**Proposition 27.** Let $m_1, m_2$ be positive integers or half-integers and let $m_3, \dots, m_n$ be non-negative integers ($n \ge 3$). Then, there is a bijection between the set of planar tight maps with $n$ boundaries labeled from 1 to $n$ with respective lengths $2m_1, 2m_2, \dots, 2m_n$, and the set of pairs $(\mathbf{m}_{12}, \mathbf{s})$ such that there exists $k \in \{0, \dots, n-3\}$ for which:

* $\mathbf{m}_{12}$ is a planar tight two-face map, with two faces of respective degrees $2m_1$ and $2m_2$, and $k+1$ distinct marked vertices, one of them distinguished,

* $\mathbf{s} = (\mathbf{s}_1, \dots, \mathbf{s}_{k+1})$ is a $(k+1)$-tuple of slices such that:

- for each $j = 1, \dots, k+1$, $\mathbf{s}_j$ is either the marked empty slice or a tight bipartite elementary slice with at least one inner face, whose inner faces and marked vertices are labeled by integers in $\{3, \dots, n\}$,

- each $i \in \{3, \dots, n\}$ appears in exactly one $\mathbf{s}_j$ and labels an inner face of degree $2m_i$ for $m_i > 0$, or a marked vertex for $m_i = 0$,

- the label 3 appears in the first slice $\mathbf{s}_1$.

Before proving this proposition, let us see how it implies Theorems 4 and 9. We start with the former: our purpose is to enumerate the pairs $(\mathbf{m}_{12}, \mathbf{s})$ of the proposition when $m_1$ and $m_2$ are integers. For a fixed $k \in \{0, \dots, n-3\}$, the number of possible maps $\mathbf{m}_{12}$ is equal to $p_k(m_1, m_2)$ by Proposition 7, which we proved in Section 4.2. As for the number of possible $\mathbf{s}$, it is given by a slight variant of the reasoning in Section 4.3. Indeed, by recursively decomposing each slice $\mathbf{s}_j$ into a tree of lattice paths, we see that the set of possible $(k+1)$-tuples $\mathbf{s}$ is in bijection with the set $\mathcal{F}_{k+1}(m_3, \dots, m_n)$ defined as the set of pairs $(\mathbf{f}, (L_i)_{3 \le i \le n})$ where:

* $\mathbf{f}$ is a plane forest with $k+1$ connected components, i.e. a $(k+1)$-tuple of rooted plane trees, whose vertices are labeled by $\{3, \dots, n\}$, the label 3 appearing in the first component,

* denoting by $k_i$ the number of children of the vertex labeled $i$ in $\mathbf{f}$:

- if $m_i > 0$ then $L_i$ is a lattice path from $(0,0)$ to $(2m_i-1, 1)$ with $k_i$ marked up steps, such that all up steps immediately following a down step are marked,

- if $m_i = 0$ then $L_i$ is the trivial path $\{(0,0)\}$, in which case necessarily $k_i = 0$.

Note that $\mathcal{F}_1(m_3, \dots, m_n)$ is nothing but the set $\mathcal{T}(m_3, \dots, m_n)$ as defined in Section 4.3. By Proposition 44 of Appendix A below, for a given $n$-uple $(k_3, \dots, k_n)$, if $k_3 + \dots + k_n = n-3-k$ then there are exactly $(n-3)!$ plane forests on the vertex set $\{3, \dots, n\}$ with $k+1$ components, the first of which contains the label 3, and such that vertex $i$ has $k_i$ children for all $i=3, \dots, n$, and there are no such forests otherwise. For any such forest $\mathbf{f}$, there are $\prod_{i=3}^n q_{k_i}(m_i)$ possible choices of marked lattice paths $L_i$, $3 \le i \le n$ such that $(\mathbf{f}, (L_i)_{3 \le i \le n})$ belongs to $\mathcal{F}_{k+1}(m_3, \dots, m_n)$. This gives

$$ \begin{align} \mathrm{Card} (\mathcal{F}_{k+1}(m_3, \dots, m_n)) &= (n-3)! \sum_{\substack{k_3, \dots, k_n \ge 0 \\ k_3 + \dots + k_n = n-3-k}} \prod_{i=3}^{n} q_{k_i}(m_i) \tag{47} \\ &= (n-3)! q_{n-3-k}(m_3, \dots, m_n). \end{align} $$
<figure><img src="image_15.png" /><figcaption>Figure 4.10: Sketch of an annular map $\mathfrak{m}$ (left) and of its universal cover $\tilde{\mathfrak{m}}$ (right). The external face 1 and the central face 2 of $\mathfrak{m}$ lift respectively to the lower face $\tilde{1}$ and to the upper face $\tilde{2}$ of $\tilde{\mathfrak{m}}$, which have infinite degrees (marked corners are represented by arrows). All the other faces form the grey region. The innermost minimal separating cycle $\gamma$ lifts to a biinfinite geodesic $\tilde{\gamma}$.</figcaption></figure>

Thus, multiplying by $p_k(m_1, m_2)$, summing over $k$ and using (2.5), we get

$$ 
\begin{aligned}
N_{0,n}(2m_1, 2m_2, 2m_3, \dots, 2m_n) &= (n-3)! \sum_{k=0}^{n-3} p_k(m_1, m_2) q_{n-3-k}(m_3, \dots, m_n) \quad (4.6) \\
&= (n-3)! p_{n-3}(m_1, m_2, m_3, \dots, m_n)
\end{aligned}
$$

as wanted. This concludes the proof of Theorem 2.3 assuming Proposition 4.7.

If we now assume that $m_1, m_2$ are half-integers, the only change we have to do in the above reasoning is that, for a fixed $k$, the number of possible maps $\mathfrak{m}_{12}$ is now equal to $\tilde{p}_k(m_1, m_2)$ by Proposition 2.9 (also proved in Section 4.2). Thus, by (2.5) and (2.20), we now have

$$ 
\begin{aligned}
N_{0,n}(2m_1, 2m_2, 2m_3, \dots, 2m_n) &= (n-3)! \sum_{k=0}^{n-3} \tilde{p}_k(m_1, m_2) q_{n-3-k}(m_3, \dots, m_n) \quad (4.7) \\
&= (n-3)! \tilde{p}_{n-3}(m_1, m_2; m_3, \dots, m_n)
\end{aligned}
$$

which establishes Theorem 2.8.

The remainder of this section is devoted to the proof of Proposition 4.7. It uses the slice decomposition of annular maps which was introduced in [BG14, Section 9.3], see also [Bou19, Section 2.2] for another exposition. Here we will give yet another, modernized, exposition which, following [BGM22], makes use of the key notion of *Busemann function*.

**Decomposing an annular map into a pair of paths decorated with elementary slices.** Let $\mathfrak{m}$ be an *annular map*, that is a planar map with two distinguished faces labeled 1 and 2. For now, we do not assume that $\mathfrak{m}$ is a tight map, but we assume that every face other than 1 and 2 has even degree. Denoting by $2m_1$ and $2m_2$ the respective degrees of 1 and 2, $\mathfrak{m}$ is either bipartite when $m_1$ and $m_2$ are integers, or quasi-bipartite when they are half-integers.
<figure><img src="image_14.png" /><figcaption>Figure 13: Sketch of an annular map **m** (left) and of its universal cover **\tilde{m}** (right). The external face 1 and the central face 2 of **m** lift respectively to the lower face **1** and to the upper face **2** of **\tilde{m}**, which have infinite degrees (marked corners are represented by arrows). All the other faces form the grey region. The innermost minimal separating cycle **γ** lifts to a biinfinite geodesic **\tilde{\gamma}$**.</figcaption></figure>

Thus, multiplying by $p_k(m_1, m_2)$, summing over $k$ and using (6), we get

$$ 
\begin{aligned}
N_{0,n}(2m_1, 2m_2, 2m_3, \dots, 2m_n) &= (n-3)! \sum_{k=0}^{n-3} p_k(m_1, m_2) q_{n-3-k}(m_3, \dots, m_n) \\
&= (n-3)! p_{n-3}(m_1, m_2, m_3, \dots, m_n)
\end{aligned} 
\quad (48) $$

as wanted. This concludes the proof of Theorem 4 assuming Proposition 27.

If we now assume that $m_1, m_2$ are half-integers, the only change we have to do in the above reasoning is that, for a fixed $k$, the number of possible maps $m_{12}$ is now equal to $\tilde{p}_k(m_1, m_2)$ by Proposition 10 (also proved in Section 4.2). Thus, by (6) and (20), we now have

$$ 
\begin{aligned}
N_{0,n}(2m_1, 2m_2, 2m_3, \dots, 2m_n) &= (n-3)! \sum_{k=0}^{n-3} \tilde{p}_k(m_1, m_2) q_{n-3-k}(m_3, \dots, m_n) \\
&= (n-3)! \tilde{p}_{n-3}(m_1, m_2; m_3, \dots, m_n)
\end{aligned} 
\quad (49) $$

which establishes Theorem 9.

The remainder of this section is devoted to the proof of Proposition 27. It uses the slice decomposition of annular maps which was introduced in [BG14, Section 9.3], see also [Bou19, Section 2.2] for another exposition. Here we will give yet another, modernized, exposition which, following [BGM21], makes use of the key notion of *Busemann function*.

**Decomposing an annular map into a pair of paths decorated with elementary slices.**
Let **m** be an *annular* map, that is a planar map with two distinguished faces labeled 1 and 2. For now, we do not assume that **m** is a tight map, but we assume that every face
Let us choose a representation of **m** in the complex plane such that face 1 is the unbounded face, and such that the origin (point of affix zero) lies in face 2. We then consider the preimage $\tilde{\mathfrak{m}}$ of **m** by the mapping $z \mapsto \exp(2i\pi z)$: it is an infinite map which we call the *universal cover* of **m** (upon viewing 0 and $\infty$ as “punctures”). We refer to [BGM22] for a detailed discussion of the properties of the universal cover of a map drawn on the triply-punctured sphere, it can be adapted without difficulty to the simpler case considered here of a map drawn on the doubly-punctured sphere. For our purposes, we simply note that the faces 1 and 2 lift in $\tilde{\mathfrak{m}}$ to unique faces $\tilde{1}$ and $\tilde{2}$, which have infinite degrees, while all the other faces and vertices of **m** lift to infinitely many preimages in $\tilde{\mathfrak{m}}$ with the same finite degree. In particular, $\tilde{\mathfrak{m}}$ is bipartite. See Figure 4.10 for an illustration.

Let $s$ denote the *separating girth* of **m**, that is the minimal length (number of edges) of a closed path in **m** winding around the origin. We call such path a *separating cycle* and say that it is *minimal* if it has length $s$. Let then $\gamma$ be the *innermost* minimal separating cycle, which we define as follows. Consider the interiors of all minimal separating cycles. It is straightforward to check that their intersection is a simply connected region containing the origin, and that its boundary oriented clockwise is still a minimal separating cycle: this is the innermost minimal separating $\gamma$ that we are looking for. We consider the path $\tilde{\gamma} = (\tilde{\gamma}(t))_{t \in \mathbb{Z}}$ obtained by following $\gamma$ counterclockwise infinitely many times and lifting this biinfinite path in $\tilde{\mathfrak{m}}$. The parametrization of $\tilde{\gamma}$ depends on a choice of a vertex $\tilde{\gamma}(0)$ whose projection belongs to $\gamma$, but we will see that the outcome of our construction does not depend on it. By [CdVE10, Proposition 2.5], $\tilde{\gamma}$ is a *biinfinite geodesic* in $\tilde{\mathfrak{m}}$, which by definition means that

$$ \tilde{d}(\tilde{\gamma}(t), \tilde{\gamma}(t')) = |t - t'| \qquad (4.8) $$

for any $t, t' \in \mathbb{Z}$, where $\tilde{d}$ denotes the graph distance in $\tilde{\mathfrak{m}}$. We now define the *Busemann function* $B_{\tilde{\gamma}}$ from the vertex set $\tilde{V}$ of $\tilde{\mathfrak{m}}$ to $\mathbb{Z}$ by

$$ B_{\tilde{\gamma}}(v) := \lim_{t \to \infty} \tilde{d}(v, \tilde{\gamma}(t)) - t, \quad v \in \tilde{V}. \qquad (4.9) $$

By the triangle inequality, one easily checks that the limit indeed exists and is attained for $t$ large enough (for instance $B_{\tilde{\gamma}}(\tilde{\gamma}(s)) = -s$ for every $s$ in $\mathbb{Z}$). Furthermore, for any two adjacent vertices $v, v'$ we have $B_{\tilde{\gamma}}(v) - B_{\tilde{\gamma}}(v') = \pm 1$ (since $\tilde{\mathfrak{m}}$ is bipartite), and every vertex $v$ has a neighbor $v'$ for which this difference is $+1$ (i.e. $B_{\tilde{\gamma}}$ has no local minimum). Let us denote by $T$ the automorphism of $\tilde{\mathfrak{m}}$ corresponding to the translation $z \mapsto z + 1$ of the complex plane (it corresponds to making one turn counterclockwise around the origin in **m**): we then have $B_{\tilde{\gamma}}(Tv) = B_{\tilde{\gamma}}(v) - s$ as a consequence of the relation $T\tilde{\gamma}(t) = \tilde{\gamma}(t+s)$.

Now, let us denote by $(c_i)_{i \in \mathbb{Z}}$ and $(c'_i)_{i \in \mathbb{Z}}$ the successive corners incident to the faces $\tilde{1}$ and $\tilde{2}$, respectively, as we follow their contours walking with the face at hand to the right. This depends on a choice for the corners $c_0$ and $c'_0$, whose influence will be discussed later. We have $Tc_i = c_{i+2m_1}$ and $Tc'_i = c'_{i-2m_2}$ for all $i$.

We then set, for any $i \in \mathbb{Z}$, $\ell_i := B_{\tilde{\gamma}}(c_0) - B_{\tilde{\gamma}}(c_i)$ and $\ell'_i := B_{\tilde{\gamma}}(c'_0) - B_{\tilde{\gamma}}(c'_i)$ (these differences do not depend on the choice of $\tilde{\gamma}(0)$, as claimed). Observe that the definition of $\ell_i$ is analogous to that used in Section 4.3 for the decomposition of a slice into a path decorated with elementary slices: the distance to the apex $d(\cdot, A)$ is just replaced by the Busemann function
other than 1 and 2 has even degree. Denoting by $2m_1$ and $2m_2$ the respective degrees of 1 and 2, $\mathbf{m}$ is either bipartite when $m_1$ and $m_2$ are integers, or quasi-bipartite when they are half-integers.

Let us choose a representation of $\mathbf{m}$ in the complex plane such that face 1 is the unbounded face, and such that the origin (point of affix zero) lies in face 2. We then consider the preimage $\tilde{\mathbf{m}}$ of $\mathbf{m}$ by the mapping $z \mapsto \exp(2i\pi z)$: it is an infinite map which we call the *universal cover* of $\mathbf{m}$ (upon viewing 0 and $\infty$ as “punctures”). We refer to [BGM21] for a detailed discussion of the properties of the universal cover of a map drawn on the triply-punctured sphere, it can be adapted without difficulty to the simpler case considered here of a map drawn on the doubly-punctured sphere. For our purposes, we simply note that the faces 1 and 2 lift in $\tilde{\mathbf{m}}$ to unique faces $\tilde{1}$ and $\tilde{2}$, which have infinite degrees, while all the other faces and vertices of $\mathbf{m}$ lift to infinitely many preimages in $\tilde{\mathbf{m}}$ with the same finite degree. In particular, $\tilde{\mathbf{m}}$ is bipartite. See Figure 13 for an illustration.

Let $s$ denote the *separating girth* of $\mathbf{m}$, that is the minimal length (number of edges) of a closed path in $\mathbf{m}$ winding around the origin. We call such path a *separating cycle* and say that it is *minimal* if it has length $s$. Let then $\gamma$ be the *innermost minimal separating cycle*, which is defined as the contour of the intersection of the interiors of all minimal separating cycles. We consider the path $\tilde{\gamma} = (\tilde{\gamma}(t))_{t \in \mathbb{Z}}$ obtained by following $\gamma$ counterclockwise infinitely many times and lifting this biinfinite path in $\tilde{\mathbf{m}}$. The parametrization of $\tilde{\gamma}$ depends on a choice of a vertex $\tilde{\gamma}(0)$ whose projection belongs to $\gamma$, but we will see that the outcome of our construction does not depend on it. By [CE10, Proposition 2.5], $\tilde{\gamma}$ is a *biinfinite geodesic* in $\tilde{\mathbf{m}}$, i.e. we have

$$ \tilde{d}(\gamma(t), \gamma(t')) = |t - t'| \qquad (50) $$

for any $t, t' \in \mathbb{Z}$, where $\tilde{d}$ denotes the graph distance in $\tilde{\mathbf{m}}$. We now define the *Busemann function* $B_{\tilde{\gamma}}$ from the vertex set $\tilde{V}$ of $\tilde{\mathbf{m}}$ to $\mathbb{Z}$ by

$$ B_{\tilde{\gamma}}(v) := \lim_{t \to \infty} \tilde{d}(v, \gamma(t)) - t, \quad v \in \tilde{V}. \qquad (51) $$

By the triangle inequality, one easily checks that the limit indeed exists and is attained for $t$ large enough. Furthermore, for any two adjacent vertices $v, v'$ we have $B_{\tilde{\gamma}}(v) - B_{\tilde{\gamma}}(v') = \pm 1$ (since $\tilde{\mathbf{m}}$ is bipartite), and every vertex $v$ has a neighbor $v'$ for which this difference is $+1$ (i.e. $B_{\tilde{\gamma}}$ has no local minimum). Let us denote by $T$ the automorphism of $\tilde{\mathbf{m}}$ corresponding to the translation $z \mapsto z+1$ of the complex plane (it corresponds to making one turn counterclockwise around the origin in $\mathbf{m}$): we then have $B_{\tilde{\gamma}}(Tv) = B_{\tilde{\gamma}}(v) - s$ as a consequence of the relation $T\tilde{\gamma}(t) = \tilde{\gamma}(t+s)$.

Now, let us denote by $(c_i)_{i \in \mathbb{Z}}$ and $(c'_i)_{i \in \mathbb{Z}}$ the successive corners incident to the faces $\tilde{1}$ and $\tilde{2}$, respectively, as we follow their contours walking with the face at hand to the right. This depends on a choice for the corners $c_0$ and $c'_0$, whose influence will be discussed later. We have $Tc_i = c_{i+2m_1}$ and $Tc'_i = c'_{i-2m_2}$ for all $i$.

We then set, for any $i \in \mathbb{Z}$, $\ell_i := B_{\tilde{\gamma}}(c_0) - B_{\tilde{\gamma}}(c_i)$ and $\ell'_i := B_{\tilde{\gamma}}(c'_0) - B_{\tilde{\gamma}}(c'_i)$ (these differences do not depend on the choice of $\tilde{\gamma}(0)$, as claimed). Observe that the definition of $\ell_i$ is analogous to that used in Section 4.3 for the decomposition of a slice into a path
<figure><img src="image_16.png" /><figcaption>Figure 4.11: Sketch of the decomposition of the universal cover of an annular map into slices. The leftmost infinite geodesics (shown in dotted lines) eventually merge with the biinfinite geodesic $\tilde{\gamma}$ (shown in dashed lines). These geodesics delimit slices $\sigma_i$ and $\sigma'_i$, $i \in \mathbb{Z}$. In the situation displayed here, the slices $\sigma_2, \sigma_7, \sigma'_1$, etc, are trivial slices (reduced to an edge).</figcaption></figure>

$B_{\tilde{\gamma}}(\cdot)$). The sequences $\ell = (\ell_i)_{i \in \mathbb{Z}}$ and $\ell' = (\ell'_i)_{i \in \mathbb{Z}}$ form infinite lattice paths (with increments $\pm 1$) which are periodic: indeed, we have $\ell_{i+2m_1} = \ell_i + s$ and $\ell'_{i+2m_2} = \ell'_i - s$ for all $i$. Thus, these sequences are entirely determined by their data in a fundamental domain: $L := (\ell_0, \ell_1, \dots, \ell_{2m_1})$ and $L' := (\ell'_0, \ell'_1, \dots, \ell'_{2m_2})$ form lattice paths connecting $(0, 0)$ to respectively $(2m_1, s)$ and $(2m_2, -s)$.

With each corner $c_i$, we associate the leftmost infinite geodesic $\gamma_i = (\gamma_i(t))_{t \ge 0}$ defined inductively as follows. We let $\gamma_i(0)$ be the vertex incident to $c_i$ and, assuming that $\gamma_i(t)$ is known, we let $\gamma_i(t+1)$ be the leftmost vertex such that $B_{\tilde{\gamma}}(\gamma_i(t+1)) = B_{\tilde{\gamma}}(\gamma_i(t)) - 1$, using the edge $\gamma_i(t-1)\gamma_i(t)$ (or the corner $c_i$ for $t=0$) as a reference. We define in the same way the leftmost infinite geodesic $\gamma'_i$ starting at $c'_i$. We may check that each of these leftmost geodesics eventually merges with $\tilde{\gamma}$ (for $\gamma'_i$, this uses the fact that we chose $\gamma$ to be the innermost minimal separating cycle).

Let $\sigma_i$ (resp. $\sigma'_i$) be the map delimited by $\gamma_{i-1}$ and $\gamma_i$ (resp. $\gamma'_{i-1}$ and $\gamma'_i$), which we stop at their first common vertex $v_i$ (resp. $v'_i$). It is an elementary bipartite slice with base $c_{i-1}c_i$ (resp. $c'_{i-1}c'_i$). As in Section 4.3, the slice is trivial whenever $i$ corresponds to a down step of $\ell$ (resp. $\ell'$), and is non-trivial otherwise (but may be empty). By periodicity, we have $\sigma_i = \sigma_{i+2m_1}$ and $\sigma'_i = \sigma'_{i-2m_2}$ (in the sense of equality as maps). Note that every finite face, edge and vertex of $\tilde{m}$ belongs to exactly one slice $\sigma_i$ or $\sigma'_i$ deprived of its right boundary.

Let us finally discuss the roles of the reference corners $c_0$ and $c'_0$. Changing the reference corner $c_0$ amounts to translating the lattice path $\ell$, and to reparametrizing the sequence $\sigma = (\sigma_i)_{i \in \mathbb{Z}}$ by a translation of $i$. In particular, both $\ell$ and $\sigma$ are invariant if we change $c_0$ into one of its translates $T^k c_0$, thus they only depend on the choice of a corner incident to face 1 in $\mathfrak{m}$. Similarly, $\ell'$ and $\sigma' = (\sigma'_i)_{i \in \mathbb{Z}}$ only depend on the choice of a corner incident to face 2 in $\mathfrak{m}$.

Let us now assume that the map $\mathfrak{m}$ has a third distinguished element (face or vertex) labeled 3, which is indeed the case in the setting of Proposition 4.7. Then it is possible to choose corners incident to faces 1 and 2 in a canonical way. Namely, we pick a preimage $\tilde{3}$ of 3 in $\tilde{\mathfrak{m}}$: $\tilde{3}$ belongs
<figure><img src="image_15.png" /><figcaption>Figure 14: Sketch of the decomposition of the universal cover of an annular map into slices. The leftmost infinite geodesics (shown in dotted lines) eventually merge with the biinfinite geodesic $\tilde{\gamma}$ (shown in dashed lines). These geodesics delimit slices $\sigma_i$ and $\sigma'_i$, $i \in \mathbb{Z}$. In the situation displayed here, the slices $\sigma_2, \sigma_7, \sigma'_1$, etc, are trivial slices (reduced to an edge).</figcaption></figure>

decorated with elementary slices: the distance to the apex $d(\cdot, A)$ is just replaced by the Busemann function $B_{\tilde{\gamma}}(\cdot)$. The sequences $\ell = (\ell_i)_{i \in \mathbb{Z}}$ and $\ell' = (\ell'_i)_{i \in \mathbb{Z}}$ form infinite lattice paths (with increments $\pm 1$) which are periodic: indeed, we have $\ell_{i+2m_1} = \ell_i + s$ and $\ell'_{i+2m_2} = \ell'_i - s$ for all $i$. Thus, these sequences are entirely determined by their data in a fundamental domain: $L := (\ell_0, \ell_1, \dots, \ell_{2m_1})$ and $L' := (\ell'_0, \ell'_1, \dots, \ell'_{2m_2})$ form lattice paths connecting $(0,0)$ to respectively $(2m_1, s)$ and $(2m_2, -s)$.

With each corner $c_i$, we associate the leftmost infinite geodesic $\gamma_i = (\gamma_i(t))_{t \ge 0}$ defined inductively as follows. We let $\gamma_i(0)$ be the vertex incident to $c_i$ and, assuming that $\gamma_i(t)$ is known, we let $\gamma_i(t+1)$ be the leftmost vertex such that $B_{\tilde{\gamma}}(\gamma_i(t+1)) = B_{\tilde{\gamma}}(\gamma_i(t)) - 1$, using the edge $\gamma_i(t-1)\gamma_i(t)$ (or the corner $c_i$ for $t=0$) as a reference. We define in the same way the leftmost infinite geodesic $\gamma'_i$ starting at $c'_i$. We may check that each of these leftmost geodesics eventually merges with $\tilde{\gamma}$ (for $\gamma'_i$, this uses the fact that we chose $\gamma$ to be the innermost minimal separating cycle).

Let $\sigma_i$ (resp. $\sigma'_i$) be the map delimited by $\gamma_{i-1}$ and $\gamma_i$ (resp. $\gamma'_{i-1}$ and $\gamma'_i$), which we stop at their first common vertex $v_i$ (resp. $v'_i$). It is an elementary bipartite slice with base $c_{i-1}c_i$ (resp. $c'_{i-1}c'_i$). As in Section 4.3, the slice is trivial whenever $i$ corresponds to a down step of $\ell$ (resp. $\ell'$), and is non-trivial otherwise (but may be empty). By periodicity, we have $\sigma_i = \sigma_{i+2m_1}$ and $\sigma'_i = \sigma'_{i-2m_2}$ (in the sense of equality as maps). Note that every finite face, edge and vertex of $\tilde{m}$ belongs to exactly one slice $\sigma_i$ or $\sigma'_i$ deprived of its right boundary.

Let us finally discuss the roles of the reference corners $c_0$ and $c'_0$. Changing the reference corner $c_0$ amounts to translating the lattice path $\ell$, and to reparametrizing the sequence $\sigma = (\sigma_i)_{i \in \mathbb{Z}}$ by a translation of $i$. In particular, both $\ell$ and $\sigma$ are invariant if we change $c_0$ into one its translates $T^k c_0$, thus they only depend on the choice of a corner incident
to precisely one slice $\sigma_i$ or $\sigma'_i$ deprived of its right boundary. If $\tilde{3}$ belongs to a $\sigma_i$, then by changing the reference corner $c_0$ we can ensure that it belongs to $\sigma_1$. Then, we choose $c'_0$ in such way that $B_{\tilde{\gamma}}(c'_0) = B_{\tilde{\gamma}}(c_0)$: this is possible because the function $B_{\tilde{\gamma}}(\cdot)$ has $\pm 1$ increments along the contour of $\tilde{2}$ and decreases by $s$ over a translation $T$, hence it is surjective. There might exist several such $c'_0$, in which case we pick the last one, so that $B_{\tilde{\gamma}}(c'_i) \neq B_{\tilde{\gamma}}(c'_0)$ for every $i > 0$. If $\tilde{3}$ belongs to a $\sigma'_i$, we proceed in the same way upon exchanging the roles of $c_0$ and $c'_0$.

