1 Introduction
All graphs considered here are finite, undirected, and simple, i.e., contain no loops nor multiple edges. The fractional arboricity of a graph (sometimes also denoted by ) is a classical measure of density, introduced by Payan [13]. The famous NashWilliams Theorem [12] says that the edges of any graph can be decomposed into forests. While the number of forests cannot be reduced, it might still be possible to improve the decomposition by imposing a low maximum degree on one forest.
For positive integers , a graph is called decomposable if its edges decompose into forests, one of maximum degree at most . A weaker notion than decomposable is the following: A graph is called decomposable if its edges decompose into forests and one subgraph of maximum degree at most . Let us also define the following notion that is stronger than decomposability: A graph is called decomposable if its edges decompose into forests and one tree of maximum degree at most .
Affirming a conjecture of Balogh et al. [1], Gonçalves [5] proves that every planar graph is decomposable, while large enough planar trees are not decomposable [1].
Our Results.
In this paper we consider decompositions of maximally planar graphs, which we will call planar triangulations, or just triangulations for short. First, we improve the classical result of Whitney [16] that every connected triangulation is Hamiltonian. Recall that a graph is connected if and for any triple of vertices the graph is connected. Moreover, a Hamiltonian path in is a simple path in that contains all vertices of , and is Hamiltonian if it contains at least one Hamiltonian path.
Theorem 1.1.
Every connected planar triangulation decomposes into two trees and one Hamiltonian path. In particular, is decomposable.
Combining this result with recursive decompositions of nonconnected triangulations, we obtain decompositions of Hamiltonian triangulations.
Theorem 1.2.
Every Hamiltonian planar triangulation decomposes into two trees and a spanning tree of maximum degree . In particular, is decomposable.
Furthermore, our methods give a new proof of (a slight strengthening of) Gonçalves’ result.
Theorem 1.3.
Every planar triangulation decomposes into two trees and a spanning tree of maximum degree . In particular, is decomposable.
Finally, we show that all our results are bestpossible, where in the last case we provide a family of examples that is richer than just planar trees, as given in [1].
Theorem 1.4.
Each of the following holds.

Some connected planar triangulations are not decomposable.

Some Hamiltonian planar triangulations are not decomposable.

