If $ k(x)$ is the least cycle for $ x$, we have $ f^{k(x)}(x) = x$, then we have $ p(x)a - q(x)b = 0$ where $ p(x)$ is the number of "$ + a$" moves and $ q(x)$ the number of "$ - b$" moves for the least cycle (and so $ p(x) + q(x) = k(x)$) And so the least values possible for $ p(x)$ and $ q(x)$ are $ p(x) = \frac {b}{\gcd(a,b)}$ and $ q(x) = \frac {a}{\gcd(a,b)}$ So $ k(x)\geq \frac {a + b}{\gcd(a,b)}$ Now, if $ x = 0$, $ f^i(0)$ can only take values in $ ([\frac {a + b}{2}] - b,[\frac {a + b}{2}] + a]$, all multiples of $ \gcd(a,b)$ (since the starting value $ 0$ is such a multiple), so $ \frac {a + b}{\gcd(a,b)}$ values. So a cycle is encountered in the first $ \frac {a + b}{\gcd(a,b)}$ moves. As a consequence, this cycle includes $ 0$ (else we would have a cycle whose length would be $ < \frac {a + b}{\gcd(a,b)}$ And so the result : $ k(0) = \frac {a + b}{\gcd(a,b)}$ (Remark : with some other departure points, the cycle may not involve the departure point and so $ k(x) > \frac {a + b}{\gcd(a,b)}$)