**Lemma 4.8.** *The above construction is a bijection between:*

• the set of maps $\mathbf{m}$ with two marked faces of degrees $2m_1, 2m_2 \in \mathbb{Z}_{>0}$ and all other faces of even degree, with a third distinguished face or vertex, and with separating girth $s \in \mathbb{Z}_{>0}$,

• the set of quadruples $(L, L', \boldsymbol{\sigma}, \boldsymbol{\sigma}')$, where $L$ and $L'$ are lattice paths connecting $(0, 0)$ to respectively $(2m_1, s)$ and $(2m_2, -s)$, where $\boldsymbol{\sigma}$ and $\boldsymbol{\sigma}'$ are sequences of elementary slices of respective periods $2m_1$ and $2m_2$, such that for every $i = 1, \dots, 2m_1$ (resp. $i = 1, \dots, 2m_2$) the tilt of $\boldsymbol{\sigma}_i$ and (resp. $\boldsymbol{\sigma}'_i$) is equal to $\ell_i - \ell_{i-1}$ (resp. $\ell'_i - \ell'_{i-1}$), where either $\boldsymbol{\sigma}_1$ or $\boldsymbol{\sigma}'_1$ carries a distinguished face or vertex, and where $L'$ (resp. $L$) does not return to height 0 if $\boldsymbol{\sigma}_1$ (resp. $\boldsymbol{\sigma}'_1$) carries the distinguished element.

To prove this lemma, we now describe the inverse procedure which consists in assembling an annular map $\mathbf{m}$ from a quadruple $(L, L', \boldsymbol{\sigma}, \boldsymbol{\sigma}')$.

**Assembling an annular map from a pair of paths decorated with elementary slices.** Using the lattice path $L$ and the sequence of slices $\boldsymbol{\sigma}$ restricted to a period, we may use the inverse of the decomposition described in Section 4.3 to obtain a slice $\boldsymbol{\Sigma}$ of width $2m_1$ and tilt $s$. We then perform the operation described in [BG14, Section 7.1] and called *wrapping* in [Bou19, Section 2.2]. It consists in gluing “isometrically” the left and right boundaries of $\boldsymbol{\Sigma}$ together, but unlike the gluing performed in Section 4.3 to obtain a pointed rooted map, we now start from the two endpoints of the base of $\boldsymbol{\Sigma}$ which we identify together, and perform the gluing from there (see the illustrations in the aforementioned references). As the tilt $s$ is positive, the $s$ edges of the left boundary closest to the apex remain unglued. This produces an annular map $\mathbf{m}_1$ whose two distinguished faces have respective degrees $2m_1$ and $s$, the latter resulting from the unglued edges, the former resulting from the base, whose two identified endpoints yield a distinguished corner denoted $c$.

Similarly, the lattice path $L'$ and the sequence of slices $\boldsymbol{\sigma}'$ yield a slice $\boldsymbol{\Sigma}'$ of width $2m_2$ and tilt $-s$, whose wrapping produces an annular map $\mathbf{m}_2$ with two distinguished faces of respective degrees $2m_2$ and $s$ (since the tilt of $\boldsymbol{\Sigma}'$ is negative, it is now the $s$ edges closest to the apex on the right boundary that form this latter face), the former having a distinguished corner denoted $c'$.

The annular map $\mathbf{m}$ is then obtained by gluing the two annular maps $\mathbf{m}_1$ and $\mathbf{m}_2$ together along the contours of their distinguished faces of degree $s$. The contour then becomes a separating cycle $\gamma$, which can be shown to have minimal length. Note that there are a priori $s$ ways to glue $\mathbf{m}_1$ and $\mathbf{m}_2$ together, however only one way is compatible with the prescription that, in the universal cover $\tilde{\mathbf{m}}$ of $\mathbf{m}$, the corners $c$ and $c'$ admit respective lifts $c_0$ and $c'_0$ such that $B_{\tilde{\gamma}}(c_0) = B_{\tilde{\gamma}}(c'_0)$ where $\tilde{\gamma}$ is the infinite geodesic constructed by lifting $\gamma$.
to face 1 in **m**. Similarly, $\ell'$ and $\sigma' = (\sigma'_i)_{i \in \mathbb{Z}}$ only depend on the choice of a corner incident to face 2 in **m**.

If we assume that the map **m** has a third distinguished element (face, edge or vertex) labeled 3, which will be the case in our application, then it is possible to choose corners incident to faces 1 and 2 in a canonical way. Namely, we pick a preimage $\tilde{3}$ of 3 in $\tilde{\mathbf{m}}$: $\tilde{3}$ belongs to precisely one slice $\boldsymbol{\sigma}_i$ or $\boldsymbol{\sigma}'_i$ deprived of its right boundary. If $\tilde{3}$ belongs to a $\boldsymbol{\sigma}_i$, then by changing the reference corner $c_0$ we can ensure that it belongs to $\boldsymbol{\sigma}_1$. Then, we choose $c'_0$ in such way that $B_{\tilde{\gamma}}(c'_0) = B_{\tilde{\gamma}}(c_0)$: this is possible because the function $B_{\tilde{\gamma}}(\cdot)$ has $\pm 1$ increments along the contour of $\tilde{2}$ and decreases by $s$ over a translation $T$, hence it is surjective. There might exist several such $c'_0$, in which case we may pick, say, the "rightmost" one. If $\tilde{3}$ belongs to a $\boldsymbol{\sigma}'_i$, we proceed in the same way upon exchanging the roles of $c_0$ and $c'_0$.

**Lemma 28.** The above construction is a bijection between:

* the set of maps **m** with two marked faces of degrees $2m_1, 2m_2 \in \mathbb{Z}_{>0}$ and all other faces of even degree, and separating girth $s \in \mathbb{Z}_{>0}$,

* the set of quadruples $(L, L', \boldsymbol{\sigma}, \boldsymbol{\sigma}')$, where $L$ and $L'$ are lattice paths connecting $(0, 0)$ to respectively $(2m_1, s)$ and $(2m_2, -s)$, where $\boldsymbol{\sigma}$ and $\boldsymbol{\sigma}'$ are sequences of elementary slices of respective periods $2m_1$ and $2m_2$, such that for every $i = 1, \dots, 2m_1$ (resp. $i = 1, \dots, 2m_2$) the tilt of $\boldsymbol{\sigma}_i$ and (resp. $\boldsymbol{\sigma}'_i$) is equal to $\ell_i - \ell_{i-1}$ (resp. $\ell'_i - \ell'_{i-1}$).

To prove this lemma, we now describe the inverse procedure which consists in assembling an annular map **m** from a quadruple $(L, L', \boldsymbol{\sigma}, \boldsymbol{\sigma}')$.

**Assembling an annular map from a pair of paths decorated with elementary slices.**

Using the lattice path $L$ and the sequence of slices $\boldsymbol{\sigma}$ restricted to a period, we may use the inverse of the decomposition described in Section 4.3 to obtain a slice $\boldsymbol{\Sigma}$ of width $2m_1$ and tilt $s$. We then perform the operation described in [BG14, Section 7.1] and called *wrapping* in [Bou19, Section 2.2]. It consists in gluing “isometrically” the left and right boundaries of $\boldsymbol{\Sigma}$ together, but unlike the gluing performed in Section 4.3 to obtain a pointed rooted map, we now start from the two endpoints of the base of $\boldsymbol{\Sigma}$ which we identify together, and perform the gluing from there (see the illustrations in the aforementioned references). As the tilt $s$ is positive, the $s$ edges of the left boundary closest to the apex remain unglued. This produces an annular map $\mathbf{m}_1$ whose two distinguished faces have respective degrees $2m_1$ and $s$, the latter resulting from the unglued edges, the former resulting from the base, whose two identified endpoints yield a distinguished corner denoted $c$.

Similarly, the lattice path $L'$ and the sequence of slices $\boldsymbol{\sigma}'$ yield a slice $\boldsymbol{\Sigma}'$ of width $2m_2$ and tilt $-s$, whose wrapping produces an annular map $\mathbf{m}_2$ with two distinguished faces of respective degrees $2m_2$ and $s$ (since the tilt of $\boldsymbol{\Sigma}'$ is negative, it is now the $s$ edges closest to the apex on the *right* boundary that form this latter face), the former having a distinguished corner denoted $c'$.

The annular map **m** is then obtained by gluing the two annular maps $\mathbf{m}_1$ and $\mathbf{m}_2$ together along the contours of their distinguished faces of degree $s$. The contour then
We may check that the assembling procedure is indeed the inverse of the decomposition described above. The most subtle point is to show that any quadruple $(L, L', \sigma, \sigma')$ is left invariant if we assemble it then decompose the resulting annular map. This requires to check that the boundaries of the slices become, in the annular map, precisely the leftmost geodesics that we use in the decomposition. More details can be found in [BG14, Section 7].

**Application to tight maps and end of the proof of Proposition 4.7.** So far, all our discussion holds without the assumption that $\mathbf{m}$ is tight. Adding this constraint amounts to two restrictions on the corresponding quadruple $(L, L', \sigma, \sigma')$, which are similar to those encountered in Section 4.3. First, every $\sigma_i$ and $\sigma'_i$ must be a tight elementary slice. Here, we use the natural convention that the mark of a marked vertex is transferred to its copy in the unique slice deprived of its right boundary that contains it (in particular, marked vertices lying on the separating cycle $\gamma$ have their marks transferred to $\sigma_i$'s and not to $\sigma'_i$'s). Second, the lattice paths $\ell$ and $\ell'$ (which are the periodic extensions of $L$ and $L'$) must be such that every up step immediately following a down step cannot be decorated with an unmarked empty slice.

To complete the proof of Proposition 4.7, we assume that $\mathbf{m}$ has $n \ge 3$ boundaries, labeled 1 to $n$. By the slice decomposition, the labels 3, ..., $n$ get distributed among the $\sigma_i$ and $\sigma'_i$. Let $i_1 \le \dots \le i_j$ be the indices $i$ between 1 and $2m_1$ such that $\sigma_i$ is neither the unmarked empty slice nor the trivial slice, i.e. contains at least one label. Let $i'_1 \le \dots \le i'_j$ be similarly the indices $i$ between 1 and $2m_2$ such that $\sigma'_i$ contains at least one label. We set $\bar{k} = j + j' - 1$ and note that $0 \le k \le n-3$ since there are $n-2$ labels in total to distribute among the slices. Recall that, by the aforementioned prescription for choosing the reference corners, the label 3 is either in $\sigma_1$ or $\sigma'_1$. If it is in $\sigma_1$, we set $(\mathbf{s}_1, \dots, \mathbf{s}_{k+1}) := (\sigma_{i_1=1}, \dots, \sigma_{i_j}, \sigma'_{i'_1}, \dots, \sigma'_{i'_j})$. Otherwise, we set $(\mathbf{s}_1, \dots, \mathbf{s}_{k+1}) := (\sigma'_{i'_1}=1, \dots, \sigma'_{i'_j'}, \sigma_{i_1}, \dots, \sigma_{i_j})$. This defines the $(k+1)$-tuple of slices $s$ of the proposition.

As for the two-face map $\mathbf{m}_{12}$, it is obtained as follows. Let $\omega$ and $\omega'$ be the sequences obtained from respectively $\sigma$ and $\sigma'$ by replacing every slice different from the unmarked empty slice and from the trivial slice with the marked empty slice. Then, $\mathbf{m}_{12}$ is the annular map obtained by assembling the quadruple $(L, L', \omega, \omega')$. It is by construction a tight two-face map with $k+1$ marked vertices, and we distinguish the marked vertex coming from the marked empty slice which replaces the slice containing the label 3.

We check that the mapping $\mathbf{m} \mapsto (\mathbf{m}_{12}, \mathbf{s})$ is a bijection by exhibiting the inverse bijection. Let $(\mathbf{m}_{12}, \mathbf{s})$ be a pair as in the proposition, and let $(L, L', \omega, \omega')$ be the quadruple obtained by decomposing the two-face map $\mathbf{m}_{12}$, its distinguished marked vertex playing the role of the third distinguished element labeled 3 used in the construction of page 38. By construction the sequences $\omega$ and $\omega'$ consist only of empty or trivial slices, with a number $k+1$ of marked empty slices. Replacing these marked empty slices with $\mathbf{s}_1, \dots, \mathbf{s}_{k+1}$, we obtain two sequences $\sigma$ and $\sigma'$ such that the assembling of the quadruple $(L, L', \sigma, \sigma')$ gives the tight map $\mathbf{m}$ we are looking for. This ends the proof of Proposition 4.7. $\square$

**Remark 4.9.** It might seem more direct to attempt to enumerate the quadruples $(L, L', \sigma, \sigma')$. But, since the paths $L$ and $L'$ have a height variation depending on the separating girth $s$, this leads to an expression involving a sum over $s$. The trick of “recombining” $L$ and $L'$ into a two-face map allows to circumvent this issue.
becomes a separating cycle $\gamma$, which can be shown to have minimal length. Note that there are a priori $s$ ways to glue $\mathbf{m}_1$ and $\mathbf{m}_2$ together, however only one way is compatible with the prescription that, in the universal cover $\tilde{\mathfrak{m}}$ of $\mathfrak{m}$, the corners $c$ and $c'$ admit respective lifts $c_0$ and $c'_0$ such that $B_{\tilde{\gamma}}(c_0) = B_{\tilde{\gamma}}(c'_0)$ where $\tilde{\gamma}$ is the infinite geodesic constructed by lifting $\gamma$.

We may check that the assembling procedure is indeed the inverse of the decomposition described above. The most subtle point is to show that any quadruple $(L, L', \sigma, \sigma')$ is left invariant if we assemble it then decompose the resulting annular map. This requires to check that the boundaries of the slices become, in the annular map, precisely the leftmost geodesics that we use in the decomposition. More details can be found in [BG14, Section 7].

**Application to tight maps and end of the proof of Proposition 27.** So far, all our discussion holds without the assumption that $\mathfrak{m}$ is tight. Adding this constraint amounts to two restrictions on the corresponding quadruple $(L, L', \sigma, \sigma')$, which are similar to those encountered in Section 4.3. First, every $\sigma_i$ and $\sigma'_i$ must be a tight elementary slice. Here, we use the natural convention that the mark of a marked vertex is transferred to its copy in the unique slice deprived of its right boundary that contains it (in particular, marked vertices lying on the separating cycle $\gamma$ have their marks transferred to $\sigma_i$'s and not to $\sigma'_i$'s). Second, the lattice paths $\ell$ and $\ell'$ (which are the periodic extensions of $L$ and $L'$) must be such that every up step immediately following a down step cannot be decorated with an unmarked empty slice.

To complete the proof of Proposition 27, we assume that $\mathfrak{m}$ has $n \ge 3$ boundaries, labeled 1 to $n$. By the slice decomposition, the labels $3, \dots, n$ get distributed among the $\sigma_i$ and $\sigma'_i$. Let $i_1 \le \dots \le i_j$ be the indices $i$ between 1 and $2m_1$ such that $\sigma_i$ is neither the unmarked empty slice nor the trivial slice, i.e. contains at least one label. Let $i'_1 \le \dots \le i'_{j'}$ be similarly the indices $i$ between 1 and $2m_2$ such that $\sigma'_i$ contains at least one label. We set $k = j + j' - 1$ and note that $0 \le k \le n - 3$ since there are $n-2$ labels in total to distribute among the slices. Recall that, by the aforementioned prescription for choosing the reference corners, the label 3 is either in $\sigma_1$ or $\sigma'_1$. If it is in $\sigma_1$, we set $(\mathbf{s}_1, \dots, \mathbf{s}_{k+1}) := (\sigma_{i_1=1}, \dots, \sigma_{i_j}, \sigma'_{i'_1}, \dots, \sigma'_{i'_{j'}})$. Otherwise, we set $(\mathbf{s}_1, \dots, \mathbf{s}_{k+1}) := (\sigma'_{i'_1=1}, \dots, \sigma'_{i'_{j'}}, \sigma_{i_1}, \dots, \sigma_{i_j})$. This defines the $(k+1)$-tuple of slices $\mathbf{s}$ of the proposition.

As for the two-face map $\mathbf{m}_{12}$, it is obtained as follows. Let $\omega$ and $\omega'$ be the sequences obtained from respectively $\sigma$ and $\sigma'$ by replacing every slice different from the unmarked empty slice and from the trivial slice with the marked empty slice. Then, $\mathbf{m}_{12}$ is the annular map obtained by assembling the quadruple $(L, L', \omega, \omega')$. It is by construction a tight two-face map with $k+1$ marked vertices, and we distinguish the marked vertex coming from the marked empty slice which replaces the slice containing the label 3.

We check that the mapping $\mathfrak{m} \mapsto (\mathbf{m}_{12}, \mathbf{s})$ is a bijection by exhibiting the inverse bijection. Let $(\mathbf{m}_{12}, \mathbf{s})$ be a pair as in the proposition, and let $(L, L', \omega, \omega')$ be the quadruple obtained by decomposing the two-face map $\mathbf{m}_{12}$, its distinguished marked vertex playing the role of the third distinguished element labeled 3 used in the construction of page 40.
## 5. Bijective proofs for non necessarily bipartite maps

In this section, we explain how the bijective approach of Section 4 may be extended so as to enumerate planar tight maps which are not necessarily bipartite.

Recall from Section 2.2 that a *petal* is a face of degree one. The key idea to extend our construction to the non bipartite case is to realize that petals play a role very similar to marked vertices. Mimicking the organization of Section 4, we will first enumerate tight maps with a single non-petal face in Section 5.1, then tight maps with just two non-petal faces in Section 5.2, and finally tight maps with an arbitrary number of non-petal faces in Section 5.4. A prerequisite to this latter enumeration will be that of tight non necessarily bipartite slices in Section 5.3.

Before we start our discussion, let us recall from Section 2.2 the definition of the polynomials

$$p_{k,e}(m) := \frac{1}{(k!)^2} \prod_{i=1}^{k} \left( m^2 - \left( i - \frac{e}{2} \right)^2 \right) = \binom{m + \frac{e}{2} - 1}{k} \binom{m - \frac{e}{2} + k}{k} \quad (5.1)$$

for $k \in \mathbb{Z}_{\ge 0}$. For $m - \frac{e}{2} \in \mathbb{Z}$, we may interpret $p_{k,e}(m)$ as counting words of the form (4.1) where the $a_i$ and $b_i$ are nonnegative integers such that $a_1 + a_2 + \dots + a_{k+1} = m + \frac{e}{2} - k - 1$ and $b_1 + b_2 + \dots + b_{k+1} = m - \frac{e}{2}$ (i.e. the word has length $2m$, $k+1$ occurrences of $D_0$ and $e$ more $D$'s than $U$'s). This interpretation holds a priori only for $m \ge \max(k + 1 - \frac{e}{2}, \frac{e}{2})$ but this domain may be extended to $m \ge \max(1 - \frac{e}{2}, \frac{e}{2} - k)$ since $p_{k,e}(m)$ vanishes in the additional domain.

**Remark 5.1.** Even though we will not use it in the sequel, let us mention that, by a variation of the arguments of Section 4.1, we may show that, for $m$ as above, $p_{k,e}(m)$ is the number of two-face tight maps with one face of degree $2m$, one simple face of degree $|e|$ and $k+1$ marked vertices, one of them distinguished, with the condition that for $e < 0$ no marked vertex is incident to the face of degree $e$. Note that such maps are closely related with the notion of $(a, b)^*$-forests of Definition 4.3: for $e > 0$, $p_{k,e}(m)$ counts $(0, e)^*$-forests with size $2m$ and $k+1$ marked vertices including the distinguished vertex.

### 5.1. Petal trees

We recall that a *petal tree* is a planar map having an *exterior* face of arbitrary degree, and such that every other face is a petal. A tight petal tree is just a petal tree with marked vertices, which is tight as a map, see again Figure 2.1.

We also recall, for $r, s \in \mathbb{Z}_{\ge 0}$, $\epsilon \in \mathbb{Z}$ and $m \in \mathbb{Z}/2$, the definition of the quasi-polynomial

$$\pi_{r,s}^{(\epsilon)}(m) := \begin{cases} \binom{r+s}{s} p_{r+s,s+1+\epsilon}(m) & \text{if } m - \frac{s+1+\epsilon}{2} \in \mathbb{Z}, \\ 0 & \text{otherwise.} \end{cases} \quad (5.2)$$

Our goal in this section is to prove Proposition 2.14, showing that, for $m \in \mathbb{Z}_{>0}/2$ and $\epsilon \in \{-1, 0, 1\}$, $\pi_{r,s}^{(\epsilon)}(m)$ enumerates tight petal trees with one exterior face of degree $2m$, $s+1+\epsilon$ petals (excluding the exterior face when $m = 1/2$), $1+\epsilon$ of which are distinguished, and $r+1-\epsilon$ marked vertices, $1-\epsilon$ of which are distinguished. The reason for the $\epsilon$-dependence in the
By construction the sequences $\omega$ and $\omega'$ consist only of empty or trivial slices, with a number $k+1$ of marked empty slices. Replacing these marked empty slices with $\mathbf{s}_1, \dots, \mathbf{s}_{k+1}$, we obtain two sequences $\sigma$ and $\sigma'$ such that the assembling of the quadruple $(L, L', \sigma, \sigma')$ gives the tight map $\mathbf{m}$ we are looking for. This ends the proof of Proposition 27. $\square$

*Remark 29.* It might seem a priori more direct to attempt to enumerate the quadruples $(L, L', \sigma, \sigma')$. But, since the paths $L$ and $L'$ have a height variation depending on the separating girth $s$, this leads to an expression involving a sum over $s$. The trick of “recombining” $L$ and $L'$ into a two-face map allows to circumvent this issue.

# 5. Bijective proofs for non necessarily bipartite maps

In this section, we explain how the bijective approach of Section 4 may be extended so as to enumerate planar tight maps which are not necessarily bipartite.

Recall from Section 2.2 that a *petal* is a face of degree one. The key idea to extend our construction to the non bipartite case is to realize that petals play a role very similar to marked vertices. Mimicking the organization of Section 4, we will first enumerate tight maps with a single non-petal face in Section 5.1, then tight maps with just two non-petal faces in Section 5.2, and finally tight maps with an arbitrary number of non-petal faces in Section 5.4. A prerequisite to this latter enumeration will be that of tight non necessarily bipartite slices in Section 5.3.

Before we start our discussion, let us recall from Section 2.2 the definition of the polynomials

$$p_{k,e}(m) := \frac{1}{(k!)^2} \prod_{i=1}^{k} \left( m^2 - \left( i - \frac{e}{2} \right)^2 \right) = \left( m + \frac{e}{2} - 1 \right) \left( m - \frac{e}{2} + k \right) \quad (52)$$

for $k \in \mathbb{Z}_{\ge 0}$. For $m - \frac{e}{2} \in \mathbb{Z}$, we may interpret $p_{k,e}(m)$ as counting words of the form (43) where the $a_i$ and $b_i$ are nonnegative integers such that $a_1 + a_2 + \dots + a_{k+1} = m + \frac{e}{2} - k - 1$ and $b_1 + b_2 + \dots + b_{k+1} = m - \frac{e}{2}$ (i.e. the word has length $2m$, $k+1$ occurrences of $D_o$ and $e$ more $D$'s than $U$'s). This interpretation holds a priori only for $m \ge \max(k+1 - \frac{e}{2}, \frac{e}{2})$ but this domain may be extended to $m \ge \max(1 - \frac{e}{2}, \frac{e}{2} - k)$ since $p_{k,e}(m)$ vanishes in the additional domain.

*Remark 30.* Even though we will not use it in the sequel, let us mention that, by a variation of the arguments of Section 4.1, we may show that, for $m$ as above, $p_{k,e}(m)$ is the number of two-face tight maps with one face of degree $2m$, one simple face of degree $|e|$ and $k+1$ marked vertices, one of them distinguished, with the condition that for $e < 0$ no marked vertex is incident to the face of degree $e$. Note that such maps are closely related with the notion of $(a, b)^*$-forests of Definition 23: for $e > 0$, $p_{k,e}(m)$ counts $(0, e)^*$-forests with size $2m$ and $k+1$ marked vertices including the distinguished vertex.

## 5.1. Petal trees

We recall that a *petal tree* is a planar map having an *exterior* face of arbitrary degree, and such that every other face is a petal. A tight petal tree is just a petal tree with
statement is that we must distinguish two elements among marked vertices and petals. For this reason, there are three situations to consider: we may distinguish two marked vertices ($\epsilon = -1$), two petals ($\epsilon = 1$), or one of each type ($\epsilon = 0$).

We note that, upon “labeling” the undistinguished elements, Proposition 2.14 is equivalent to the more symmetric statement:

**Proposition 5.2.** For $0 \le e \le k + 2$ and $m \in \mathbb{Z}_{>0}/2$, the number of tight petal trees with an exterior face of degree $2m$, with $e$ petals and $k + 2 - e$ marked vertices, all labeled, is given by:

$$N_{0,k+3}(2m, \underbrace{1, \dots, 1}_{e}, \underbrace{0, \dots, 0}_{k+2-e}) = \begin{cases} k! p_{k,e}(m) & \text{if } m - \frac{e}{2} \in \mathbb{Z}, \\ 0 & \text{otherwise.} \end{cases} \qquad (5.3)$$

More precisely, Proposition 5.2 is recovered from Proposition 2.14 by setting $k = r + s$ and $e = s + 1 + \epsilon$ (hence $k + 2 - e = r + 1 - \epsilon$). The petal trees considered in Proposition 2.14 have $s$ (respectively $r$) non-distinguished petals (respectively marked vertices) which we may label in $s!$ (respectively $r!$) ways. Using $s!r!(\binom{r+s}{s}) = (r+s)! = k!$, we obtain (5.3) for $1 + \epsilon \le e \le k + 1 + \epsilon$, hence for the whole range $0 \le e \le k + 2$ by letting in $\epsilon$ vary in $\{-1, 0, 1\}$.

*Proof of Proposition 2.14.* Let us first discuss how we may code petal trees using words, or equivalently lattice paths, as we did in Section 4.1 for ordinary trees. Note first that petal trees have two types of edges: “tree-type” edges whose both sides are incident to the exterior face and “petal-type” edges with one edge side incident to a petal and the other to the exterior face. In particular, the tree-type edges have distinct endpoints and they form a plane tree (i.e. a map with a single face), which we call the “wood” of the petal tree. Here the coding that we shall use applies to *rooted* petal trees, i.e. petal trees where we distinguish a corner (the root corner) in the exterior face (note that this also induces a rooting of the wood tree). Starting from this root corner and following the contour of the exterior face going counterclockwise around the tree (i.e. with the exterior face on the right), we record a letter $U$ (respectively $D$) for each tree-type edge visited while going away from (respectively towards) the root and a letter $E$ for each visited petal-type edge. The obtained three-letter word may alternatively be visualized as a lattice path with three types of elementary steps: up, down and horizontal, associated respectively to the letters $U$, $D$ and $E$. This path is a *Motzkin path* of length equal to the degree $2m'$ of the exterior face, i.e. it goes from $(0,0)$ to $(2m',0)$ and *stays above the x-axis*. Indeed, the ordinates of the path are *non-negative* since they record graph distances to the *root vertex* incident to the root corner. The number $s'$ of occurrences of the letter $E$ in the coding word is nothing but the number of petals, and is such that $m' - \frac{s'}{2}$, which is the number of $U$'s (or equivalently of $D$'s), is a nonnegative integer.

Let us now consider a *rooted* petal tree with marked vertices, where the root vertex is *unmarked*, and such that all non-root leaves are marked. Every non-root vertex being bijectively associated with its parent tree-type edge in the wood tree, hence with a letter $D$, we record the markings as in Section 4.1 by replacing each $D$ by a $D_\circ$ if the associated vertex is marked and by a $D_\bullet$ otherwise. We end up with a word made of four letters, $U$, $D_\circ$, $D_\bullet$ and $E$, associated with a “dressed” Motzkin path, with $k'$ occurrences of $D_\circ$ and $m' - \frac{s'}{2} - k'$ occurrences of $D_\bullet$.
marked vertices, which is tight as a map.

