---
abstract: |
    We consider planar cubic maps, i.e. connected cubic graphs embedded into plane, with marked spanning tree and marked directed edge (not in this tree). The number of such objects with $2n$ vertices is $C_{2n}\cdot
    C_{n+1}$, where $C_k$ is Catalan number.
author:
- Yury Kochetkov
title: 'On enumeration of tree-rooted planar cubic maps'
---

Introduction
============

Plane triangulation is a planar map, where the perimeter of each face is three. The corresponding dual graph is *cubic*, i.e. the degree of each vertex is three. A plane triangulation will be called *proper*, if each edge is incident to exactly *two* faces. Otherwise it will be called *improper*.

$$\begin{picture}(265,70) \put(15,5){\small proper triangulation}
\put(165,5){\small improper triangulation} \put(25,20){\line(1,0){60}}
\put(25,20){\circle*{2}} \put(85,20){\circle*{2}} \put(55,65){\circle*{2}}
\put(25,20){\line(2,3){30}} \put(85,20){\line(-2,3){30}}
\put(215,45){\oval(40,40)} \put(180,45){\circle*{2}} \put(195,45){\circle*{2}}
\put(210,45){\circle*{2}} \put(180,45){\line(1,0){30}}
\end{picture}$$ The corresponding dual graphs are presented below: $$\begin{picture}(160,50) \put(20,25){\oval(40,40)} \put(20,5){\circle*{2}}
\put(20,45){\circle*{2}} \put(20,5){\line(0,1){40}} \put(60,22){¨}
\put(100,25){\oval(30,30)} \put(160,25){\oval(30,30)} \put(115,25){\circle*{2}}
\put(145,25){\circle*{2}} \put(115,25){\line(1,0){30}}
\end{picture}$$

A connected graph with marked directed edge will be called *edge-rooted*. Proper edge-rooted triangulations where enumerated by Tutte in the work [@Tu]: the number $T_n$ of proper planar triangulations with $2n$ faces and marked directed edge is $$T_n=\frac{2\,(4n-3)!}{n!\,(3n-1)!}.$$

A combinatorial proof of Tutte formula see in [@PS] (see also [@AP]).

Let $F_n$ be the number of planar edge-rooted cubic graphs with $2n$ vertices, i.e. the number of planar edge-rooted triangulations (proper and improper) with $2n$ faces. Let us define numbers $f_n$, $n\geqslant -1$, in the following way:

-   $f_{-1}=1/2$;

-   $f_0=2$;

-   $f_n=(3n+2)F_n$, $n>0$.

In [@GJ] a recurrent relation for numbers $f_n$ was proposed: $$f_n=\frac{4(3n+2)}{n+1}\sum_{\scriptsize \begin{array}{c} i\geqslant -1,\, j\geqslant
-1\\ i+j=n-2\end{array}} f(i)f(j).\eqno(1)$$

From (1) it follows that $F_1=4$. Indeed, there are four ways to choose a root edge in a planar cubic map with two vertices: $$\begin{picture}(340,40) \put(15,20){\oval(30,30)}  \put(15,5){\vector(0,1){30}}
\put(70,20){\oval(30,30)} \put(120,20){\oval(30,30)}
\put(85,20){\vector(1,0){20}} \put(175,20){\oval(30,30)}
\put(225,20){\oval(30,30)} \put(190,20){\line(1,0){20}}
\put(210,15){\vector(0,1){5}} \put(275,20){\oval(30,30)}
\put(290,20){\line(1,0){20}} \put(325,20){\oval(30,30)}
\put(310,25){\vector(0,-1){5}} \end{picture}$$ Also we have that $F_2=32$. Indeed, there are six cubic maps with 4 vertices (and 6 edges): $$\begin{picture}(300,50) \put(40,25){\oval(40,40)} \put(40,5){\line(0,1){40}}
\put(60,25){\line(1,0){20}} \put(95,25){\oval(30,30)} \put(5,20){\small 1)}

\put(155,25){\oval(30,30)} \put(215,25){\oval(30,30)}
\qbezier(155,40)(185,55)(215,40) \qbezier(155,10)(185,-5)(215,10)
\put(125,20){\small 2)}

