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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.

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.