Some planar triangulations are not decomposable.
Related Work.
While the present paper is focused on planar triangulations and the cases , , let us give an account of the history of decomposability and decomposability for general graphs and general integers .
One motivation for decomposability are applications to bounding the (incidence) gamechromatic number [6, 11, 2] and the spectral radius of a graph [4]. However, most of the research in this field was inspired by the famous Nine Dragon Tree Conjecture of Montassier, Ossona de Mendez, Raspaud, and Zhu [10], which states that if the difference between and is large, then decomposes into forests, where the maximum degree of one forest can be bounded. More precisely, if for positive integers , then is decomposable. The Nine Dragon Tree Conjecture was proved for several special cases [8, 10, 9, 3] before it was confirmed in full generality by Jiang and Wang in 2016 [7].
The original motivation for the Nine Dragon Tree Conjecture in [10] comes as a generalization of decomposition results in sparse planar graphs: Planar graphs are known to be decomposable, i.e., decompose into a forest and a matching, when they have girth at least [15, 10], while some planar graphs of girth are not decomposable [10]. For , He et al. [6] show that planar graphs of girth at least are decomposable, which was improved to decomposability by Gonçalves [5], while some planar graphs of girth are not decomposable [10]. In [6] it is further shown that planar graphs of girth at least are decomposable. All these decomposition results follow immediately from the Nine Dragon Tree Theorem [7] as for every planar graph of girth at least we have , and thus these decompositions rely purely on the low fractional arboricity of graphs.
However, for planar triangulations the fractional arboricity tends to as the number of vertices tends to infinity. Thus the Nine Dragon Tree Theorem does not give decomposability of all planar graphs for any fixed . Hence, for the following results (as well as our results in the present paper) the structure of planar graphs had to be exploited on a different level: In [6] it is shown that planar graphs are decomposable, which is strengthened to decomposability in [1]. Moreover, Balogh et al. [1] show that Hamiltonian and consequently connected planar graphs are decomposable. Finally Gonçalves [5] improved these results to decomposability of all planar graphs, which is bestpossible [1].
Organization of the Paper.
In Section 2 we prove our key lemma, Lemma 2.2, which is crucial for all our decomposition results. In Section 3 we decompose any planar triangulation along its separating triangles and introduce triangle assignments. Here we also combine Lemma 2.2 to obtain decompositions for of planar triangulations admitting socalled assignments, c.f. Proposition 3.2. In Section 4 we prove our main decomposition results, namely Theorems 1.1–1.3. To this end, we show that connected (respectively Hamiltonian and general) triangulations admit assignments (respectively assignments and assignments) and use the decompositions given by Proposition 3.2 in Section 3. In Section 5 we show that our results are bestpossible by constructing connected (respectively Hamiltonian and general) triangulations that are not decomposable (respectively decomposable and decomposable); In other words, we prove Theorem 1.4. Finally, we conclude the paper in Section 6.
2 The Key Lemma
This section is devoted to the proof of Lemma 2.2, which is a central element of the proofs of all our Theorems.
Let be a plane embedded graph with a simple outer cycle . Moreover, let be inner triangulated, that is, every inner face of is a triangle. For two outer vertices of we denote by the path from to along the outer cycle in counterclockwise direction, and define . If then consists of only one vertex and . A filled triangle in is a triple of pairwise adjacent vertices, such that at least one vertex of lies inside this triangle. A separating triangle in is a triple of pairwise adjacent vertices, such that at least one vertex of lies inside this triangle and at least one vertex lies outside this triangle. It is wellknown and easy to see that a planar triangulation is connected if and only if it does not have^{1}^{1}1Using planar triangulation instead of plane triangulation is justified as the existence of separating triangles is independent of the chosen plane embedding. any separating triangles.
Definition 2.1.
A plane inner triangulated graph with simple outer cycle is a Whitney graph with respect to if are outer vertices of with , such that:

contains no filled triangle.

appear in this counterclockwise order around .

, , are induced paths in .

If , then is not an edge of .
Note that the outer face of a Whitney graph with at least four vertices cannot be a triangle. Moreover, any inner triangulated connected graph with some outer vertices in this counterclockwise order is a Whitney graph with respect to .
Lemma 2.2.
If is a Whitney graph with respect to , then the edges of can be oriented and colored black, red, and blue, such that each of the following holds.

The black edges form a directed Hamiltonian path from to in .

Every inner vertex has precisely one outgoing red edge and one outgoing blue edge.

Every vertex on has precisely one outgoing blue edge and no outgoing red edge.

Every vertex on has precisely one outgoing red edge and no outgoing blue edge.

Every vertex on has precisely one outgoing blue edge and no outgoing red edge.

Neither nor has outgoing red edges nor outgoing blue edges.

If , then vertex has precisely one outgoing blue edge which is if it exists and no outgoing red edge. If , then has no outgoing blue and no outgoing red edge.

There is no monochromatic directed cycle in .
Proof.
Recall that our proof is based on the decomposition of planar connected triangulations by Whitney [16]. We do induction on the number of vertices of . If then is a triangle with vertices and . We define to be the path and orient from to and color it blue. It is easy to see that a–h are satisfied. If we distinguish seven cases, which we go through in this order, i.e., when considering Case we sometimes make use of the fact that Case does not apply for .
 Case 1: .