\put(280,25){\oval(40,40)} \put(280,25){\line(0,1){20}} \put(245,20){\small 3)}
\put(280,25){\line(3,-2){17}} \put(280,25){\line(-3,-2){17}}  \end{picture}$$ $$\begin{picture}(310,50) \put(20,35){\circle{10}} \put(20,5){\circle{10}}
\put(40,20){\line(-4,3){16}} \put(40,20){\line(-4,-3){16}}
\put(40,20){\line(1,0){20}} \put(65,20){\circle{10}} \put(0,17){\small 4)}

\put(110,20){\oval(20,20)} \put(120,20){\line(1,0){20}}
\put(150,20){\oval(20,20)} \put(160,20){\line(1,0){20}}
\put(190,20){\oval(20,20)} \put(85,17){\small 5)}

\put(240,20){\oval(20,20)} \put(250,20){\line(1,0){20}}
\put(290,20){\oval(40,40)} \put(290,20){\oval(20,20)}
\put(300,20){\line(1,0){10}} \put(215,17){\small 6)}
\end{picture}$$

[Figure 1]{}

Group of automorphisms of the first map is trivial, of the second has order 4, of the third has order 12, of the forth has order 3, of the fifth and the sixth has order 2. Thus, there are 12 ways to choose a root edge in the first map, 3 — in the second, 1 — in the third, 4 — in the forth, 6 — in the fifth and the sixth. All this gives us 32 edge-rooted maps.

However, this formula does not seem to have a geometrical/combinatorial explanation.

In [@Mu] a nice formula was proposed for the number tree-rooted planar maps, i.e. edge-rooted planar maps with distinguished spanning tree: the number of such maps with $n$ edges is $C_n\cdot
C_{n+1}$, where $C_k$ is $k$-th Catalan number. An elegant proof of this formula see in [@Be].

There are four planar maps with two edges: $$\begin{picture}(215,40) \multiput(0,25)(15,0){3}{\circle*{3}}
\put(0,25){\line(1,0){30}} \put(13,2){\small 1}

\put(70,25){\oval(20,20)} \put(80,25){\circle*{3}}
\put(95,25){\circle*{3}} \put(80,25){\line(1,0){15}}
\put(68,2){\small 2}

\put(125,25){\circle*{3}} \put(145,25){\circle*{3}}
\put(135,25){\oval(20,20)} \put(134,2){\small 3}

\put(195,25){\circle*{3}} \put(185,25){\oval(20,20)}
\put(205,25){\oval(20,20)} \put(194,2){\small 4}
\end{picture}$$

-   There is one way to choose a spanning tree in the first map and two ways to choose a directed edge.

-   There is one way to choose a spanning tree in the second map and four ways to choose a directed edge.

-   There is one way to choose a spanning tree in the third map and two ways to choose a directed edge.

-   There is no spanning trees in the forth map and two ways to choose a directed edge.

Thus we have $10=C_2\cdot C_3$ tree rooted planar maps with two edges.

We will study tree-rooted cubic maps with additional property: a root edge *does not* belong to the spanning tree.

**Theorem.** *The number of such tree-rooted cubic maps with $2n$ vertices is $C_{2n}\cdot
C_{n+1}$, where $C_k$ is $k$-th Catalan number.*

The main construction: from map to curve
========================================

By tree-rooted plane cubic map we will understand a cubic graph imbedded into plane (sphere) with

-   marked spanning tree;

-   marked directed edge that *does not* belong to the spanning tree.

