Solution

Take

I_{0} = [x_{0}, x_{2}], I_{1} = [x_{2}, x_{1}], I_{2} = [x_{1}, x_{3}] .

Now

I_{0} ⟶ I_{2}, I_{1} ⟶ I_{1}, I_{2}, I_{2} ⟶ I_{0},

A = 001010110 .

The 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. Also

tr (A^{3}) = tr A = 1,

so this ordering forces no 3-cycle.

The bound is attained by the piecewise-linear function

F (x) = ⎩ ⎨ ⎧ x + \frac{2}{3}, \frac{5}{3} - 2 x, 1 - x, 0 \leq x \leq \frac{1}{3}, \frac{1}{3} \leq x \leq \frac{2}{3}, \frac{2}{3} \leq x \leq 1.

Indeed,

0 \mapsto \frac{2}{3} \mapsto \frac{1}{3} \mapsto 1 \mapsto 0

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

and

F (x) - \frac{5}{9} = - 2 (x - \frac{5}{9}),

so 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

0

Solved by gpt-5.6-sol high.

 Ancestors (12)