2. Surface triangulations, flip-graphs, and contractions

Throughout the paper, we consider a finite-type orientable surface $\Sigma$ , possibly with boundary curves and with a non-empty set P of marked points. In order to ensure that P can be the vertex set of a triangulation of $\Sigma$ , each boundary curve must contain at least one marked point. Marked points that lie in the interior of the surface we call punctures.

An arc is a subset of $\Sigma$ that is obtained as the continuous image of a closed interval, which is an embedding on its interior, and with the endpoints of the interval sent to P. We are interested in arcs up to the homotopies that preserve P and we only consider arcs which are not homotopic to a point (which can only occur if the two endpoints are the same). Two arcs are said to be disjoint if they are disjoint in their interiors and a triangulation of $\Sigma$ is a set of pairwise disjoint arcs that is maximal with respect to inclusion.

We call the number of arcs in a triangulation (that does not depend on the choice of the triangulation) the arc complexity of $\Sigma$ and denote it by $\kappa(\Sigma)$ .

Fig. 2. The flip-graph of a 3-punctured sphere.

We also consider the unoriented graph $\mathcal{F}(\Sigma)$ whose vertices are the triangulations of $\Sigma$ and whose edges link two triangulations that differ by a single arc. In other words, two triangulations T and T ^′ of $\Sigma$ are the endpoints of an edge of $\mathcal{F}(\Sigma)$ when T ^′ can be obtained from T by replacing an arc with a different one. This operation is called a flip and the graph $\mathcal{F}(\Sigma)$ the flip-graph of $\Sigma$ . We will think of $\mathcal{F}(\Sigma)$ as a metric space whose elements are the triangulations of $\Sigma$ where the distance d(T, T ^′) between two triangulations T and T ^′ in $\mathcal{F}(\Sigma)$ is the minimal number of flips needed to transform T into T ^′.

A minimal length path between two triangulations is called geodesic. For any two triangulations T and T ^′ of $\Sigma$ , we denote by $\#(T,T')$ the number of geodesic paths between T and T ^′ in $\mathcal{F}(\Sigma)$ . For any k in $\mathbb{N}$ , we consider the quantity

\begin{align*}\Delta_k(\Sigma)=\max \Bigl\{ \#(T,T')\;:\;(T,T')\in\mathcal{F}(\Sigma)^2,\,d(T,T')=k\Bigr\}\mbox{.}\end{align*}

Note that $\Delta_k(\Sigma)$ is always well-defined and finite. This is because the number of possible flips in a triangulation of $\Sigma$ is bounded by $\kappa(\Sigma)$ and hence the number of possible sequences of k flips that start at a given triangulation is always at most $\kappa(\Sigma)^k$ . Since a flip in a geodesic path cannot be immediately followed by the inverse flip, we get the following.

Throughout the paper, p will denote the number of punctures of $\Sigma$ , g its genus, and b the number of its boundary components. It will be important to remember that p only counts punctures (marked points in the interior of $\Sigma$ ) but ignores the number of marked points of P that lie on the boundary curves of $\Sigma$ . If p and b are simultaneously equal to 0, then $\mathcal{F}(\Sigma)$ is empty and for this reason, we will assume that their sum is positive. If b and g are both equal to 0, then $\Sigma$ is a sphere. If in addition, p is equal to 1 or 2, there are not enough marked points to build a triangulation of $\Sigma$ (at best, one can decompose $\Sigma$ into bigons when p is equal to 2). Therefore, we assume that if b and g are both equal to 0, then p is at least 3.

Let us also recall the three cases when $\mathcal{F}(\Sigma)$ is a finite graph. These three cases are marked with a triangle in Figure 1. The first case is the 3-times punctured sphere (when b and g are equal to 0 and p to 3). In that case $\mathcal{F}(\Sigma)$ is finite and the full graph is depicted in Figure 2. The second case is the disk (when g and p are equal to 0 and b is equal to 1). In that case, $\mathcal{F}(\Sigma)$ is empty when there are less than 3 marked points on the boundary of the disk and non-empty otherwise. This particular flip-graph is a graph made up of the vertices and edges of a polytope called the associahedron [ Reference Lee6, Reference Stasheff15–Reference Tamari17 ], whose number of vertices is counted by the Catalan numbers. The third case where $\mathcal{F}(\Sigma)$ is finite is the once-punctured disk (that is when b and p are both equal to 1 and g to 0) [ Reference Parlier and Pournin10 ]. In this case $\mathcal{F}(\Sigma)$ is never empty and its size depends on the number of marked points on the boundary curve.

We distinguish two kinds of arcs in a triangulation T of $\Sigma$ : if an arc in T can be homotoped to the boundary of $\Sigma$ then we call that arc a boundary arc of T. Any arc in T that is not a boundary arc will be called an interior arc. Observe that all the triangulations of $\Sigma$ have the same set of boundary arcs. For this reason, we will also refer to these arcs as the boundary arcs of $\Sigma$ . When a boundary component $\gamma$ of $\Sigma$ contains a single point p from P, then p splits $\gamma$ into a single boundary arc $\alpha$ of $\Sigma$ . In that case, $\alpha$ is twice incident to p and we call this boundary arc a boundary loop. In the remainder of the paper, we denote the number of boundary arcs of $\Sigma$ by n. As each boundary component of $\Sigma$ contains at least one point from P, it gives rise to at least one boundary arc and therefore n is not less than b. Note that when n is equal to b, all the boundary arcs of $\Sigma$ are boundary loops.

Let us now describe an operation on $\Sigma$ and its triangulations that will be useful throughout the paper. Consider a boundary arc $\alpha$ of $\Sigma$ and assume that $\alpha$ is not a boundary loop. In that case, we can build a new surface $\Sigma{{\backslash\!\backslash}}\alpha$ by removing one of the vertices of $\alpha$ from the marked points of $\Sigma$ . In other words, $\Sigma{{\backslash\!\backslash}}\alpha$ is identical to $\Sigma$ except that its set of marked points is $P\mathord{\setminus}\{a\}$ instead of P, where a is a vertex of $\alpha$ . Note that the two surfaces obtained from this construction using, for a, one or the other vertex of $\alpha$ are homeomorphic. For this reason, we will identify $\Sigma{{\backslash\!\backslash}}\alpha$ with any of these surfaces.