Let $G$ be a tree-rooted pane cubic map with $2n$ vertices. We draw triangles, one triangle for each vertex, in such way that:

-   triangles are disjoint;

-   each vertex is inside the corresponding triangle;

-   each side of triangle intersect one outgoing edge of corresponding vertex.

$$\begin{picture}(260,60) \qbezier[25](0,30)(0,50)(20,50)
\qbezier[25](0,30)(0,10)(20,10) \qbezier[25](20,10)(40,10)(40,30)
\qbezier[20](60,30)(60,45)(75,45)
\qbezier[20](60,30)(60,15)(75,15)
\qbezier[20](75,45)(90,45)(90,30)
\qbezier[20](75,15)(90,15)(90,30) \linethickness{0.5mm}
\put(20,10){\line(0,1){40}} \put(40,30){\line(1,0){20}}
\qbezier(20,50)(40,50)(40,30) \put(20,7){$\to$}
\put(110,28){$\Rightarrow$} \thinlines
\qbezier[25](140,30)(140,50)(160,50)
\qbezier[25](140,30)(140,10)(160,10)
\qbezier[25](160,10)(180,10)(180,30)
\qbezier[20](200,30)(200,45)(215,45)
\qbezier[20](200,30)(200,15)(215,15)
\qbezier[20](215,45)(230,45)(230,30)
\qbezier[20](215,15)(230,15)(230,30) \linethickness{0.5mm}
\put(160,10){\line(0,1){40}} \put(180,30){\line(1,0){20}}
\qbezier(160,50)(180,50)(180,30) \thinlines
\put(152,42){\line(1,0){16}} \put(152,42){\line(1,2){8}}
\put(168,42){\line(-1,2){8}} \put(152,18){\line(1,0){16}}
\put(152,18){\line(1,-2){8}} \put(168,18){\line(-1,-2){8}}
\put(188,22){\line(0,1){16}} \put(188,22){\line(-2,1){16}}
\put(188,38){\line(-2,-1){16}} \put(192,22){\line(0,1){16}}
\put(192,22){\line(2,1){16}} \put(192,38){\line(2,-1){16}}
\put(160,7){$\longrightarrow$}\end{picture}$$ Thick lines above mark spanning tree and an arrow indicates the direction of the root edge.

Two triangles will be called adjacent, if the corresponding vertices are adjacent and the edge, that connects them, belongs to the spanning tree. The sides of adjacent triangles that intersect this edge also will be called adjacent. We construct a polygon $P$ by glewing adjacent triangles by adjacent sides. This polygon has $2n+2$ sides and is divided into $2n$ triangles. Each edge of the cubic map, that does not belong to the spanning tree, intersects two sides of $P$ and we will say that these sides constitute a pair. Polygon $P$ has a marked side: the marked edge of the cubic map intersects it in direction from inside $P$ to outside.

Continuation of Example. $$\begin{picture}(160,200) \put(20,70){\line(2,-3){40}}
\put(20,70){\line(1,0){80}} \put(20,70){\line(2,3){80}}
\put(60,10){\line(2,3){80}} \put(60,130){\line(2,-3){40}}
\put(60,130){\line(1,0){80}} \put(100,190){\line(2,-3){40}}
\linethickness{0.6mm} \put(60,50){\line(0,1){40}}
\qbezier(60,90)(80,100)(100,110) \put(100,110){\line(0,1){40}}
\thinlines \put(60,50){\line(-2,-1){40}}
\put(60,50){\vector(2,-1){40}} \put(60,90){\line(-2,1){40}}
\put(100,110){\line(2,-1){40}} \put(100,150){\line(2,1){40}}
\put(100,150){\line(-2,1){40}} \put(65,5){\small A}
\put(105,65){\small B} \put(145,125){\small C} \put(10,68){\small
F} \put(50,128){\small E} \put(90,188){\small D}
\end{picture}$$ Here $EF$ and $FA$, $AB$ and $BC$, $CD$ and $DE$ are pairs and $AB$ is the marked edge. If we identify sides that are in pairs (i.e. $EF$ with $FA$, $AB$ with $BC$ and $CD$ with $DE$), then we will obtain a triangulated genus 0 curve.

The main construction: from curve to map
========================================

Let $P$ be a $2n$-gon with marked side $M$ and triangulated by non-intersecting diagonals into $2n-2$ triangles. Sides of $P$ are divided into pairs in such way, that the identification of sides in each pair gives us a genus 0 curve. We will construct a plane tree-rooted cubic map with root edge (not in the spanning tree) in the following way.

-   We put a vertex $v_i$ inside each triangle $\triangle_i$ and connect vertices in adjacent triangles — the spanning tree is constructed.

