1 Introduction
For a positive integer , a interval representation of a graph is a set where is the union of at most intervals on the real line representing vertex , such that () is an edge of if and only if . In 1979 Trotter and Harary [16] introduced the interval number of , denoted by , as the smallest such that has a interval representation. In 1983 Scheinerman and West [13] constructed planar graphs with interval number at least and proposed a proof for the following:
Theorem 1.1 (Scheinerman, West [13]).
If is planar, then .
In the present paper, we show that the original proof has a flaw and give a different proof of Theorem 1.1.
Related work.
A graph with is called interval graph. A concept closely related to the interval number is the track number of , denoted by , which is the smallest such that is the union of interval graphs. Equivalently, is the smallest such that admits a interval representation that is the union of interval representations, each on a different copy of the real line (called a track) and each containing one interval per vertex. More recently, Knauer and Ueckerdt [11] defined the local track number of , denoted by , to be the smallest such that is the union of interval graphs, for some , where every vertex of is contained in at most of them. It is easy to see that for every graph we have .
Gonçalves [7] proved that the track number of planar graphs is at most , which is bestpossible [8]. It is an open problem whether there is a planar graph with local track number [11]. Balogh et al. [2] show that any planar graph of maximum degree at most has interval number at most . For further recent results about interval numbers of other classes of planar graphs and general graphs, let us refer to [3] and [6], respectively.
2 Preliminaries
All graphs considered here are finite, simple, undirected, and nonempty. If is a graph and is a interval representation of for some , we say that a subset of the real line is intersected or covered by some if . A vertex has a broken end if is an endpoint of an interval in and is not covered by any other for . We say that a vertex is displayed if contains a portion (i.e., a nonempty open interval) not intersected by any other for . An edge is displayed if contains a portion not intersected by any other for .
The depth of the representation is the largest integer such that every point of the real line is covered by at most sets with . For , the depth interval number of , denoted by , is the smallest such that admits a interval representation of depth . Scheinerman and West [13] proved that there are planar graphs with , while Gonçalves [7] proved that for all planar graphs . In Theorem 4.2 we obtain that for any connected planar graph it holds . Finally, we shall prove here that for all planar graphs , which is also the original claim of Scheinerman and West.
3 The approach of Scheinerman and West
To explain the proof strategy of Scheinerman and West, we introduce some more terminology from their paper. For a plane embedded graph let denote the (outerplanar) subgraph induced by its external vertices, i.e., those lying on the unbounded, outer face. The edges of that bound the outer face are the external edges, while those that bound two inner faces are the chords. The graph is considered together with its decomposition into its inclusionmaximal connected subgraphs, called blocks. Furthermore, we fix for each component of a noncutvertex as the root of that component and say that is the root of its corresponding block . For each remaining block of let be the cutvertex of in that is closest to the root of the corresponding component. For a chord of contained in block with root and a third vertex of , say that is on the side of if and lie in different connected components of .
The argument of Scheinerman and West proceeds by induction on the number of vertices in with a stronger induction hypothesis on how , i.e., the subgraph of induced by the external vertices, is represented. Every edge of shall be represented in such a way that for a possible future vertex which is adjacent to and we can define only one interval that has at least two of the following three properties:
If is displayed, it is easy to find an interval for having properties P1 and P2. Moreover if , respectively , has a broken end, it is easy to satisfy P1, respectively P2, and P3. However if is not displayed and neither nor has a broken end, Scheinerman and West propose to alter the existing representation of another vertex whose interval covers an endpoint of . As their modification, which is depicted in Figure 1, splits an interval of into two, it is necessary that appears at most twice so far, meaning that consists of at most two intervals. This shall be achieved by specifying the current representation of quite precisely.
Given a rooted plane embedded graph , i.e., with a fixed root for each block of , a representation is called Pspecial if every vertex is displayed and each of the following holds:

Each root is represented by one interval and every other external vertex is represented by at most two intervals.

For each block of all edges incident to are displayed.

Each nondisplayed edge of any block of is assigned to an endpoint of or , say , such that the following hold.

If is an external edge, the endpoint is a broken end.

If is a chord, the endpoint is a broken end, or is covered by for only one other vertex , where additionally is on the side of and is an external edge of . An endpoint satisfying this condition is called reusable for edge .

For each endpoint there is at most one edge assigned to .

Each vertex covers at most one endpoint which has an edge assigned.