We also recall, for $r, s \in \mathbb{Z}_{\ge 0}$, $\epsilon \in \mathbb{Z}$ and $m \in \mathbb{Z}/2$, the definition of the quasi-polynomial

$$ \pi_{r,s}^{(\epsilon)}(m) := \begin{cases} \binom{r+s}{s} p_{r+s,s+1+\epsilon}(m) & \text{if } m - \frac{s+1+\epsilon}{2} \in \mathbb{Z}, \\ 0 & \text{otherwise.} \end{cases} \qquad (53) $$

Our goal in this section is to prove Proposition 15, showing that, for $m \in \mathbb{Z}_{>0}/2$ and $\epsilon \in \{-1, 0, 1\}$, $\pi_{r,s}^{(\epsilon)}(m)$ enumerates tight petal trees with one exterior face of degree $2m$, $s+1+\epsilon$ petals (excluding the exterior face when $m=1/2$), $1+\epsilon$ of which are distinguished, and $r+1-\epsilon$ marked vertices, $1-\epsilon$ of which are distinguished. The reason for the $\epsilon$-dependence in the statement is that we must distinguish two elements among marked vertices and petals. For this reason, there are three situations to consider: we may distinguish two marked vertices ($\epsilon = -1$), two petals ($\epsilon = 1$), or one of each type ($\epsilon = 0$).

We note that, upon “labeling” the undistinguished elements, Proposition 15 is equivalent to the more symmetric statement:

**Proposition 31.** For $0 \le e \le k+2$ and $m \in \mathbb{Z}_{>0}/2$, the number of tight petal trees with an exterior face of degree $2m$, with $e$ petals and $k+2-e$ marked vertices, all labeled, is given by:

$$ N_{0,k+3}(2m, \underbrace{1, \dots, 1}_{e}, \underbrace{0, \dots, 0}_{k+2-e}) = \begin{cases} k! p_{k,e}(m) & \text{if } m - \frac{e}{2} \in \mathbb{Z}, \\ 0 & \text{otherwise.} \end{cases} \qquad (54) $$

More precisely, Proposition 31 is recovered from Proposition 15 by setting $k = r + s$ and $e = s+1+\epsilon$ (hence $k+2-e = r+1-\epsilon$). The petal trees considered in Proposition 15 have $s$ (respectively $r$) non-distinguished petals (respectively marked vertices) which we may label in $s!$ (respectively $r!$) ways. Using $s!r!(\binom{r+s}{s}) = (r+s)! = k!$, we obtain (54) for $1+\epsilon \le e \le k+1+\epsilon$, hence for the whole range $0 \le e \le k+2$ by letting in $\epsilon$ vary in $\{-1, 0, 1\}$.

*Proof of Proposition 15.* Let us first discuss how we may code petal trees using words, or equivalently lattice paths, as we did in Section 4.1 for ordinary trees. Note first that petal trees have two types of edges: “tree-type” edges whose both sides are incident to the exterior face and “petal-type” edges with one edge side incident to a petal and the other to the exterior face. In particular, the tree-type edges have distinct endpoints and they form a plane tree (i.e. a map with a single face), which we call the “wood” of the petal tree. Here the coding that we shall use applies to *rooted* petal trees, i.e. petal trees where we distinguish a corner (the root corner) *in the exterior face* (note that this also induces a rooting of the wood tree). Starting from this root corner and following the contour of the exterior face going counterclockwise around the tree (i.e. with the exterior face on the right), we record a letter $U$ (respectively $D$) for each tree-type edge visited while going away from (respectively towards) the root and a letter $E$ for each visited petal-type edge. The obtained three-letter word may alternatively be visualized as a lattice path with three types of elementary steps: up, down and horizontal, associated
<figure><img src="image_17.png" /><figcaption>Figure 5.1: Schematic picture of the decomposition of a lattice path of length $2m-1$, height difference $-\epsilon$, and minimal height $-\ell$ into a sequence of $2\ell + 1 - \epsilon$ Motzkin paths, where the squares indicate that the underlying descending step may be marked (if associated with a $D_\circ$) or not (if associated with a $D_\bullet$). For $j = 1, 2, \dots, \ell$, the $j$-th Motzkin path is followed by the first elementary down step reaching height $-j$. These $\ell$ Motzkin paths code for an ordered set $\mathcal{F}^\square$ of $\ell$ rooted petal trees whose root vertex are marked or not according to the nature of the following elementary down step. For $j = \ell+2, \ell+3, \dots, 2\ell+1-\epsilon$, the $j$-th Motzkin path is preceded by the last elementary up step reaching height $j-1-2\ell$. Together with the $(\ell+1)$-th Motzkin path, these Motzkin paths code for an ordered set $\mathcal{F}_\times$ of $\ell+1-\epsilon$ rooted petal trees whose root vertex is unmarked. The ordered sets $\mathcal{F}^\square$ and $\mathcal{F}_\times$ are drawn here with additional edges connecting the successive roots of their petal tree components, represented schematically by grey blobs with a black boundary, with a root vertex drawn as a square if markable and a cross if not.</figcaption></figure>

if the petal tree has $k'$ marked vertices. As before, requiring that all non-root leaves be marked simply amounts to forbidding the sequence $UD_\bullet$ in the coding words.

Returning to the setting of Proposition 2.14 and assuming that $m - \frac{s+1+\epsilon}{2} \in \mathbb{Z}$, recall that $p_{r+s,s+1+\epsilon}(m)$ counts three-letter words of the form (4.1) with $r+s+1$ occurrences of $D_\circ$, $m-r - \frac{s+1-\epsilon}{2}$ occurrences of $D_\bullet$ and $m - \frac{s+1+\epsilon}{2}$ occurrences of $U$, so that the height difference is $-(s+1+\epsilon)$. As already mentioned, the property holds for the extended range $m \ge \max(1-\frac{s+1+\epsilon}{2}, 1-r-\frac{s+1-\epsilon}{2})$, hence for all positive $m$ in the current setting where $r, s \ge 0$ and $\epsilon \in \{-1, 0, 1\}$. We now transform the three-letter word into a four-letter word coding for a sequence of petal trees as follows: we first remove the first letter $D_\circ$, which leaves us with $r+s$ occurrences of $D_\circ$, and pick $s$ of them that we transform into $E$'s. This transformation can be done in $\binom{r+s}{s}$ ways, hence the obtained four-letter words are counted by $\binom{r+s}{s}p_{r+s,s+1+\epsilon}(m)$. By construction, these words have $s$ occurrences of $E$, $r$ occurrences of $D_\circ$, $m - \frac{s+1+\epsilon}{2}$ occurrences of $U$ and $m-r - \frac{s+1-\epsilon}{2}$ occurrences of $D_\bullet$. These correspond to lattice paths of length $2m-1$ and height difference $-\epsilon$, say from $(0,0)$ to $(2m-1, -\epsilon)$, hence with minimal height $-\ell$ for some $\ell \ge \max(0, \epsilon)$. Any such path is canonically decomposed into a sequence of $2\ell+1-\epsilon$ dressed Motzkin paths (possibly of length $0$), obtained by cutting out the first elementary down
respectively to the letters $U, D$ and $E$. This path is a Motzkin path of length equal to the degree $2m'$ of the exterior face, i.e. it goes from $(0,0)$ to $(2m',0)$ and stays above the $x$-axis. Indeed, the ordinates of the path record graph distances to the root vertex incident to the root corner. The number $s'$ of occurrences of the letter $E$ in the coding word is nothing but the number of petals, and is such that $m' - \frac{s'}{2}$, which is the number of $U$'s (or equivalently of $D$'s), is a nonnegative integer.

Let us now consider a rooted petal tree with marked vertices, where the root vertex is unmarked, and such that all non-root leaves are marked. Every non-root vertex being bijectively associated with its parent tree-type edge in the wood tree, hence with a letter $D$, we record the markings as in Section 4.1 by replacing each $D$ by a $D_\circ$ if the associated vertex is marked and by a $D_\bullet$ otherwise. We end up with a word made of four letters, $U, D_\circ, D_\bullet$ and $E$, associated with a “dressed” Motzkin path, with $k'$ occurrences of $D_\circ$ and $m' - \frac{s'}{2} - k'$ occurrences of $D_\bullet$ if the petal tree has $k'$ marked vertices. As before, requiring that all non-root leaves be marked simply amounts to forbidding the sequence $UD_\bullet$ in the coding words.

Returning to the setting of Proposition 15 and assuming that $m - \frac{s+1+\epsilon}{2} \in \mathbb{Z}$, recall that $p_{r+s,s+1+\epsilon}(m)$ counts three-letter words of the form (43) with $r+s+1$ occurrences of $D_\circ$, $m-r - \frac{s+1-\epsilon}{2}$ occurrences of $D_\bullet$ and $m - \frac{s+1+\epsilon}{2}$ occurrences of $U$. As already mentioned, the property holds for the extended range $m \ge \max(1-\frac{s+1+\epsilon}{2}, 1-r-\frac{s+1-\epsilon}{2})$, hence for all positive $m$ in the current setting where $r,s \ge 0$ and $\epsilon \in \{-1,0,1\}$. We now transform the three-letter word into a four-letter word coding for a sequence of petal trees as follows: we first remove the first letter $D_\circ$, which leaves us with $r+s$ occurrences of $D_\circ$, and pick $s$ of them that we transform into $E$'s. This transformation can be done in $\binom{r+s}{s}$ ways, hence the obtained four-letter words are counted by $\binom{r+s}{s}p_{r+s,s+1+\epsilon}(m)$. By construction, these words have $s$ occurrences of $E$, $r$ occurrences of $D_\circ$, $m - \frac{s+1+\epsilon}{2}$ occurrences of $U$ and $m-r-\frac{s+1-\epsilon}{2}$ occurrences of $D_\bullet$. These correspond to lattice paths of length $2m-1$ and height difference $-\epsilon$, say from $(0,0)$ to $(2m-1,-\epsilon)$, hence with minimal height $-\ell$ for some $\ell \ge \max(0,\epsilon)$. Any such path is canonically decomposed into a sequence of $2\ell+1-\epsilon$ dressed Motzkin paths (possibly of length 0), obtained by cutting out the first elementary down steps reaching height $-j$ for $j = 1,2,\dots, \ell$ and the last elementary up steps reaching height $-j'+1$ for $j' = \ell, \ell-1, \dots, \epsilon+1$, see Figure 15. Each Motzkin path component codes for a marked petal tree whose root vertex is unmarked, and we decide to mark it if this Motzkin path is followed by a $D_\circ$ in the original path. By doing so, we both ensure that the total number of marked vertices in the petal tree sequence is $r$ and that the four-letter word can be recovered bijectively from the petal tree sequence (since we know the nature $D_\bullet, D_\circ$ or $U$ of all the removed steps). Note that by construction, only the first $\ell$ petal trees may have their root vertex marked. All in all, $\binom{r+s}{s}p_{r+s,s+1+\epsilon}(m)$ enumerates pairs $(\mathcal{F}^\square, \mathcal{F}_\times)$ made of a sequence $\mathcal{F}^\square$ of $\ell$ petal trees with marked vertices whose root vertex is markable, and a sequence $\mathcal{F}_\times$ of $\ell+1-\epsilon$ petal trees with marked vertices whose root vertex is unmarked, for some arbitrary integer $\ell \ge \max(0,\epsilon)$, with a total of $s$ petals, $r$ marked vertices, and $m-\ell - \frac{s+1-\epsilon}{2}$ tree-type edges, and with all the non-root leaves in the petal trees marked. As a final step, the pairs $(\mathcal{F}^\square, \mathcal{F}_\times)$ are transformed into the desired tight petal trees by attaching the roots of the $\ell$ (respectively $\ell+1-\epsilon$) petal trees in the sequence $\mathcal{F}^\square$ (respectively $\mathcal{F}_\times$) by
steps reaching height $-j$ for $j = 1, 2, \dots, \ell$ and the last elementary up steps reaching height $-j' + 1$ for $j' = \ell, \ell-1, \dots, \epsilon+1$, see Figure 5.1. Each Motzkin path component codes for a marked petal tree whose root vertex is unmarked, and we decide to mark it if this Motzkin path is followed by a $D_\bullet$ in the original path. By doing so, we both ensure that the total number of marked vertices in the petal tree sequence is $r$ and that the four-letter word can be recovered bijectively from the petal tree sequence (since we know the nature $D_\bullet, D_\circ$ or $U$ of all the removed steps). Note that by construction, only the first $\ell$ petal trees may have their root vertex marked. All in all, $\binom{r+s}{s} p_{r+s,s+1+\epsilon}(m)$ enumerates pairs $(\mathcal{F}_\square, \mathcal{F}_\times)$ made of a sequence $\mathcal{F}_\square$ of $\ell$ petal trees with marked vertices whose root vertex is markable, and a sequence $\mathcal{F}_\times$ of $\ell+1-\epsilon$ petal trees with marked vertices whose root vertex is unmarked, for some arbitrary integer $\ell \ge \max(0, \epsilon)$, with a total of $s$ petals, $r$ marked vertices, and $m-\ell-\frac{s+1-\epsilon}{2}$ tree-type edges, and with all the non-root leaves in the petal trees marked. As a final step, the pairs $(\mathcal{F}_\square, \mathcal{F}_\times)$ are transformed into the desired tight petal trees by attaching the roots of the $\ell$ (respectively $\ell+1-\epsilon$) petal trees in the sequence $\mathcal{F}_\square$ (respectively $\mathcal{F}_\times$) by additional edges and gluing them along a spine of length $\ell-\epsilon$ as displayed in Figure 5.2. Note that a markable root vertex is always matched to an unmarked one: the markings of root vertices may thus be transferred without ambiguity after gluing, with no risk of double markings along the spine. Let us detail the three possibilities $\epsilon = -1, 0, 1$.

For $\epsilon = -1$, the sequence $\mathcal{F}_\times$ has two more petal trees than $\mathcal{F}_\square$ so that the root vertices of its two extremal petal trees remain unmatched. We decide to mark these vertices and distinguish them as vertices 1 and 2: the resulting object is a tight petal tree with $s$ petals and $r+2$ marked vertices, two of which distinguished. The construction is clearly reversible by cutting along the branch between vertices 1 and 2, and holds for any $\ell \ge 0$ (for $\ell = 0$, vertices 1 and 2 are incident). The degree of the exterior face is easily found equal to $2(m-\ell-\frac{s+2}{2})+s+2(\ell+1) = 2m$ as wanted.

For $\epsilon = 0$, the sequence $\mathcal{F}_\times$ has one more petal tree than $\mathcal{F}_\square$ so that the root vertex of its first petal tree remains unmatched. We decide to mark this vertex and to add an additional, distinguished petal to mark the corner inbetween the last two glued petal trees at the end of the spine: the resulting object is a tight petal tree with $s+1$ petals, one of which distinguished, and $r+1$ marked vertices, one of which distinguished. The construction is clearly reversible by cutting along the branch between the distinguished marked vertex and the distinguished petal, eventually removing this petal, and holds for any $\ell \ge 0$ (for $\ell = 0$, the distinguished vertex is incident to the distinguished petal). The degree of the exterior face is $2(m-\ell-\frac{s+1}{2})+s+1+2\ell = 2m$ as wanted.

Finally, for $\epsilon = 1$, the sequences $\mathcal{F}_\times$ and $\mathcal{F}_\square$ have the same number $\ell \ge 1$ of petal tree components and their gluing is made reversible by adding a petal inbetween the glued petal trees at each extremity of the spine, that we distinguish as petals 1 and 2: the resulting object is a tight petal tree with $s+2$ petals, two of which distinguished, and $r$ marked vertices. The construction is clearly reversible by cutting along the branch between petals 1 and 2, eventually removing these petals, and holds for any $\ell \ge 1$ (for $\ell = 1$, the distinguished petals are incident to the same vertex). The degree of the exterior face is $2(m-\ell-\frac{s}{2})+s+2+2(\ell-1) = 2m$ as wanted. This ends the proof of Proposition 2.14. $\square$
<figure><img src="image_16.png" /><figcaption>Figure 15: Schematic picture of the decomposition of a lattice path of length $2m-1$, height difference $-\epsilon$, and minimal height $-\ell$ into a sequence of $2\ell + 1 - \epsilon$ Motzkin paths, where the squares indicate that the underlying descending step may be marked (if associated with a $D_\bullet$) or not (if associated with a $D_\circ$). For $j = 1, 2, \dots, \ell$, the $j$-th Motzkin path is followed by the first elementary down step reaching height $-j$. These $\ell$ Motzkin paths code for an ordered set $\mathcal{F}^\square$ of $\ell$ rooted petal trees whose root vertex are marked or not according to the nature of the following elementary down step. For $j = \ell+2, \ell+3, \dots, 2\ell+1-\epsilon$, the $j$-th Motzkin path is preceded by the last elementary up step reaching height $j-1-2\ell$. Together with the $(\ell+1)$-th Motzkin path, these Motzkin paths code for an ordered set $\mathcal{F}_\times$ of $\ell+1-\epsilon$ rooted petal trees whose root vertex is unmarked. The ordered sets $\mathcal{F}^\square$ and $\mathcal{F}_\times$ are drawn here with additional edges connecting the successive roots of their petal tree components, represented schematically by grey blobs with a black boundary, with a root vertex drawn as a square if markable and a cross if not.</figcaption></figure>

additional edges and gluing them along a spine of length $\ell - \epsilon$ as displayed in Figure 16. Note that a markable root vertex is always matched to an unmarked one: the markings of root vertices may thus be transferred without ambiguity after gluing, with no risk of double markings along the spine. Let us detail the three possibilities $\epsilon = -1, 0, 1$.

For $\epsilon = -1$, the sequence $\mathcal{F}_\times$ has two more petal trees than $\mathcal{F}^\square$ so that the root vertices of its two extremal petal trees remain unmatched. We decide to mark these vertices and distinguish them as vertices 1 and 2: the resulting object is a tight petal tree with $s$ petals and $r+2$ marked vertices, two of which distinguished. The construction is clearly reversible by cutting along the branch between vertices 1 and 2, and holds for any $\ell \ge 0$ (for $\ell = 0$, vertices 1 and 2 are incident). The degree of the exterior face is easily found equal to $2(m - \ell - \frac{s+2}{2}) + s + 2(\ell + 1) = 2m$ as wanted.

For $\epsilon = 0$, the sequence $\mathcal{F}_\times$ has one more petal tree than $\mathcal{F}^\square$ so that the root vertex of its first petal tree remains unmatched. We decide to mark this vertex and to add an
<figure><img src="image_18.png" /><figcaption>Figure 5.2: The gluing of a sequence $\mathcal{F}^\square$ of $\ell$ petal trees with markable roots and a sequence $\mathcal{F}_\times$ of $\ell+1-\epsilon$ petal trees with unmarked roots into a tight petal tree with $1-\epsilon$ distinguished marked vertices (represented by circles) and $1+\epsilon$ distinguished petals (represented by small empty loops), for $\epsilon = -1$ (left), $0$ (center) and $1$ (right).</figcaption></figure>

## 5.2. Petal necklaces

We now consider *petal necklaces*, namely planar maps having two distinguished faces of arbitrary degrees, and such that any other face is a petal. Again, a tight petal necklace is just a petal necklace with marked vertices, which is tight as a map. We have the following enumeration result:

**Proposition 5.3.** For $r, s$ nonnegative integers not both zero and $m_1, m_2 \in \mathbb{Z}_{>0}/2$, let

$$ \Pi_{r,s}(m_1, m_2) := N_{0,r+s+2}(2m_1, 2m_2, \underbrace{0, \dots, 0}_{r}, \underbrace{1, \dots, 1}_{s}) \quad (5.4) $$

be the number of tight petal necklaces with two distinguished faces of degrees $2m_1$ and $2m_2$, with $r$ marked vertices and $s$ petals, all labeled. Then, for $r \ge 1$ we have

$$ \Pi_{r,s}(m_1, m_2) = (r-1)!s! \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} (\pi_{r_1,s_1}^{(-1)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2) + \pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2)) \quad (5.5) $$

and, for $s \ge 1$,

$$ \Pi_{r,s}(m_1, m_2) = r!(s-1)! \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} (\pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2) + \pi_{r_1,s_1}^{(1)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2)) \quad (5.6) $$
<figure><img src="image_17.png" /><figcaption>Figure 16: The gluing of a sequence $\mathcal{F}^\square$ of $\ell$ petal trees with markable roots and a sequence $\mathcal{F}_\times$ of $\ell+1-\epsilon$ petal trees with unmarked roots into a tight petal tree with $1-\epsilon$ distinguished marked vertices (represented by circles) and $1+\epsilon$ distinguished petals (represented by small empty loops), for $\epsilon = -1$ (left), $0$ (center) and $1$ (right).</figcaption></figure>

additional, distinguished petal to mark the corner inbetween the last two glued petal trees at the end of the spine: the resulting object is a tight petal tree with $s+1$ petals, one of which distinguished, and $r+1$ marked vertices, one of which distinguished. The construction is clearly reversible by cutting along the branch between the distinguished marked vertex and the distinguished petal, eventually removing this petal, and holds for any $\ell \ge 0$ (for $\ell=0$, the distinguished vertex is incident to the distinguished petal). The degree of the exterior face is $2(m-\ell-\frac{s+1}{2})+s+1+2\ell=2m$ as wanted.

Finally, for $\epsilon = 1$, the sequences $\mathcal{F}_\times$ and $\mathcal{F}^\square$ have the same number $\ell \ge 1$ of petal tree components and their gluing is made reversible by adding a petal inbetween the glued petal trees at each extremity of the spine, that we distinguish as petals 1 and 2: the resulting object is a tight petal tree with $s+2$ petals, two of which distinguished, and $r$ marked vertices. The construction is clearly reversible by cutting along the branch between petals 1 and 2, eventually removing these petals, and holds for any $\ell \ge 1$ (for $\ell=1$, the distinguished petals are incident to the same vertex). The degree of the exterior face is $2(m-\ell-\frac{s}{2})+s+2+2(\ell-1)=2m$ as wanted. This ends the proof of Proposition 15. □
<figure><img src="image_19.png" /><figcaption>Figure 5.3: The partial and mutual gluings of the segments $S_1$ (oriented from left to right) and $S_2$ (oriented from right to left) connecting the petal tree components of the sequences $\mathcal{F}_1$ and $\mathcal{F}_2$ respectively, here for $\ell_2 \le \ell_1$. The vertex identifications are indicated by arrows. Top: in the case $\epsilon_1 = -1$, $\epsilon_2 = +1$, the partial gluing of $S_1$ amounts to pulling in its ($\ell_1 + 1$)-th vertex, creating a branch of length $\ell_1 - \ell_2 + 1$ in $f_1$ with a distinguished vertex 1 at its end. Bottom: in the case $\epsilon_1 = 0$, $\epsilon_2 = +1$, the partial gluing of $S_1$ amounts to pulling in its $\ell_1$-th edge, creating a branch of length $\ell_1 - \ell_2$ in $f_1$ with a distinguished petal 1 at its end. In both cases, the mutual gluing generates a loop of even length $2\ell_2$.</figcaption></figure>
## 5.2. Petal necklaces

We now consider *petal necklaces*, namely planar maps having two distinguished faces of arbitrary degrees, and such that any other face is a petal. Again, a tight petal necklace is just a petal necklace with marked vertices, which is tight as a map. We have the following enumeration result:

**Proposition 32.** For $r, s$ nonnegative integers not both zero and $m_1, m_2 \in \mathbb{Z}_{>0}/2$, let

$$ \Pi_{r,s}(m_1, m_2) := N_{0,r+s+2}(2m_1, 2m_2, \underbrace{0, \dots, 0}_{r}, \underbrace{1, \dots, 1}_{s}) \quad (55) $$

be the number of tight petal necklaces with two distinguished faces of degrees $2m_1$ and $2m_2$, with $r$ marked vertices and $s$ petals, all labeled. Then, for $r \ge 1$ we have

$$ \Pi_{r,s}(m_1, m_2) = (r-1)!s! \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} \left( \pi_{r_1,s_1}^{(-1)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2) + \pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2) \right) \quad (56) $$

and, for $s \ge 1$,

$$ \Pi_{r,s}(m_1, m_2) = r!(s-1)! \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} \left( \pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2) + \pi_{r_1,s_1}^{(1)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2) \right) \quad (57) $$

where $\pi_{r,s}^{(\epsilon)}$ is the univariate quasi-polynomial defined in (26).

This proposition admits a “partially unlabeled” equivalent, which will be useful later on.

**Proposition 33.** Given $r_0, s_0 \in \mathbb{Z}_{\ge 0}$ and $m_1, m_2 \in \mathbb{Z}_{>0}/2$, the number of tight petal necklaces with two distinguished labeled faces of degrees $2m_1, 2m_2$, one distinguished marked vertex, $r_0$ other unlabeled marked vertices, and $s_0$ unlabeled petals, is equal to

$$ \pi_{r_0, s_0}^{(0)}(m_1, m_2) := \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r_0 \\ s_1+s_2=s_0}} \left( \pi_{r_1, s_1}^{(-1)}(m_1) \pi_{r_2, s_2}^{(1)}(m_2) + \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) \right), \quad (58) $$

while the number of tight petal necklaces with two distinguished labeled faces of degrees $2m_1, 2m_2$, one distinguished petal, $s_0$ other unlabeled petals, and $r_0$ unlabeled marked vertices, is equal to

$$ \pi_{r_0, s_0}^{(1)}(m_1, m_2) := \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r_0 \\ s_1+s_2=s_0}} \left( \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(1)}(m_2) + \pi_{r_1, s_1}^{(1)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) \right). \quad (59) $$
where $\pi_{r,s}^{(\epsilon)}$ is the univariate quasi-polynomial defined in (2.26).

This proposition admits a “partially unlabeled” equivalent, which will be useful later on.

**Proposition 5.4.** Given $r_0, s_0 \in \mathbb{Z}_{\ge 0}$ and $m_1, m_2 \in \mathbb{Z}_{>0}/2$, the number of tight petal necklaces with two distinguished labeled faces of degrees $2m_1, 2m_2$, one distinguished marked vertex, $r_0$ other unlabeled marked vertices, and $s_0$ unlabeled petals, is equal to

$$ \pi_{r_0, s_0}^{(0)}(m_1, m_2) := \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r_0 \\ s_1+s_2=s_0}} \left( \pi_{r_1, s_1}^{(-1)}(m_1) \pi_{r_2, s_2}^{(1)}(m_2) + \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) \right), \quad (5.7) $$

while the number of tight petal necklaces with two distinguished labeled faces of degrees $2m_1, 2m_2$, one distinguished petal, $s_0$ other unlabeled petals, and $r_0$ unlabeled marked vertices, is equal to

$$ \pi_{r_0, s_0}^{(1)}(m_1, m_2) := \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r_0 \\ s_1+s_2=s_0}} \left( \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(1)}(m_2) + \pi_{r_1, s_1}^{(1)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) \right). \quad (5.8) $$

Propositions 5.3 and 5.4 are indeed equivalent to one another, since the petal necklaces considered in the latter have three distinguished boundaries hence no symmetries. Namely, we pass from (5.7) to (5.5) by taking $r = r_0 + 1$ and $s = s_0$, and from (5.8) to (5.6) by taking $r = r_0$ and $s = s_0 + 1$. In both cases, we multiply by $r_0! s_0!$ to label the unlabeled marked vertices/petals in all possible ways. For later use we record the following:

**Remark 5.5.** Given a petal necklace $m_{12}$ with a distinguished marked vertex/petal as considered in Proposition 5.4, it is possible to label the marked vertices and petals in a canonical way from 1 to $r_0 + s_0 + 1$, with the label 1 assigned to the distinguished marked vertex/petal. The precise labeling procedure is irrelevant, as long as it is deterministic<fnref n="9" />.

