordered sequence repeats, giving displaced pulses and rounded exponential rises and falls.
Every event can be constructed without a numerical differential-equation solver: a rising variable starting at $v_0<h$ takes $\log[(1-v_0)/(1-h)]$ to reach $h$, while a decaying variable starting at $v_0>h$ takes $\log(v_0/h)$. Advance all three variables with their current production terms to the earliest such event, switch the affected production term, and continue. The original plot uses the explicitly illustrative choice $h=1/2$. If $h\ge1$, the unrepressed levels cannot cross the threshold and this oscillatory sequence does not follow.
\Image[/past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2015-repressilator.png]
{title=Original <logic approximation for gene regulation> trajectory of the <repressilator>, with threshold one half of the unrepressed level}
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