-   Let sides $L$ and $L'$ be in pair. $L$ and $L'$ are sides of triangles $\triangle_i$ and $\triangle_j$, respectively (these triangles may coincide). We draw an arc that connect $v_i$ and $v_j$ in the following way: going from $v_i$ the arc intersects $L$. Its next part lies in the exterior of $P$ and connects $L$ and $L'$. After intersecting $L'$ the arc goes to $v_j$.

-   An arc, that intersects $M$ will be the root edge. At intersection point it is directed from inside $P$ to outside.

$$\begin{picture}(320,120) \put(10,40){\line(0,1){40}}
\put(10,40){\line(1,-1){30}} \put(10,40){\line(1,1){70}}
\put(10,80){\line(1,1){30}} \qbezier(10,40)(25,75)(40,110)
\put(40,110){\line(1,0){40}} \put(40,10){\vector(1,0){40}}
\put(40,10){\line(2,5){40}} \qbezier(40,10)(75,25)(110,40)
\put(80,110){\line(1,-1){30}} \qbezier(80,110)(95,75)(110,40)
\put(110,40){\line(0,1){40}} \put(80,10){\line(1,1){30}}
\put(0,36){\small H} \put(0,76){\small G} \put(36,0){\small A}
\put(36,113){\small F} \put(79,0){\small B} \put(79,113){\small E}
\put(115,36){\small C} \put(115,76){\small D}
\put(150,55){$\Rightarrow$}

\put(200,40){\line(0,1){40}} \put(200,40){\line(1,-1){30}}
\put(200,40){\line(1,1){70}} \put(200,80){\line(1,1){30}}
\qbezier(200,40)(215,75)(230,110) \put(230,110){\line(1,0){40}}
\put(230,10){\vector(1,0){40}} \put(230,10){\line(2,5){40}}
\qbezier(230,10)(265,25)(300,40) \put(270,110){\line(1,-1){30}}
\qbezier(270,110)(285,75)(300,40) \put(300,40){\line(0,1){40}}
\put(270,10){\line(1,1){30}} \put(190,36){\small H}
\put(190,76){\small G} \put(226,0){\small A} \put(226,113){\small
F} \put(269,0){\small B} \put(269,113){\small E}
\put(305,36){\small C} \put(305,76){\small D}

\linethickness{0.5mm} \put(210,80){\line(1,0){20}}
\qbezier(230,80)(235,65)(240,50) \put(240,50){\line(1,0){30}}
\put(270,50){\line(0,-1){30}} \qbezier(270,50)(280,65)(290,80)

\thinlines \put(270,20){\vector(-3,-2){30}}
\put(270,20){\line(2,-1){20}} \put(290,80){\line(2,-3){20}}
\put(290,80){\line(0,1){20}} \put(240,50){\line(-1,-1){35}}
\put(230,80){\line(2,3){25}} \put(210,80){\line(-2,-3){20}}
\put(210,80){\line(0,1){25}}
\end{picture}$$ Here sides $AB$ and $FG$, $BC$ and $CD$, $DE$ and $EF$, $GH$ and $HA$ constitute pairs and $AB$ is the marked side. Thus, we must connect the arc that intersects $AB$ with the arc that intersects $FG$, the arc that intersects $BC$ with the arc that intersects $CD$, the arc that intersects $DE$ with the arc that intersects $EF$ and the arc that intersects $GH$ with the arc that intersects $HA$. An arrow in the arc that intersects $AB$ indicates the direction of the root edge of the cubic graph. The cubic graph itself and its “simplification” are presented in the figure below. $$\begin{picture}(300,100) \linethickness{0.6mm} \put(0,30){\line(1,0){80}}
\put(60,30){\line(0,1){10}} \thinlines \put(20,30){\line(0,1){10}}
\put(20,30){\oval(40,40)[b]} \put(50,30){\oval(100,100)[t]}
\put(40,40){\oval(40,40)[t]} \put(60,30){\oval(40,20)[tr]}
\put(90,30){\oval(20,20)[b]} \put(78,23){$\downarrow$}
\put(140,50){$\Rightarrow$} \linethickness{0.6mm}
\put(200,20){\line(0,1){60}} \put(200,20){\line(1,0){80}}
\put(280,20){\line(0,1){30}} \qbezier(280,20)(310,50)(280,80)
\thinlines \qbezier(200,20)(170,50)(200,80)
\put(200,80){\line(1,0){80}} \put(200,50){\line(1,0){80}}
\put(280,50){\line(0,1){30}} \put(280,80){\vector(-1,0){15}}
\end{picture}$$

We can draw above mentioned arcs in such way, that they do not intersect in the exterior of $P$.