**Proof of Proposition 5.4.** Our proof is a direct generalization of that of Proposition 2.6 given in Section 4.2. Still, rather than recoursing to $(a, b)$- and $(a, b)^*$-forests or generalizations thereof, we will instead use the related notion of pairs of petal tree sequences<fnref n="10" />, as introduced in the proof of Proposition 2.14. More precisely, we have seen there that, for $r_1, s_1$ nonnegative integers, $m_1 \in \mathbb{Z}_{>0}/2$, and $\epsilon_1 = -1, 0, 1$, the quantity $\pi_{r_1, s_1}^{(\epsilon_1)}(m_1)$ defined in (2.26) enumerates pairs $(\mathcal{F}_{\square}, \mathcal{F}_{\times})$ of sequences of respectively $\ell_1$ and $\ell_1 + 1 - \epsilon_1$ rooted petal trees for some arbitrary integer $\ell_1 \ge \max(0, \epsilon_1)$. The petal trees have a total of $r_1$ marked vertices, with all their non-root leaves marked, a total of $s_1$ petals, and a total of $m_1 - \ell_1 - \frac{s_1+1-\epsilon_1}{2}$ tree-type edges. Finally,

<footnote n="9">⁹For instance, calling $f_1$ and $f_2$ the two distinguished faces, a possible algorithm consists in picking the vertex $v$ on the boundary between $f_1$ and $f_2$ lying on the same branch (possibly of length 0) in the petal necklace as the distinguished marked vertex/petal. Following the contour of $f_1$ clockwise, starting from the leftmost corner at $v$ in this face, and doing then the same for $f_2$, we label the marked vertices and petals, including the distinguished element, by successive integers from 1 to $r_0 + s_0 + 1$ at the first visit of such marked vertex (encountered via an incident corner) or petal (encountered via a petal-type edge). We then perform a cyclic permutation of the labels so that the distinguished element receives the label 1.</footnote>

<footnote n="10">¹⁰There is indeed a clear correspondence between $(a, b)$-forests and collections of trees with $a$ unmarkable roots and $b$ markable ones, and this could be generalized to our setting, with petal trees replacing trees.</footnote>
<figure><img src="image_18.png" /><figcaption>Figure 17: The partial and mutual gluings of the segments $S_1$ (oriented from left to right) and $S_2$ (oriented from right to left) connecting the petal tree components of the sequences $\mathcal{F}_1$ and $\mathcal{F}_2$ respectively, here for $\ell_2 \le \ell_1$. The vertex identifications are indicated by arrows. Top: in the case $\epsilon_1 = -1$, $\epsilon_2 = +1$, the partial gluing of $S_1$ amounts to pulling in its ($\ell_1 + 1$)-th vertex, creating a branch of length $\ell_1 - \ell_2 + 1$ in $f_1$ with a distinguished vertex 1 at its end. Bottom: in the case $\epsilon_1 = 0$, $\epsilon_2 = +1$, the partial gluing of $S_1$ amounts to pulling in its $\ell_1$-th edge, creating a branch of length $\ell_1 - \ell_2$ in $f_1$ with a distinguished petal 1 at its end. In both cases, the mutual gluing generates a loop of even length $2\ell_2$.</figcaption></figure>
only the petal trees in $\mathcal{F}_{\square}$ may have their root vertex marked. The enumeration statement above holds even if $m_1 - \frac{s+1+\epsilon_1}{2}$ is a half-integer, as the number of sequence pairs is trivially 0 in this case, and so is $\pi_{r_1,s_1}^{(\epsilon_1)}(m_1)$ by definition.

To construct a petal necklace, we first merge the two sequences $\mathcal{F}_{\square}$ and $\mathcal{F}_{\times}$ into a single sequence $\mathcal{F}_1$ of $2\ell_1 + 1 - \epsilon_1$ petal trees, which we transform into a single connected object by adding edges connecting the successive roots of the petal tree components and forming a linear segment $S_1$ of length $2\ell_1 - \epsilon_1$ – see Figure 5.3 for a schematic representation. This first connected object will code for the face of degree $2m_1$ of our necklace. The coding of the face of degree $2m_2$ is performed via a second sequence $\mathcal{F}_2$, taken now in the set enumerated by $\pi_{r_2,s_2}^{(\epsilon_2)}(m_2)$. The sequence $\mathcal{F}_2$ has the same properties, mutatis mutandis, as the sequence $\mathcal{F}_1$, with now $2\ell_2 + 1 - \epsilon_2$ petal trees, and a total of $r_2$ marked vertices and $s_2$ petals. Again $\mathcal{F}_2$ is transformed into a connected object by adding a linear segment $S_2$ of $2\ell_2 - \epsilon_2$ edges connecting its successive petal tree roots. To obtain the desired necklace, we will simply, by some “partial gluing” process reminiscent of that for (a, b)-forests, squeeze the largest of the two segments $S_1$ and $S_2$ so as to get two segments of the same length $\min(2\ell_1 - \epsilon_1, 2\ell_2 - \epsilon_2)$ which can then be glued together, head to tail, into a map with a spine that, after closing by some additional edge, forms the cycle separating the two distinguished faces of the necklace. Let us discuss in detail how we perform the partial and mutual gluing processes. An important property of both processes is that we always identify a markable vertex to an unmarked one. By transferring the possible marking of their markable copy, this guarantees that all the vertices obtained by gluing are markable and marked or not without ambiguity (and without double markings).

Consider first the case where, say $\epsilon_1 = -1$ and $\epsilon_2 = +1$ so that $S_2$ has length $2\ell_2 - 1$, with some $\ell_2 \ge 1$ and $S_1$ has length $2\ell_1 + 1$ for some $\ell_1 \ge 0$. Assume first that $\ell_2 \le \ell_1$: we may then squeeze $S_1$ by “pulling in” its $(\ell_1+1)$-th vertex, namely by gluing the $(\ell_1+1-i)$-th (markable) vertex along $S_1$ to the $(\ell_1+1+i)$-th (unmarked) one for $i = 1, \dots, \ell_1 - \ell_2 + 1$. This results in a segment $S_1'$ of the same length as $S_2$, with $\ell_2$ markable vertices followed by $\ell_2$ unmarked ones, together with a branch (attached to the $\ell_2$-th vertex of $S_1'$) of length $\ell_1 - \ell_2 + 1$ with all its vertices markable but the last one, corresponding to the former $(\ell_1+1)$-th vertex along $S_1$. This vertex was unmarked and we decide to mark and distinguish it, say with the label 1. We finally glue the two segments $S_1'$ and $S_2$, head to tail, into a linear spine that we close with an additional edge into a cycle of length $2\ell_2$ separating two faces $f_1$ and $f_2$ – see Figure 5.3-top. Again every markable vertex of $S_2$ is matched to an unmarked vertex of $S_1'$ and vice versa, so that all the vertices along the cycle are markable. It is easily checked that $f_1$ and $f_2$ have respective degrees $2m_1$ and $2m_2$ and we end up with a petal necklace with an additional marked vertex incident to $f_1$ but not to $f_2$. For $\ell_2 \ge \ell_1 + 2$, a similar construction consisting now in squeezing the segment $S_2$ by pulling in its $(\ell_2+1)$-th vertex generates a necklace where the separating cycle has length $2(\ell_1+1)$ and where the additional marked vertex 1 is incident to $f_2$ but not to $f_1$. Finally, if $\ell_2 = \ell_1+1$, $S_1$ and $S_2$ have the same length and no squeezing is necessary: the faces $f_1$ and $f_2$ are then separated by a cycle of length $2\ell_2 = 2(\ell_1+1)$ and the vertex 1 (obtained by gluing the $(\ell_1+1)$-th vertex of $S_1$ to the $(\ell_2+1)$-th vertex of $S_2$) lies along this cycle, i.e. is incident to both faces. The construction is clearly reversible by a cutting procedure along the separating cycle and along the branch leading from this cycle to vertex 1. We deduce that $\pi_{r_1,s_1}^{(-1)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2)$ counts petal necklaces with $s_1$ unlabeled petals and $r_1$ unlabeled marked vertices in its distinguished face $f_1$.
Propositions 32 and 33 are indeed equivalent to one another, since the petal necklaces considered in the latter have three distinguished boundaries hence no symmetries. Namely, we pass from (58) to (56) by taking $r = r_0 + 1$ and $s = s_0$, and from (59) to (57) by taking $r = r_0$ and $s = s_0 + 1$. In both cases, we multiply by $r_0!s_0!$ to label the unlabeled marked vertices/petals in all possible ways. For later use we record the following:

**Remark 34.** Given a petal necklace $\mathbf{m}_{12}$ with a distinguished marked vertex/petal as considered in Proposition 33, it is possible to label the marked vertices and petals in a *canonical* way from 1 to $r_0+s_0+1$, with the label 1 assigned to the distinguished marked vertex/petal. The precise labeling procedure is irrelevant, as long as it is deterministic$^7$.

*Proof of Proposition 33.* Our proof is a direct generalization of that of Proposition 7 given in Section 4.2. Still, rather than recoursing to $(a, b)$- and $(a, b)^*$-forests or generalizations thereof, we will instead use the related notion of pairs of petal tree sequences, as introduced in the proof of Proposition 15. More precisely, we have seen there that, for $r_1, s_1$ nonnegative integers, $m_1 \in \mathbb{Z}_{>0}/2$, and $\epsilon_1 = -1, 0, 1$, the quantity $\pi_{r_1,s_1}^{(\epsilon_1)}(m_1)$ defined in (26) enumerates pairs $(\mathcal{F}_{\square}, \mathcal{F}_{\times})$ of sequences of respectively $\ell_1$ and $\ell_1 + 1 - \epsilon_1$ rooted petal trees for some arbitrary integer $\ell_1 \ge \max(0, \epsilon_1)$. The petal trees have a total of $r_1$ marked vertices, with all their non-root leaves marked, a total of $s_1$ petals, and a total of $m_1 - \ell_1 - \frac{s_1+1-\epsilon_1}{2}$ tree-type edges. Finally, only the petal trees in $\mathcal{F}_{\square}$ may have their root vertex marked. The enumeration statement above holds even if $m_1 - \frac{s_1+1-\epsilon_1}{2}$ is a half-integer, as the number of sequence pairs is trivially 0 in this case, and so is $\pi_{r_1,s_1}^{(\epsilon_1)}(m_1)$ by definition.

To construct a petal necklace, we first merge the two sequences $\mathcal{F}_{\square}$ and $\mathcal{F}_{\times}$ into a single sequence $\mathcal{F}_1$ of $2\ell_1 + 1 - \epsilon_1$ petal trees, which we transform into a single connected object by adding edges connecting the successive roots of the petal tree components and forming a linear segment $S_1$ of length $2\ell_1 - \epsilon_1$ – see Figure 17 for a schematic representation. This first connected object will code for the face of degree $2m_1$ of our necklace. The coding of the face of degree $2m_2$ is performed via a second sequence $\mathcal{F}_2$, taken now in the set enumerated by $\pi_{r_2,s_2}^{(\epsilon_2)}(m_2)$. The sequence $\mathcal{F}_2$ has the same properties, mutatis mutandis, as the sequence $\mathcal{F}_1$, with now $2\ell_2 + 1 - \epsilon_2$ petal trees, and a total of $r_2$ marked vertices and $s_2$ petals. Again $\mathcal{F}_2$ is transformed into a connected object by adding a linear segment $S_2$ of $2\ell_2 - \epsilon_2$ edges connecting its successive petal tree roots. To obtain the desired necklace, we will simply, by some “partial gluing” process reminiscent of that for $(a, b)$-forests, squeeze the largest of the two segments $S_1$ and $S_2$ so as to get two segments of the same length $\min(2\ell_1 - \epsilon_1, 2\ell_2 - \epsilon_2)$ which can then be glued together, head to tail, into a

<footnote>$^7$For instance, calling $f_1$ and $f_2$ the two distinguished faces, a possible algorithm consists in picking the vertex $v$ on the boundary between $f_1$ and $f_2$ lying on the same branch (possibly of length 0) in the petal necklace as the distinguished marked vertex/petal. Following the contour of $f_1$ clockwise, starting from the leftmost corner at $v$ in this face, and doing then the same for $f_2$, we label the marked vertices and petals, including the distinguished element, by successive integers from 1 to $r_0 + s_0 + 1$ at the *first visit* of such marked vertex (encountered via an incident corner) or petal (encountered via a petal-type edge). We then perform a cyclic permutation of the labels so that the distinguished element receives the label 1.</footnote>
of degree $2m_1$, $s_2$ unlabeled petals and $r_2$ unlabeled marked vertices in its distinguished face $f_2$ of degree $2m_2$, and with an additional marked vertex (distinct from all the others) distinguished as vertex 1 and lying anywhere in the map (the map hence has a total of $r_1+r_2+1 \ge 1$ marked vertices). The necklaces are easily seen to be tight, since the petal tree components have no unmarked leaves and no unmarked leaf was created in the process. Still, as apparent from the above discussion, a restriction applies to these necklaces since, by construction, the length of the cycle separating their distinguished faces is always even. It is easily seen that the missing set of necklaces is enumerated by $\pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2)$: indeed, repeating the above construction on the corresponding pairs of petal tree sequences $\mathcal{F}_1$ (with $2\ell_1+1$ petal trees) and $\mathcal{F}_2$ (with $2\ell_2+1$ petal trees), we get necklaces whose separating cycle now has the odd length $2\min(\ell_1, \ell_2)+1$ (for some arbitrary $\ell_1, \ell_2 \ge 0$), again with an additional marked vertex distinguished as vertex 1. We finally obtain (5.7) by summing over $r_1$ and $r_2$ with $r_1+r_2=r_0$, and over $s_1$ and $s_2$ with $s_1+s_2=s_0$, to account for the dispatching of the marked vertices and petals among $f_1$ and $f_2$.

An alternative construction consists in starting from $\epsilon_1 = 0$ and $\epsilon_2 = +1$ so that $S_2$ still has length $2\ell_2 - 1$, with some $\ell_2 \ge 1$ while $S_1$ has length $2\ell_1$ for some $\ell_1 \ge 0$. As before, we assume first that $\ell_2 \le \ell_1$: we may then squeeze $S_1$, now by “pulling in” its $\ell_1$-th connecting edge, namely by gluing the $(\ell_1 + 1 - i)$-th (markable) vertex along $S_1$ to the $(\ell_1 + i)$-th (unmarked) one for $i = 1, \dots, \ell_1 - \ell_2 + 1$. This results in a segment $S'_1$ of the same length as $S_2$, with $\ell_2$ markable vertices followed by $\ell_2$ unmarked ones, together with a branch (attached to the $\ell_2$-th vertex of $S'_1$) of length $\ell_1 - \ell_2$ with all its vertices markable, having now a petal at its end corresponding to the former $\ell_1$-th edge along $S_1$. This newly created petal is adjacent to $f_1$ but not to $f_2$ and we decide to distinguish it with the label 1. Gluing $S'_1$ and $S_2$ and closing, we end up with a cycle of even length $2\ell_2$ separating two faces $f_1$ and $f_2$, see the bottom of Figure 5.3. For $\ell_2 \ge \ell_1 + 1$, squeezing now the segment $S_2$ by pulling in its $\ell_2$-th edge generates a necklace with a separating cycle of odd length $2\ell_1 + 1$ and where the additional petal is now adjacent to $f_2$. To get the missing necklaces, namely those with an odd separating cycle and a distinguished petal adjacent to $f_1$, and those with an even separating cycle and a distinguished petal adjacent to $f_2$, we have, as clear by symmetry, to supplement the above family of necklaces enumerated by $\pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2)$ by that enumerated by $\pi_{r_1,s_1}^{(1)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2)$. We finally obtain (5.8) by summing over $r_1$ and $r_2$ with $r_1+r_2=r_0$, and over $s_1$ and $s_2$ with $s_1+s_2=s_0$, to account for the dispatching of the marked vertices and petals among $f_1$ and $f_2$. $\square$

**Remark 5.6.** The consistency of the two expressions (5.5) and (5.6) for $\Pi_{r,s}(m_1, m_2)$ when $r, s \ge 1$ may be checked directly by setting $s! = (s-1)!(s_1+s_2)$ in (5.5), $r! = (r-1)!(r_1+r_2)$ in (5.6), and upon using the identifications

$$ \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_1 \pi_{r_1,s_1}^{(-1)}(m_1) \pi_{r_2,s_2}^{(1)}(m_2) = \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} r_1 \pi_{r_1,s_1}^{(0)}(m_1) \pi_{r_2,s_2}^{(1)}(m_2) \quad (5.9) $$

$$ \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_2 \pi_{r_1,s_1}^{(-1)}(m_1) \pi_{r_2,s_2}^{(1)}(m_2) \stackrel{*}{=} \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_2 \pi_{r_1,s_1}^{(1)}(m_1) \pi_{r_2,s_2}^{(-1)}(m_2) \quad (5.10) $$
map with a spine that, after closing by some additional edge, forms the cycle separating
the two distinguished faces of the necklace. Let us discuss in detail how we perform the
partial and mutual gluing processes. An important property of both processes is that
we always identify a markable vertex to an unmarked one. By transferring the possible
marking of their markable copy, this guarantees that all the vertices obtained by gluing
are markable and marked or not without ambiguity (and without double markings).

Consider first the case where, say $\epsilon_1 = -1$ and $\epsilon_2 = +1$ so that $S_2$ has length $2\ell_2 - 1$, with some $\ell_2 \ge 1$ and $S_1$ has length $2\ell_1 + 1$ for some $\ell_1 \ge 0$. Assume first that $\ell_2 \le \ell_1$: we may then squeeze $S_1$ by "pulling in" its $(\ell_1 + 1)$-th vertex, namely by gluing the $(\ell_1 + 1 - i)$-th (markable) vertex along $S_1$ to the $(\ell_1 + 1 + i)$-th (unmarked) one for $i = 1, \dots, \ell_1 - \ell_2 + 1$. This results in a segment $S_1'$ of the same length as $S_2$, with $\ell_2$ markable vertices followed by $\ell_2$ unmarked ones, together with a branch (attached to the $\ell_2$-th vertex of $S_1'$) of length $\ell_1 - \ell_2 + 1$ with all its vertices markable but the last one, corresponding to the former $(\ell_1 + 1)$-th vertex along $S_1$. This vertex was unmarked and we decide to mark and distinguish it, say with the label 1. We finally glue the two segments $S_1'$ and $S_2$, head to tail, into a linear spine that we close with an additional edge into a cycle of length $2\ell_2$ separating two faces $f_1$ and $f_2$ – see Figure 17-top. Again every markable vertex of $S_2$ is matched to an unmarked vertex of $S_1'$ and vice versa, so that all the vertices along the cycle are markable. It is easily checked that $f_1$ and $f_2$ have respective degrees $2m_1$ and $2m_2$ and we end up with a petal necklace with an additional marked vertex incident to $f_1$ but not to $f_2$. For $\ell_2 \ge \ell_1 + 2$, a similar construction consisting now in squeezing the segment $S_2$ by pulling in its $(\ell_2 + 1)$-th vertex generates a necklace where the separating cycle has length $2(\ell_1 + 1)$ and where the additional marked vertex 1 is incident to $f_2$ but not to $f_1$. Finally, if $\ell_2 = \ell_1 + 1$, $S_1$ and $S_2$ have the same length and no squeezing is necessary: the faces $f_1$ and $f_2$ are then separated by a cycle of length $2\ell_2 = 2(\ell_1 + 1)$ and the vertex 1 (obtained by gluing the $(\ell_1 + 1)$-th vertex of $S_1$ to the $(\ell_2 + 1)$-th vertex of $S_2$) lies along this cycle, i.e. is incident to both faces. The construction is clearly reversible by a cutting procedure along the separating cycle and along the branch leading from this cycle to vertex 1. We deduce that $\pi_{r_1,s_1}^{(-1)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2)$ counts petal necklaces with $s_1$ unlabeled petals and $r_1$ unlabeled marked vertices in its distinguished face $f_1$ of degree $2m_1$, $s_2$ unlabeled petals and $r_2$ unlabeled marked vertices in its distinguished face $f_2$ of degree $2m_2$, and with an additional marked vertex (distinct from all the others) distinguished as vertex 1 and lying anywhere in the map (the map hence has a total of $r_1+r_2+1 \ge 1$ marked vertices). The necklaces are easily seen to be tight, since the petal tree components have no unmarked leaves and no unmarked leaf was created in the process. Still, as apparent from the above discussion, a restriction applies to these necklaces since, by construction, the length of the cycle separating their distinguished faces is always even. It is easily seen that the missing set of necklaces is enumerated by $\pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2)$: indeed, repeating the above construction on the corresponding pairs of petal tree sequences $F_1$ (with $2\ell_1 + 1$ petal trees) and $F_2$ (with $2\ell_2 + 1$ petal trees), we get necklaces whose separating cycle now has the odd length $2\min(\ell_1, \ell_2) + 1$ (for some arbitrary $\ell_1, \ell_2 \ge 0$), again with an additional marked vertex distinguished as vertex 1. We finally obtain (58) by summing
$$ = \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} r_2 \pi_{r_1, s_1}^{(1)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) \quad (5.11) $$

$$ \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_1 \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) = \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} r_1 \pi_{r_1, s_1}^{(1)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) \quad (5.12) $$

$$ \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_2 \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) = \sum_{\substack{r_1, r_2, s_1, s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} r_2 \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(1)}(m_2) . \quad (5.13) $$

Here all equalities follow from the identity $s\pi_{r,s}^{(\epsilon)}(m) = (r+1)\pi_{r+1,s-1}^{(1+\epsilon)}(m)$ for $s \ge 1$, apart from that marked with a (*) which follows from the identity $\pi_{r,s}^{(1)}(m) = \pi_{r,s}^{(-1)}(m) + \pi_{r-1,s}^{(-1)}(m)$ for any $r \ge 0$ (with the convention $\pi_{-1,s}^{(-1)}(m) = 0$), see (B.10)-(B.13) in Appendix B for similar so-called transmutation relations.

## 5.3. Tight slices

As in the case of bipartite maps, a key ingredient in the derivation of a general formula for the number of tight general non-bipartite maps is the enumeration of tight slices. Recall the basic definitions pertaining to (tight) slices from Section 4.3, and note that we do not assume anymore that the face degrees be even integers. In this general context, it is still true that the only elementary slice of tilt $-1$ is the trivial slice, but a major difference is that the set of elementary slices of tilt $0$ is now non-empty.

As discussed in Section 4.3, a (tight) slice $s$ with width $w \ge 1$ and tilt $t$ can be decomposed into a collection of $w$ (tight) elementary slices. This discussion applies without change in our general context: we cut the slice along the leftmost geodesics started from the consecutive corners $c_0, c_1, \dots, c_w$ incident to the base, and record the lattice path $L = (\ell_0, \ell_1, \dots, \ell_w)$ where $\ell_i = d(c_0, A) - d(c_i, A)$, and $A$ is the apex of $s$. The only difference is that the tilt $t_i = \ell_i - \ell_{i-1}$ of the slice $s'_i$ with base $c_{i-1}c_i$ can take any value $\{-1, 0, 1\}$ rather than only $\{-1, 1\}$. As before, the tightness condition requires that, for $i \in \{2, \dots, w\}$, if $t_i = +1$ and $t_{i-1} = -1$, then the slice $s'_i$ is either the marked empty slice, or a tight elementary slice with at least one inner face. For $i \in \{1, \dots, w\}$, if $t_i = -1$ then the slice $s'_i$ is necessarily trivial, while if $t_i = 0$ then the slice $s'_i$ automatically has at least one inner face of odd degree. We may forget the redundant information of all down steps, as well as up steps decorated with unmarked empty slices, by letting $s_j, 1 \le j \le k$ be the sequence $(s'_i, 1 \le i \le w)$ whose trivial and unmarked empty elements have been removed. This gives:

**Proposition 5.7.** There is a one-to-one correspondence between tight slices of width $w$ and tilt $t$ on the one hand, and pairs of the form $(L, (s_j, 1 \le j \le k))$ on the other hand, where:

* $L = (\ell_0, \dots, \ell_w)$ is a lattice path from $(0, 0)$ to $(w, t)$ with increments in $\{-1, 0, 1\}$, that has $k$ marked steps $1 \le i_1 < i_2 < \dots < i_k \le w$ with $\ell_{i_j} - \ell_{i_{j-1}} \in \{0, 1\}$, in such a way that every up step immediately following a down step is marked, and every horizontal step is marked,
over $r_1$ and $r_2$ with $r_1 + r_2 = r_0$, and over $s_1$ and $s_2$ with $s_1 + s_2 = s_0$, to account for
the dispatching of the marked vertices and petals among $f_1$ and $f_2$.

An alternative construction consists in starting from $\epsilon_1 = 0$ and $\epsilon_2 = +1$ so that $S_2$
still has length $2\ell_2 - 1$, with some $\ell_2 \ge 1$ while $S_1$ has length $2\ell_1$ for some $\ell_1 \ge 0$.
As before, we assume first that $\ell_2 \le \ell_1$: we may then squeeze $S_1$, now by "pulling in" its
$\ell_1$-th connecting edge, namely by gluing the $(\ell_1 + 1 - i)$-th (markable) vertex along $S_1$
to the $(\ell_1 + i)$-th (unmarked) one for $i = 1, \dots, \ell_1 - \ell_2 + 1$. This results in a segment
$S_1'$ of the same length as $S_2$, with $\ell_2$ markable vertices followed by $\ell_2$ unmarked ones,
together with a branch (attached to the $\ell_2$-th vertex of $S_1'$) of length $\ell_1 - \ell_2$ with all
its vertices markable, having now a petal at its end corresponding to the former $\ell_1$-th
edge along $S_1$. This newly created petal is adjacent to $f_1$ but not to $f_2$ and we decide
to distinguish it with the label 1. Gluing $S_1'$ and $S_2$ and closing, we end up with a
cycle of even length $2\ell_2$ separating two faces $f_1$ and $f_2$, see the bottom of Figure 17.
For $\ell_2 \ge \ell_1 + 1$, squeezing now the segment $S_2$ by pulling in its $\ell_2$-th edge generates a
necklace with a separating cycle of odd length $2\ell_1 + 1$ and where the additional petal is
now adjacent to $f_2$. To get the missing necklaces, namely those with an odd separating
cycle and a distinguished petal adjacent to $f_1$, and those with an even separating cycle
and a distinguished petal adjacent to $f_2$, we have, as clear by symmetry, to supplement
the above family of necklaces enumerated by $\pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2)$ by that enumerated by
$\pi_{r_1,s_1}^{(1)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2)$. We finally obtain (59) by summing over $r_1$ and $r_2$ with $r_1+r_2 = r_0$,
and over $s_1$ and $s_2$ with $s_1+s_2 = s_0$, to account for the dispatching of the marked vertices
and petals among $f_1$ and $f_2$.
$\square$

Remark 35. The consistency of the two expressions (56) and (57) for $\Pi_{r,s}(m_1, m_2)$ when $r, s \ge 1$ may be checked directly by setting $s! = (s-1)!(s_1+s_2)$ in (56), $r! = (r-1)!(r_1+r_2)$ in (57), and upon using the identifications

$$
\sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_1 \pi_{r_1,s_1}^{(-1)}(m_1) \pi_{r_2,s_2}^{(1)}(m_2) = \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} r_1 \pi_{r_1,s_1}^{(0)}(m_1) \pi_{r_2,s_2}^{(1)}(m_2) \quad (60)
$$

$$
\sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_2 \pi_{r_1,s_1}^{(-1)}(m_1) \pi_{r_2,s_2}^{(1)}(m_2) \stackrel{(*)}{=} \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_2 \pi_{r_1,s_1}^{(1)}(m_1) \pi_{r_2,s_2}^{(-1)}(m_2) \quad (61)
$$

$$
= \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} r_2 \pi_{r_1,s_1}^{(1)}(m_1) \pi_{r_2,s_2}^{(0)}(m_2) \quad (62)
$$

$$
\sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_1 \pi_{r_1,s_1}^{(0)}(m_1) \pi_{r_2,s_2}^{(0)}(m_2) = \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} r_1 \pi_{r_1,s_1}^{(1)}(m_1) \pi_{r_2,s_2}^{(0)}(m_2) \quad (63)
$$

$$
\sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r-1 \\ s_1+s_2=s}} s_2 \pi_{r_1,s_1}^{(0)}(m_1) \pi_{r_2,s_2}^{(0)}(m_2) = \sum_{\substack{r_1,r_2,s_1,s_2 \ge 0 \\ r_1+r_2=r \\ s_1+s_2=s-1}} r_2 \pi_{r_1,s_1}^{(0)}(m_1) \pi_{r_2,s_2}^{(1)}(m_2) . \quad (64)
$$
* for $1 \le j \le k$, $s_j$ is a tight elementary slice of tilt $\ell_{i_j} - \ell_{i_{j-1}}$, and in the case where this tilt is 1, $s_j$ is either the marked empty slice, or has at least one inner face.