Let and be the neighbor of on and , respectively. Then is a Whitney graph with respect to . Indeed the vertices on are neighbors of and hence a chord in would give a separating triangle in . Hence is an induced path in . Moreover, and are subsets of the induced paths and , respectively, and thus induced, too.
By induction there is a Hamiltonian path from to in and an orientation and coloring of the edges in , such that b–h are satisfied. We extend by the edge , i.e., , and the coloring/orientation by orienting all incident edges at (except ) towards and coloring them red. See Figure 1 for an illustration.
We need to argue that together with the coloring/orientation satisfies b–h. Indeed b follows from b, c with respect to and the coloring/orientation of the edges incident to , for c there is nothing to show, d–f follow from d–f with respect to , and g follows again from the coloring/orientation of the edges incident to . Finally, we need to show h, namely that there is no directed monochromatic cycle in . By h with respect to such a cycle would contain , which has not incoming and outgoing edges of the same color.
 Case 2: There is an edge with .

We choose to be the vertex that is a neighbor of and is closest to on . The three illustrations in Figure 2 display the three subcases that we sometimes have to treat differently along the construction: , and , and .
Define to be the inner triangulated subgraph of with outer cycle . Then is a Whitney graph with respect to . We define to be the graph . If both, and , are edges in , then is just a single edge, which we put into the Hamiltonian path. Otherwise is a Whitney graph with respect to (or rather in case ), where is the neighbor of on if . However, if the embedding of needs to be flipped so that appear in counterclockwise order. Indeed is induced since it consists solely of neighbors of , and similarly is induced in case . We apply induction to both, and (if is not just an edge), and concatenate the obtained Hamiltonian paths in and to a Hamiltonian path from to in .
It remains to color and orient the edges incident to , but not contained in . Additionally, we also define a color and orientation for the edge , disregarding its color and orientation given by induction on . (Note that is certainly not in the Hamiltonian path of .) If we orient these edges, as well as the edge , towards and color them red, except for which we color blue and orient to . On the other hand if we swap the colors red and blue in the coloring for , but keep the orientation the same. Moreover, we color the edges incident to blue and orient them towards , except for which is oriented towards . See Figure 2 for an illustration.
It is straightforward to check that b–g follow from b–g with respect to and and the coloring/orientation of the edges incident to . Moreover, by h with respect to and , every directed monochromatic cycle has to contain . The only case in which has incoming and outgoing edges of the same color is when . There the only cycle would have to be blue and go through which has no blue outgoing edges by g for and since red and blue were swapped in . Thus there is no such directed monochromatic cycle in , i.e., h is satisfied.
 Case 3: There is an edge with and .

We choose to be the vertex that has a neighbor on and is closest to on , and to be the neighbor of on that is closest to on . Note that , but possibly . The three illustrations in Figure 2 display the three subcases that we sometimes have to treat differently along the construction: , and , and . We define to be the inner triangulated subgraph of with outer cycle . Then is a Whitney graph with respect to . Indeed, in is induced since in is and by the choice of there is no edge between and . Next consider and the neighbor of on . If both, and , are edges in , then is just the edge and we put this edge into the Hamiltonian path. Otherwise is a Whitney graph with respect to , since, by the choice of , there is no edge from to a vertex on between and . However, the embedding of needs to be flipped so that appear in counterclockwise order.
We apply induction to both, (if is not just an edge) and , and concatenate the obtained Hamiltonian paths in and to a Hamiltonian path from to in . We orient the edges incident to towards and color them blue. Note that if then the vertices on that are contained in have no outgoing red edge, since we flipped the embedding of . Similarly the neighbors of in have no outgoing blue edge within .
It is again straightforward to check that b–g follow from b–g with respect to and and the coloring/orientation of the edges incident to and . It remains to show that h is satisfied, i.e., there is no monochromatic directed cycle in . By h with respect to and , such a cycle would contain edges from to and pass through , but has no outgoing edges towards .
 Case 4: .

Let be the neighbor of in . Note that possibly , which is illustrated second in Figure 4. Moreover is not an edge since Case 3 does not apply. Now is a Whitney graph with respect to , since the vertices in are neighbors of and hence is induced. We apply induction to and extend the obtained Hamiltonian path by the edge and orient from to and color it blue. We orient the remaining edges incident to towards and color them red. See Figure 4 for an illustration.
Now b follows from b, e with respect to and the coloring/orientation of the edges incident to , c, d and f follow from c, d and f (and g in case ) with respect to , and e and g follow again from the orientation/coloring of the edges at . Finally, h follows from h with respect to and the fact that has no outgoing and incoming edges of the same color.
 Case 5: .