Let us connect midpoints of all sides in pairs by segments inside $P$. As the identification of sides in pairs generates a genus zero curve, then these segments do not intersect. The polygon $P$ is embedded into sphere, so we can interchange its interior and exterior domains.

Main statement
==============

The number of tree-rooted cubic maps with $2n$ vertices and a marked edge, that does not belong to the spanning tree, is $C_{2n}\cdot C_{n+1}$, where $C_k$ is $k$-th Catalan number.

Our theorem follows from two statements.

1.  A convex $n$-gon with a marked side can be divided into triangles by non-intersecting diagonals in $C_{n-2}$ ways [@St].

2.  There are $C_n$ ways to define a pairwise identification of sides of a convex $2n$-gon with a marked side to obtain a genus 0 curve [@LZ].

According to theorem, we have $C_4\cdot C_3=70$ tree-rooted cubic maps with $4$ vertices. In what follows a map with a marked spanning tree will be called *t-map*. The first cubic map in Figure 1 generates six t-maps. $$\begin{picture}(330,50) \qbezier[20](0,25)(0,45)(20,45)
\qbezier[20](0,25)(0,5)(20,5) \qbezier[20](60,25)(60,40)(75,40)
\qbezier[20](60,25)(60,10)(75,10)
\qbezier[20](75,40)(90,40)(90,25)
\qbezier[20](75,10)(90,10)(90,25) \qbezier[30](20,5)(20,25)(20,45)

\qbezier[20](180,25)(180,40)(195,40)
\qbezier[20](180,25)(180,10)(195,10)
\qbezier[20](195,40)(210,40)(210,25)
\qbezier[20](195,10)(210,10)(210,25)
\qbezier[20](120,25)(120,45)(140,45)
\qbezier[20](120,25)(120,5)(140,5)
\qbezier[20](140,5)(160,5)(160,25)

\qbezier[20](300,25)(300,40)(315,40)
\qbezier[20](300,25)(300,10)(315,10)
\qbezier[20](315,40)(330,40)(330,25)
\qbezier[20](315,10)(330,10)(330,25)
\qbezier[20](260,5)(280,5)(280,25)
\qbezier[30](260,5)(260,25)(260,45)

\linethickness{0.5mm} \qbezier(20,45)(40,45)(40,25)
\qbezier(20,5)(40,5)(40,25) \put(40,25){\line(1,0){20}}
\put(160,25){\line(1,0){20}} \put(280,25){\line(1,0){20}}
\qbezier(140,45)(160,45)(160,25) \put(140,5){\line(0,1){40}}
\qbezier(240,25)(240,45)(260,45) \qbezier(240,25)(240,5)(260,5)
\qbezier(260,45)(280,45)(280,25)
\end{picture}$$

$$\begin{picture}(210,50) \qbezier[20](0,25)(0,45)(20,45)
\qbezier[20](0,25)(0,5)(20,5) \qbezier[20](60,25)(60,40)(75,40)
\qbezier[20](60,25)(60,10)(75,10)
\qbezier[20](75,40)(90,40)(90,25)
\qbezier[20](75,10)(90,10)(90,25)
\qbezier[20](20,45)(40,45)(40,25)

\qbezier[20](180,25)(180,40)(195,40)
\qbezier[20](180,25)(180,10)(195,10)
\qbezier[20](195,40)(210,40)(210,25)
\qbezier[20](195,10)(210,10)(210,25)
\qbezier[20](140,45)(160,45)(160,25) \put(140,5){\line(0,1){40}}

\qbezier[20](140,5)(160,5)(160,25)
\qbezier[30](140,5)(140,25)(140,45)

\linethickness{0.5mm} \put(20,5){\line(0,1){40}}
\put(40,25){\line(1,0){20}} \qbezier(20,5)(40,5)(40,25)
\qbezier(120,25)(120,45)(140,45) \qbezier(120,25)(120,5)(140,5)
\put(160,25){\line(1,0){20}} \qbezier(140,5)(160,5)(160,25)
\end{picture}$$ In each case we have six ways to choose a marked edge, that does not belong to the tree. Thus, the first map generates $30$ tree-rooted maps.