We obtain the following generalization of Corollary 4.5.

**Corollary 5.8.** For $m \in \mathbb{Z}_{>0}/2$, there is a one-to-one correspondence between non-empty tight elementary slices of tilt $\epsilon \in \{0, 1\}$, whose inner face incident to the base edge has degree $2m$ on the one hand, and pairs of the form $(L, (s_j, 1 \le j \le k))$ on the other hand, where $k$ is a non-negative integer, and:

* $L = (\ell_0, \dots, \ell_{2m-1})$ is a lattice path from $(0, 0)$ to $(2m-1, \epsilon)$ with increments in $\{-1, 0, 1\}$, that has $k$ marked steps $1 \le i_1 < i_2 < \dots < i_k \le 2m-1$ with $\ell_{i_j} - \ell_{i_{j-1}} \in \{0, 1\}$, in such a way that every up step immediately following a down step is marked, and every horizontal step is marked,

* for $1 \le j \le k$, $s_j$ is a tight elementary slice of tilt $\ell_{i_j} - \ell_{i_{j-1}}$, and in the case where this tilt is 1, $s_j$ is either the marked empty slice, or has at least one inner face.

By iterating the decomposition of this corollary, we obtain an encoding of tight elementary slices by plane trees, where every vertex may be associated with a slice of tilt in $\{0, 1\}$. Let $m_1, \dots, m_n$ be integers or half integers, not all equal to 0, and $\epsilon \in \{0, 1\}$. Let $T_{\epsilon}(m_1, \dots, m_n)$ be the family of pairs $(t, (L_i)_{1 \le i \le n})$, satisfying the following conditions.

* $t$ is a rooted plane tree with $n$ vertices labeled by $\{1, 2, \dots, n\}$.

* Each vertex receives a type in $\{0, 1\}$, and the root has type $\epsilon$. We let $\epsilon_i \in \{0, 1\}$ be the type of vertex $i$ and $(w_1^{(i)}, \dots, w_{k_i}^{(i)})$ be the types of the $k_i$ children of vertex $i$ in $t$ in planar order.

* For $i \in \{1, 2, \dots, n\}$ such that $m_i > 0$, $L_i = (\ell_0^{(i)}, \dots, \ell_{2m_i-1}^{(i)})$ is a lattice path from $(0, 0)$ to $(2m_i - 1, \epsilon_i)$ with increments in $\{-1, 0, 1\}$, and with $k_i$ marked steps $1 \le j_1 < \dots < j_{k_i} \le 2m_i - 1$, such that $\ell_{j_r}^{(i)} - \ell_{j_{r-1}}^{(i)} = w_r^{(i)}$ for every $r \in \{1, \dots, k_i\}$. In particular, only up or horizontal steps may be marked. We further require that every up step of $L_i$ immediately following a down step is marked, and every horizontal step of $L_i$ is marked.

* For $i \in \{1, 2, \dots, n\}$ such that $m_i = 0$, we have $\epsilon_i = 1$, $k_i = 0$ and $L_i$ is the trivial path $\{(0, 0)\}$.

This leads to the following generalization of Corollary 4.6.

**Corollary 5.9.** For $\epsilon \in \{0, 1\}$ and every choice of integers or half-integers $m_1, \dots, m_n$ not all equal to 0, the iterated slice decomposition yields a bijection between the set of elementary tight slices with $n$ labeled boundaries of respective lengths $2m_1, \dots, 2m_n$ with tilt $\epsilon$ and the set $T_{\epsilon}(m_1, \dots, m_n)$.
Here all equalities follow from the identity $s\pi_{r,s}^{(\epsilon)}(m) = (r+1)\pi_{r+1,s-1}^{(1+\epsilon)}(m)$ for $s \ge 1$, apart from that marked with a (*) which follows from the identity $\pi_{r,s}^{(1)}(m) = \pi_{r,s}^{(-1)}(m) + \pi_{r-1,s}^{(-1)}(m)$ for any $r \ge 0$ (with the convention $\pi_{-1,s}^{(-1)}(m) = 0$), see (95) in Appendix B for similar so-called transmutation relations.

### 5.3. Tight slices

As in the case of bipartite maps, a key ingredient in the derivation of a general formula for the number of tight general non-bipartite maps is the enumeration of tight slices. Recall the basic definitions pertaining to (tight) slices from Section 4.3, and note that we do not assume anymore that the face degrees be even integers. In this general context, it is still true that the only elementary slice of tilt $-1$ is the trivial slice, but a major difference is that the set of elementary slices of tilt 0 is now non-empty.

As discussed in Section 4.3, a (tight) slice $\mathbf{s}$ with width $w \ge 1$ and tilt $t$ can be decomposed into a collection of $w$ (tight) elementary slices. This discussion applies without change in our general context: we cut the slice along the leftmost geodesics started from the consecutive corners $c_0, c_1, \dots, c_w$ incident to the base, and record the lattice path $L = (\ell_0, \ell_1, \dots, \ell_w)$ where $\ell_i = d(c_0, A) - d(c_i, A)$, and $A$ is the apex of $\mathbf{s}$. The only difference is that the tilt $t_i = \ell_i - \ell_{i-1}$ of the slice $\mathbf{s}'_i$ with base $c_{i-1}c_i$ can take any value $\{-1, 0, 1\}$ rather than only $\{-1, 1\}$. As before, the tightness condition requires that, for $i \in \{2, \dots, w\}$, if $t_i = +1$ and $t_{i-1} = -1$, then the slice $\mathbf{s}'_i$ is either the marked empty slice, or a tight elementary slice with at least one inner face. For $i \in \{1, \dots, w\}$, if $t_i = -1$ then the slice $\mathbf{s}'_i$ is necessarily trivial, while if $t_i = 0$ then the slice $\mathbf{s}'_i$ automatically has at least one inner face of odd degree. We may forget the redundant information of all down steps, as well as up steps decorated with unmarked empty slices, by letting $\mathbf{s}_j, 1 \le j \le k$ be the sequence $(\mathbf{s}'_i, 1 \le i \le w)$ whose trivial and unmarked empty elements have been removed. This gives:

**Proposition 36.** There is a one-to-one correspondence between tight slices of width $w$ and tilt $t$ on the one hand, and pairs of the form $(L, (\mathbf{s}_j, 1 \le j \le k))$ on the other hand, where:

* $L = (\ell_0, \dots, \ell_w)$ is a lattice path from $(0,0)$ to $(w,t)$ with increments in $\{-1,0,1\}$, that has $k$ marked steps $1 \le i_1 < i_2 < \dots < i_k \le w$ with $\ell_{i_j} - \ell_{i_{j-1}} \in \{0,1\}$, in such a way that every up step immediately following a down step is marked, and every horizontal step is marked,

* for $1 \le j \le k$, $\mathbf{s}_j$ is a tight elementary slice of tilt $\ell_{i_j} - \ell_{i_{j-1}}$, and in the case where this tilt is 1, $\mathbf{s}_j$ is either the marked empty slice, or has at least one inner face.

We obtain the following generalization of Corollary 25.

**Corollary 37.** For $m \in \mathbb{Z}_{>0}/2$, there is a one-to-one correspondence between non-empty tight elementary slices of tilt $\epsilon \in \{0,1\}$, whose inner face incident to the base edge has degree $2m$ on the one hand, and pairs of the form $(L, (\mathbf{s}_j, 1 \le j \le k))$ on the other hand, where $k$ is a non-negative integer, and:
It turns out that, in order to prove Theorem 2.12 in Section 5.4, we will need an extension of this corollary associating sequences of slices with sequences of trees, namely forests. Still, as a warm-up, let us first consider the enumeration of $\mathcal{T}_\epsilon(m_1, \dots, m_n)$, which is done based on that of two-type trees, obtained as a special case of Proposition A.3 of Appendix A. Note that the types 1 and 0 considered in the present section correspond to types A and B respectively in the notation of the appendix. We count the elements $(\mathbf{t}, (L_i)_{1\le i\le n})$ of $\mathcal{T}_\epsilon(m_1, \dots, m_n)$ by fixing the types $\epsilon_1, \dots, \epsilon_n$ of the vertices $1, \dots, n$ of $\mathbf{t}$ (so that $\epsilon_i = w_0^{(i)}$ in the notation of Appendix A) with the constraint that the root vertex should have type $\epsilon$, as well as the sequence of types $w_1^{(i)}, \dots, w_{k_i}^{(i)}$ of the consecutive children (in planar order) of vertex $i$. These sequences must satisfy the consistency conditions (A.1) and (A.2) of Appendix A, which in this context are rewritten as

$$ \sum_{i=1}^{n} \epsilon_i = \epsilon + \sum_{i=1}^{n} r_i \quad \text{and} \quad \sum_{i=1}^{n} \bar{\epsilon}_i = \bar{\epsilon} + \sum_{i=1}^{n} s_i, \qquad (5.14) $$

where we let $\bar{\eta} = 1 - \eta$ for $\eta \in \{0, 1\}$, and where $r_i = \sum_{j=1}^{k_i} w_j^{(i)}$ and $s_i = \sum_{j=1}^{k_i} \bar{w}_j^{(i)}$ are the numbers of type-1 and type-0 children of vertex $i$. These consistency relations simply express in two different ways the number of type 1 (resp. 0) vertices in the tree $\mathbf{t}$. Note that the number of type-0 (resp. type-1) vertices that are the children of type-1 (resp. type-0) vertices (those are respectively the numbers $b^A$ and $a^B$ considered in Appendix A) is given by

$$ \sum_{i=1}^{n} \epsilon_i s_i \quad \left( \text{resp.} \quad \sum_{i=1}^{n} \bar{\epsilon}_i r_i \right). \qquad (5.15) $$

Finally, by Proposition A.3 (in the notation therein, we have $a^O = \epsilon$ and $b^O = \bar{\epsilon}$, expressing the fact that we are counting “forests” made of only one tree with root of type $\epsilon$), we see that the number of possible trees $\mathbf{t}$ contributing to $\mathcal{T}_\epsilon(m_1, \dots, m_n)$, with types given by $\epsilon_1, \dots, \epsilon_n$ and given a consistent type array $(w_j^{(i)})_{j=1,\dots,k_i}^{i=1,\dots,n}$, is equal to

$$ \left( \epsilon \sum_{i=1}^{n} \epsilon_i s_i + \bar{\epsilon} \sum_{i=1}^{n} \bar{\epsilon}_i r_i \right) \left( \sum_{i=1}^{n} \epsilon_i - 1 \right)! \left( \sum_{i=1}^{n} \bar{\epsilon}_i - 1 \right)! \qquad (5.16) $$

where $\sum_{i=1}^{n} \epsilon_i$ and $\sum_{i=1}^{n} \bar{\epsilon}_i$ are the numbers of vertices of types 1 and 0 respectively. In the case where either one of these numbers is equal to 0, in accordance with (A.6), we should replace the whole formula by $(n-2)!$.

It now remains to enumerate, for a given consistent array $(w_j^{(i)})_{j=1,\dots,k_i}^{i=1,\dots,n}$, and for a given tree $\mathbf{t}$ as above, the number of possible lattice paths $L_i$, $1 \le i \le n$ so that $(\mathbf{t}, (L_i)_{1\le i\le n})$ is an element of $\mathcal{T}_\epsilon(m_1, \dots, m_n)$. As before, we let $r_i = \sum_{j=1}^{k_i} w_j^{(i)}$ and $s_i = \sum_{j=1}^{k_i} \bar{w}_j^{(i)} = k_i - r_i$. If $m_i = 0$ then by definition of $\mathcal{T}_\epsilon(m_1, \dots, m_n)$, this requires $L_i = \{(0, 0)\}$, $r_i = s_i = 0$ and $\epsilon_i = 1$: by (2.27), this is precisely counted by $\pi_{r_i, s_i}^{(\epsilon_i)}(0) = \delta_{r_i, 0}\delta_{s_i, 0}\delta_{\epsilon_i, 1}$. If $m_i > 0$, $L_i$ should be a lattice path from $(0, 0)$ to $(2m_i - 1, \epsilon_i)$, with increments in $\{-1, 0, 1\}$, in which $k_i$ up or horizontal steps are marked, say $1 \le j_1 < \dots < j_{k_i} \le 2m_i - 1$, with increments at these steps respectively given by $w_1^{(i)}, \dots, w_{k_i}^{(i)}$, and in such a way that all horizontal steps are marked, as well as all the up steps immediately following a down step. As explained in the proof of Proposition 2.14 (and
• $L = (\ell_0, \dots, \ell_{2m-1})$ is a lattice path from $(0,0)$ to $(2m-1, \epsilon)$ with increments in $\{-1, 0, 1\}$, that has $k$ marked steps $1 \le i_1 < i_2 < \dots < i_k \le 2m-1$ with $\ell_{i_j} - \ell_{i_{j-1}} \in \{0, 1\}$, in such a way that every up step immediately following a down step is marked, and every horizontal step is marked,

• for $1 \le j \le k$, $s_j$ is a tight elementary slice of tilt $\ell_{i_j} - \ell_{i_{j-1}}$, and in the case where this tilt is 1, $s_j$ is either the marked empty slice, or has at least one inner face.

By iterating the decomposition of this corollary, we obtain an encoding of tight elementary slices by plane trees, where every vertex may be associated with a slice of tilt in $\{0, 1\}$. Let $m_1, \dots, m_n$ be integers or half integers, not all equal to 0, and $\epsilon \in \{0, 1\}$. Let $\mathcal{T}_\epsilon(m_1, \dots, m_n)$ be the family of pairs $(\mathbf{t}, (L_i)_{1\le i\le n})$, satisfying the following conditions.

• $\mathbf{t}$ is a rooted plane tree with $n$ vertices labeled by $\{1, 2, \dots, n\}$.

• Each vertex receives a type in $\{0, 1\}$, and the root has type $\epsilon$. We let $\epsilon_i \in \{0, 1\}$ be the type of vertex $i$ and $(w_1^{(i)}, \dots, w_{k_i}^{(i)})$ be the types of the $k_i$ children of vertex $i$ in $\mathbf{t}$ in planar order.

• For $i \in \{1, 2, \dots, n\}$ such that $m_i > 0$, $L_i = (\ell_0^{(i)}, \dots, \ell_{2m_i-1}^{(i)})$ is a lattice path from $(0, 0)$ to $(2m_i - 1, \epsilon_i)$ with increments in $\{-1, 0, 1\}$, and with $k_i$ marked steps $1 \le j_1 < \dots < j_{k_i} \le 2m_i - 1$, such that $\ell_{j_r}^{(i)} - \ell_{j_{r-1}}^{(i)} = w_r^{(i)}$ for every $r \in \{1, \dots, k_i\}$. In particular, only up or horizontal steps may be marked. We further require that every up step of $L_i$ immediately following a down step is marked, and every horizontal step of $L_i$ is marked.

• For $i \in \{1, 2, \dots, n\}$ such that $m_i = 0$, we have $\epsilon_i = 1$, $k_i = 0$ and $L_i$ is the trivial path $\{(0, 0)\}$.

This leads to the following generalization of Corollary 26.

**Corollary 38.** For $\epsilon \in \{0, 1\}$ and every choice of integers or half-integers $m_1, \dots, m_n$ not all equal to 0, the iterated slice decomposition yields a bijection between the set of elementary tight slices with $n$ labeled boundaries of respective lengths $2m_1, \dots, 2m_n$ with tilt $\epsilon$ and the set $\mathcal{T}_\epsilon(m_1, \dots, m_n)$.

It turns out that, in order to prove Theorem 13 in Section 5.4, we will need an extension of this corollary associating sequences of slices with sequences of trees, namely forests. Still, as a warm-up, let us first consider the enumeration of $\mathcal{T}_\epsilon(m_1, \dots, m_n)$, which is done based on that of two-type trees, obtained as a special case of Proposition 46 of Appendix A. Note that the types 1 and 0 considered in the present section correspond to types A and B respectively in the notation of the appendix. We count the elements $(\mathbf{t}, (L_i)_{1\le i\le n})$ of $\mathcal{T}_\epsilon(m_1, \dots, m_n)$ by fixing the types $\epsilon_1, \dots, \epsilon_n$ of the vertices 1, \dots, n of $\mathbf{t}$ (so that $\epsilon_i = w_0^{(i)}$ in the notation of Appendix A) with the constraint that the root vertex should have type $\epsilon$, as well as the sequence of types $w_1^{(i)}, \dots, w_{k_i}^{(i)}$ of the consecutive children (in planar
up to changing the up steps into down steps and vice versa), the number of such paths is equal to $p_{r_i+s_i, s_i+1+\epsilon_i}(m_i)$ if $m_i - \frac{s_i+1+\epsilon_i}{2} \in \mathbb{Z}$, and to zero otherwise. In particular, this number depends on the array $(w_j^{(i)})_{j=1,...,n}$ only through the values of $r_i, s_i$. Noting that for a given value of $r_i, s_i$, there are $\binom{r_i+s_i}{s_i}$ possible choices of $(w_j^{(i)}, 1 \le j \le k_i)$ inducing these values, and recalling that $\pi_{r_i, s_i}^{(\epsilon_i)} = \binom{r_i+s_i}{s_i} p_{r_i+s_i, s_i+1+\epsilon_i}(m_i)$, we finally obtain the following result.

**Proposition 5.10.** For $\epsilon = 0, 1$ and for every choice of integers or half-integers $m_1, ..., m_n$ not all equal to 0, the cardinality of $\mathcal{T}_\epsilon(m_1, ..., m_n)$, and hence the number of elementary tight slices with $n$ labeled boundaries of respective lengths $2m_1, ..., 2m_n$ and tilt $\epsilon$, is given by

$$ \mathrm{Card}(\mathcal{T}_\epsilon(m_1, \dots, m_n)) = \sum_{\substack{\epsilon_1, \dots, \epsilon_n=0,1 \\ r_1, \dots, r_n, s_1, \dots, s_n \ge 0 \\ \epsilon_1+\dots+\epsilon_n=\epsilon+r_1+\dots+r_n \\ \bar{\epsilon}_1+\dots+\bar{\epsilon}_n=\bar{\epsilon}+s_1+\dots+s_n}} \left( \epsilon \sum_{i=1}^{n} \epsilon_i s_i + \bar{\epsilon} \sum_{i=1}^{n} \bar{\epsilon}_i r_i \right) \left( \sum_{i=1}^{n} \epsilon_i - 1 \right)! \left( \sum_{i=1}^{n} \bar{\epsilon}_i - 1 \right)! \prod_{i=1}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i), \quad (5.17) $$

where the first three factors in the sum should be replaced by $(n-2)!$ whenever the argument in either of the two factorials equals $-1$.

This proposition has interesting consequences for certain evaluations of lattice count polynomials. Proceeding similarly to Section 4.3, we may indeed associate bijectively tight maps with elementary tight slices of tilt in $\{0, 1\}$. Precisely, starting from an elementary tight slice of tilt 0, we may glue isometrically the left and right boundaries to obtain a tight map with one marked petal (delimited by the base edge of the slice) and one marked vertex (given by the apex). This construction can be inverted by cutting along the leftmost geodesic from the marked petal to the pointed vertex. Similarly, starting from an elementary tight slice of tilt 1, we may glue isometrically the left and right boundaries starting from the base<fnref n="11" /> to obtain a tight map with two marked petals (one being delimited by the base edge of the slice, and the other one by the edge of the left boundary incident to the apex). The inverse construction is more involved, but is in fact a particular case of the slice decomposition of annular maps.

These bijections imply that for every $m_1, ..., m_n \in \mathbb{Z}_{\ge 0}/2$ not all equal to zero, and for $\epsilon \in \{0, 1\}$, we have

$$ N_{0,n}(2m_1, \dots, 2m_n, 1, \epsilon) = \mathrm{Card}(\mathcal{T}_\epsilon(m_1, \dots, m_n)). \quad (5.18) $$

Finally, from the discussion of [BG12, Appendix A], we obtain the identity

$$ N_{0,n}(2m_1, \dots, 2m_n, 2, 0) = \mathrm{Card}(\mathcal{T}_1(m_1, \dots, m_n)) + \sum_{\{1\} \subset I \subsetneq \{1, \dots, n\}} \mathrm{Card}(\mathcal{T}_0((m_i)_{i \in I})) \mathrm{Card}(\mathcal{T}_0((m_i)_{i \notin I})). \quad (5.19) $$

<footnote n="11">¹¹This means in particular that the two endpoints of the base are glued together. If we proceed as in Section 4.3 and glue the left and right boundaries of a slice of tilt 1 starting from the apex, we do obtain a pointed rooted tight map, but this is not the most general such map, since by construction the two ends of the root edge (inherited from the base) are at different distances from the distinguished vertex (inherited from the apex).</footnote>
order) of vertex $i$. These sequences must satisfy the consistency conditions (80) and (81) of Appendix A, which in this context are rewritten as

$$ \sum_{i=1}^{n} \epsilon_i = \epsilon + \sum_{i=1}^{n} r_i \quad \text{and} \quad \sum_{i=1}^{n} \bar{\epsilon}_i = \bar{\epsilon} + \sum_{i=1}^{n} s_i, \qquad (65) $$

where we let $\bar{\eta} = 1 - \eta$ for $\eta \in \{0, 1\}$, and where $r_i = \sum_{j=1}^{k_i} w_j^{(i)}$ and $s_i = \sum_{j=1}^{k_i} \bar{w}_j^{(i)}$ are the numbers of type-1 and type-0 children of vertex $i$. These consistency relations simply express in two different ways the number of type 1 (resp. 0) vertices in the tree $\mathbf{t}$. Note that the number of type-0 (resp. type-1) vertices that are the children of type-1 (resp. type-0) vertices (those are respectively the numbers $b^A$ and $a^B$ considered in Appendix A) is given by

$$ \sum_{i=1}^{n} \epsilon_i s_i \quad \left( \text{resp.} \quad \sum_{i=1}^{n} \bar{\epsilon}_i r_i \right). \qquad (66) $$

Finally, by Proposition 46 (in the notation therein, we have $a^O = \epsilon$ and $b^O = \bar{\epsilon}$, expressing the fact that we are counting “forests” made of only one tree with root of type $\epsilon$), we see that the number of possible trees $\mathbf{t}$ contributing to $\mathcal{T}_{\epsilon}(m_1, \dots, m_n)$, with types given by $\epsilon_1, \dots, \epsilon_n$ and given a consistent type array $(w_j^{(i)})_{j=1,\dots,n}^{i=1,\dots,n}$, is equal to

$$ \left( \epsilon \sum_{i=1}^{n} \epsilon_i s_i + \bar{\epsilon} \sum_{i=1}^{n} \bar{\epsilon}_i r_i \right) \left( \sum_{i=1}^{n} \epsilon_i - 1 \right)! \left( \sum_{i=1}^{n} \bar{\epsilon}_i - 1 \right)! \qquad (67) $$

where $\sum_{i=1}^{n} \epsilon_i$ and $\sum_{i=1}^{n} \bar{\epsilon}_i$ are the numbers of vertices of types 1 and 0 respectively. In the case where either one of these numbers is equal to 0, we should replace the whole formula by $(n-2)!$.

It now remains to enumerate, for a given consistent array $(w_j^{(i)})_{j=1,\dots,n}^{i=1,\dots,n}$, and for a given tree $\mathbf{t}$ as above, the number of possible lattice paths $L_i$, $1 \le i \le n$ so that $(\mathbf{t}, (L_i)_{1 \le i \le n})$ is an element of $\mathcal{T}_{\epsilon}(m_1, \dots, m_n)$. As before, we let $r_i = \sum_{j=1}^{k_i} w_j^{(i)}$ and $s_i = \sum_{j=1}^{k_i} \bar{w}_j^{(i)} = k_i - r_i$. If $m_i = 0$ then by definition of $\mathcal{T}_{\epsilon}(m_1, \dots, m_n)$, this requires $L_i = \{(0,0)\}$, $r_i = s_i = 0$ and $\epsilon_i = 1$: by (27), this is precisely counted by $\pi_{r_i,s_i}(0) = \delta_{r_i,0}\delta_{s_i,0}\delta_{\epsilon_i,1}$. If $m_i > 0$, $L_i$ should be a lattice path from $(0,0)$ to $(2m_i - 1, \epsilon_i)$, with increments in $\{-1, 0, 1\}$, in which $k_i$ up or horizontal steps are marked, say $1 \le j_1 < \dots < j_{k_i} \le 2m_i - 1$, with increments at these steps respectively given by $w_1^{(i)}, \dots, w_{k_i}^{(i)}$, and in such a way that all horizontal steps are marked, as well as all the up steps immediately following a down step. As explained in the proof of Proposition 15 (and up to changing the up steps into down steps and vice versa), the number of such paths is equal to $p_{r_i+s_i,s_i+1+\epsilon_i}(m_i)$ if $m_i - \frac{s_i+1+\epsilon_i}{2} \in \mathbb{Z}$, and to zero otherwise. In particular, this number depends on the array $(w_j^{(i)})_{j=1,\dots,n}^{i=1,\dots,n}$ only through the values of $r_i, s_i$. Noting that for a given value of $r_i, s_i$, there are $\binom{r_i+s_i}{s_i}$ possible choices of $(w_j^{(i)}, 1 \le j \le k_i)$ inducing these values, and recalling that $\pi_{r_i,s_i}^{(\epsilon_i)} = \binom{r_i+s_i}{s_i} p_{r_i+s_i,s_i+1+\epsilon_i}(m_i)$, we finally obtain the following result.
5.4. General maps

We are now ready to get a general enumeration formula for planar tight maps. Our approach will be parallel to that of Section 4.4, and we start by stating the following analog of Proposition 4.7:

**Proposition 5.11.** Let $m_1, \dots, m_n$ be non-negative integers or half-integers ($n \ge 3$) with $m_1, m_2 > 0$. Then, there is a bijection between the set of planar tight maps with $n$ boundaries labeled from 1 to $n$ with respective lengths $2m_1, 2m_2, \dots, 2m_n$, and the set of pairs $(\mathbf{m}_{12}, \mathbf{s})$ such that there exist $r_0, s_0 \ge 0$ with $r_0 + s_0 \in \{0, \dots, n-3\}$ for which the following holds.

• $\mathbf{m}_{12}$ is a tight petal necklace with two distinguished faces of degrees $2m_1, 2m_2$, with one extra distinguished element being either a marked vertex or a petal, and with $r_0$ other marked vertices and $s_0$ other petals.

• $\mathbf{s} = (\mathbf{s}_1, \dots, \mathbf{s}_{r_0+s_0+1})$ is a $(r_0+s_0+1)$-tuple of slices such that:

- each $\mathbf{s}_j$ ($j = 1, \dots, r_0+s_0+1$) is a tight elementary slice of tilt 0 or 1 containing at least one inner face or marked vertex,

- the inner faces and marked vertices of these slices are labeled by integers in $\{3, \dots, n\}$,

- each $i \in \{3, \dots, n\}$ appears in exactly one $\mathbf{s}_j$ and labels an inner face of degree $2m_i$ for $m_i > 0$, or a marked vertex for $m_i = 0$,

- the label 3 appears in the first slice $\mathbf{s}_1$.

• $\mathbf{m}_{12}$ and $\mathbf{s}$ are compatible in the sense that, if we label the marked vertices and petals of $\mathbf{m}_{12}$ in a canonical way from 1 to $r_0+s_0+1$ as in Remark 5.5, and for $j = 1, \dots, r_0+s_0+1$ we set $w_j^{(0)} = 1$ (resp. $w_j^{(0)} = 0$) if label $j$ is on a marked vertex (resp. petal), then $\mathbf{s}_j$ has tilt $w_j^{(0)}$. Note that $r_0 = \sum_{j=2}^{r_0+s_0+1} w_j^{(0)}$.

