%display latex 

load("../specifier.py")

We consider the class $\mathrm{Av}(132)$. It contains no simple permutation. Hence we obtain its specification as follows:

S = Specification([],[[1,3,2]])

S

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array} \mathcal T_{0}= &\{ \bullet \} \uplus \oplus[\mathcal T_{1},\mathcal T_{2}]\uplus \ominus[\mathcal T_{3},\mathcal T_{0}]\\ \mathcal T_{1}= &\{ \bullet \} \uplus \ominus[\mathcal T_{3},\mathcal T_{0}]\\ \mathcal T_{2}= &\{ \bullet \} \uplus \oplus[\mathcal T_{4},\mathcal T_{2}]\\ \mathcal T_{3}= &\{ \bullet \} \uplus \oplus[\mathcal T_{1},\mathcal T_{2}]\\ \mathcal T_{4}= &\{ \bullet \} \end{array}$$

In the specification algorithm, the families $\mathcal T_0, \ldots, \mathcal T_d$ are families of constrained permutations. We can see which families exactly as follows:

S.concrete()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array} \mathcal C = &\mathrm{Subst}(\oplus, \ominus)\\ \mathcal T_{0} = &C\langle 132\rangle \\ \mathcal T_{1} = &C^+\langle 132\rangle \\ \mathcal T_{2} = &C\langle 21\rangle \\ \mathcal T_{3} = &C^-\langle 132\rangle \\ \mathcal T_{4} = &C^+\langle 21\rangle \end{array}$$

S.solve()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[-\frac{2 \, z}{2 \, z - \sqrt{-4 \, z + 1} - 1}, \frac{2 \, z^{2} + z {\left(\sqrt{-4 \, z + 1} - 3\right)} - \sqrt{-4 \, z + 1} + 1}{2 \, z}, -\frac{z}{z - 1}, -\frac{1}{2} \, \sqrt{-4 \, z + 1} + \frac{1}{2}, z\right]$$

We see that $T_0,T_1,T_3$ have radius of convergence $1/4$, while the others have strictly larger radius. Hence the critical families are $\mathcal T_0,\mathcal T_1,\mathcal T_3$. Hypothesis (SC) (strong connectivity among critical families) holds, and the specification is branching (see for instance the term $\ominus[\mathcal T_{3},\mathcal T_{0}]$).

Since the series are convergent, we may perform Boltzmann sampling right at the radius of convergence:

C = S.sampler(0.25, formal=True)

perm = C.run(10)

list_plot(perm,aspect_ratio=1)

The limiting permuton seems to be the antidiagonal. This is confirmed by our results: indeed, we are in the essentially branching case, and the substitutions containing two critical series all have a $\ominus$ at the root (it is always $\ominus[\mathcal T_0,\mathcal T_3]$).