Now consider a triangulation T of $\Sigma$ and denote by t the triangle of T incident to $\alpha$ . Further denote by a and a ^′ the vertices of $\alpha$ as shown on the left of Figure 3. We can obtain a triangulation $T{{\backslash\!\backslash}}\alpha$ of $\Sigma{{\backslash\!\backslash}}\alpha$ from T as follows. First displace a and a ^′ within $\alpha$ until they merge into a point that we will still denote by a ^′. During that motion, the arcs in T incident to a and a ^′ should be kept incident to the moving vertices as shown in Figure 3. This makes $\alpha$ disappear and transforms t into the bigon that is striped in the center of Figure 3. In particular, the set of arcs resulting from that operation is not a triangulation. However, this set of arcs can be turned into the announced triangulation $T{{\backslash\!\backslash}}\alpha$ of $\Sigma{{\backslash\!\backslash}}\alpha$ by removing one of the arcs bounding the bigon. The only thing that remains of the triangle t is the other arc bounding the bigon, that is labelled by $\beta$ on the right of Figure 3. This operation is referred to as the contraction of $\alpha$ within either $\Sigma$ or T. It will be important to keep in mind that this contraction operation can only be performed when $\alpha$ is not a boundary loop.

Fig. 3. The contraction of $\alpha$ in a triangulation of $\Sigma$ .

The contraction operation was first used in [ Reference Pournin11 ] in order to compute the diameter of the flip-graph of a disk without punctures and later in [ Reference Parlier and Pournin8–Reference Parlier and Pournin10 ] to do the same for the modular flip-graphs of more general surfaces up to homeomorphism. It is also used more generally in [ Reference Pournin and Wang12 ] to compute distances within the flip-graph of a disk without punctures. Let us recall a property of the contraction operation that is instrumental in all of these articles and will also turn up later here. This property is related to the following definition.

In other words, a flip is incident to $\alpha$ when it exchanges the diagonals of a quadrilateral which has $\alpha$ as one of its sides. Now recall that all the triangulations of $\Sigma$ contain $\alpha$ and assume that $\alpha$ is not a boundary loop. In that case, for any two triangulations T and T ^′ that are related by a flip, $T{{\backslash\!\backslash}}\alpha$ and $T'{{\backslash\!\backslash}}\alpha$ are two triangulations of $\Sigma{{\backslash\!\backslash}}\alpha$ that either coincide or are still related by a flip. More precisely $T{{\backslash\!\backslash}}\alpha$ and $T'{{\backslash\!\backslash}}\alpha$ coincide precisely when the flip that relates T and T ^′ is incident to $\alpha$ . We therefore obtain the following.

Lemma 7. Consider a non-loop boundary arc $\alpha$ of $\Sigma$ . If $\nu$ flips are incident to $\alpha$ along some geodesic path in $\mathcal{F}(\Sigma)$ between two triangulations T and T^′, then

\begin{align*}d(T,T')\geq{d(T{{\backslash\!\backslash}}\alpha,T'{{\backslash\!\backslash}}\alpha)}+\nu\mbox{.}\end{align*}

3. The exponential regime

Denote by $\Gamma$ the surface with two boundary components, genus zero, no puncture, and a single marked point on each boundary component. In this section, we prove Theorem 1: our aim is to show that, except for a few cases, $\Delta_k(\Sigma)$ is at least an exponential function of k when k goes to infinity. In order to do that, we are going to exhibit strongly convex subgraphs of $\mathcal{F}(\Sigma)$ that are isomorphic to the cartesian product $\mathcal{F}(\Gamma)\mathord{\times}\mathcal{F}(\Gamma)$ . Indeed, we shall see that the largest possible number of geodesic paths between two vertices of that cartesian product is an exponential function of their distance. We recall that a subgraph G of $\mathcal{F}(\Sigma)$ is strongly convex when the geodesic paths in $\mathcal{F}(\Sigma)$ between two vertices of G are all entirely contained in G. Strong convexity in flip-graphs has been studied in different disguises for instance in [ Reference Ceballos and Pilaud3, Reference Disarlo and Parlier4, Reference Pournin and Wang12, Reference Sleator, Tarjan and Thurston14 ]. Further recall that $\Sigma^\star$ denotes the surface obtained from $\Sigma$ by removing, on each boundary component, all the marked points except one.

Proof. Denote by $\alpha_1$ to $\alpha_b$ the boundary components of $\Sigma$ . As there is at least a marked point in any of them, we can pick one of these marked points, say $a_i$ in $\alpha_i$ . Cutting $\alpha_i$ at $a_i$ results in a boundary loop $\alpha_i^\star$ twice incident to $a_i$ that contains in its interior all the marked points in $\alpha_i$ but $a_i$ . Now consider a sequence $\beta_1$ to $\beta_b$ arcs in $\Sigma$ such that $\beta_i$ is twice incident to $a_i$ and the portion of $\Sigma$ bounded by $\alpha_i\cup\beta_i$ is a topological disk $\Sigma_i$ . By construction, $\Sigma^\star$ is homeomorphic to the surface obtained by cutting the disks $\Sigma_1$ to $\Sigma_b$ out from $\Sigma$ . We will therefore identify $\Sigma^\star$ with this surface.

Now consider a sequence $T_1$ to $T_b$ of triangulations of $\Sigma_1$ to $\Sigma_b$ , respectively and observe that, for any triangulation $T^\star$ of $\Sigma^\star$ ,

\begin{align*}\left(\bigcup_{i=1}^bT_i\right)\cup{T^\star}\end{align*}

is a triangulation of $\Sigma$ . It immediately follows that $\mathcal{F}(\Sigma)$ contains a copy of $\mathcal{F}(\Sigma^\star)$ as an induced subgraph. According to [ Reference Disarlo and Parlier4 ], that subgraph is strongly convex in $\mathcal{F}(\Sigma)$ .

Using Lemma 8, we can state sufficient conditions on $\Sigma$ under which $\mathcal{F}(\Sigma)$ has a strongly convex subgraph isomorphic to $\mathcal{F}(\Gamma)$ .

