Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 33B / b

LetThe covering relations areThus the covering graph has adjacency matrixThe two closed length-three walksgive a horseshoe from two closed covering walks for $F^3$, so $F$ must be chaotic. Moreover,By counting cycles in an interval covering graph, the number of primitive length-three cyclic itineraries isHence $F$ must have at least, and the connect-the-dots realization shows that it need have only, $\boxed{2}$ distinct 3-cycles.

SetTheir covering graph iswith adjacency matrixThe closed walks $I_1\to I_1\to I_1$ and $I_1\to I_0\to I_1$ give a horseshoe from two closed covering walks for $F^2$, so $F$ must be chaotic. Sincecounting cycles in an interval covering graph givesdistinct 3-cycles in the minimum case, attained by the corresponding connect-the-dots interval map.

TakeNowsoThe recurrent pieces of this covering graph are the two-cycle $I_0\leftrightarrow I_2$ and the loop at $I_1$; neither contains two competing closed routes. Alsoso this ordering forces no 3-cycle.

The bound is attained by the piecewise-linear functionIndeed,is a 4-cycle with the required spatial order. On $I_0\cup I_2$, the fourth iterate is the identity function. Within $I_1$, the fixed point is $5/9$ andso every other point eventually leaves $I_1$ and enters $I_0\cup I_2$. The map therefore has no horseshoe in any iterate and is not chaotic. The minimum number of distinct 3-cycles is consequently $\boxed{0}$.