The second cubic map in Figure 1 generates four t-maps. $$\begin{picture}(310,80) \put(0,37){\small 1)} \qbezier[20](40,60)(60,60)(60,40)
\qbezier[20](40,20)(60,20)(60,40)
\qbezier[20](120,60)(100,60)(100,40)
\qbezier[20](120,20)(100,20)(100,40)
\qbezier[20](120,60)(140,60)(140,40)
\qbezier[20](120,20)(140,20)(140,40) \put(40,20){\circle*{3}}
\put(40,60){\circle*{3}} \put(120,20){\circle*{3}}
\put(120,60){\circle*{3}}

\put(170,37){\small 2)} \qbezier[20](210,60)(230,60)(230,40)
\qbezier[20](210,20)(230,20)(230,40)
\qbezier[20](290,60)(270,60)(270,40)
\qbezier[20](290,20)(270,20)(270,40) \put(210,20){\circle*{3}}
\put(210,60){\circle*{3}} \put(290,20){\circle*{3}}
\put(290,60){\circle*{3}} \qbezier[60](210,60)(250,90)(290,60)

\linethickness{0.5mm} \qbezier(40,60)(20,60)(20,40)
\qbezier(40,20)(20,20)(20,40) \qbezier(40,20)(80,-10)(120,20)
\qbezier(40,60)(80,90)(120,60) \qbezier(210,60)(190,60)(190,40)
\qbezier(210,20)(190,20)(190,40) \qbezier(210,20)(250,-10)(290,20)

\qbezier(290,60)(310,60)(310,40) \qbezier(290,20)(310,20)(310,40)
\end{picture}$$

$$\begin{picture}(310,80) \put(0,37){\small 3)} \qbezier[20](40,60)(60,60)(60,40)
\qbezier[20](40,20)(60,20)(60,40)
\qbezier[20](120,60)(140,60)(140,40)
\qbezier[20](120,20)(140,20)(140,40)  \put(40,20){\circle*{3}}
\put(40,60){\circle*{3}} \put(120,20){\circle*{3}}
\put(120,60){\circle*{3}} \qbezier[60](40,60)(80,90)(120,60)

\put(170,37){\small 4)} \qbezier[20](210,60)(190,60)(190,40)
\qbezier[20](210,20)(190,20)(190,40)
\qbezier[20](290,60)(270,60)(270,40)
\qbezier[20](290,20)(270,20)(270,40) \put(210,20){\circle*{3}}
\put(210,60){\circle*{3}} \put(290,20){\circle*{3}}
\put(290,60){\circle*{3}} \qbezier[60](210,60)(250,90)(290,60)

\linethickness{0.5mm} \qbezier(40,60)(20,60)(20,40)
\qbezier(40,20)(20,20)(20,40) \qbezier(40,20)(80,-10)(120,20)
\qbezier(120,60)(100,60)(100,40) \qbezier(120,20)(100,20)(100,40)

\qbezier(210,60)(230,60)(230,40) \qbezier(210,20)(230,20)(230,40)
\qbezier(210,20)(250,-10)(290,20) \qbezier(290,60)(310,60)(310,40)
\qbezier(290,20)(310,20)(310,40)
\end{picture}$$ The first two t-maps have trivial groups of automorphisms. Thus, they generate six tree-rooted cubic maps each. But the group of automorphisms of the third and the of forth t-maps has order two. Thus, they generate three tree-rooted cubic maps each and the second cubic map generates $18$ tree-rooted maps.