Proof. Assume that $\Sigma$ has at least one boundary curve and at least two punctures. Denote by $\beta$ one of the boundary curves of $\Sigma$ and by x one of its punctures. Pick an arc $\alpha$ in $\Sigma$ twice incident to x that separates the surface into a cylinder $\Gamma'$ without punctures bounded by $\alpha$ and $\beta$ and a surface $\Sigma'$ that contains all the remaining topology of $\Sigma$ as shown on the left of Figure 4. Note that $\alpha$ is not contractible because $\Sigma'$ has at least one puncture. Therefore, $\Gamma'$ is indeed an essential cylinder. Pick a triangulation T ^′ of $\Sigma'$ and consider the subgraph $\mathcal{G}$ of $\mathcal{F}(\Sigma)$ induced by the triangulations that contain all the arcs in T ^′.

Fig. 4. The constructions in Propositions 9, 10 and 11.

By construction, $\mathcal{G}$ is isomorphic to $\mathcal{F}(\Gamma')$ and according to [ Reference Disarlo and Parlier4 ], $\mathcal{G}$ is a strongly convex subgraph of $\mathcal{F}(\Sigma)$ . Therefore, $\mathcal{F}(\Sigma)$ has a strongly convex subgraph isomorphic to $\mathcal{F}(\Gamma')$ . This does not immediately provide the desired result because $\mathcal{F}(\Gamma')$ is not necessarily isomorphic to $\mathcal{F}(\Gamma)$ . Indeed, $\beta$ can contain several marked points and, in that case, $\Gamma'$ is not homeomorphic to $\Gamma$ . However, $[\Gamma']^\star$ is homeomorphic to $\Gamma$ and, by Lemma 8, $\mathcal{F}(\Gamma')$ has a strongly convex subgraph isomorphic to $\mathcal{F}(\Gamma)$ , which proves the proposition.

Using the above three propositions, we can exhibit strongly convex subgraphs of $\mathcal{F}(\Sigma)$ isomorphic to $\mathcal{F}(\Gamma)\mathord{\times}\mathcal{F}(\Gamma)$ except in a select number of cases.

Proof. Assume that $\Sigma$ has genus one and at least two punctures but no boundary. Consider a non-separating arc $\alpha$ in $\Sigma$ that is twice incident to one of the punctures. Cutting $\Sigma$ along $\alpha$ results in a cylinder $\Sigma'$ with at least one puncture. As in the above proofs, the subgraph of $\mathcal{F}(\Sigma)$ induced by the triangulations that admit $\alpha$ as an arc is strongly convex and isomorphic to $\mathcal{F}(\Sigma')$ . Hence, the result is immediately obtained from Lemma 14.

Fig. 5. The flip-graph of a cylinder without punctures and a marked point on each boundary.

We are now ready to prove Theorem 1. The proof relies on the above lemmas and on the fact that $\mathcal{F}(\Gamma)$ is a bi-infinite simple path. More precisely, as already observed in [ Reference Parlier and Pournin8 ], $\mathcal{F}(\Gamma)$ is the infinite connected graph whose vertices each are incident to exactly two edges, as shown in Figure 5. In other words, $\mathcal{F}(\Gamma)$ can be modeled as the graph whose vertices are the integers in $\mathbb{Z}$ and whose edges connect any two consecutive integers. In turn, the cartesian product $\mathcal{F}(\Gamma)\mathord{\times}\mathcal{F}(\Gamma)$ is isomorphic to the graph $\mathcal{G}$ whose vertices are the points from $\mathbb{Z}^2$ and whose edges connect two points that coincide in one coordinate and differ by 1 in the other. For any two positive integers i and j, the distance in the graph $\mathcal{G}$ between the origin (0,0) of $\mathbb{Z}^2$ and the point (i, j) is precisely $i+j$ and the geodesics starting the the former point and ending in the latter are made of edges along which one coordinate increases by 1 while the other coordinate does not change. As a consequence, there are

\begin{align*}\frac{(i+j)!}{i!j!}\end{align*}

geodesic paths in $\mathcal{G}$ between these two points of $\mathbb{Z}^2$ . As a consequence, if k is any integer greater than or equal to 2, then taking

\begin{align*}\left\{\begin{array}{l}i=\!\left\lfloor\frac{k}{2}\right\rfloor\!\mbox{ and}\\[5pt] j=\!\left\lceil\frac{k}{2}\right\rceil\\\end{array}\right.\end{align*}

and using Stirling’s approximation of the factorial, one obtains that there are two vertices of $\mathcal{F}(\Gamma)\mathord{\times}\mathcal{F}(\Gamma)$ whose distance is equal to k such that the number of geodesics between these vertices in $\mathcal{F}(\Gamma)\mathord{\times}\mathcal{F}(\Gamma)$ is greater than

\begin{align*}\left(1-\frac{1}{k}\right)\!\sqrt{\frac{2}{k\pi}}2^ke^{-\frac{1}{4(k-1)}}\mbox{.}\end{align*}

With this estimate, we can prove the following theorem that gives an explicit exponential lower bound on $\Delta_k(\Sigma)$ and therefore implies Theorem 1.

Theorem 18. If b and p are not both equal to 0 and $2g+p+b$ is at least 3, then for every integer k greater than or equal to 2,

\begin{align*}\Delta_k(\Sigma) \gt \!\left(1-\frac{1}{k}\right)\!\sqrt{\frac{2}{k\pi}}2^ke^{-\frac{1}{4(k-1)}}\end{align*}

except possibly when:

1. (i) $\Sigma$ is the 4-punctured sphere (g and b are equal to 0 and p is equal to 4);

2. (ii) $\Sigma$ is a genus one surface with a unique boundary component and no puncture (g and b are equal to 1 and p is equal to 0); or

3. (iii) $\Sigma$ is a 2-punctured disk (g is equal to 0, b to 1, and p to 2).

4. General estimates

For any triangulation T of $\Sigma$ and any non-negative integer k, we consider the set $B_\Sigma(T,k)$ of all the triangulations in $\mathcal{F}(\Sigma)$ at distance at most k from T. Equivalently, $B_\Sigma(T,k)$ is the ball of radius k centered at T in $\mathcal{F}(\Sigma)$ . Recall that the number of triangulations contained in $B_\Sigma(T,k)$ is bounded above by $\kappa(\Sigma)^k$ . We will denote by $\Lambda_k(\Sigma)$ the largest possible number of triangulations in any such ball:

\begin{align*}\Lambda_k(\Sigma)=\max \Bigl\{ |B_\Sigma(T,k)|:T\in\mathcal{F}(\Sigma)\Bigr\}\mbox{.}\end{align*}

We also denote by $\widetilde{\Delta}_k(\Sigma)$ the largest value of $\Delta_i(\Sigma)$ when $1\leq{i}\leq{k}$ :

\begin{align*}\widetilde{\Delta}_k(\Sigma)=\mathrm{max}\bigl\{\Delta_i(\Sigma):1\leq{i}\leq{k}\bigr\}\mbox{.}\end{align*}

The aim of this section is to prove the following theorem.

