= Solution
Let
$$
I_0=[x_0,x_1],\qquad I_1=[x_1,x_2],\qquad I_2=[x_2,x_3].
$$
The <interval covering relation>[covering relations] are
$$
I_0\longrightarrow I_1,\qquad
I_1\longrightarrow I_2,\qquad
I_2\longrightarrow I_0,I_1,I_2.
$$
Thus the <directed covering graph of an interval map>[covering graph] has <adjacency matrix of a directed graph>[adjacency matrix]
$$
A=\begin{pmatrix}0&1&0\\0&0&1\\1&1&1\end{pmatrix}.
$$
The two closed length-three walks
$$
I_2\to I_2\to I_2\to I_2,
\qquad
I_2\to I_0\to I_1\to I_2
$$
give a <horseshoe from two closed covering walks> for $F^3$, so $F$ must be <Glendinning chaos>[chaotic]. Moreover,
$$
\operatorname{tr}A=1,
\qquad
\operatorname{tr}(A^3)=7.
$$
By <counting cycles in an interval covering graph>, the number of primitive length-three cyclic itineraries is
$$
\frac{7-1}{3}=2.
$$
Hence $F$ must have at least, and the connect-the-dots realization shows that it need have only, $\boxed{2}$ distinct 3-cycles.
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