Now Scheinerman and West propose to show by induction on the number of vertices in that every rooted plane embedded graph admits a Pspecial interval representation. They remove a small set of carefully chosen external vertices from , induct on the smaller instance to obtain a Pspecial representation, create intervals for the removed vertex/vertices, extend and/or alter the existing representation, and argue that the result is a Pspecial interval representation of .
The problem lies in the last step. Some removed vertex may cover an edge that is external in , but a chord in . But it may be that some nondisplayed chord of is assigned to an endpoint of that is reusable and covered by . This assignment cannot be kept, since is no longer external, which however is required for reusability as it is defined in cb above. Thus, the invariants of the induction cannot be maintained.
Let us explain in more detail below (discussing only the relevant cases) on basis of a small example graph that the above problem can indeed occur, and why some straightforward attempt to fix it does not work. To this end, consider the plane embedded vertex graph in the topleft of Figure 2. The thick edges highlight the subgraph , which has only one block , and let us pick vertex to be the root of that block, i.e., . The base case of the inductive construction is an independent set of vertices. Otherwise, we identify a set of external vertices that we want to remove as follows. Consider a leafblock of , i.e., one that contains no root of another block, its root , and distinguish the following cases.
 Case I: Every chord of is incident to .

In this case label the vertices of as , , in this cyclic order around the outer face of . If , let , where is the smallest integer greater than or equal to for which has degree in .
 Case II: Some chord of is not incident to .

Consider an inner face of that does not contain and is bounded by only one chord , and distinguish further.
 Case IIa: is a triangle.

Let , where is the third vertex in .
 Case IIb: has at least four vertices.

In this case label the vertices of as , , in this cyclic order around , and let .
Figure 2 shows how our example graph is reduced to an independent set according to these rules. Next, we consider these steps in reverse order and construct an interval representation following the case distinction as proposed in [13]. Let us refer to Figure 3 for a stepbystep illustration of this construction. The problem arises in Step 5.
 Step 1: base case.

Represent with a single interval.
 Step 2: Case I, .

Insert intervals for into the displayed interval for . Place overlapping, displayed intervals for in an unused portion of the real line.
 Step 3: Case IIa, .

As the chord is displayed, called subcase (1), add a displayed interval for and place the second interval available for in the displayed portion for . Assign one broken end of to each of the external edges and .
 Step 4: Case IIa, .

As the chord is not displayed, but assigned to a broken end of , called subcase (2), add the displayed interval for so that it overlaps at . Add a second interval for in the displayed portion of . Assign the chord to the reusable endpoint , which is here covered by with being an external edge.
 Step 5: Case IIa, .

As the chord is displayed, we are in subcase (1) again. We add a displayed interval for and place a second interval for in the displayed portion for . Now the problem is that the chord can no longer be assigned to the endpoint , since the edge is no longer external.
One might be tempted to assume that the problem can be fixed by relaxing the definition of reusability as follows. Instead of being external, maybe it suffices to require that appears at most twice so far. Let us call this new concept almost reusability. However, the assumption that is external is crucial for Case IIb, as we will demonstrate in Steps 6 and 7 below.
 Step 6: Case IIb, .

Assign an interval in the displayed interval for , and assign an interval in the displayed interval for . As the chord is not displayed, but assigned to an almost reusable endpoint of , we are in subcase (2c) of [13]. Here it is concluded in [13] that the vertex covering must be due to the externality of edge . Hence it would suffice to place overlapping displayed intervals for and in an unused portion of the line and add intervals for the inner neighbors of in the displayed portion for , , and , according to the subset of these vertices that they are adjacent to.
However, in our example we have , which at least appears only twice so far, but which is an inner neighbor of and , and thus has to spend its third interval in the displayed portion of . We end up with chord still assigned to the endpoint that is covered by , but neither is external, nor is appearing only twice so far. In particular, we are not prepared to insert a future vertex adjacent to and , as we illustrate in Step 7.
 Step 7: Case IIa, .