Theorem 20. For every positive integer k,

(1) \begin{equation}\Delta_k(\Sigma^\star)\leq\Delta_k(\Sigma)\leq\bigl(2\kappa(\Sigma)-4\bigr)^{(n-b)r}\Lambda_k(\Sigma^\star)^{2(r-1)}\widetilde{\Delta}_k(\Sigma^\star)^r\end{equation}

where r is equal to $\bigl(2\kappa(\Sigma)\bigr)^{n-b}$ .

The first inequality in (1) is easily obtained. Recall that a subgraph G ^′ of a graph G is strongly convex when all the geodesic paths in G between any two vertices of G ^′ remain in G ^′. Hence, we immediately get the inequality $\Delta_k(\Sigma^\star)\leq\Delta_k(\Sigma)$ from Lemma 8.

Hence, we turn our attention to proving the second inequality in (1). As a first step, we show that when a boundary arc of $\Sigma$ is not a boundary loop, the number of flips incident to this arc along any geodesic path of length k in $\mathcal{F}(\Sigma)$ can be bounded above independently from k. In order to show this, consider a boundary arc $\alpha$ of $\Sigma$ that is not a boundary loop and a vertex x of $\alpha$ . Let $\beta$ be the boundary arc of $\Sigma$ other than $\alpha$ that admits x as a vertex. Consider the only triangle t contained in $\Sigma$ that is incident to both $\alpha$ and $\beta$ and observe that t exists precisely because $\alpha$ is not a boundary loop. Now denote by $\varepsilon$ the edge of t that is distinct from $\alpha$ and $\beta$ . We will denote by $\mathcal{F}_x(\Sigma)$ the subgraph induced in $\mathcal{F}(\Sigma)$ by the triangulations that admit $\varepsilon$ as an arc. Equivalently, $\mathcal{F}(\Sigma)$ is made up of all the triangulations of $\Sigma$ that do not contain any interior arc incident to x. Observe that the map $\psi_{x,\alpha}:\mathcal{F}_x(\Sigma)\rightarrow\mathcal{F}(\Sigma{{\backslash\!\backslash}}\alpha)$ that sends a triangulation T in $\mathcal{F}_x(\Sigma)$ to $T{{\backslash\!\backslash}}\alpha$ is an isomorphism (and therefore, since we think of flip-graphs as metric spaces, an isometry). In the sequel, we denote by $\mathrm{deg}_T(x)$ the number of incidences between x and the arcs of T. It will be important to keep in mind that any arc in T that is twice incident to x contributes 2 to $\mathrm{deg}_T(x)$ .

Lemma 21. Consider a boundary arc $\alpha$ of $\Sigma$ and a vertex x of $\alpha$ . If $\alpha$ is not a loop then, for any triangulation T of $\Sigma$ ,

\begin{align*}d\bigl(T,\psi_{x,\alpha}^{-1}(T{{\backslash\!\backslash}}\alpha)\bigr)\leq\mathrm{deg}_T(x)-2\mbox{.}\end{align*}

Proof. Consider a triangulation T of $\Sigma$ and the triangle t of T incident to $\alpha$ . Denote

\begin{align*}i(T)=\left\{\begin{array}{l}\mathrm{deg}_T(x)\text{ if }\textit{t} \text{ does not have an edge twice incident to }\textit{x} \text{ and}\\\mathrm{deg}_T(x)-1\text{ if }\textit{t} \text{ has an edge twice incident to }\textit{x}.\\ \end{array}\right.\end{align*}

Under the assumption that $\alpha$ is not a loop, i(T) is necessarily at least 2. Indeed, in that case, x must be incident to two boundary arcs of $\Sigma$ (none of which is a boundary loop).

We will prove by induction on i(T) that

(2) \begin{equation}d\bigl(T,\psi_{x,\alpha}^{-1}(T{{\backslash\!\backslash}}\alpha)\bigr)\leq{i(T)-2}\mbox{.}\end{equation}

As i(T) is at most $\mathrm{deg}_T(x)$ , the desired result will follow.

Observe that, if i(T) is equal to 2, then T belongs to $\mathcal{F}_x(\Sigma)$ . As a consequence, T coincides with $\psi_{x,\alpha}^{-1}(T{{\backslash\!\backslash}}\alpha)$ , which establishes the base case for the induction. Now assume that i(T) is at least 3. Denote by y the vertex of $\alpha$ distinct from x and by $\gamma$ the edge of t that is not incident to y. Note that $\gamma$ is a flippable interior arc of T. Moreover, $\gamma$ is the only edge of t that can possibly be twice incident to x because y is distinct from x. Denote by T ^′ the triangulation of $\Sigma$ obtained by flipping $\gamma$ in T. Since T and T ^′ are related by a flip incident to $\alpha$ , the triangulations $T'{{\backslash\!\backslash}}\alpha$ and $T{{\backslash\!\backslash}}\alpha$ coincide. In particular,

(3) \begin{equation}d\bigl(T,\psi_{x,\alpha}^{-1}(T{{\backslash\!\backslash}}\alpha)\bigr)\leq{d\bigl(T',\psi_{x,\alpha}^{-1}(T'{{\backslash\!\backslash}}\alpha)\bigr)+1}\mbox{.}\end{equation}

