= Solution
Set
$$
I_0=[x_0,x_1],\qquad I_1=[x_1,x_3],\qquad I_2=[x_3,x_2].
$$
Their covering graph is
$$
I_0\longrightarrow I_1,I_2,\qquad
I_1\longrightarrow I_0,I_1,I_2,\qquad
I_2\longrightarrow I_0,I_1,
$$
with <adjacency matrix of a directed graph>[adjacency matrix]
$$
A=\begin{pmatrix}0&1&1\\1&1&1\\1&1&0\end{pmatrix}.
$$
The 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 <Glendinning chaos>[chaotic]. Since
$$
\operatorname{tr}A=1,
\qquad
\operatorname{tr}(A^3)=13,
$$
<counting cycles in an interval covering graph> gives
$$
\frac{13-1}{3}=\boxed4
$$
distinct 3-cycles in the minimum case, attained by the corresponding <connect-the-dots interval map>.
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