Let be the neighbor of in . Note that since Case 4 does not apply. Now is a Whitney graph with respect to . However, the embedding of needs to be flipped so that appear in counterclockwise order. Indeed is an induced path since all its vertices are neighbors of . We apply induction to and extend the obtained Hamiltonian path by the edge . We orient the remaining edges incident to towards and color them blue. See Figure 5 for an illustration.
 Case 6: There is an edge with .

We choose to be any neighbor of on . Note that since Case 5 does not apply. We define to be the inner triangulated subgraph of with outer cycle . Then is a Whitney graph with respect to . However, the embedding of needs to be flipped so that appear in counterclockwise order. Indeed, in is induced since in is so and by the choice of there is no edge between and . Next consider and the neighbor of on . The three illustrations in Figure 6 display the three subcases that we sometimes have to treat differently along the construction: , and , and . If both, and , are edges in , then is just the edge and we put this edge into the Hamiltonian path. Otherwise is a Whitney graph with respect to . However, the embedding of needs to be flipped so that appear in counterclockwise order.
We apply induction to both, (if is not just an edge) and , and concatenate the obtained Hamiltonian paths in and to a Hamiltonian path from to in . We swap the colors red and blue in the coloring for , but keep the orientation the same. We orient the edges incident to towards and color them blue. Note that if then the vertices on that are contained in have no outgoing red edge, since we flipped the embedding of . Similarly the neighbors of in have no outgoing blue edge within .
It is again straightforward to check that b–g follow from b–g with respect to and and the coloring/orientation of the edges incident to and . By f there is no outgoing edge at towards . Thus no directed cycle contains and hence from h with respect to and follows h for , i.e., there is no directed monochromatic cycle in .
 Case 7: None of Case 1 – Case 6 applies.

Let be the edge between a vertex and a vertex farthest away from . If no such edge exists set and let be the neighbor of and on . Then and , since otherwise an earlier case would apply.
Let be the neighbor of on . Since earlier cases do not apply, we have . Consider the subgraph of induced by all vertices that are not on but have at least one neighbor in . Since none of Case 2, Case 3 and Case 6 applies, and are the only outer vertices of that are contained in . Let be a shortest path in . (Clearly is connected since is inner triangulated.) We define to be the inner triangulated subgraph of with outer cycle . Then is a Whitney graph with respect to . However, the embedding of needs to be flipped so that appear in counterclockwise order. Indeed, the path is induced since it is a shortest path.
Moreover, if define to be the inner triangulated subgraph of with outer cycle , which clearly is a Whitney graph with respect to . If , then is the neighbor of on and we set to be the edge .
Furthermore we define to be the blocks of , where the numbering is according to their appearance along . For every the graph contains two vertices and that lie on and have a neighbor on . Let be the one that is closer to on . If and have a common neighbor on , then is either just an edge or a Whitney graph with respect to . Otherwise there is a unique vertex in that has two neighbors on , and is a Whitney graph with respect to . Whenever is not just an edge, we flip the embedding of so that appear in counterclockwise order.
We color and orient all the Whitney graphs by induction, where if , we put directed from to into the black path, and whenever , we put directed from to into the black path. In we swap the colors red and blue.
The remaining edges emanating from outside are oriented as follows. All such edges at a vertex on are blue and oriented towards it until the last edge from , which is blue but oriented towards . If there are more edges to the vertex they are red and oriented towards it. See Figure 7 for an illustration.
It is maybe a bit tedious but again straightforward to check that b–g follow from b–g with respect to after swapping red and blue in .
In order to see h, first note that no red edges leave and thus any monochromatic cycle would have to be blue. Moreover, vertices on have no blue outgoing edges towards , i.e., a blue cycle has to use and the explicitly colored edges. The latter form a set of rooted trees, where the only vertices with ingoing blue edges in are the vertices , since these have no outgoing blue edges within the respective any directed blue path continues to the right until eventually reaching and then possibly , which has no blue outgoing edge. This gives h and concludes the proof.∎
3 Triangle Assignments
In this section we will establish the notion of triangle assignments and under which conditions they are sufficient to apply Lemma 2.2 to recursive decompositions of planar triangulations.
Let be a plane triangulation and let and denote the set of filled triangles in and inner faces in , respectively. In particular, is a partition of the triangles in , we have if and only if is just a triangle, and we have if and only if is connected. The set is naturally endowed with a partial order , where whenever the interior of is strictly contained in the interior of . Since any two filled triangles that both contain a third filled triangle (i.e., and ) are necessarily contained in each other ( or ), we have that the partially ordered set has the structure of a rooted tree whose root is the outer triangle and where if and only if lie on a roottoleaf path in with being closer to the root than . Tree is sometimes called the separation tree of , see [14] for an early appearance of the term.
For each triangle , let denote the subgraph of induced by and all vertices inside . We further define to be the plane triangulated graph obtained from by removing for each all inner vertices of . Note that if and only if is connected. For the graphs and with we have that