We shall now prove that

(4) \begin{equation}i(T')=i(T)-1\mbox{.}\end{equation}

Denote by t ^′ the triangle of T ^′ incident to $\alpha$ and by $\gamma'$ the edge of t ^′ that is not incident to y. The flip between T and T ^′ is shown in Figure 6 depending on whether $\gamma$ and $\gamma'$ are twice incident to x or just once. For instance, on the left of the figure, neither $\gamma$ nor $\gamma'$ are twice incident to x. In that case, i(T) coincides with $\mathrm{deg}_T(x)$ and i(T ^′) with $\mathrm{deg}_{T'}(x)$ . Hence, (4) follows from the observation that $\mathrm{deg}_{T'}(x)$ is less than $\mathrm{deg}_T(x)$ by exactly one. The situation depicted next on Figure 6 is when $\gamma$ is not twice incident to x but $\gamma'$ is. Therefore, i(T) is equal to $\mathrm{deg}_T(x)$ and i(T ^′) to $\mathrm{deg}_{T'}(x)-1$ . However, here $\mathrm{deg}_{T'}(x)$ is equal to $\mathrm{deg}_T(x)$ and we recover (4) as well. In the situation depicted third on the figure, $\gamma$ is twice incident to x and $\gamma'$ only once. Hence, i(T) is equal to $\mathrm{deg}_T(x)-1$ and i(T ^′) to $\mathrm{deg}_{T'}(x)$ . As in this case, $\mathrm{deg}_{T'}(x)$ is less than $\mathrm{deg}_T(x)$ by two, (4) holds again. In the situation shown last, $\gamma$ and $\gamma'$ are both twice incident to x and $i(T')-i(T)$ is equal to $\mathrm{deg}_{T'}(x)-\mathrm{deg}_{T'}(x)$ . As $\mathrm{deg}_{T'}(x)$ is less than $\mathrm{deg}_T(x)$ by one, this proves (4), as announced. Therefore, by induction,

(5) \begin{equation}d\bigl(T',\psi_{a,\alpha}^{-1}(T'{{\backslash\!\backslash}}\alpha)\bigr)\leq{i(T')-2}\mbox{.}\end{equation}

Combining (3), (4) and (5) proves (2), as desired.

Fig. 6. The flip between T and T ^′.

Proof. Let x and y be the vertices of $\alpha$ and consider two triangulations T and T ^′ of $\Sigma$ . As $\alpha$ is not a loop, x and y are distinct and

\begin{align*}\mathrm{deg}_T(x)+\mathrm{deg}_T(y)+\mathrm{deg}_{T'}(x)+\mathrm{deg}_{T'}(y)\leq4\kappa(\Sigma)\mbox{.}\end{align*}

We can therefore assume without loss of generality that

(6) \begin{equation}\mathrm{deg}_T(x)+\mathrm{deg}_{T'}(x)\leq2\kappa(\Sigma)\end{equation}

by, if needed, exchanging the labels of x and y. Now consider a geodesic path between T and T ^′ in $\mathcal{F}(\Sigma)$ and denote by $\nu$ the number of flips along it that are incident to $\alpha$ . We shall prove that $\nu\leq2\kappa(\Sigma)-4$ . According to Lemma 7,

\begin{align*}d(T{{\backslash\!\backslash}}\alpha,T'{{\backslash\!\backslash}}\alpha)+\nu\leq{d(T,T')}\mbox{.}\end{align*}

Since $\psi_{x,\alpha}$ is an isometry, this can be rewritten as

(7) \begin{equation}d\bigl(\psi^{-1}_{x,\alpha}(T{{\backslash\!\backslash}}\alpha),\psi^{-1}_{x,\alpha}(T'{{\backslash\!\backslash}}\alpha)\bigr)+\nu\leq{d(T,T')}\mbox{.}\end{equation}

According to Lemma 21,

\begin{align*}\left\{\begin{array}{l}d\bigl(T,\psi_{x,\alpha}^{-1}(T{{\backslash\!\backslash}}\alpha)\bigr)\leq\mathrm{deg}_T(x)-2\mbox{ and}\\ d\bigl(T',\psi_{x,\alpha}^{-1}(T'{{\backslash\!\backslash}}\alpha)\bigr)\leq\mathrm{deg}_{T'}(x)-2\mbox{.}\\\end{array}\right.\end{align*}

Combining these inequalities with (6) yields

\begin{align*}d\bigl(T,\psi_{x,\alpha}^{-1}(T{{\backslash\!\backslash}}\alpha)\bigr)+d\bigl(T',\psi_{x,\alpha}^{-1}(T'{{\backslash\!\backslash}}\alpha)\bigr)\leq2\kappa(\Sigma)-4\mbox{.}\end{align*}

Summing this with (7), one obtains that $2\kappa(\Sigma)-4-\nu$ is at least

\begin{align*}d\bigl(T,\psi_{x,\alpha}^{-1}(T{{\backslash\!\backslash}}\alpha)\bigr)+d\bigl(\psi^{-1}_{x,\alpha}(T{{\backslash\!\backslash}}\alpha),\psi^{-1}_{x,\alpha}(T'{{\backslash\!\backslash}}\alpha)\bigr)+d\bigl(T',\psi_{x,\alpha}^{-1}(T'{{\backslash\!\backslash}}\alpha)\bigr)-d(T,T')\end{align*}

and it follows from the triangle inequality that $\nu\leq2\kappa(\Sigma)-4$ .

Recall that n is the number of the boundary arcs of $\Sigma$ and b is the number of its boundary components. The second ingredient for the proof of the second inequality in (1) is the following lemma according to which $\Lambda_k(\Sigma)$ only differs from $\Lambda_k(\Sigma^\star)$ by a factor that does not depend on k but only on $\kappa(\Sigma)$ , n, and b.

Proof. According to Lemma 8, $\mathcal{F}(\Sigma)$ admits a copy of $\mathcal{F}(\Sigma^\star)$ as a strongly convex subgraph. We obtain as an immediate consequence that

\begin{align*}\Lambda_k(\Sigma^\star)\leq\Lambda_k(\Sigma)\mbox{.}\end{align*}

The other inequality is proven by induction on n. The base case is immediate as $\Sigma$ coincides with $\Sigma^\star$ when n is equal to b. Assume that n is greater than b. In that case, $\Sigma$ admits a boundary arc $\alpha$ that is not a loop. By Lemma 7, for any two triangulations T and T ^′ of $\Sigma$ ,

\begin{align*}d(T{{\backslash\!\backslash}}\alpha,T'{{\backslash\!\backslash}}\alpha)\leq{d(T,T')}\mbox{.}\end{align*}