The chord is not displayed, but assigned to the endpoint of , which is covered by with however being internal and appearing three times already. Thus, if we apply the proposed modification as illustrated in Figure 1, we split one interval of into two, causing to consist of four intervals.
This concludes our example. Let us remark that the fourth interval for in Step 7 of Figure 3 might seem superfluous, but is actually needed to give a displayed portion. This in turn is necessary as there might be a neighbor of of degree . For example, with inside face , it would have been inserted between Steps 5 and 6, according to Case I with , spending one interval in the displayed portion of .
Also note that in this particular example other ways of assigning endpoints in Step 5 would have allowed the process to continue. However, the first author together with Daniel Gonçalves tried some time to obtain new invariants to fix this induction but did not succeed.
In his thesis [12], Scheinerman proposes a different argumentation for handling the subcase (2c) of Case IIb, which however also works only if (corresponding to edge in our example) is an external edge of .
4 An alternative proof of Theorem 1.1
In this section we give a new proof for Theorem 1.1, i.e., that every planar graph has interval number at most . A triangulation is a plane embedded graph in which every face is bounded by a triangle. As every planar graph is an induced subgraph of some triangulation and the interval number is monotone under taking induced subgraphs, we may assume without loss of generality that is a triangulation.
A triangle in is nonempty if its interior contains at least one vertex of . We shall construct a interval representation of by recursively splitting along its nonempty triangles. This leaves us with the task to represent connected triangulations, i.e., triangulations whose only nonempty triangle is the outer triangle, and to “glue” those representation along the nonempty triangles of . More precisely, we shall roughly proceed as follows:

Consider a nonempty triangle with inclusionminimal interior and the set of all vertices in its interior.

Call induction on the graph , obtaining a interval representation of with additional properties on how inner faces are represented.

Since the subgraph of induced by is connected, we can utilize a recent result of the second and fourth author to decompose into a path and two forests.

Using this decomposition, we define a interval representation of that coincides with on .
We remark that the essentials of the construction in step d can be already found in [1]. Also, the decomposition of a triangulation along its nonempty triangles is a common method in the field of intersection graphs; see e.g., [15, 4, 5]. Hence the key to our new proof is the most recent [10] decomposition used in step c, which we shall state next.
Let be a connected triangulation with outer vertices . We denote by the unique inner vertex of adjacent to and (if there were several, would not be connected), and call the vertex the vertex opposing . Similarly, the vertex opposing and the vertex opposing are defined. Note that since is connected we have that all coincide if , and are pairwise distinct if . We use the following recent result of the second and fourth author (Lemma 3.1 in [10]), which we have adapted here to the case of connected triangulations (see Figure 4 for an illustration):
Lemma 4.1 (Knauer, Ueckerdt [10]).
Let be a plane connected triangulation with outer triangle and corresponding opposing vertices . Then 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 .
Note that the conditions on the decomposition in Lemma 4.1 imply that the edge is in and the edge is in . We are now ready to give our new proof of Theorem 1.1.
Proof of Theorem 1.1.
We have to show that every planar graph admits a interval representation of depth at most , in particular that . As every planar graph is an induced subgraph of some planar triangulation and the interval number is monotone under taking induced subgraphs, we may assume without loss of generality that is a triangulation.
We proceed by induction on the number of vertices in , showing that admits a interval representation of depth with the additional invariant that (I1) every vertex is displayed, and (I2) every inner face contains at least one displayed edge.
The base case is , i.e., is a triangle with vertices . In this case, it is easy to define a interval representation of with invariants (I1) and (I2). For example, take , , and , and note that edges and are displayed.
Now assume that , i.e., contains at least one nonempty triangle. Let be a nonempty triangle in with inclusionminimal interior among all nonempty triangles in . Let be the (nonempty) set of vertices in the interior of , be the triangulation induced by , and be the connected triangulation induced by . By induction hypothesis there exists a interval representation of of depth satisfying the invariant (I1) and (I2). In particular, the three vertices of , which now form an inner face of , are already assigned to up to three intervals each. We shall now extend this representation to the vertices in .
By invariant (I2) at least one edge of , say , is a displayed edge. Consider the vertices opposing in , respectively, and let , , be the decomposition of given by Lemma 4.1. For convenience, if , let additionally have vertex as a onevertex component. We define, based on this decomposition, three intervals for each vertex as follows.

Represent the path on an unused portion of the real line by using one interval per vertex, and in such a way that every vertex and every edge is displayed. Let this representation be denoted by .

Consider to be the root of the tree , and for each vertex in consider the father of in , i.e., the neighbor of on the to path in . Create a new interval for strictly inside the displayed portion of . If , such a portion exists in as (I1) holds for , and if , such a portion exists in . Let these new intervals be denoted by .