is a connected triangulation for every ,

every separating triangle of is an inner face of for exactly one ,

the graphs form an edgepartition of .
Properties a–c will enable us to find decompositions of for small , based on decompositions of , , given by the following lemma. Let be the outer triangle of . An outer vertex of different from is called special vertex of , and is denoted by , when it is adjacent to and (indices modulo ). We also say that is the special vertex opposing . Note that are pairwise distinct when , and pairwise coincide when , i.e., . An immediate consequence of Lemma 2.2 is the following.
Lemma 3.1.
Let be a plane triangulation with outer triangle and corresponding special vertices . Then for any the edges of can be partitioned into three forests such that

is a Hamiltonian path of going from to ,

is a spanning tree of ,

is a spanning forest of consisting of two trees, one containing and one containing , unless . In this case .
Proof.
If has only vertices, i.e., , then the decomposition for all has the desired properties. So assume that has at least vertices. Since is connected, is a Whitney graph with respect to . By rotating and flipping we can assume without loss of generality that . Then by Lemma 2.2 the edges of can be oriented and colored black, red and blue with the following properties. The black edges form a Hamiltonian path from to , which we extend by the edge to obtain . Every inner vertex has exactly one outgoing red edge and one outgoing blue edge, every vertex on has an outgoing blue but no outgoing red edge, every vertex on has an outgoing red but no outgoing blue edge, and neither nor have outgoing blue or red edges. Since there are no directed monochromatic cycles. This implies that red and blue edges form a forest each, where the roots of the red trees correspond to the vertices on and the roots of the blue trees . Thus, coloring all edges except from to or red gives and coloring all edges from to blue gives . See Figure 8 for an illustration. We have obtained the desired decomposition of the edges of .
∎
We define for a assignment of a plane triangulation with outer triangle to be a map which satisfies for every and
Proposition 3.2.
Let be a plane triangulation with outer triangle such that or is in no separating triangle. If admits a assignment , , such that , then can be decomposed into two trees and one spanning tree with maximum degree at most , i.e., is decomposable.
Proof.
Without loss of generality assume that is in no separating triangle.
We will prove that decomposes into three trees , where is spanning and has maximum degree , spans , and spans . Even stronger, for every vertex we shall have if or is a special vertex opposing for some , and otherwise. This will be done by induction along the separation tree of .
If is just the triangle we set , , and , and we are done.
Let now be an inclusionminimal filled triangle of . Hence is connected and we have . We apply induction to the graph obtained from by removing all vertices inside , together with the assignment obtained from by restricting to , and obtain three trees with the desired properties. We have to include the edges of into our decomposition.
Note that at most two vertices of also appear in but by the choice of these cannot be and simultaneously. Without loss of generality let and be such that . Since is in no separating triangle, we may also assume that . Let be a partition of the edges in as given by Lemma 3.1 for , and . (Recall that .) We include into , into , and into . Clearly, by c we obtain a set of trees with the desired properties, except that it remains to bound the degrees of vertices in .
First consider any inner vertex of . If is the vertex in opposing , then has one incident edge in and no incident edge in , and thus with . If is an inner vertex of different from , then has two incident edges in and no incident edge in , and thus with .