Therefore, the map $T'\mapsto{T'{{\backslash\!\backslash}}\alpha}$ sends $B_\Sigma(T,k)$ to a subset of $B_{\Sigma{{\backslash\!\backslash}}\alpha}(T{{\backslash\!\backslash}}\alpha,k)$ . Now consider a triangulation T ^′ in $B_\Sigma(T,k)$ . Recall that, when $\alpha$ is contracted within T ^′, then its two vertices are merged into a single vertex a ^′ and only one of the edges of the triangle t of T ^′ incident to $\alpha$ remains in $T'{{\backslash\!\backslash}}\alpha$ as shown in Figure 3 where that edge is labelled $\beta$ . We can recover all the triangulations T ^′′ in $B_\Sigma(T,k)$ such that $T''{{\backslash\!\backslash}}\alpha$ is equal to $T'{{\backslash\!\backslash}}\alpha$ by reversing the process shown in Figure 3. It suffices to pick any arc $\beta$ of $T'{{\backslash\!\backslash}}\alpha$ incident to a ^′, to insert another arc $\beta'$ homotopic to $\beta$ , thus creating a bigon bounded by $\beta\cup\beta'$ and then changing this bigon into a triangle by splitting its boundary at vertex a ^′ and connecting the two obtained copies of a ^′ with the boundary arc $\alpha$ . Note that if $\beta$ is twice incident to a ^′, then the bigon can be completed into a triangle in two different ways because its boundary can be split at both of its vertices. The number of triangulations in $B_\Sigma(T,k)$ that can be obtained by this reversed procedure is therefore the number of arcs in $T'{{\backslash\!\backslash}}\alpha$ incident to a ^′ where the arcs twice incident to a ^′ are counted twice. This number is precisely $\mathrm{deg}_{T'{{\backslash\!\backslash}}\alpha}(a')$ which is, in turn, not greater than $2\kappa(\Sigma{{\backslash\!\backslash}}\alpha)$ .

This shows that the map $T'\mapsto{T'{{\backslash\!\backslash}}\alpha}$ sends at most $2\kappa(\Sigma{{\backslash\!\backslash}}\alpha)$ triangulations from $B_\Sigma(T,k)$ to the same triangulation of $\Sigma{{\backslash\!\backslash}}\alpha$ and as a consequence,

\begin{align*}|B_\Sigma(T,k)|\leq2\kappa(\Sigma{{\backslash\!\backslash}}\alpha)|B_{\Sigma{{\backslash\!\backslash}}\alpha}(T{{\backslash\!\backslash}}\alpha,k)|\mbox{.}\end{align*}

Assume that T is a triangulation of $\Sigma$ such that $|B_\Sigma(T,k)|$ is equal to $\Lambda_k(\Sigma)$ . In that case, the previous inequality can be rewritten as

\begin{align*}\Lambda_k(\Sigma)\leq2\kappa(\Sigma{{\backslash\!\backslash}}\alpha)|B_{\Sigma{{\backslash\!\backslash}}\alpha}(T{{\backslash\!\backslash}}\alpha,k)|\mbox{.}\end{align*}

Now observe that $\kappa(\Sigma{{\backslash\!\backslash}}\alpha)$ is equal to $\kappa(\Sigma)-2$ . As in addition $|B_{\Sigma{{\backslash\!\backslash}}\alpha}(T{{\backslash\!\backslash}}\alpha,k)|$ is not greater than $\Lambda_k(\Sigma{{\backslash\!\backslash}}\alpha)$ , we obtain that

(8) \begin{equation}\Lambda_k(\Sigma)\leq\bigl(2\kappa(\Sigma)-4\bigr)\Lambda_k(\Sigma{{\backslash\!\backslash}}\alpha)\mbox{.}\end{equation}

However $\Sigma{{\backslash\!\backslash}}\alpha$ has one boundary arc less than $\Sigma$ . Therefore, by induction,

(9) \begin{equation}\Lambda_k(\Sigma{{\backslash\!\backslash}}\alpha)\leq\bigl(2\kappa(\Sigma{{\backslash\!\backslash}}\alpha)-4\bigr)^{n-1-b}\Lambda_k(\Sigma^\star)\mbox{.}\end{equation}

As $\kappa(\Sigma{{\backslash\!\backslash}}\alpha)\leq\kappa(\Sigma)$ , combining (8) and (9) completes the proof.

Remark 24. Note that, according to the proof of Lemma 23, the upper bound on $\Lambda_k(\Sigma)$ provided by this lemma can immediately be improved to

\begin{align*}2^{n-b}\Lambda_k(\Sigma^\star)\prod_{i=1}^{n-b}\bigl(\kappa(\Sigma)-2i\bigr)\end{align*}

when $n-b$ is greater than 1 but we will not make use of this in the sequel.

Just as Lemma 23, the second inequality in (1) will be proven by induction on n. The third ingredient in this proof is the induction step, provided by the following lemma.

Lemma 25. If a boundary arc $\alpha$ of $\Sigma$ is not a loop, then for any positive k,

\begin{align*}\Delta_k(\Sigma)\leq\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)^{2\kappa(\Sigma)}\Lambda_k(\Sigma)^{2\kappa(\Sigma)}\mbox{.}\end{align*}

Proof. Let $\alpha$ be a non-loop boundary arc of $\Sigma$ . Consider two triangulations T and T ^′ of $\Sigma$ whose distance in $\mathcal{F}(\Sigma)$ is equal to k and such that $\#(T,T')$ is equal to $\Delta_k(\Sigma)$ . Denote

\begin{align*}S=\!\!\!\bigcup_{m=0}^{2\kappa(\Sigma)-1}\!\!\!B_\Sigma(T,k)^m\mbox{.}\end{align*}

We associate an element of S to each geodesic in $\mathcal{F}(\Sigma)$ between T and T ^′ as follows. Denote by $T_i$ the triangulation of $\Sigma$ at distance i from T along such a geodesic. Denote by $\phi(1)$ to $\phi(m)$ the indices such that the flip between $T_{\phi(i)}$ and $T_{\phi(i)+1}$ is incident to $\alpha$ . We assume without loss of generality that $\phi$ is an increasing function of i. Observe that m might be equal to 0 if there is no flip incident to $\alpha$ along the considered geodesic. Moreover, it follows from Lemma 22 that m is less than $2\kappa(\Sigma)$ . If $m=0$ , then the element s of S that we associate with the considered geodesic is the empty set (which is the only element of $B_\Sigma(T,k)^0$ ). Otherwise, we associate the geodesic with

(10) \begin{equation}s=(T_{\phi(1)}, T_{\phi(2)},\ldots,T_{\phi(m)})\mbox{.}\end{equation}

Observe that some elements of S may not be associated to any geodesic between T and T ^′. For instance, if T and T ^′ do not contain the same triangle incident to $\alpha$ , then the empty set canot be associated to any geodesic between T and T ^′ as all of them must contain a flip incident to $\alpha$ . Now let us bound the number of geodesics between T and T ^′ that are associated to s. If s is the empty set, then the triangle incident to s is the same in T and in T ^′. Moreover, all the triangulations along all the geodesics associated to s contain that triangle. Therefore, the number of geodesics associated to s is at most the number of geodesics between $T{{\backslash\!\backslash}}\alpha$ and $T'{{\backslash\!\backslash}}\alpha$ which, in turn is at most $\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)$ as, by Lemma 7, $d(T{{\backslash\!\backslash}}\alpha,T'{{\backslash\!\backslash}}\alpha)\leq{k}$ .