The proof of this proposition follows exactly the same lines as that of Proposition 4.7 in Section 4.4. In particular, Lemma 4.8 admits a direct non-bipartite extension in which the maps $\mathbf{m}$ may have all their faces of arbitrary degrees, the lattice paths $L, L'$ may contain horizontal steps, and consistently the sequences $\boldsymbol{\sigma}, \boldsymbol{\sigma}'$ may contain slices of tilt 0. Further details are left to the reader. We deduce the following enumerative result:

**Proposition 5.12.** For $n \ge 3$ and for non-negative integers or half-integers $m_1, \dots, m_n$ with, say, $m_1, m_2 > 0$, we have

$$
\begin{align}
N_{0,n}(2m_1, \dots, 2m_n) = & \sum_{\substack{\epsilon_3, \dots, \epsilon_n = 0, 1 \\ r_1, \dots, r_n, s_1, \dots, s_n \ge 0 \\ \sum_{i=3}^n \epsilon_i = \sum_{i=1}^n r_i + 1 \\ \sum_{i=3}^n (1-\epsilon_i) = \sum_{i=1}^n s_i}} \left(\sum_{i=1}^n r_i\right)! \left(\epsilon_3(s_1 + s_2) + \sum_{i=3}^n \epsilon_i s_i\right) \left(\sum_{i=1}^n s_i - 1\right)! \tag{5.20} \\
& \times \left(\pi_{r_1, s_1}^{(-1)}(m_1) \pi_{r_2, s_2}^{(1)}(m_2) + \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2)\right) \prod_{i=3}^n \pi_{r_i, s_i}^{(\epsilon_i)}(m_i) \nonumber
\end{align}
$$
**Proposition 39.** For $\epsilon = 0, 1$ and for every choice of integers or half-integers $m_1, \dots, m_n$ not all equal to 0, the cardinality of $\mathcal{T}_\epsilon(m_1, \dots, m_n)$, and hence the number of elementary tight slices with $n$ labeled boundaries of respective lengths $2m_1, \dots, 2m_n$ and tilt $\epsilon$, is given by

$$ \text{Card}(\mathcal{T}_\epsilon(m_1, \dots, m_n)) = \sum_{\substack{\epsilon_1, \dots, \epsilon_n = 0, 1 \\ r_1, \dots, r_n, s_1, \dots, s_n \ge 0 \\ \epsilon_1 + \dots + \epsilon_n = \epsilon + r_1 + \dots + r_n \\ \bar{\epsilon}_1 + \dots + \bar{\epsilon}_n = \bar{\epsilon} + s_1 + \dots + s_n}} \left( \epsilon \sum_{i=1}^{n} \epsilon_i s_i + \bar{\epsilon} \sum_{i=1}^{n} \bar{\epsilon}_i r_i \right) \left( \sum_{i=1}^{n} \epsilon_i - 1 \right)! \left( \sum_{i=1}^{n} \bar{\epsilon}_i - 1 \right)! \prod_{i=1}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i), \tag{68} $$

where the first three factors in the sum should be replaced by $(n-2)!$ whenever the argument in either of the two factorials equals $-1$.

This proposition has interesting consequences for certain evaluations of lattice count polynomials. Proceeding similarly to Section 4.3, we may indeed associate bijectively tight maps with elementary tight slices of tilt in $\{0, 1\}$. Precisely, starting from an elementary tight slice of tilt 0, we may glue isometrically the left and right boundaries to obtain a tight map with one marked petal (delimited by the base edge of the slice) and one marked vertex (given by the apex). This construction can be inverted by cutting along the leftmost geodesic from the marked petal to the pointed vertex. Similarly, starting from an elementary tight slice of tilt 1, we may glue isometrically the left and right boundaries starting from the base (rather than from the apex as was done in Section 4.3) to obtain a tight map with two marked petals (one being delimited by the base edge of the slice, and the other one by the edge of the left boundary incident to the apex). The inverse construction is more involved, but is in fact a particular case of the slice decomposition of annular maps.

These bijections imply that for every $m_1, \dots, m_n \in \mathbb{Z}_{\ge 0}/2$ not all equal to zero, and for $\epsilon \in \{0, 1\}$, we have

$$ N_{0,n}(2m_1, \dots, 2m_n, 1, \epsilon) = \text{Card}(\mathcal{T}_\epsilon(m_1, \dots, m_n)). \tag{69} $$

Finally, from the discussion of [BG12, Appendix A], we obtain the identity

$$ N_{0,n}(2m_1, \dots, 2m_n, 2, 0) = \text{Card}(\mathcal{T}_1(m_1, \dots, m_n)) + \sum_{\{1\} \subset I \subset \{1, \dots, n\}} \text{Card}(\mathcal{T}_0((m_i)_{i \in I})) \text{Card}(\mathcal{T}_0((m_i)_{i \notin I})). \tag{70} $$

## 5.4. General maps

We are now ready to get a general enumeration formula for planar tight maps. Our approach will be parallel to that of Section 4.4, and we start by stating the following analog of Proposition 27:
$$
+ \sum_{\substack{\epsilon_3, \dots, \epsilon_n = 0, 1 \\ r_1, \dots, r_n, s_1, \dots, s_n \ge 0 \\ \sum_{i=3}^n \epsilon_i = \sum_{i=1}^n r_i \\ \sum_{i=3}^n (1-\epsilon_i) r_i = \sum_{i=1}^n s_i + 1}} \left( \sum_{i=1}^n s_i \right)! \left( (1-\epsilon_3)(r_1+r_2) + \sum_{i=3}^n (1-\epsilon_i)r_i \right) \left( \sum_{i=1}^n r_i - 1 \right)! \\
\times \left( \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(1)}(m_2) + \pi_{r_1, s_1}^{(1)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) \right) \prod_{i=3}^n \pi_{r_i, s_i}^{(\epsilon_i)}(m_i)
$$

where $\pi_{r,s}^{(\epsilon)}$ is the univariate quasi-polynomial defined in (2.26) and where it is understood that
$(\epsilon_3(s_1+s_2) + \sum_{i=3}^n \epsilon_i s_i)(\sum_{i=1}^n s_i - 1)!$ is equal to 1 if all the $s_i$ are zero, and that
$((1-\epsilon_3)(r_1+r_2) + \sum_{i=3}^n (1-\epsilon_i)r_i)(\sum_{i=1}^n r_i - 1)!$ is equal to 1 if all the $r_i$ are zero.

Proof. We need to enumerate the compatible pairs ($\mathbf{m}_{12}$, $\mathbf{s}$) of Proposition 5.11. Note first that, given $r_0$ and $s_0$, the number of possible petal necklaces $\mathbf{m}_{12}$ is given by Proposition 5.4. Turning now to the number of possible $\mathbf{s}$ compatible with a given petal necklace $\mathbf{m}_{12}$, it is given by a direct extension of Corollary 5.9 as follows: by decomposing recursively each elementary tight slice $\mathbf{s}_j$ (with tilt $w_j^{(0)}$) into a tree of lattice paths, we see that the set of possible $(r_0 + s_0 + 1)$-tuples $\mathbf{s}$ is in bijection with the set $\mathcal{F}_{(w_1^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})}(m_3, \dots, m_n)$ defined as the set of pairs $(\mathbf{f}, (L_i)_{1 \le i \le n})$ satisfying the following conditions.

• $\mathbf{f}$ is a plane forest with $r_0 + s_0 + 1$ connected components, i.e. a $(r_0 + s_0 + 1)$-tuple of rooted plane trees, and with a total of $n - 2$ vertices labeled by $\{3, \dots, n\}$, the label 3 appearing in the first component.

* Each vertex receives a type in {0, 1}, and the root vertex of the j-th tree component in the forest has type $w_j^{(0)}$. For $i \in \{3, \dots, n\}$, we let $\epsilon_i \in \{0, 1\}$ be the type of vertex $i$, $k_i$ be its number of children, and $(w_1^{(i)}, \dots, w_{k_i}^{(i)})$ be the types of these children (numbered in planar order in the rooted tree component at hand). We also set $r_i := \sum_{j=1}^{k_i} w_j^{(i)}$ the numbers of those children which are of type 1 and $s_i := k_i - r_i$ the numbers of those children which are of type 0.

• For $i \in \{3, \dots, n\}$:

- if $m_i > 0$, $L_i = (\ell_0^{(i)}, \dots, \ell_{2m_i-1}^{(i)})$ is a lattice path from $(0,0)$ to $(2m_i-1, \epsilon_i)$ with increments in $\{-1, 0, 1\}$, and with $k_i$ marked steps $1 \le j_1 < \dots < j_{k_i} \le 2m_i-1$, such that $\ell_{j_r}^{(i)} - \ell_{j_r-1}^{(i)} = w_r^{(i)}$ for every $r \in \{1, \dots, k_i\}$. In particular, only up or horizontal steps may be marked, and we further require that every horizontal step of $L_i$ is marked, as well as every up step immediately following a down step,

- if $m_i = 0$, we have $\epsilon_i = 1$, $k_i = r_i = s_i = 0$, and $L_i$ is the trivial path $\{(0,0)\}$.

We may now obtain the number of elements of $\mathcal{F}_{(w_1^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})}(m_3, \dots, m_n)$ from the results of Appendix A for the enumeration of two-type forests, where the types $A$ and $B$ therein correspond to types 1 and 0 in the present setting. For fixed $\epsilon_i$ and $k_i$, $i = 3, \dots, n$ and for fixed $(w_j^{(i)})_{j=1,\dots,k_i}^{i=3,\dots,n}$,
**Proposition 40.** Let $m_1, \dots, m_n$ be non-negative integers or half-integers ($n \ge 3$) with $m_1, m_2 > 0$. Then, there is a bijection between the set of planar tight maps with $n$ boundaries labeled from 1 to $n$ with respective lengths $2m_1, 2m_2, \dots, 2m_n$, and the set of pairs $(\mathbf{m}_{12}, \mathbf{s})$ such that there exist $r_0, s_0 \ge 0$ with $r_0 + s_0 \in \{0, \dots, n-3\}$ for which the following holds.

* $\mathbf{m}_{12}$ is a tight petal necklace with two distinguished faces of degrees $2m_1, 2m_2$, with one extra distinguished element being either a marked vertex or a petal, and with $r_0$ other marked vertices and $s_0$ other petals.

* $\mathbf{s} = (\mathbf{s}_1, \dots, \mathbf{s}_{r_0+s_0+1})$ is a $(r_0+s_0+1)$-tuple of slices such that:

- each $\mathbf{s}_j$ ($j = 1, \dots, r_0+s_0+1$) is a tight elementary slice of tilt 0 or 1 containing at least one inner face or marked vertex,

- the inner faces and marked vertices of these slices are labeled by integers in $\{3, \dots, n\}$,

- each $i \in \{3, \dots, n\}$ appears in exactly one $\mathbf{s}_j$ and labels an inner face of degree $2m_i$ for $m_i > 0$, or a marked vertex for $m_i = 0$,

- the label 3 appears in the first slice $\mathbf{s}_1$.

* $\mathbf{m}_{12}$ and $\mathbf{s}$ are compatible in the sense that, if we label the marked vertices and petals of $\mathbf{m}_{12}$ in a canonical way from 1 to $r_0+s_0+1$ as in Remark 34, and for $j = 1, \dots, r_0+s_0+1$ we set $w_j^{(0)} = 1$ (resp. $w_j^{(0)} = 0$) if label $j$ is on a marked vertex (resp. petal), then $\mathbf{s}_j$ has tilt $w_j^{(0)}$. Note that $r_0 = \sum_{j=2}^{r_0+s_0+1} w_j^{(0)}$.

The proof of this proposition follows exactly the same lines as that of Proposition 27 in Section 4.4. In particular, Lemma 28 admits a direct non-bipartite extension in which the maps $\mathbf{m}$ may have all their faces of arbitrary degrees, the lattice paths $L, L'$ may contain horizontal steps, and consistently the sequences $\boldsymbol{\sigma}, \boldsymbol{\sigma}'$ may contain slices of tilt 0. Further details are left to the reader. We deduce the following enumerative result:

**Proposition 41.** For $n \ge 3$ and for non-negative integers or half-integers $m_1, \dots, m_n$ with, say, $m_1, m_2 > 0$, we have

$$N_{0,n}(2m_1, \dots, 2m_n) = \sum_{\substack{\epsilon_3, \dots, \epsilon_n = 0, 1 \\ \sum_{i=3}^{n} \epsilon_i = \sum_{i=1}^{n} r_i + 1 \\ \sum_{i=3}^{n} (1-\epsilon_i) = \sum_{i=1}^{n} s_i}} \left(\sum_{i=1}^{n} r_i\right)! \left(\epsilon_3(s_1 + s_2) + \sum_{i=3}^{n} \epsilon_i s_i\right) \left(\sum_{i=1}^{n} s_i - 1\right)! \times \left(\pi_{r_1,s_1}^{(-1)}(m_1)\pi_{r_2,s_2}^{(1)}(m_2) + \pi_{r_1,s_1}^{(0)}(m_1)\pi_{r_2,s_2}^{(0)}(m_2)\right) \prod_{i=3}^{n} \pi_{r_i,s_i}^{(\epsilon_i)}(m_i) \quad (71)$$
the number of forests f is non-zero only if the two consistency relations (A.1) and (A.2) are satisfied, namely

$$ \sum_{i=3}^{n} \epsilon_i = w_1^{(0)} + r_0 + \sum_{i=3}^{n} r_i \quad \text{and} \quad \sum_{i=3}^{n} \bar{\epsilon}_i = \bar{w}_1^{(0)} + s_0 + \sum_{i=3}^{n} s_i \qquad (5.21) $$

where, as before, we use the shorthand notation $\bar{\eta} := 1 - \eta$ for $\eta \in \{0, 1\}$. Again, these identities simply express in two different ways the number of vertices of type 1 (first identity) and of type 0 (second identity), corresponding respectively to the quantities denoted by a and b in Appendix A. When these conditions are satisfied, we may use the constrained enumeration result (A.6), with the correspondence $a^O = w_1^{(0)} + r_0$, $b^O = \bar{w}_1^{(0)} + s_0$, $a^B = \sum_{i=3}^n \bar{\epsilon}_i r_i$ and $b^A = \sum_{i=3}^n \epsilon_i s_i$: if the vertex 3 is of type 1, i.e. $\epsilon_3 = 1$, the number of forests f is given by

$$ \left\{ \begin{array}{ll} \left( s_0 + \sum_{i=3}^{n} \epsilon_i s_i \right) \left( r_0 + \sum_{i=3}^{n} r_i \right)! \left( s_0 + \sum_{i=3}^{n} s_i - 1 \right)! & \text{if } w_1^{(0)} = 1, \\ \left( \sum_{i=3}^{n} \bar{\epsilon}_i r_i \right) \left( r_0 + \sum_{i=3}^{n} r_i - 1 \right)! \left( s_0 + \sum_{i=3}^{n} s_i \right)! & \text{if } w_1^{(0)} = 0, \end{array} \right. \qquad (5.22) $$

with, in the first line, the convention $(s_0 + \sum_{i=3}^n \epsilon_i s_i) \times (s_0 + \sum_{i=3}^n s_i - 1) = 1$ if $s_0 + \sum_{i=3}^n s_i = 0$, i.e. when $s_0$ and all the $s_i$'s for $i \ge 3$ are zero. Note that, in the second line, the quantity $r_0 + \sum_{i=3}^n r_i - 1$ is always nonnegative since, from the first consistency relation, it is equal to $\sum_{i=4}^n \epsilon_i$. By symmetry, if the vertex 3 is of type 0, i.e. $\epsilon_3 = 0$, the number of forests f is given by

$$ \left\{ \begin{array}{ll} \left( r_0 + \sum_{i=3}^{n} \bar{\epsilon}_i r_i \right) \left( s_0 + \sum_{i=3}^{n} s_i \right)! \left( r_0 + \sum_{i=3}^{n} r_i - 1 \right)! & \text{if } w_1^{(0)} = 0, \\ \left( \sum_{i=3}^{n} \epsilon_i s_i \right) \left( s_0 + \sum_{i=3}^{n} s_i - 1 \right)! \left( r_0 + \sum_{i=3}^{n} r_i \right)! & \text{if } w_1^{(0)} = 1, \end{array} \right. \qquad (5.23) $$

with, in the first line, the convention $(r_0 + \sum_{i=3}^n \bar{\epsilon}_i r_i) \times (r_0 + \sum_{i=3}^n r_i - 1) = 1$ if $r_0 + \sum_{i=3}^n r_i = 0$, i.e. when $r_0$ and all the $r_i$'s for $i \ge 3$ are zero. Again, in the second line, the quantity $s_0 + \sum_{i=3}^n s_i - 1$ is always nonnegative since, from the second consistency relation, it is equal to $\sum_{i=4}^n \bar{\epsilon}_i$. Both cases $\epsilon_3 = 1$ or 0 above may be summarized into the enumeration formulas

$$ \left\{ \begin{array}{ll} \left( \epsilon_3 s_0 + \sum_{i=3}^{n} \epsilon_i s_i \right) \left( s_0 + \sum_{i=3}^{n} s_i - 1 \right)! \left( r_0 + \sum_{i=3}^{n} r_i \right)! & \text{if } w_1^{(0)} = 1, \\ \left( \bar{\epsilon}_3 r_0 + \sum_{i=3}^{n} \bar{\epsilon}_i r_i \right) \left( r_0 + \sum_{i=3}^{n} r_i - 1 \right)! \left( s_0 + \sum_{i=3}^{n} s_i \right)! & \text{if } w_1^{(0)} = 0, \end{array} \right. \qquad (5.24) $$

with the conventions that $(\epsilon_3 s_0 + \sum_{i=3}^n \epsilon_i s_i)(s_0 + \sum_{i=3}^n s_i - 1) = 1$ if $s_0$ and all the $s_i$'s for $i \ge 3$ are zero and that $(\bar{\epsilon}_3 r_0 + \sum_{i=3}^n \bar{\epsilon}_i r_i)(r_0 + \sum_{i=3}^n r_i - 1) = 1$ if $r_0$ and all the $r_i$'s for $i \ge 3$ are zero.

It remains to enumerate, for a given array $(w_j^{(i)})_{j=1,...,n}$ satisfying the consistency relations (5.21), and for a given two-type forest f as above, the number of families of lattice paths $(L_i)_{3 \le i \le n}$ so that $(f, (L_i)_{3 \le i \le n})$ is an element of $\mathcal{F}_{(w_1^{(0)},...,w_{r_0+s_0+1}^{(0)})}(m_3, ..., m_n)$. Repeating the counting
$$ + \sum_{\substack{\epsilon_3, \dots, \epsilon_n = 0, 1 \\ r_1, \dots, r_n, s_1, \dots, s_n \ge 0 \\ \sum_{i=3}^n \epsilon_i = \sum_{i=1}^n r_i \\ \sum_{i=3}^n (1-\epsilon_i) = \sum_{i=1}^n s_i + 1}} \left( \sum_{i=1}^n s_i \right)! \left( (1-\epsilon_3)(r_1+r_2) + \sum_{i=3}^n (1-\epsilon_i)r_i \right) \left( \sum_{i=1}^n r_i - 1 \right)! \\ \times \left( \pi_{r_1, s_1}^{(0)}(m_1) \pi_{r_2, s_2}^{(1)}(m_2) + \pi_{r_1, s_1}^{(1)}(m_1) \pi_{r_2, s_2}^{(0)}(m_2) \right) \prod_{i=3}^n \pi_{r_i, s_i}^{(\epsilon_i)}(m_i) $$

where $\pi_{r,s}^{(\epsilon)}$ is the univariate quasi-polynomial defined in (26) and where it is understood that $(\epsilon_3(s_1 + s_2) + \sum_{i=3}^n \epsilon_i s_i)(\sum_{i=1}^n s_i - 1)!$ is equal to 1 if all the $s_i$ are zero, and that $((1 - \epsilon_3)(r_1 + r_2) + \sum_{i=3}^n (1 - \epsilon_i)r_i)(\sum_{i=1}^n r_i - 1)!$ is equal to 1 if all the $r_i$ are zero.

**Proof.** We need to enumerate the compatible pairs $(\mathbf{m}_{12}, \mathbf{s})$ of Proposition 40. Note first that, given $r_0$ and $s_0$, the number of possible petal necklaces $\mathbf{m}_{12}$ is given by Proposition 33. Turning now to the number of possible $\mathbf{s}$ compatible with a given petal necklace $\mathbf{m}_{12}$, it is given by a direct extension of Corollary 38 as follows: by decomposing recursively each elementary tight slice $\mathbf{s}_j$ (with tilt $w_j^{(0)}$) into a tree of lattice paths, we see that the set of possible $(r_0 + s_0 + 1)$-tuples $\mathbf{s}$ is in bijection with the set $\mathcal{F}_{(w_1^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})}(m_3, \dots, m_n)$ defined as the set of pairs $(\mathbf{f}, (L_i)_{1 \le i \le n})$ satisfying the following conditions.

* $\mathbf{f}$ is a plane forest with $r_0 + s_0 + 1$ connected components, i.e. a $(r_0 + s_0 + 1)$-tuple of rooted plane trees, and with a total of $n - 2$ vertices labeled by $\{3, \dots, n\}$, the label 3 appearing in the first component.

* Each vertex receives a type in $\{0, 1\}$, and the root vertex of the $j$-th tree component in the forest has type $w_j^{(0)}$. For $i \in \{3, \dots, n\}$, we let $\epsilon_i \in \{0, 1\}$ be the type of vertex $i$, $k_i$ be its number of children, and $(w_1^{(i)}, \dots, w_{k_i}^{(i)})$ be the types of these children (numbered in planar order in the rooted tree component at hand). We also set $r_i := \sum_{j=1}^{k_i} w_j^{(i)}$ the numbers of those children which are of type 1 and $s_i := k_i - r_i$ the numbers of those children which are of type 0.

* For $i \in \{3, \dots, n\}$:

- if $m_i > 0$, $L_i = (\ell_0^{(i)}, \dots, \ell_{2m_i-1}^{(i)})$ is a lattice path from $(0, 0)$ to $(2m_i - 1, \epsilon_i)$ with increments in $\{-1, 0, 1\}$, and with $k_i$ marked steps $1 \le j_1 < \dots < j_{k_i} \le 2m_i - 1$, such that $\ell_{j_r}^{(i)} - \ell_{j_r-1}^{(i)} = w_r^{(i)}$ for every $r \in \{1, \dots, k_i\}$. In particular, only up or horizontal steps may be marked, and we further require that every horizontal step of $L_i$ is marked, as well as every up step immediately following a down step,

- if $m_i = 0$, we have $\epsilon_i = 1$, $k_i = r_i = s_i = 0$, and $L_i$ is the trivial path $\{(0, 0)\}$.

We may now obtain the number of elements of $\mathcal{F}_{(w_1^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})}(m_3, \dots, m_n)$ from the results of Appendix A for the enumeration of two-type forests, where the types A and B
argument in the proof of Proposition 5.10, this number is simply equal to $\prod_{i=3}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i)$. Combining with (5.24), and recalling the consistency relations (5.21), we obtain in the case $w_1^{(0)} = 1$

$$ \text{Card} \left( \mathcal{F}_{(w_1^{(0)}=1, w_2^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})} (m_3, \dots, m_n) \right) = \\ \sum_{\substack{\epsilon_3, \dots, \epsilon_n = 0, 1 \\ r_3, \dots, r_n, s_3, \dots, s_n \ge 0 \\ \sum_{i=3}^{n} \epsilon_i = r_0 + \sum_{i=3}^{n} r_i + 1 \\ \sum_{i=3}^{n} \bar{\epsilon}_i = s_0 + \sum_{i=3}^{n} s_i}} \left( r_0 + \sum_{i=3}^{n} r_i \right)! \left( \epsilon_3 s_0 + \sum_{i=3}^{n} \epsilon_i s_i \right) \left( s_0 + \sum_{i=3}^{n} s_i - 1 \right)! \prod_{i=3}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i), \tag{5.25} $$

and in the case $w_1^{(0)} = 0$

$$ \text{Card} \left( \mathcal{F}_{(w_1^{(0)}=0, w_2^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})} (m_3, \dots, m_n) \right) = \\ \sum_{\substack{\epsilon_3, \dots, \epsilon_n = 0, 1 \\ r_3, \dots, r_n, s_3, \dots, s_n \ge 0 \\ \sum_{i=3}^{n} \epsilon_i = r_0 + \sum_{i=3}^{n} r_i \\ \sum_{i=3}^{n} \bar{\epsilon}_i = s_0 + \sum_{i=3}^{n} s_i + 1}} \left( s_0 + \sum_{i=3}^{n} s_i \right)! \left( \bar{\epsilon}_3 r_0 + \sum_{i=3}^{n} \bar{\epsilon}_i r_i \right) \left( r_0 + \sum_{i=3}^{n} r_i - 1 \right)! \prod_{i=3}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i). \tag{5.26} $$

Note that these expressions do not depend on the precise sequence $(w_1^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})$ but only on the values of $r_0 = \sum_{j=2}^{r_0+s_0+1} w_j^{(0)}$, of $s_0 = \sum_{j=2}^{r_0+s_0+1} \bar{w}_j^{(0)}$ and of $w_1^{(0)}$. This is expected since permuting the terms of the sequence $(w_j^{(0)})_{j=2, \dots, n}$ at fixed $r_0$ and $s_0$ simply amounts to changing the order of the last $r_0 + s_0$ trees in the forest $f$, a harmless operation as far as enumeration is concerned. We may therefore use the counting formulas (5.7) and (5.8) for petal necklaces $m_{12}$ with fixed $r_0, s_0$ and with a fixed value $w_1^{(0)} = 1$ and $w_1^{(0)} = 0$ respectively. The expression (5.20) for $N_{0,n}(2m_1, \dots, 2m_n)$ is obtained by inserting (5.25) into (5.7) and (5.26) into (5.8), adding these two contributions and finally summing over $r_0$ and $s_0$, which trivially amounts to replacing in (5.25) and (5.26) each occurrence of $r_0$ by $r_1 + r_2$ and each occurrence of $s_0$ by $s_1 + s_2$, and summing over $r_1, r_2, s_1, s_2 \ge 0$. This ends the proof of Proposition 5.12. $\square$

Even though it is not apparent, the right-hand side of (5.20) turns out to be, as expected, symmetric upon permuting the $m_i$'s for $i$ in the whole set $\{1, \dots, n\}$. This property is shown in Appendix B where, after some algebraic manipulations, this quantity is given a manifestly symmetric form, see for instance (B.4) or (B.17). Using this result, we arrive at the following unified theorem encompassing all the main theorems of the paper (Theorems 2.3, 2.8 and 2.12):