Now consider any vertex of . Note that is a special vertex opposing for some if and only if that is also the case for and . If , then and , as has exactly one incident edge in , which implies the desired degree of in for this case. Finally, if is any vertex of different from , then and , as has no incident edge in , which implies the desired degree of in also for this case. ∎
4 Proofs of Decomposition Results
In this section, we show that all the classes of concern admit triangle assignments qualifying for Proposition 3.2 with respect to the claimed parameters. This will suffice to prove the Theorems of this paper.
Now Proposition 3.2 immediately implies Theorem 1.1, namely that every connected triangulation is decomposable.
Proof of Theorem 1.1.
Clearly, if is a connected plane triangulation, then consists only of the outer face and is a assignment of . Since contains no separating triangles, applying Proposition 3.2 yields the result. ∎
We remark that some plane triangulations (such as the one in the left of Figure 9) are not connected and still admit a assignment qualifying for Proposition 3.2.
Next, we shall turn our attention to general plane triangulations. Recall that denotes the set of all inner faces of a plane triangulation .
Lemma 4.1.
Let be a plane triangulation on at least four vertices and be a special vertex of . Then there is a map with for every and
Proof.
Let be two outer vertices adjacent to the special vertex and be the inner face formed by , and . Consider a plane straightline embedding of in which the only horizontal edge is . For every inner face , let be the vertex in with the middle coordinate. Moreover, let . It is easily seen that has the desired properties. ∎
Lemma 4.2.
Every plane triangulation admits a assignment.
Proof.
Let be any plane triangulation with outer triangle . By b we have that . Similarly, every inner vertex of is an inner vertex of for exactly one . By Lemma 4.1 we have maps mapping the inner faces of to inner vertices of such that every inner vertex is hit twice, except for a fixed special vertex of , which is hit three times. From this collection of maps we obtain a desired assignment by setting for some outer vertex of and whenever . ∎
We can now give an alternative proof of Gonçalves’s result [5] that every planar graph is decomposable and indeed a slight strengthening thereof, since in [5] the forests of the decomposition are not necessarily connected. That is, we prove Theorem 1.3 stating that every planar triangulation decomposes into two trees and one spanning tree of maximum degree .
Proof of Theorem 1.3.
Lemma 4.2 gives a assignment for a planar triangulation for any prescribed outer triangle , where moreover the choice of is arbitrary. Since clearly any planar triangulation contains an edge , that is not contained in a separating triangle we can choose an outer triangle containing . Use Lemma 4.2 to get a assignment such that is a vertex of and apply Proposition 3.2 to get the desired decomposition. ∎
Next, we shall turn our attention to Hamiltonian planar triangulations.
Lemma 4.3.
Let be a plane triangulation with outer triangle .

If admits a Hamiltonian path, then admits a assignment with and

If admits a Hamiltonian path, then admits a assignment with and

If admits a Hamiltonian path, then admits a assignment with and
Proof.
Let be a plane triangulation with outer triangle and let be a fixed Hamiltonian path in for items a,b or in for item c. We shall prove all three items by induction on , the number of filled triangles in . If , i.e., is connected, then a desired assignment is given by for a,b and for c, respectively.
So assume that , i.e., has at least one separating triangle. Let denote the set of all inclusionmaximal separating triangles in . For each , let denote the inclusionmaximal subpath of consisting only of edges in . Then has distinct endpoints, which both lie on . Note that is a Hamiltonian path either in , or in for one . Moreover, for each and thus for , by c.
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