Now, if s is given by (10), then observe that along any geodesic associated with s, the triangle incident to $\alpha$ in $T_{\phi(i)+1}$ and $T_{\phi(i+1)}$ must be the same. In particular, the flip that relates $T_{\phi(i)}$ and $T_{\phi(i)+1}$ is prescribed by s when $i \lt m$ and by T ^′ when $i=m$ . Therefore, the number of geodesics in $\mathcal{F}(\Sigma)$ between T and T ^′ that are associated to s is precisely

\begin{align*}\#\bigl(T,T_{\phi(1)}\bigr)\!\left[\prod_{i=1}^{m-1}\#\bigl(T_{\phi(i)+1},T_{\phi(i+1)}\bigr)\right]\!\#\bigl(T_{\phi(m)},T'\bigr)\mbox{.}\end{align*}

However, since the triangle incident to $\alpha$ is not modified along the portion between $T_{\phi(i)+1}$ and $T_{\phi(i+1)}$ of any geodesic associated to s,

\begin{align*}\#\bigl(T_{\phi(i)+1},T_{\phi(i+1)}\bigr)\leq\#\bigl(T_{\phi(i)+1}{{\backslash\!\backslash}}\alpha,T_{\phi(i+1)}{{\backslash\!\backslash}}\alpha\bigr)\mbox{.}\end{align*}

By Lemma 7, $T_{\phi(i)+1}{{\backslash\!\backslash}}\alpha$ and $T_{\phi(i+1)}{{\backslash\!\backslash}}\alpha$ have distance at most k in $\mathcal{F}(\Sigma{{\backslash\!\backslash}}\alpha)$ . Hence,

\begin{align*}\#\bigl(T_{\phi(i)+1}{{\backslash\!\backslash}}\alpha,T_{\phi(i+1)}{{\backslash\!\backslash}}\alpha\bigr)\leq\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)\end{align*}

and in turn,

\begin{align*}\#\bigl(T_{\phi(i)+1},T_{\phi(i+1)}\bigr)\leq\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)\end{align*}

By the same argument, we obtain that $\#\bigl(T,T_{\phi(1)}\bigr)$ and $\#\bigl(T_{\phi(m)},T'\bigr)$ are also both bounded above by $\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)$ . Therefore, the number of geodesics associated to s is at most $\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)^{m+1}$ . Hence, any element of $B_\Sigma(T,k)^m$ is associated with at most this many geodesics, and we can bound the number of geodesics in $\mathcal{F}(\Sigma)$ between T and T ^′ as

\begin{align*}\Delta_k(\Sigma)\leq\!\!\!\sum_{m=0}^{2\kappa(\Sigma)-1}\!\!\!|B_\Sigma(T,k)|^{m}\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)^{m+1}\end{align*}

The right-hand side of this inequality can be rewritten as

\begin{align*}\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)\frac{|B_\Sigma(T,k)|^{2\kappa(\Sigma)}\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)^{2\kappa(\Sigma)}-1}{|B_\Sigma(T,k)|\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)-1}\end{align*}

Observe that, under the assumption that k is positive, $\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)$ is also positive and $|B_\Sigma(T,k)|$ is at least 2. In particular, not only is the denominator of the above fraction non-zero, but it is also least $\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)$ . As a consequence,

\begin{align*}\Delta_k(\Sigma)\leq|B_\Sigma(T,k)|^{2\kappa(\Sigma)}\widetilde{\Delta}_k(\Sigma{{\backslash\!\backslash}}\alpha)^{2\kappa(\Sigma)}\end{align*}

Since $|B_\Sigma(T,k)|$ is at most $\Lambda_k(\Sigma)$ , this completes the proof.

We are now ready to prove Theorem 20.

Proof of Theorem 20. By Lemma 8, we only need to prove the second inequality from (1). As announced, we will prove it by induction on n. Note that the base case, when n is equal to b, is immediate. Assume that n is greater than b. In that case, $\Sigma$ has a boundary arc $\alpha$ that is not a loop. According to Lemma 25,

\begin{align*}\Delta_k(\Sigma)\leq\Delta_i(\Sigma{{\backslash\!\backslash}}\alpha)^{2\kappa(\Sigma)}\Lambda_k(\Sigma)^{2\kappa(\Sigma)},\end{align*}

where i denotes an integer satisfying $1\leq{i}\leq{k}$ . In turn, Lemma 23 makes it possible to bound $\Lambda_k(\Sigma)$ above in terms of $\Lambda_k(\Sigma^\star)$ and we obtain

(11) \begin{equation}\Delta_k(\Sigma)\leq\bigl(2\kappa(\Sigma)-4\bigr)^{2(n-b)\kappa(\Sigma)}\Delta_i(\Sigma{{\backslash\!\backslash}}\alpha)^{2\kappa(\Sigma)}\Lambda_k(\Sigma^\star)^{2\kappa(\Sigma)}.\end{equation}

By induction,

(12) \begin{equation}\Delta_i(\Sigma{{\backslash\!\backslash}}\alpha)\leq\bigl(2\kappa(\Sigma{{\backslash\!\backslash}}\alpha)-4\bigr)^{(n-b-1)r'}\Lambda_i(\Sigma^\star)^{2(r'-1)}\widetilde{\Delta}_i(\Sigma^\star)^{r'},\end{equation}

where

\begin{align*}r'=\bigl(2\kappa(\Sigma{{\backslash\!\backslash}}\alpha)\bigr)^{n-b-1}\mbox{.}\end{align*}