**Theorem 5.13.** For $n \ge 3$ and $m_1, m_2, \dots, m_n \in \mathbb{Z}_{\ge 0}/2$, not all equal to zero, the number $N_{0,n}(2m_1, 2m_2, \dots, 2m_n)$ of planar tight maps with $n$ boundaries labeled from 1 to $n$ with re-
therein correspond to types 1 and 0 in the present setting. For fixed $\epsilon_i$ and $k_i$, $i = 3, \dots, n$ and for fixed $(w_j^{(i)})_{j=1, \dots, n}^{i=3, \dots, n}$, the number of forests $f$ is non-zero only if the two consistency relations (80) and (81) are satisfied, namely

$$ \sum_{i=3}^{n} \epsilon_i = w_1^{(0)} + r_0 + \sum_{i=3}^{n} r_i \quad \text{and} \quad \sum_{i=3}^{n} \bar{\epsilon}_i = \bar{w}_1^{(0)} + s_0 + \sum_{i=3}^{n} s_i \qquad (72) $$

where, as before, we use the shorthand notation $\bar{\eta} := 1 - \eta$ for $\eta \in \{0, 1\}$. Again, these identities simply express in two different ways the number of vertices of type 1 (first identity) and of type 0 (second identity), corresponding respectively to the quantities denoted by $a$ and $b$ in Appendix A. When these conditions are satisfied, we may use the constrained enumeration result (85), with the correspondence $a^O = w_1^{(0)} + r_0$, $b^O = \bar{w}_1^{(0)} + s_0$, $a^B = \sum_{i=3}^n \bar{\epsilon}_i r_i$ and $b^A = \sum_{i=3}^n \epsilon_i s_i$: if the vertex 3 is of type 1, i.e. $\epsilon_3 = 1$, the number of forests $f$ is given by

$$ \left\{ \begin{array}{ll} \left( s_0 + \sum_{i=3}^{n} \epsilon_i s_i \right) \left( r_0 + \sum_{i=3}^{n} r_i \right)! \left( s_0 + \sum_{i=3}^{n} s_i - 1 \right)! & \text{if } w_1^{(0)} = 1, \\ \left( \sum_{i=3}^{n} \bar{\epsilon}_i r_i \right) \left( r_0 + \sum_{i=3}^{n} r_i - 1 \right)! \left( s_0 + \sum_{i=3}^{n} s_i \right)! & \text{if } w_1^{(0)} = 0, \end{array} \right. \qquad (73) $$

with, in the first line, the convention $(s_0 + \sum_{i=3}^n \epsilon_i s_i) \times (s_0 + \sum_{i=3}^n s_i - 1)! = 1$ if $s_0 + \sum_{i=3}^n s_i = 0$, i.e. when $s_0$ and all the $s_i$'s for $i \ge 3$ are zero. Note that, in the second line, the quantity $r_0 + \sum_{i=3}^n r_i - 1$ is always nonnegative since, from the first consistency relation, it is equal to $\sum_{i=4}^n \epsilon_i$. By symmetry, if the vertex 3 is of type 0, i.e. $\epsilon_3 = 0$, the number of forests $f$ is given by

$$ \left\{ \begin{array}{ll} \left( r_0 + \sum_{i=3}^{n} \bar{\epsilon}_i r_i \right) \left( s_0 + \sum_{i=3}^{n} s_i \right)! \left( r_0 + \sum_{i=3}^{n} r_i - 1 \right)! & \text{if } w_1^{(0)} = 0, \\ \left( \sum_{i=3}^{n} \epsilon_i s_i \right) \left( s_0 + \sum_{i=3}^{n} s_i - 1 \right)! \left( r_0 + \sum_{i=3}^{n} r_i \right)! & \text{if } w_1^{(0)} = 1, \end{array} \right. \qquad (74) $$

with, in the first line, the convention $(r_0 + \sum_{i=3}^n \bar{\epsilon}_i r_i) \times (r_0 + \sum_{i=3}^n r_i - 1)! = 1$ if $r_0 + \sum_{i=3}^n r_i = 0$, i.e. when $r_0$ and all the $r_i$'s for $i \ge 3$ are zero. Again, in the second line, the quantity $s_0 + \sum_{i=3}^n s_i - 1$ is always nonnegative since, from the second consistency relation, it is equal to $\sum_{i=4}^n \bar{\epsilon}_i$. Both cases $\epsilon_3 = 1$ or $0$ above may be summarized into the enumeration formulas

$$ \left\{ \begin{array}{ll} \left( \epsilon_3 s_0 + \sum_{i=3}^{n} \epsilon_i s_i \right) \left( s_0 + \sum_{i=3}^{n} s_i - 1 \right)! \left( r_0 + \sum_{i=3}^{n} r_i \right)! & \text{if } w_1^{(0)} = 1, \\ \left( \bar{\epsilon}_3 r_0 + \sum_{i=3}^{n} \bar{\epsilon}_i r_i \right) \left( r_0 + \sum_{i=3}^{n} r_i - 1 \right)! \left( s_0 + \sum_{i=3}^{n} s_i \right)! & \text{if } w_1^{(0)} = 0, \end{array} \right. \qquad (75) $$

with the conventions that $(\epsilon_3 s_0 + \sum_{i=3}^n \epsilon_i s_i)(s_0 + \sum_{i=3}^n s_i - 1)! = 1$ if $s_0$ and all the $s_i$'s for $i \ge 3$ are zero and that $(\bar{\epsilon}_3 r_0 + \sum_{i=3}^n \bar{\epsilon}_i r_i)(r_0 + \sum_{i=3}^n r_i - 1)! = 1$ if $r_0$ and all the $r_i$'s for $i \ge 3$ are zero.
spective lengths $2m_1, 2m_2, \dots, 2m_n$ is given by the symmetric quasi-polynomial

$$
\begin{align}
N_{0,n}(2m_1, 2m_2, \dots, 2m_n) = & \sum_{\left(\begin{smallmatrix} \epsilon_1, \dots, \epsilon_n \\ r_1, \dots, r_n \\ s_1, \dots, s_n \end{smallmatrix}\right) \in I_n} \left(\sum_{i=1}^n r_i\right)! \left(\sum_{i=1}^n \epsilon_i s_i\right) \left(\sum_{i=1}^n s_i - 1\right)! \prod_{i=1}^n \pi_{r_i, s_i}^{(\epsilon_i)}(m_i) \nonumber \\
& + (n-3)! \sum_{\substack{(r_1, \dots, r_n) \in \mathbb{Z}^n \\ r_1 + \dots + r_n = n-3}} \sum_{1 \le j < \ell \le n} \pi_{r_j, 0}^{(0)}(m_j) \pi_{r_\ell, 0}^{(0)}(m_\ell) \prod_{\substack{i=1 \\ i \ne j, \ell}}^n \pi_{r_i, 0}^{(1)}(m_i) \tag{5.27} \\
& + (n-3)! \sum_{\substack{(r_1, \dots, r_n) \in \mathbb{Z}^n \\ r_1 + \dots + r_n = n-3}} \pi_{r_1, 0}^{(-1)}(m_1) \prod_{i=2}^n \pi_{r_i, 0}^{(1)}(m_i). \nonumber
\end{align}
$$

with $\pi_{r,s}^{(\epsilon)}(m)$ as in (2.26) and $I_n$ as in (2.29). Note that the sum in the last line is equal to the sym-
metric polynomial $(n-3)!p_{n-3}(m_1, \dots, m_n)$ when all $m_i$ are integers, and to zero otherwise,
hence it is symmetric.

Proof. Since $N_{0,n}(2m_1, 2m_2, \dots, 2m_n)$ is (by definition) symmetric in $m_1, m_2, \dots, m_n$ and so is the right-hand side of (5.27), we may assume without loss of generality that $m_1 > 0$ and either $m_2 > 0$ or $m_2 = \dots = m_n = 0$. In the first case, Proposition 5.12 and the identification of the right-hand sides of (5.20) and (5.27) proved in Proposition B.1 of Appendix B allows to conclude. In the second case, both sides of the equation reduce to the univariate quasi-polynomial $(n-3)!\pi_{n-3,0}^{(-1)}(m_1)$, equal to $(n-3)!p_{n-3}(m_1)$ if $m_1$ is a positive integer, and to zero otherwise. $\square$

Let us now explain how to recover Theorems 2.3, 2.8 and 2.12 from Theorem 5.13. As in Remark 2.13, let us denote by $k$ the number of half-integers among $m_1, \dots, m_n$. From the general property that $\pi_{r_i, s_i}^{(\epsilon_i)}(m_i)$ is non-zero only if $m_i - \frac{s_i+1+\epsilon_i}{2} \in \mathbb{Z}$, we see that the sum in the third line of the right-hand side of (5.27) is non-zero if and only if all the $m_i$'s are integers, i.e. the map is bipartite ($k=0$), in which case it reduces precisely to $(n-3)!p_{n-3}(m_1, \dots, m_n)$. Similarly, the sum in the second line is non-zero if and only if exactly two of the $m_i$'s, say for $i=i_1$ and $i=i_2$ ($1 \le i_1 < i_2 \le n$), are half-integers, i.e. the map is quasi-bipartite ($k=2$), in which case it reduces to $(n-3)!\tilde{p}_{n-3}(m_{i_1}, m_{i_2}; m_1, \dots, \tilde{m}_{i_1}, \dots, \tilde{m}_{i_2}, \dots, m_n)$ where $\tilde{m}_{i_\ell}$ means that the argument $m_{i_\ell}$ is omitted. Therefore, for $k=1$ and $k \ge 3$, the only possibly non-zero term is the sum in the first line, which matches precisely the right hand side of (2.28). This proves Theorem 2.12 (where in practice, only even values of $k$ yield a non-zero result, as it should). In order to recover Theorems 2.3 and 2.8, it only remains to check that the sum in the first line of the right-hand side of (5.27) vanishes in the bipartite and quasi-bipartite cases. Note that if $m_i$ is an integer, $\pi_{r_i, s_i}^{(\epsilon_i)}(m_i)$ is non-zero only if $s_i$ and $1-\epsilon_i$ have the same parity, a property which, for $s_i \ge 0$ and $1-\epsilon_i \in \{0, 1\}$ implies that $s_i \ge 1-\epsilon_i$. If all the $m_i$'s are integers, the required constraint $\sum_{i=1}^n (1-\epsilon_i) = \sum_{i=1}^n s_i + 2$ in the definition of the set $I_n$ cannot be fulfilled, hence the sum vanishes. If exactly two of the $m_i$'s, say for $i=i_1$ and $i=i_2$, are
It remains to enumerate, for a given array $(w_j^{(i)})_{j=1,...,n}^{i=3,...,n}$ satisfying the consistency relations (72), and for a given two-type forest $f$ as above, the number of families of lattice paths $(L_i)_{3\le i \le n}$ so that $(f, (L_i)_{3\le i \le n})$ is an element of $\mathcal{F}_{(w_1^{(0)},...,w_{r_0+s_0+1}^{(0)})}(m_3, ..., m_n)$. Repeating the counting argument in the proof of Proposition 39, this number is simply equal to $\prod_{i=3}^n \pi_{r_i, s_i}^{(\epsilon_i)}(m_i)$. Combining with (75), and recalling the consistency relations (72), we obtain in the case $w_1^{(0)} = 1$

$$ \text{Card} \left( \mathcal{F}_{(w_1^{(0)}=1, w_2^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})}(m_3, \dots, m_n) \right) = \\ \sum_{\substack{\epsilon_3, \dots, \epsilon_n = 0, 1 \\ r_3, \dots, r_n, s_3, \dots, s_n \ge 0 \\ \sum_{i=3}^n \epsilon_i = r_0 + \sum_{i=3}^n r_i + 1 \\ \sum_{i=3}^n \bar{\epsilon}_i = s_0 + \sum_{i=3}^n s_i}} \left( r_0 + \sum_{i=3}^n r_i \right)! \left( \epsilon_3 s_0 + \sum_{i=3}^n \epsilon_i s_i \right) \left( s_0 + \sum_{i=3}^n s_i - 1 \right)! \prod_{i=3}^n \pi_{r_i, s_i}^{(\epsilon_i)}(m_i), \tag{76} $$

and in the case $w_1^{(0)} = 0$

$$ \text{Card} \left( \mathcal{F}_{(w_1^{(0)}=0, w_2^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})}(m_3, \dots, m_n) \right) = \\ \sum_{\substack{\epsilon_3, \dots, \epsilon_n = 0, 1 \\ r_3, \dots, r_n, s_3, \dots, s_n \ge 0 \\ \sum_{i=3}^n \epsilon_i = r_0 + \sum_{i=3}^n r_i \\ \sum_{i=3}^n \bar{\epsilon}_i = s_0 + \sum_{i=3}^n s_i + 1}} \left( s_0 + \sum_{i=3}^n s_i \right)! \left( \bar{\epsilon}_3 r_0 + \sum_{i=3}^n \bar{\epsilon}_i r_i \right) \left( r_0 + \sum_{i=3}^n r_i - 1 \right)! \prod_{i=3}^n \pi_{r_i, s_i}^{(\epsilon_i)}(m_i). \tag{77} $$

Note that these expressions do not depend on the precise sequence $(w_1^{(0)}, \dots, w_{r_0+s_0+1}^{(0)})$ but only on the values of $r_0 = \sum_{j=2}^{r_0+s_0+1} w_j^{(0)}$, of $s_0 = \sum_{j=2}^{r_0+s_0+1} \bar{w}_j^{(0)}$ and of $w_1^{(0)}$. This is expected since permuting the terms of the sequence $(w_j^{(0)})_{j=2,...,n}$ at fixed $r_0$ and $s_0$ simply amounts to changing the order of the last $r_0+s_0$ trees in the forest $f$, a harmless operation as far as enumeration is concerned. We may therefore use the counting formulas (58) and (59) for petal necklaces $m_{12}$ with fixed $r_0, s_0$ and with a fixed value $w_1^{(0)} = 1$ and $w_1^{(0)} = 0$ respectively. The expression (71) for $N_{0,n}(2m_1, \dots, 2m_n)$ is obtained by inserting (76) into (58) and (77) into (59), adding these two contributions and finally summing over $r_0$ and $s_0$, which trivially amounts to replacing in (76) and (77) each occurrence of $r_0$ by $r_1 + r_2$ and each occurrence of $s_0$ by $s_1 + s_2$, and summing over $r_1, r_2, s_1, s_2 \ge 0$. This ends the proof of Proposition 41. $\square$

Even though it is not apparent, the right-hand side of (71) turns out to be, as expected, symmetric upon permuting the $m_i$'s for $i$ in the whole set $\{1, \dots, n\}$. This property is shown in Appendix B where, after some algebraic manipulations, this quantity is given a manifestly symmetric form, see for instance (89) or (99). Using this result, we arrive at the following unified theorem encompassing all the main theorems of the paper (Theorems 4, 9 and 13):
half-integers, this same constraint imposes that $\epsilon_{i_1} = \epsilon_{i_2} = 0$ and $s_{i_1} = s_{i_2} = 0$ while $s_i = 1 - \epsilon_i$ for all $i \neq i_1, i_2$, hence $(\sum_{i=1}^n \epsilon_i s_i) = \sum_{i \neq i_1, i_2} (1 - \epsilon_i) \epsilon_i = 0$ and the first sum again vanishes. To summarize, for an even value of $k$, exactly one of the three lines in the right-hand side of (5.27) is not identically zero. Each line corresponds to one of the mutually exclusive situations $k \ge 4$ (first line), $k = 2$ (second line) and $k = 0$ (third line), corresponding to Theorem 2.12, 2.8 and 2.3 respectively.

# 6. Conclusion

In this paper, we have provided an explicit expression for the planar lattice count quasi-polynomial $N_{0,n}(2m_1, \dots, 2m_n)$, by extending the slice decomposition to the case of planar tight maps. Note that, in contrast with previous work such as [BG14, Bud22b], our approach is not based on generating functions but rather on direct bijective enumeration techniques.

We note that our most general formula, stated in Theorem 5.13, still involves a complicated sum over a no less complicated set $I_n$. As discussed in the Appendix B, there are in fact many identities that can be used to rewrite it, and it is not impossible that it admits a significantly simpler expression yet to be unveiled. A similar remark applies to our extension of Tutte's slicings formula given in Theorem 3.2.

We believe that the methodology of Sections 4 and 5 is quite robust and may be adapted to other map enumeration problems. A first problem one may think of is a model of maps with continuous edge lengths, whose set can be associated with a natural volume measure. The volumes of such measures have been considered by Kontsevich [Kon92], and, as noted by Norbury, they correspond to the homogeneous top degree part of the lattice count polynomials, see [Nor10, Theorem 3]. Combining with our Theorem 2.3, we obtain the following result (see Norbury's paper for the definition of $V_{0,n}$).

**Proposition 6.1.** *The genus zero volume polynomial $V_{0,n}$ reads explicitly*

$$V_{0,n}(b_1, \dots, b_n) = \frac{(n-3)!}{2^{2n-7}} \sum_{\substack{k_1, \dots, k_n \ge 0 \\ k_1 + \dots + k_n = n-3}} \left( \frac{b_1^{k_1} \dots b_n^{k_n}}{k_1! \dots k_n!} \right)^2. \qquad (6.1)$$

This amounts to the known expression for the genus zero intersection numbers, see for instance [LZ04, Proposition 4.6.10] and [Nor10, Corollary 1]. Still, a direct construction of maps with continuous edge lengths, using continuous paths and mimicking the above discrete paths encoding for slices, should also be possible and interesting, and we plan to investigate this question in future work. A second extension is motivated by the work of Budd [Bud22b] who generalized Norbury's results to the case of *irreducible maps*, i.e. maps with a girth constraint. The slice decomposition of these maps was discussed in [BG14] in the non-tight case, and we expect it to be adaptable to the tightness constraint. Finally, combining these two ideas, namely, passing to maps with continuous edge lengths and adding an irreducibility constraint, one obtains *irreducible metric maps* which have been considered by Budd [Bud22a], who showed that their volumes are related to the Weil-Petersson volumes of hyperbolic surfaces. We plan to investigate the slice decomposition of these maps, which might shed new light on these questions.
**Theorem 42.** For $n \ge 3$ and $m_1, m_2, \dots, m_n \in \mathbb{Z}_{\ge 0}/2$, not all equal to zero, the number $N_{0,n}(2m_1, 2m_2, \dots, 2m_n)$ of planar tight maps with $n$ boundaries labeled from 1 to $n$ with respective lengths $2m_1, 2m_2, \dots, 2m_n$ is given by the symmetric quasi-polynomial

$$ N_{0,n}(2m_1, 2m_2, \dots, 2m_n) = \sum_{\substack{\left(\begin{smallmatrix} \epsilon_1, \dots, \epsilon_n \\ r_1, \dots, r_n \\ s_1, \dots, s_n \end{smallmatrix}\right) \in I_n \\ (r_1+\dots+r_n=n-3}} \left(\sum_{i=1}^n r_i\right)! \left(\sum_{i=1}^n \epsilon_i s_i\right) \left(\sum_{i=1}^n s_i - 1\right)! \prod_{i=1}^n \pi_{r_i,s_i}^{(\epsilon_i)}(m_i) \\ + (n-3)! \sum_{\substack{\left(\begin{smallmatrix} r_1, \dots, r_n \end{smallmatrix}\right) \in \mathbb{Z}_{\ge 0}^n \\ r_1+\dots+r_n=n-3}} \sum_{1 \le j < \ell \le n} \pi_{r_j,0}^{(0)}(m_j) \pi_{r_\ell,0}^{(0)}(m_\ell) \prod_{\substack{i=1 \\ i \ne j, \ell}}^n \pi_{r_i,0}^{(1)}(m_i) \\ + (n-3)! \sum_{\substack{\left(\begin{smallmatrix} r_1, \dots, r_n \end{smallmatrix}\right) \in \mathbb{Z}_{\ge 0}^n \\ r_1+\dots+r_n=n-3}} \pi_{r_1,0}^{(-1)}(m_1) \prod_{i=2}^n \pi_{r_i,0}^{(1)}(m_i). \tag{78} $$

with $\pi_{r,s}^{(\epsilon)}(m)$ as in (26) and $I_n$ as in (29). Note that the sum in the third line is equal to the symmetric polynomial $(n-3)!p_{n-3}(m_1, \dots, m_n)$ when all $m_i$ are integers, and to zero otherwise, hence it is symmetric.

*Proof.* Since $N_{0,n}(2m_1, 2m_2, \dots, 2m_n)$ is (by definition) symmetric in $m_1, m_2, \dots, m_n$ and so is the right-hand side of (78), we may assume without loss of generality that $m_1 > 0$ and either $m_2 > 0$ or $m_2 = \dots = m_n = 0$. In the first case, Proposition 41 and the identification the right-hand sides of (71) and (78) proved in Proposition 48 of Appendix B allows to conclude. In the second case, both sides of the equation reduce to the univariate quasi-polynomial $(n-3)!\pi_{n-3,0}^{(-1)}(m_1)$, equal to $(n-3)!p_{n-3}(m_1)$ if $m_1$ is a positive integer, and to zero otherwise. $\square$

Let us now explain how to recover Theorems 4, 9 and 13 from Theorem 42. As in Remark 14, let us denote by $k$ the number of half-integers among $m_1, \dots, m_n$. From the general property that $\pi_{r_i,s_i}^{(\epsilon_i)}(m_i)$ is non-zero only if $m_i - \frac{s_i+1+\epsilon_i}{2} \in \mathbb{Z}$, we see that the sum in the third line of (78) is non-zero if and only if all the $m_i$'s are integers, i.e. the map is bipartite ($k=0$), in which case it reduces precisely to $(n-3)!p_{n-3}(m_1, \dots, m_n)$. Similarly, the sum in the second line is non-zero if and only exactly two of the $m_i$'s, say for $i=i_1$ and $i=i_2$ ($1 \le i_1 < i_2 \le n$), are half-integers, i.e. the map is quasi-bipartite ($k=2$), in which case it reduces to $(n-3)!\tilde{p}_{n-3}(m_{i_1}, m_{i_2}; m_1, \dots, \tilde{m}_{i_1}, \dots, \tilde{m}_{i_2}, \dots, m_n)$ where $\tilde{m}_{i_\ell}$ means that the argument $m_{i_\ell}$ is omitted. Therefore, for $k=1$ and $k \ge 3$, the only possibly non-zero term is the sum in the first line, which matches precisely the right hand side of (28). This proves Theorem 13 (where in practice, only even values of $k$ yield a non-zero result, as it should). In order to recover Theorems 4 and 9, it only remains to check that the sum in the first line of (78) vanishes in the bipartite and quasi-bipartite cases. Note that if $m_i$ is an integer, $\pi_{r_i,s_i}^{(\epsilon_i)}(m_i)$ is non-zero only if $s_i$ and $1-\epsilon_i$ have the same parity, a property which, for $s_i \ge 0$ and $1-\epsilon_i \in \{0, 1\}$ implies that $s_i \ge 1-\epsilon_i$. If all the $m_i$'s are integers, the required constraint $\sum_{i=1}^n (1-\epsilon_i) = \sum_{i=1}^n s_i + 2$ in the definition
In another direction, the bijective techniques presented in this paper should pave the road to the study of continuum limits of random planar tight maps, when the number of faces tends to infinity. We believe in particular that, as soon as the face degrees are well-behaved in a certain sense, the Gromov-Hausdorff limit of appropriately renormalized planar tight maps, seen as discrete metric spaces, should be given by the *Brownian sphere*. To this purpose, the recent approach by Marzouk [Mar18, Mar22], dealing with limits of planar non necessarily tight maps with prescribed face degrees, should be particularly relevant.

Note that our approach is currently restricted to the planar (i.e. genus 0) case. This is a current limitation of the slice decomposition. We hope however that this limitation will be challenged by further investigations. A first result in this vein is the bijective study of pairs of pants (planar maps with three boundaries) done in [BGM22], and the fact that general surfaces can be decomposed into pairs of pants gives some support to our hope.

**A. Enumeration of one- and two-type labeled plane forests**

This appendix lists the forest enumeration results that we need in this paper. We consider *plane forests* (sequences of rooted plane trees) which are *labeled* (distinct labels are assigned to the vertices). We start with the case of forests with one type of vertices.

**Proposition A.1.** Let $n, k_1, \dots, k_n$ be non-negative integers such that $k_1 + \dots + k_n < n$. Then, there are exactly $(n-1)!$ plane forests with $n$ vertices labeled $\{1, \dots, n\}$, such that vertex $i$ has $k_i$ children for all $i = 1, \dots, n$ and such that vertex 1 appears in the first tree. (Such forests consist necessarily of $k_0 := n - k_1 - \dots - k_n$ trees.)

*Proof*. The case of trees ($k_0 = 1$) is given explicitly in [BM14, Section 5, Equation (18)]. It implies the general case since, for $k_0 > 1$, there is a straightforward bijection between the set of forests at hand and the set of labeled plane trees such that vertex 1 has $k_1 + k_0 - 1$ children, the number of children of the other vertices being unmodified. $\square$

*Remark A.2.* Proposition A.1 can alternatively be proved directly by exhibiting a bijection between the set of forests at hand and the set of cyclic orders on $\{1, \dots, n\}$. Such a bijection is obtained by simply listing the vertex labels of a labeled plane forest in depth-first order, giving a linear, hence a cyclic, order on $\{1, \dots, n\}$. Conversely, from a cyclic order and the data of the $k_i$'s, we construct a conjugacy class of Łukasiewicz words—see e.g. [Sta99, Section 5.3]—whose letters are labeled and which contains exactly one word coding for a plane forest having vertex 1 in the first tree.

We now turn our attention to labeled plane forests with two types of vertices, say *A* and *B*. Our purpose is to enumerate such *two-type* forests in which, for every vertex, we prescribe not only its type but also the sequence formed by the types of all its children, read in the planarity order. Similar counting problems, for an arbitrary number of types, have been previously considered in the literature—see e.g. [BS13, BM14, CL16] and references therein—but since the general formulas are quite complicated we provide a self-contained derivation for two types. Our approach is closely related with that of Chottin [Cho81] who treated the case of two-type trees.
of the set $I_n$ cannot be fulfilled, hence the sum vanishes. If exactly two of the $m_i$'s, say for $i = i_1$ and $i = i_2$, are half-integers, this same constraint imposes that $\epsilon_{i_1} = \epsilon_{i_2} = 0$ and $s_{i_1} = s_{i_2} = 0$ while $s_i = 1 - \epsilon_i$ for all $i \neq i_1, i_2$, hence $(\sum_{i=1}^n \epsilon_i s_i) = \sum_{i \neq i_1, i_2} (1 - \epsilon_i) \epsilon_i = 0$ and the first sum again vanishes. To summarize, for an even value of $k$, exactly one of the three lines in the right-hand side of (78) is not identically zero. Each line corresponds to one of the mutually exclusive situations $k \ge 4$ (first line), $k = 2$ (second line) and $k = 0$ (third line), corresponding to Theorem 13, 9 and 4 respectively.

## 6. Conclusion

In this paper, we have provided an explicit expression for the planar lattice count quasi-polynomial $N_{0,n}(2m_1, \dots, 2m_n)$, by extending the slice decomposition to the case of planar tight maps. Note that, in contrast with previous work such as [BG14; Bud20b], our approach is not based on generating functions but rather on direct bijective enumeration techniques.

We note that our most general formula, stated in Theorem 42, still involves a complicated sum over a no less complicated set $I_n$. As discussed in the Appendix B, there are in fact many identities that can be used to rewrite it, and it is not impossible that it admits a significantly simpler expression yet to be unveiled. A similar remark applies to our extension of Tutte's slicings formula given in Theorem 19.

We believe that the methodology of Sections 4 and 5 is quite robust and may be adapted to other map enumeration problems. A first problem one may think of is a model of maps with continuous edge lengths, whose set can be associated with a natural volume measure. The volumes of such measures have been considered by Kontsevich [Kon92], and, as noted by Norbury, they correspond to the homogeneous top degree part of the lattice count polynomials, see [Nor10, Theorem 3]. Combining with our Theorem 4, we immediately obtain the following result (see Norbury's paper for the definition of $V_{0,n}$).

**Corollary 43.** *The genus zero volume polynomial $V_{0,n}$ reads explicitly*

$$V_{0,n}(b_1, \dots, b_n) = \frac{(n-3)!}{2^{2n-7}} \sum_{\substack{k_1, \dots, k_n \ge 0 \\ k_1 + \dots + k_n = n-3}} \left( \frac{b_1^{k_1} \cdots b_n^{k_n}}{k_1! \cdots k_n!} \right)^2. \qquad (79)$$

Still, a direct construction of maps with continuous edge lengths, using continuous paths and mimicking the above discrete paths encoding for slices, should also be possible and interesting, and we plan to investigate this question in future work. A second extension is motivated by the work of Budd [Bud20b] who generalized Norbury's results to the case of *irreducible maps*, i.e. maps with a girth constraint. The slice decomposition of these maps was discussed in [BG14] in the non-tight case, and we expect it to be adaptable to the tightness constraint. Finally, combining these two ideas, namely, passing to maps with continuous edge lengths and adding an irreducibility constraint, one obtains *irreducible metric maps* which have been considered by Budd [Bud20a], who showed that their volumes are related to the Weil-Petersson volumes of hyperbolic surfaces. We plan
Handling forests with several components however involves an extra difficulty, which we circumvent by specializing the general multitype approach of Bacher and Schaeffer [BS13]. Note that the latter two references consider plane forests which are unlabeled, but adding vertex labels does not fundamentally change the problem since plane forests have no symmetries.

