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If a directed covering graph of an interval map has two distinct closed walks of the same length $n$, based at the same interval and with different first edges, successive use of the interval covering relation produces two subintervals with disjoint interiors that $F^n$ maps across the base interval. Hence $F^n$ has a horseshoe for an interval map.

A horseshoe for an interval map consists of two branches covering a common interval: there are an open interval $J$ and disjoint open subintervals $K_0,K_1\subseteq J$ with $f(K_0)=f(K_1)=J$. Equivalently one can work with two closed intervals with disjoint interiors whose images cover their union, and then restrict the branches. Glendinning chaos means that some positive iterate $f^m$ has such a horseshoe, rather than requiring one for $f$ itself.

For the three-cycle, use the intermediate value theorem on endpoint images. The interval covering relations are $I_a\to I_b$ and $I_b\to I_a,I_b$. Consequently both $f^2(I_a)$ and $f^2(I_b)$ cover $I_a\cup I_b$. This proves $f^2$ has a horseshoe. The directed covering graph of an interval map has exactly these forced arrows; a closed walk $a\to b\to b\to a$ has length three and a primitive itinerary. Pulling back the intervals successively and using the fixed-point property produces a point fixed by $f^3$ with this itinerary. It cannot be fixed by $f$, and a shared endpoint belongs to the already given three-cycle, so a genuine period-three orbit results.

For the increasing spatial order of the five-cycle, let $I_j=[x_{j-1},x_j]$, $1\le j\le 4$. The forced graph isThe primitive loops $I_4$, $I_3{I}_4$, $I_2{I}_3{I}_4$, and $I_1{I}_2{I}_3{I}_4$ force exact periods $\boxed{1,2,3,4}$. For periods below five the known five-cycle endpoints cannot be responsible, so the interval itineraries give distinct interior orbits. Both $g^2(I_3)$ and $g^2(I_4)$ cover $I_3\cup I_4$, so $g$ is guaranteed to be Glendinning-chaotic.

A horseshoe for $g$ itself is not guaranteed. As a counterexample put $x_j=j$ on $[0,4]$ and interpolate by $g(x)=x+1$ for $0\le x\le 3$, $g(x)=16-4x$ for $3\le x\le 4$. This has the required five-cycle. The unique fixed point is $16/5$, with $g(x)>x$ to its left and $g(x)<x$ to its right. Any nondegenerate closed interval whose image covers itself must contain this fixed point in its interior: an interval entirely on either side, or with the fixed point as an endpoint, fails to cover its extreme endpoint. Two intervals with disjoint interiors therefore cannot both cover themselves, as a horseshoe would require. Hence this example has no one-iterate horseshoe.

For the alternative spatial order, set $y_0=x_4,y_1=x_2,y_2=x_1,y_3=x_3,y_4=x_0$ and $J_j=[y_{j-1},y_j]$. Endpoint images giveThe self-loop at $J_2$, the two-cycle $J_1{J}_4$, and the four-cycle $J_2{J}_3{J}_1{J}_4$ force $\boxed{1,2,4}$; period three is not guaranteed. To see this last claim, take the connect-the-dots map at $y_j=j$ with values $(4,3,1,0,2)$. Its exact interval transitions are those displayed. Any length-three closed itinerary is the repeated self-loop in $J_2$; there the map is the affine decreasing function $5-2x$, whose third iterate has only its original fixed point. Endpoints belong to the five-cycle. Thus no exact three-cycle exists in this example.

Chaos is still guaranteed: from $J_2$ there are two different length-four closed walks, four self-loops and $J_2{J}_3{J}_1{J}_4{J}_2$. Pullbacks give disjoint branches of $g^4$ over a common interval in $J_2$, hence a horseshoe for an interval map for an iterate. A horseshoe for $g$ itself is again not guaranteed: the same connect-the-dots example is strictly decreasing on $[0,3]$ and increasing on $[3,4]$ with image $[0,2]$. Its unique fixed point is $5/3$, again with $g(x)>x$ to the left and $g(x)<x$ to the right. The same fixed-point argument rules out two self-covering intervals with disjoint interiors. These counterexamples distinguish the three separate questions about periods, chaos and a one-iterate horseshoe.

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 $\boxed{0}$.

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.

Every closed walkin a directed covering graph of an interval map has a point $x\in J_0$ with $F^j(x)\in J_j$ and $F^n(x)=x$. This follows by successively pulling $J_0$ back through the covering relations and applying the fixed-point property of a closed interval. If the itinerary has least period $n$ and avoids shared endpoints, $x$ lies on an $n$-cycle.