Recall that $\kappa(\Sigma{{\backslash\!\backslash}}\alpha)$ is at most $\kappa(\Sigma)$ . As in addition $\Lambda_i(\Sigma^\star)$ and $\widetilde{\Delta}_i(\Sigma^\star)$ both are nondecreasing functions of i, (12) yields

\begin{align*}\Delta_i(\Sigma{{\backslash\!\backslash}}\alpha)\leq\bigl(2\kappa(\Sigma)-4\bigr)^{(n-b-1)r'}\Lambda_k(\Sigma^\star)^{2(r'-1)}\widetilde{\Delta}_k(\Sigma^\star)^{r'}\mbox{.}\end{align*}

Combining this with (11), one obtains

\begin{align*}\Delta_k(\Sigma)\leq\bigl(2\kappa(\Sigma)-4\bigr)^{2(n-b+(n-b-1)r')\kappa(\Sigma)}\Lambda_k(\Sigma^\star)^{2\kappa(\Sigma)(2r'-1)}\widetilde{\Delta}_k(\Sigma^\star)^{2\kappa(\Sigma)r'}\mbox{.}\end{align*}

Now denote

\begin{align*}r=\bigl(2\kappa(\Sigma)\bigr)^{n-b}\end{align*}

and recall that $\kappa(\Sigma{{\backslash\!\backslash}}\alpha)\leq\kappa(\Sigma)$ . Hence, $2\kappa(\Sigma)r'\leq{r}$ . Moreover,

\begin{align*}2\kappa(\Sigma)(2r'-1)\leq2r-2\kappa(\Sigma)\mbox{.}\end{align*}

As $\kappa(\Sigma)$ is positive, the right-hand side of this inequality is at most $2(r-1)$ . Therefore,

(13) \begin{equation}\Delta_k(\Sigma)\leq\bigl(2\kappa(\Sigma)-4\bigr)^{2(n-b+(n-b-1)r')\kappa(\Sigma)}\Lambda_k(\Sigma^\star)^{2(r-1)}\widetilde{\Delta}_k(\Sigma^\star)^r\mbox{.}\end{equation}

As $2\kappa(\Sigma)r'\leq{r}$ , the exponent of $2\kappa(\Sigma)-4$ in (13) can be bounded as

(14) \begin{equation}2\bigl(n-b+(n-b-1)r'\bigr)\kappa(\Sigma)\leq2(n-b)\kappa(\Sigma)+(n-b-1)r\end{equation}

However, as $\kappa(\Sigma)$ and $n-b$ are both at least 1, we have

\begin{align*}n-b\leq\bigl(2\kappa(\Sigma)\bigr)^{n-b-1}\mbox{.}\end{align*}

Therefore, the first term in the right-hand side of (14) is at most r and in turn, that right-hand side is at most $(n-b)r$ . Since $2\kappa(\Sigma)-4$ is at least 2 (otherwise $\Sigma$ would not admit a triangulation), the desired upper bound on $\Delta_k(\Sigma)$ follows from (13).

5. Polynomial cases

Recall that $\Gamma$ is a cylinder without punctures and exactly one marked point on each boundary component. As already mentioned at the end of Section 3, $\mathcal{F}(\Sigma)$ is isomorphic to a bi-infinite simple path. We obtain the following statement as an immediate consequence.

When $\Sigma$ is a cylinder without punctures, $\Sigma^\star$ is equal to $\Gamma$ . Hence, Theorem 20 and Proposition 26 provide Theorem 2. More precisely, we have the following statement.

Theorem 27. If $\Sigma$ is a cylinder without punctures, then

\begin{align*}\Delta_k(\Sigma)\leq\bigl(2\kappa(\Sigma)-4\bigr)^{(n-b)r}(2k+1)^{2(r-1)},\end{align*}

where r is equal to $\bigl(2\kappa(\Sigma)\bigr)^{(n-b)}$ .

This case corresponds to the striped disk in Figure 1. Let us turn our attention to the case of the 2-punctured disk. For the remainder of the section, we assume that $\Sigma$ is a 2-punctured disk. According to Theorem 20, we only need to consider the 2-punctured disk $\Sigma^\star$ with only one marked point on the boundary. Denote by x and y the two punctures of $\Sigma^\star$ and by z the marked point on its boundary. Note that there is a unique arc with vertices x and y up to homotopy. We denote that arc by $\varepsilon$ . Denote by $\mathcal{G}$ the subgraph of $\mathcal{F}(\Sigma^\star)$ induced by the triangulations that contain $\varepsilon$ and by $\mathcal{G}'$ its subgraph induced by the triangulations that do not contain $\varepsilon$ . A portion of $\mathcal{F}(\Sigma^\star)$ is shown in Figure 7. The seven triangulations in the two top rows of the figure are in $\mathcal{G}'$ and the eight in the two bottom rows in $\mathcal{G}$ . Observe that $\mathcal{G}'$ , contains two kinds of triangulations. The first kind of triangulations have two interior arcs twice incident to z and two non-flippable interior arcs, one incident to x and the other to y. Three such triangulations are shown in the top row of Figure 7. Each of them is adjacent in $\mathcal{F}(\Sigma^\star)$ to exactly two triangulations, both in $\mathcal{G}'$ and of the second kind. Four of these second kind triangulations are represented in the second row of Figure 7. In these triangulations, there is a single interior arc twice incident to z. One of the punctures is incident to a non-flippable arc and the other two flippable arcs. These triangulations are adjacent in $\mathcal{F}(\Sigma^\star)$ to two triangulations of the first kind and to exactly one triangulation in $\mathcal{G}$ since flipping the interior arc twice incident to z introduces $\varepsilon$ . With this description, $\mathcal{G}'$ is, like $\mathcal{F}(\Gamma)$ , a bi-infinite simple path because it is a connected infinite subgraph with all vertices of degree 2. Moreover one can move from $\mathcal{G}'$ to $\mathcal{G}$ by a single possible flip in every other triangulation in that path, as shown in Figure 7 and if two triangulations are far enough apart in $\mathcal{G}'$ , it is faster to pass through $\mathcal{G}$ . We get the following as a consequence.

We can now prove Theorem 3.

Fig. 7. A portion of the flip-graph of a 2-punctured disk with one marked point in the boundary.