To state our result we need some definitions and notation. Let us consider a two-type plane forest whose vertices are labeled {$1, \dots, n$}. For every vertex $i = 1, \dots, n$, we denote by $k_i$ its number of children, and we let $w^{(i)} = (w_0^{(i)}, w_1^{(i)}, \dots, w_{k_i}^{(i)}) \in \{A, B\}^{k_i+1}$ be the sequence such that $w_0^{(i)}$ is the type of $i$ and such that, for every $j = 1, \dots, k_i$, $w_j^{(i)}$ is the type of the $j$-th child of $i$ in the planarity order. We also define a sequence $w^{(0)} = (w_0^{(0)}, w_1^{(0)}, \dots, w_{k_0}^{(0)})$ where $k_0$ is the number of trees of the forest, $w_0^{(0)}$ is a third type denoted $O$ and, for $j = 1, \dots, k_0$, $w_j^{(0)}$ is the type of the $j$-th root, i.e. of the root vertex of the $j$-th tree of the forest. In this sense, $O$ can be seen as the label of a super-root of type $O$, which is the parent of all the roots (which have the usual types $A$ or $B$).

The collection $\mathbf{w} = (w^{(0)}, w^{(1)}, \dots, w^{(n)})$ is called the type array of the forest. Denoting by $[\cdot]$ the Iverson bracket ($[P]$ is equal to 1 if $P$ is true, and to 0 otherwise), we have necessarily

$$ \sum_{i=1}^{n} [w_{0}^{(i)} = A] = \sum_{i=0}^{n} \sum_{j=1}^{k_i} [w_{j}^{(i)} = A] := a \qquad (A.1) $$

as seen by expressing in two different ways the number $a$ of type $A$ vertices. Similarly, the number $b$ of type $B$ vertices is given by

$$ \sum_{i=1}^{n} [w_{0}^{(i)} = B] = \sum_{i=0}^{n} \sum_{j=1}^{k_i} [w_{j}^{(i)} = B] := b. \qquad (A.2) $$

Note that, by adding these two equations, we obtain the relation $n = k_0 + k_1 + \dots + k_n$ already seen in the context of one-type forests in Proposition A.1. An array $\mathbf{w} = (w_j^{(i)})_{j=0,\dots,k_i}^{i=0,\dots,n}$ with $w_0^{(0)} = O$ and $w_j^{(i)} \in \{A, B\}$ for $i, j$ not both zero is said consistent if it satisfies (A.1) and (A.2).

**Proposition A.3.** Let $\mathbf{w}$ be a consistent array, and let $a$ and $b$ be as defined in (A.1) and (A.2), respectively. Define furthermore the integers

$$ a^O := \sum_{j=1}^{k_0} [w_j^{(0)} = A], \qquad a^B := \sum_{i=1}^{n} \sum_{j=1}^{k_i} [w_0^{(i)} = B] [w_j^{(i)} = A], \qquad (A.3) $$

$$ b^O := \sum_{j=1}^{k_0} [w_j^{(0)} = B], \qquad b^A := \sum_{i=1}^{n} \sum_{j=1}^{k_i} [w_0^{(i)} = A] [w_j^{(i)} = B], \qquad (A.4) $$

namely $a^O$ and $b^O$ correspond to the number of type $A$ and type $B$ roots, respectively, while $a^B$ is the number of type $A$ vertices with a type $B$ parent, and vice versa for $b^A$.

Then, we have the following enumerative formulas.
to investigate the slice decomposition of these maps, which might shed new light on these
questions.

In another direction, the bijective techniques presented in this paper should pave the
road to the study of continuum limits of random planar tight maps, when the number of
faces tends to infinity. We believe in particular that, as soon as the face degrees are well-
behaved in a certain sense, the Gromov-Hausdorff limit of appropriately renormalized
planar tight maps, seen as discrete metric spaces, should be given by the *Brownian sphere*.
To this purpose, the recent approach by Marzouk [Mar18; Mar21], dealing with limits
of planar non necessarily tight maps with prescribed face degrees, should be particularly
relevant.

Note that our approach is currently restricted to the planar (i.e. genus 0) case. This
is a current limitation of the slice decomposition. We hope however that this limitation
will be challenged by further investigations. A first result in this vein is the bijective
study of pairs of pants (planar maps with three boundaries) done in [BGM21], and the
fact that general surfaces can be decomposed into pairs of pants gives some support to
our hope.

A. Enumeration of one- and two-type labeled plane forests

This appendix lists the forest enumeration results that we need in this paper. We consider
*plane forests* (sequences of rooted plane trees) which are *labeled* (distinct labels are
assigned to the vertices). We start with the case of forests with one type of vertices.

**Proposition 44.** Let $n, k_1, \dots, k_n$ be non-negative integers such that $k_1 + \dots + k_n < n$. Then, there are exactly $(n-1)!$ plane forests with $n$ vertices labeled $\{1, \dots, n\}$, such that vertex $i$ has $k_i$ children for all $i = 1, \dots, n$ and such that vertex 1 appears in the first tree. (Such forests consist necessarily of $k_0 := n - k_1 - \dots - k_n$ trees.)

*Proof*. The case of trees ($k_0 = 1$) is given explicitly in [BM14, Section 5, Equation (18)]. It implies the general case since, for $k_0 > 1$, there is a straightforward bijection between the set of forests at hand and the set of labeled plane trees such that vertex 1 has $k_1 + k_0 - 1$ children, the number of children of the other vertices being unmodified. $\square$

*Remark 45.* Proposition 44 can alternatively be proved directly by exhibiting a bijection between the set of forests at hand and the set of cyclic orders on $\{1, \dots, n\}$. Such a bijection is obtained by simply listing the vertex labels of a labeled plane forest in depth-first order, giving a linear, hence a cyclic, order on $\{1, \dots, n\}$. Conversely, from a cyclic order and the data of the $k_i$'s, we construct a conjugacy class of Łukasiewicz words—see e.g. [Sta99, Section 5.3]—whose letters are labeled and which contains exactly one word coding for a plane forest having vertex 1 in the first tree.

We now turn our attention to labeled plane forests with two types of vertices, say *A* and *B*. Our purpose is to enumerate such *two-type* forests in which, for every vertex, we prescribe not only its type but also the sequence formed by the types of all its children, read in the planarity order. Similar counting problems, for an arbitrary number of types,
• (General enumeration) The number of two-type labeled plane forests of type array w—
thus containing a type A and b type B vertices—is equal to

$$
\left\{
\begin{array}{ll}
(a^O b^O + a^O b^A + a^B b^O) (a-1)!(b-1)! & \text{if } a > 0 \text{ and } b > 0, \\
a^O (a-1)! & \text{if } a > 0 \text{ and } b = 0, \\
b^O (b-1)! & \text{if } a = 0 \text{ and } b > 0, \\
1 & \text{if } a = 0 \text{ and } b = 0.
\end{array}
\right.
\qquad (A.5)
$$

• (Constrained enumeration) Assume that, say, $a > 1$ and $w_0^{(1)} = A$, i.e. vertex 1 has type A. Then, the number of two-type labeled plane forests of type array $w$ such that vertex 1 is in the first tree is equal to

$$
\left\{
\begin{array}{ll}
(b^O + b^A)(a-1)!(b-1)! & \text{if } b > 0 \text{ and } w_1^{(0)} = A, \\
a^B(a-1)!(b-1)! & \text{if } b > 0 \text{ and } w_1^{(0)} = B, \\
(a-1)! & \text{if } b = 0.
\end{array}
\right.
\qquad (A.6)
$$

(The first and second cases correspond to a first root of type A and B, respectively.)

Proof. Let us first note that the cases where *a* or *b* vanish follow immediately from Proposition A.1 (for the general enumeration, we perform a circular permutation of the trees to lift the constraint that vertex 1 is in the first tree, giving the extra factor *a*^0 or *b*^0). Hence, we assume from now on that *a* and *b* are both positive.

We claim that the general enumeration formula follows from the constrained one. Indeed, we
may partition the set of forests of type array *w* according to the index *j* = 1, ..., *k*₀ of the tree
containing vertex 1. Upon doing a circular permutation of the trees, we deduce from (A.6) that,
for *w*₀⁽¹⁾ = *A*, the number of forests with a given value of *j* is equal to (*b*Ο + *b*Α)(*a* − 1)!(*b* − 1)!
if *w*ⱼ⁽¹⁾ = *A*, and to *a*Β(*a* − 1)!(*b* − 1)! if *w*ⱼ⁽¹⁾ = *B*. As the first (resp. second) case occurs for
*a*Ο (resp. *b*Ο) values of *j*, summing over *j* gives the first line of (A.5). The case *w*₀⁽¹⁾ = *B* is
deduced by exchanging the roles of *A* and *B*.

It remains to prove the constrained enumeration formula. We will do so by giving an algorithm to construct any forest of type array *w*, where it will be manifest that the number of possibilities is given by (A.6). The algorithm is easier to visualize if, instead of working with labels 0, 1, ..., *n*, we relabel the vertices as *O*, *A*₁, ..., *A*ₐ, *B*₁, ..., *B*ᵢ to make their type apparent (as the type array *w* is fixed, this may be done by fixing a bijection between {*0*, 1, ..., *n*} and {*O*, *A*₁, ..., *A*ₐ, *B*₁, ..., *B*ᵢ} hence does not change the counting problem). The general idea, illustrated on Figure A.1, is to proceed in several stages, by first “assembling” the type *B* vertices together, before dealing with the types *A* and *O* vertices. Let us describe the different stages of the algorithm in detail.

I. We start with isolated vertices *O*, *A*₁, ..., *A*ₐ, *B*₁, ..., *B*ᵢ, to which we attach sequences
of *dangling A-edges* and *B-edges*: these dangling edges will be connected later to type
*A* and *B* vertices, respectively. The type array *w* tells us precisely the sequence which
we have to attach to each vertex. We define a total order on the dangling edges by listing
have been previously considered in the literature—see e.g. [BS13; BM14; CL16] and references therein—but since the general formulas are quite complicated we provide a self-contained derivation for two types. Our approach is closely related with that of Chottin [Cho81] who treated the case of two-type trees. Handling forests with several components however involves an extra difficulty, which we circumvent by specializing the general multitype approach of Bacher and Schaeffer [BS13]. Note that the latter two references consider plane forests which are unlabeled, but adding vertex labels does not fundamentally change the problem since plane forests have no symmetries.

To state our result we need some definitions and notation. Let us consider a two-type plane forest whose vertices are labeled {$1, \dots, n$}. For every vertex $i = 1, \dots, n$, we denote by $k_i$ its number of children, and we let $w^{(i)} = (w_0^{(i)}, w_1^{(i)}, \dots, w_{k_i}^{(i)}) \in \{A, B\}^{k_i+1}$ be the sequence such that $w_0^{(i)}$ is the type of $i$ and such that, for every $j = 1, \dots, k_i$, $w_j^{(i)}$ is the type of the $j$-th child of $i$ in the planarity order. We also define a sequence $w^{(0)} = (w_0^{(0)}, w_1^{(0)}, \dots, w_{k_0}^{(0)})$ where $k_0$ is the number of trees of the forest, $w_0^{(0)}$ is a third type denoted $O$ and, for $j = 1, \dots, k_0$, $w_j^{(0)}$ is the type of the $j$-th root, i.e. of the root vertex of the $j$-th tree of the forest. In this sense, $0$ can be seen as the label of a super-root of type $O$, which is the parent of all the roots (which have the usual types $A$ or $B$).

The collection $\mathbf{w} = (w^{(0)}, w^{(1)}, \dots, w^{(n)})$ is called the type array of the forest. Denoting by $[\cdot]$ the Iverson bracket ($[P]$ is equal to 1 if $P$ is true, and to 0 otherwise), we have necessarily

$$ \sum_{i=1}^{n} [w_{0}^{(i)} = A] = \sum_{i=0}^{n} \sum_{j=1}^{k_i} [w_{j}^{(i)} = A] := a \qquad (80) $$

as seen by expressing in two different ways the number $a$ of type $A$ vertices. Similarly, the number $b$ of type $B$ vertices is given by

$$ \sum_{i=1}^{n} [w_{0}^{(i)} = B] = \sum_{i=0}^{n} \sum_{j=1}^{k_i} [w_{j}^{(i)} = B] := b. \qquad (81) $$

Note that, by adding these two equations, we obtain the relation $n = k_0 + k_1 + \dots + k_n$ already seen in the context of one-type forests in Proposition 44. An array $\mathbf{w} = (w_j^{(i)})_{i=0, \dots, n}^{i=0, \dots, n}$ with $w_0^{(0)} = O$ and $w_j^{(i)} \in \{A, B\}$ for $i, j$ not both zero is said consistent if it satisfies (80) and (81).

**Proposition 46.** Let $\mathbf{w}$ be a consistent array, and let $a$ and $b$ be as defined in (80) and (81), respectively. Define furthermore the integers

$$ a^O := \sum_{j=1}^{k_0} [w_j^{(0)} = A], \qquad a^B := \sum_{i=1}^{n} \sum_{j=1}^{k_i} [w_0^{(i)} = B] [w_j^{(i)} = A], \qquad (82) $$

$$ b^O := \sum_{j=1}^{k_0} [w_j^{(0)} = B], \qquad b^A := \sum_{i=1}^{n} \sum_{j=1}^{k_i} [w_0^{(i)} = A] [w_j^{(i)} = B], \qquad (83) $$
<figure><img src="image_20.png" /><figcaption>Figure A.1: Construction of a two-type forest in the case $w_0^{(1)} = B$. See the main text for a precise description of the stages I-V. We display the type A (resp. B) vertices and dangling A-edges (resp. B-edges) with squares (resp. circles). The distinguished dangling edges are indicated with arrows.</figcaption></figure>

those incident to $O$, then those incident to $A_1$, etc. We distinguish the *first* edge incident to $O$: it has type $w_0^{(1)}$. We then distinguish another dangling edge of the opposite type:

- if $w_0^{(1)} = A$, then we distinguish a *B-edge* incident either to $O$ or to a type A vertex: there are $b^O + b^A$ possible choices,

- if $w_0^{(1)} = B$, then we distinguish an *A-edge* incident to a type B vertex: there are $a^B$ possible choices.

II. We form a plane forest with the vertices $B_1, \dots, B_b$, by attaching them together via their incident B-edges. By Proposition A.1, there are $(b-1)!$ ways to do so (with $B_1$ in the first tree). A simple computation shows that the resulting forest is made of $b^O + b^A$ trees. If $w_0^{(1)} = B$, we permute the trees circularly so that the tree containing the distinguished A-edge comes first.

III. We attach the roots of the forest constructed at stage II to the *B-edges* (in number $b^O + b^A$) dangling from $O, A_1, \dots, A_a$. Precisely, we attach the root of the first tree to the distinguished *B-edge*, and we then proceed circularly using the order on dangling edges defined at stage I. Note that all the *B-edges* have now been matched to type *B* vertices. We end up with a sequence of *supernodes*, which are trees with roots $O, A_1, \dots, A_a$, subtrees made of type *B* vertices, and dangling *A-edges*. Observe that the distinguished *A-edge* always belongs to the supernode with root *O* (precisely, the first edge incident to *O* is
namely $a^O$ and $b^O$ correspond to the number of type A and type B roots, respectively, while $a^B$ is the number of type A vertices with a type B parent, and vice versa for $b^A$. Then, we have the following enumerative formulas.

• (General enumeration) The number of two-type labeled plane forests of type array $w$—thus containing a type A and b type B vertices—is equal to

$$ \left\{ \begin{array}{ll} (a^O b^O + a^O b^A + a^B b^O)(a-1)!(b-1)! & \text{if } a > 0 \text{ and } b > 0, \\ a^O(a-1)! & \text{if } a > 0 \text{ and } b = 0, \\ b^O(b-1)! & \text{if } a = 0 \text{ and } b > 0, \\ 1 & \text{if } a = 0 \text{ and } b = 0. \end{array} \right. \qquad (84) $$

• (Constrained enumeration) Assume that, say, $a > 1$ and $w_0^{(1)} = A$, i.e. vertex 1 has type A. Then, the number of two-type labeled plane forests of type array $w$ such that vertex 1 is in the first tree is equal to

$$ \left\{ \begin{array}{ll} (b^O + b^A)(a-1)!(b-1)! & \text{if } b > 0 \text{ and } w_1^{(0)} = A, \\ a^B(a-1)!(b-1)! & \text{if } b > 0 \text{ and } w_1^{(0)} = B, \\ (a-1)! & \text{if } b = 0. \end{array} \right. \qquad (85) $$

(The first and second cases correspond to a first root of type A and B, respectively.)

*Proof.* Let us first note that the cases where $a$ or $b$ vanish follow immediately from Proposition 44 (for the general enumeration, we perform a circular permutation of the trees to lift the constraint that vertex 1 is in the first tree, giving the extra factor $a^O$ or $b^O$). Hence, we assume from now on that $a$ and $b$ are both positive.

We claim that the general enumeration formula follows from the constrained one. Indeed, we may partition the set of forests of type array $w$ according to the index $j = 1, \dots, k_0$ of the tree containing vertex 1. Upon doing a circular permutation of the trees, we deduce from (85) that, for $w_0^{(1)} = A$, the number of forests with a given value of $j$ is equal to $(b^O + b^A)(a-1)!(b-1)!$ if $w_j^{(0)} = A$, and to $a^B(a-1)!(b-1)!$ if $w_j^{(0)} = B$. As the first (resp. second) case occurs for $a^O$ (resp. $b^O$) values of $j$, summing over $j$ gives the first line of (84). The case $w_0^{(1)} = B$ is deduced by exchanging the roles of A and B.

It remains to prove the constrained enumeration formula. We will do so by giving an algorithm to construct any forest of type array $w$, where it will be manifest that the number of possibilities is given by (85). The algorithm is easier to visualize if, instead of working with labels $0, 1, \dots, n$, we relabel the vertices as $O, A_1, \dots, A_a, B_1, \dots, B_b$ to make their type apparent (as the type array $w$ is fixed, this may be done by fixing a bijection between $\{0, 1, \dots, n\}$ and $\{O, A_1, \dots, A_a, B_1, \dots, B_b\}$ hence does not change the counting problem). The general idea, illustrated on Figure 18, is to proceed in several stages, by first “assembling” the type B vertices together, before dealing with the types A and O vertices. Let us describe the different stages of the algorithm in detail.
either the distinguished A-edge when $w_0^{(1)} = A$, or leads to a subtree which contains it when $w_0^{(1)} = B$).

IV. Viewing the supernodes as compound vertices, we form a plane forest with $a^O$ trees by assembling the supernodes with roots $A_1, \dots, A_a$ via their dangling A-edges. Note that the number of “children” of a supernode is prescribed by its number of dangling A-edges. By Proposition A.1, there are $(a-1)!$ ways to do so (with $A_1$ in the first tree).

V. We then complete the construction by attaching these trees to the A-edges of the supernode with root $O$, starting with the first tree which we attach to the distinguished A-edge, and then proceeding circularly using the order of stage I. All dangling edges have now been matched, and vertex 1 is in the first tree by construction.

It is plain from stage I that we obtain a forest of type array $w$. Furthermore, each such forest is obtained in precisely one way, as we may check that it is obtained for a unique choice of the second distinguished dangling edge at stage I$^{12}$ and of the one-type forests at stages II and IV. □

*Remark A.4.* The factor $a^O b^O + a^O b^A + a^B b^O$ in (A.5) corresponds to a sum over the three Cayley trees on the set $\{O, A, B\}$. For $k$ types of vertices there would be as many terms as Cayley trees on a set with $k+1$ elements [BS13].

## B. Symmetrizing the planar lattice count quasi-polynomials

Let us fix an integer $n \ge 3$ and denote by $\Pi(m_1, \dots, m_n)$ the right-hand side of (5.20). It is manifest that it is a quasi-polynomial in $2m_1, \dots, 2m_n$ of degree $n-3$, since $\pi_{r,s}^{(\epsilon)}(m)$ is a univariate quasi-polynomial in $2m$ of degree $r+s$. The purpose of this appendix is to show that $\Pi(m_1, \dots, m_n)$ is in fact symmetric in $m_1, \dots, m_n$, which we will do by rewriting it in a manifestly symmetric form. Note that the expression (5.20) displays a symmetry in $m_4, \dots, m_n$ only. Here we assume only that $m_i \in \mathbb{Z}/2, i = 1, \dots, n$ without further restriction.

It is useful to introduce compact notations for the high-dimensional sums appearing in (5.20). Let $I$ be the finite subset of $\{0, 1\}^n \times \mathbb{Z}_{\ge 0}^n \times \mathbb{Z}_{\ge 0}^n$ defined by

$$I := \left\{ \begin{pmatrix} \epsilon_1, \dots, \epsilon_n \\ r_1, \dots, r_n \\ s_1, \dots, s_n \end{pmatrix} : \begin{array}{l} \displaystyle \sum_{i=1}^{n} \epsilon_i = \sum_{i=1}^{n} r_i + 1 \\ \displaystyle \sum_{i=1}^{n} (1 - \epsilon_i) = \sum_{i=1}^{n} s_i + 2 \end{array} \right\}. \qquad (B.1)$$

Given a tuple in $I$, we set

$$r := \sum_{i=1}^{n} r_i, \qquad s := \sum_{i=1}^{n} s_i, \qquad (B.2)$$

<footnote>$^{12}$Precisely, if $w_1^{(0)} = A$ then the distinguished B-edge corresponds to the edge closest to $B_1$ attached to a parent not of type B, and if $w_1^{(0)} = B$ then the distinguished A-edge corresponds to the parent edge of the oldest type A ancestor of $A_1$.</footnote>
<figure><img src="image_19.png" /><figcaption>Figure 18: Construction of a two-type forest in the case $w_0^{(1)} = B$. See the main text for a precise description of the stages I-V. We display the type A (resp. B) vertices and dangling A-edges (resp. B-edges) with squares (resp. circles). The distinguished dangling edges are indicated with arrows.</figcaption></figure>

I. We start with isolated vertices $O, A_1, \dots, A_a, B_1, \dots, B_b$, to which we attach sequences of dangling A-edges and B-edges: these dangling edges will be connected later to type A and B vertices, respectively. The type array $w$ tells us precisely the sequence which we have to attach to each vertex. We define a total order on the dangling edges by listing those incident to $O$, then those incident to $A_1$, etc. We distinguish the first edge incident to $O$: it has type $w_0^{(1)}$. We then distinguish another dangling edge of the opposite type:

- if $w_0^{(1)} = A$, then we distinguish a B-edge incident either to $O$ or to a type A vertex: there are $b^O + b^A$ possible choices,

- if $w_0^{(1)} = B$, then we distinguish an A-edge incident to a type B vertex: there are $a^B$ possible choices.

II. We form a plane forest with the vertices $B_1, \dots, B_b$, by attaching them together via their incident B-edges. By Proposition 44, there are $(b-1)!$ ways to do so (with $B_1$ in the first tree). A simple computation shows that the resulting forest is made of $b^O + b^A$ trees. If $w_0^{(1)} = B$, we permute the trees circularly so that the tree containing the distinguished A-edge comes first.

III. We attach the roots of the forest constructed at stage II to the B-edges (in number $b^O + b^A$) dangling from $O, A_1, \dots, A_a$. Precisely, we attach the root of the first tree to the distinguished B-edge, and we then proceed circularly using the order
and note that $r + s = n - 3$ by the definition of $I$. Let $I_{r \ge 1}$, $I_{r=0}$, $I_{s \ge 1}$, and $I_{s=0}$, be the subsets of $I$ consisting of tuples such that $r \ge 1$, $r = 0$, $s \ge 1$, and $s = 0$, respectively. Note that the set $I_n$ introduced in (2.29) corresponds in our present notations to $I_{s \ge 1}$. Note also that $I_{r=0}$ consists of tuples such that $r_i = 0$ for all $i$, exactly one $\epsilon_i$ is equal to 1, and $s_1 + \cdots + s_n = s = n - 3$. Similarly, $I_{s=0}$ consists of tuples such that $s_i = 0$ for all $i$, exactly two $\epsilon_i$ are equal to 0, and $r_1 + \cdots + r_n = r = n - 3$.

For $e_1, e_2 \in \{0, 1\}$, we let $I^{(e_1,e_2)}$ be the subset of $I$ consisting of tuples such that $\epsilon_1 = e_1$ and $\epsilon_2 = e_2$. We also allow for the value $e_1 = -1$, which means that we consider tuples such that $\epsilon_1 = -1$, keeping the sum condition in (B.1) unchanged. The notations $I_{r \ge 1}^{(e_1,e_2)}$, etc, should hopefully be self-explanatory. Note that $I_{s=0}^{(-1,1)}$ consists of tuples such that $s_i = 0$ for all $i$, $\epsilon_1 = -1$ and $\epsilon_i = 1$ for all $i \ge 2$, and $r_1 + \cdots + r_n = r = n - 3$. We finally use, for $\epsilon \in \{0, 1\}$, the shorthand notation $\bar{\epsilon} := 1 - \epsilon$ (which we shall never use for $\epsilon = -1$).

Armed with all these notations, we can rewrite $\Pi(m_1, \dots, m_n)^{13}$ as

$$ \Pi(m_1, \dots, m_n) = \sum_{I_{s \ge 1}^{(-1,1)} \cup I_{s \ge 1}^{(0,0)}} r!(s-1)! \left( \epsilon_3(s_1+s_2) + \sum_{j=3}^{n} \epsilon_j s_j \right) \prod_{i=1}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i) \\ + \sum_{I_{r \ge 1}^{(0,1)} \cup I_{r \ge 1}^{(1,0)}} (r-1)!s! \left( \bar{\epsilon}_3(r_1+r_2) + \sum_{j=3}^{n} \bar{\epsilon}_j r_j \right) \prod_{i=1}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i) \tag{B.3} \\ + \sum_{I_{s=0}^{(-1,1)} \cup I_{s=0}^{(0,0)} \cup I_{r=0}^{(0,1)} \cup I_{r=0}^{(1,0)}} r!s! \prod_{i=1}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i). $$

Here, the first (resp. second) sum corresponds to the first (resp. second) sum in (5.20) when at least one of the $s_i$'s (resp. one of the $r_i$'s) is non zero, hence when $s \ge 1$ (resp. $r \ge 1$); the third sum accounts for the conventional values in (5.20): $(\epsilon_3(s_1 + s_2) + \sum_{i=3}^n \epsilon_i s_i)(\sum_{i=1}^n s_i - 1)! \to 1 = s!$ when all the $s_i$'s are zero (or equivalently $s = 0$) and $(\bar{\epsilon}_3(r_1 + r_2) + \sum_{i=3}^n \bar{\epsilon}_i r_i)(\sum_{i=1}^n r_i - 1)! \to 1 = r!$ when all the $r_i$'s are zero (or equivalently $r = 0$).

The main result of this appendix is:

**Proposition B.1.** The quasi-polynomial $\Pi(m_1, \dots, m_n)$ is symmetric in $m_1, \dots, m_n$ and admits the expression

$$ \Pi(m_1, \dots, m_n) = \sum_{I_{s \ge 1}} r!(s-1)! \left( \sum_{j=1}^{n} \epsilon_j s_j \right) \prod_{i=1}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i) + \sum_{I_{s=0}^{(-1,1)} \cup I_{s=0}} r! s! \prod_{i=1}^{n} \pi_{r_i, s_i}^{(\epsilon_i)}(m_i). \tag{B.4} $$

Note that the sum over $I_{s=0}^{(-1,1)}$ is equal to $(n-3)! \sum_{r_1+\cdots+r_n=n-3} \pi_{r_1,0}^{(-1)}(m_1) \prod_{i=2}^{n} \pi_{r_i,0}^{(1)}(m_i)$ which is

<footnote n="13">¹³Here we use the shorthand notation $\sum_J (\cdot)$ for $\sum_{\binom{\epsilon_1, \dots, \epsilon_n}{r_1, \dots, r_n} \in J} (\cdot)$.</footnote>