Consider the three trees that are the components of , that is, of after removing the edge . Say contains vertex (and possibly no other vertex), contains , and contains . Consider , respectively and , to be the root of , respectively and . For each vertex in , respectively and , consider the father of in , respectively and , and create a new interval for in the displayed portion of . (Again, if , such a portion exists in as (I1) holds for , and if , such a portion exists in .) Let these new intervals be denoted by .

Finally, create a new interval in the displayed portion of edge .
Defining for each , it is straightforward to check that is a interval representation of of depth . Moreover, as the edge , every inner edge of except for and , and every vertex of is displayed, this representation satisfies our invariants (I1) and (I2), which concludes the proof. ∎
As already mentioned above, Axenovich et al. [1] essentially used the construction above to prove that every graph whose edges decompose into forests and another forest of maximum degree admits a interval representation of depth at most . Note that if is a connected triangulation with outer triangle , the decomposition of given by Lemma 4.1 can be easily extended to a decomposition of all edges in into two forests and a path, which immediately gives the following.
Theorem 4.2.
If is planar and connected, then .
Given that the largest interval number among all planar graphs is , while the largest track number is , it remains open to determine whether the largest local track number among all planar graphs is or , c.f., [11, Question 19]. We feel that the gluing of several triangulations along separating triangles is likely to be possible along the lines discussed in [11] for planar trees. However, finding a local track representation of a connected planar triangulation, strengthening Theorem 4.2, seems to be more difficult. What is the the largest local track number among all planar graphs?
Acknowledgements.
We would like to thank Maria Axenovich and Daniel Gonçalves for fruitful discussions. Moreover, we thank Ed Scheinerman and Douglas West for their helpful comments on an earlier version of this manuscript and for providing us a copy of Ed’s PhD thesis. The second author was partially supported by ANR projects ANR16CE40000901 and ANR17CE400015.
References
 [1] M. Axenovich, A. Beveridge, J. P. Hutchinson, and D. B. West. Visibility number of directed graphs. SIAM J. Discrete Math., 27(3):1429–1449, 2013.
 [2] J. Balogh, P. Ochem, and A. Pluhár. On the interval number of special graphs. J. Graph Theory, 46(4):241–253, 2004.
 [3] J. Balogh and A. Pluhár. A sharp edge bound on the interval number of a graph. J. Graph Theory, 32(2):153–159, 1999.

[4]
J. Chalopin and D. Gonçalves.
Every planar graph is the intersection graph of segments in the
plane.
In
Proceedings of the fortyfirst annual ACM symposium on Theory of computing
, pages 631–638. ACM, 2009.  [5] S. Chaplick and T. Ueckerdt. Planar graphs as VPGgraphs. In International Symposium on Graph Drawing, pages 174–186. Springer, 2012.
 [6] A. B. De Queiroz, V. Garnero, and P. Ochem. On interval representations of graphs. Discrete Appl. Math., 202:30–36, 2016.
 [7] D. Gonçalves. Caterpillar arboricity of planar graphs. Discrete Math., 307(16):2112–2121, 2007.
 [8] D. Gonçalves and P. Ochem. On star and caterpillar arboricity. Discrete Math., 309(11):3694–3702, 2009.
 [9] M. Jiang. Recognizing dinterval graphs and dtrack interval graphs. Algorithmica, 66(3):541–563, 2013.
 [10] K. Knauer and T. Ueckerdt. Decomposing connected planar triangulations into two trees and one path. J. Comb. Theory, Ser. B (accepted).
 [11] K. Knauer and T. Ueckerdt. Three ways to cover a graph. Discrete Math., 339(2):745–758, 2016.
 [12] E. R. Scheinerman. Intersection Classes and Multiple Intersection Parameters of Graphs. PhD thesis, Princeton University, 1984.
 [13] E. R. Scheinerman and D. B. West. The interval number of a planar graph: Three intervals suffice. J. Comb. Theory, Ser. B, 35:224–239, 1983.
 [14] P. Stumpf. On covering numbers of different kinds. Bachelor thesis, Karlsruhe Institute of Technology, August 2015.
 [15] C. Thomassen. Interval representations of planar graphs. J. Comb. Theory, Ser. B, 40(1):9–20, 1986.
 [16] W. T. Trotter and F. Harary. On double and multiple interval graphs. J. Graph Theory, 3:205–211, 1979.
 [17] D. B. West and D. B. Shmoys. Recognizing graphs with fixed interval number is NPcomplete. Discrete Appl. Math., 8:295–305, 1984.
Comments
There are no comments yet.