The third cubic map in Figure 1 generates three t-maps. $$\begin{picture}(270,95) \put(0,30){\circle*{3}} \put(70,30){\circle*{3}}
\put(35,50){\circle*{3}} \put(35,90){\circle*{3}}
\qbezier[30](0,30)(35,5)(70,30) \qbezier[30](0,30)(0,70)(35,90)
\qbezier[30](70,30)(70,70)(35,90) \put(33,2){\small 1}

\put(100,30){\circle*{3}} \put(170,30){\circle*{3}}
\put(135,50){\circle*{3}} \put(135,90){\circle*{3}}
\qbezier[30](100,30)(100,70)(135,90)
\qbezier[30](135,50)(135,70)(135,90)
\qbezier[30](135,50)(152,40)(170,30) \put(133,2){\small 2}

\put(200,30){\circle*{3}} \put(270,30){\circle*{3}}
\put(235,50){\circle*{3}} \put(235,90){\circle*{3}}
\qbezier[30](270,30)(270,70)(235,90)
\qbezier[30](235,50)(235,70)(235,90)
\qbezier[30](200,30)(217,40)(235,50) \put(233,2){\small 3}

\linethickness{0.5mm} \qbezier(0,30)(17,40)(35,50)
\qbezier(35,50)(52,40)(70,30) \put(35,50){\line(0,1){40}}

\qbezier(100,30)(117,40)(135,50) \qbezier(100,30)(135,5)(170,30)
\qbezier(170,30)(170,70)(135,90)

\qbezier(200,30)(235,5)(270,30) \qbezier(200,30)(200,70)(235,90)
\qbezier(235,50)(252,40)(270,30)
\end{picture}$$ The group of automorphisms of the first of them has order $3$ and of the second and the third — order $2$. Thus they generate $2+3+3=8$ tree-rooted cubic maps.

The forth cubic map in Figure 1 generates one t-map with order three group of automorphisms. Thus, it generates $2$ tree-rooted cubic maps.

The fifth cubic map in Figure 1 also generates one t-map with trivial group of automorphisms. Thus, it generates $6$ tree-rooted cubic maps.

The sixth cubic map in Figure 1 generates two t-maps $$\begin{picture}(240,80) \put(10,40){\oval(20,20)}
\qbezier(40,40)(40,70)(70,70) \qbezier(70,70)(100,70)(100,40)
\put(70,40){\oval(20,20)}

\put(150,40){\oval(20,20)} \qbezier(180,40)(180,10)(210,10)
\qbezier(210,10)(240,10)(240,40) \put(210,40){\oval(20,20)}

\linethickness{0.6mm} \put(20,40){\line (1,0){20}}
\put(80,40){\line(1,0){20}} \qbezier(40,40)(40,10)(70,10)
\qbezier(70,10)(100,10)(100,40)

\put(160,40){\line (1,0){20}} \put(220,40){\line(1,0){20}}
\qbezier(180,40)(180,70)(210,70) \qbezier(210,70)(240,70)(240,40)
\end{picture}$$ with order two group of automorphisms each. Thus, they generate $3+3=6$ tree-rooted cubic maps.

So, we have $$30+18+8+2+6+6=70$$ tree-rooted cubic maps, as expected.

[99]{} Albenque M. and Poulalhon D., *Generic method for bijections between blossoming trees and planar maps*, arXiv: 1305.1312. Bernardi O., *Bijective counting of tree-rooted maps and shuffles of parenthesis systems*, arXiv:math/0601684. Goulden I.P. and Jackson D.M., *The KP hierarchy, branched covers, and triangulations*, arXiv: 0803.3980. Lando S. and Zvonkin A., *Graphs on surfaces and their applications. Encyclopedia of mathematical sciences* **141**, Springer-Verlag, Berlin, 2004. Mullin R.C., *On the enumeration of tree-rooted maps*, Canad. J. Math., 1967, 19, 174-183. Poulalhon D. and Schaeffer G., *Optimal coding and sampling of triangulations*, Algorithmica, 2006, 46(3), 505-527. Stanley R.P., *Enumerative combinatorics, volume 2*, Wadsworth & Brooks, 1999. Tutte W.T., *A census of planar triangulations*, Canad. J. Math., 1962, 14, 21-38.
