# Maximizing the Degree of (Geometric) Thickness-*t* Regular Graphs

## Abstract

In this paper, we show that there exist \((6t-1)\)-regular graphs with thickness *t*, by constructing such an example graph. Since all graphs of thickness *t* must have at least one node with degree less than 6*t*, this construction is optimal. We also show, by construction, that there exist 5*t*-regular graphs with geometric thickness at most *t*. Our construction for the latter builds off of a relationship between geometric thickness and the Cartesian product of two graphs.

## 1 Introduction

A straight-line drawing \(\Gamma \) of a graph \(G=(V,E)\) is a mapping of vertices to points in the plane and edges to curves between the endpoints. A drawing is planar if and only if the edges only intersect at the endpoints. For convenience, we often refer to the vertices of a given graph *G* as *V*(*G*) and the edges as *E*(*G*). The *order* of a graph, |*V*|, is the number of vertices in the graph. The *thickness* \(\theta (G)\) of a graph *G* is the minimum number of planar subgraphs whose union forms *G*. The edges of these subgraphs form a partitioning of *E*(*G*). For convenience, we identify each partition with a unique color.

The *geometric thickness* of *G*, \(\bar{\theta }(G)\), is the smallest integer *t* such that there is a *straight-line drawing* \(\Gamma (G)\) whose edges can be colored with *t* colors such that no two edges with the same color intersect, except at the endpoints. That is, each coloring (layer) is a planar drawing. We refer to such a drawing, with associated coloring, as a *t-layered planar drawing*.

When discussing a drawing \(\Gamma (G)\), it often helps to describe the grid size \(h \times w\) of the drawing. We take the convention that the vertices (points) all lie on integer coordinates. In addition, to count the width and height we use the number of grid points in the smallest axis aligned bounding box of the drawing. This convention means that a single vertex has dimension 1, as opposed to 0. See, for example, Fig. 2a, which has height 5 and width 4.

Generalized by Tutte [8], the thickness problem began as an exploration of the biplanarity of a graph, where biplanar refers to having thickness two [1, 4, 7]. Numerous research articles have also explored geometric thickness and its relationship to thickness, e.g. [2, 5]. Since the research in graph thickness is too large to summarize adequately in this technical note, we refer the interested reader to a survey by Mutzel *et al.* [6].

In [3], Durocher *et al.* explore the relationship between the colorability and minimum degree of thickness-*t* graphs and pose several questions. We investigate the question of finding the largest *k* for a given *t*, such that there exist *k*-regular graphs of (geometric) thickness *t*, as this provides a lower bound on the minimum degree. We provide a construction that shows that there exist \((6t-1)\)-regular graphs of thickness *t*, which is optimal since every graph of thickness *t* must have at least one node of degree less than 6*t*. This observation can be seen by recalling that the average degree of any planar graph is less than 6. We also show that there exist 5*t*-regular graphs of geometric thickness at most *t*, but do not claim that 5*t* is optimal.

## 2 \((6t-1)\)-Regular Thickness-*t* Graphs

To prove that there exist \((6t-1)\)-regular thickness-*t* graphs, we start by creating a graph \(\mathcal{G}\) with \(48(t-1)\) vertices of degree 6 and 48 vertices of degree 5 and a larger graph \(\mathcal{G}_C\) composed of several disjoint copies of \(\mathcal{G}\). We then create *t* layers of \(\mathcal{G}_C\) using the same vertex set but in different permutations to ensure that every vertex has degree 5 in exactly one layer, leading to the following theorem.

### **Theorem 1**

For any \(t \in \mathbb {Z}^+\), there exist \((6t-1)\)-regular graphs with thickness *t*.

### *Proof*

For \(t=1\), it is well-known that there exist 5-regular planar graphs. (See for example, Fig. 2d). Therefore, we assume that \(t > 1\).

*t*containing exactly 48 vertices of degree 5 and \(48(t-1)\) vertices of degree 6.

Let \(\mathcal{G}_C\) be the graph of order 48*tC* formed by the union of \(C=48t\) disjoint copies of \(\mathcal{G}\).^{1} We shall form our \((6t-1)\)-regular graph \(G=(V,E)\) of thickness *t* by creating *t* layers of \(\mathcal{G}_C\) where each layer uses the same vertex set, in a different permutation. That is, *V* consists of 48*tC* vertices, and *E* is the union of *t* sets, \(E_0, E_1, \dots , E_{t-1}\), with each set representing a different layer (color) of the graph. For any \(v \in V\) and \(0 \le i < t\), we refer to \(\pi _i(v)\) as the permuted vertex in \(\mathcal{G}_C\) of the *i*-th layer. For any two vertices \(v,w \in V\), the edge (*v*, *w*) is in \(E_i\) if and only if \((\pi _i(v),\pi _i(w)) \in E(\mathcal{G}_C)\), the permuted vertices have an edge in \(\mathcal{G}_C\). See Fig. 1b for an illustration using a simpler base graph, two copies of a 4-cycle.

- 1.
Every vertex gets mapped to a degree-5 vertex

*exactly once*. - 2.
If \((\pi _i(v),\pi _i(w)) \in E(\mathcal{G}_C)\) then \((\pi _j(v),\pi _j(w)) \not \in E(\mathcal{G}_C)\) for all \(j \ne i\). That is, we do not have any duplicate edges.

If we guarantee these two conditions, then we know that the degree of every vertex in *G* is exactly \(6t-1\) and given the construction we know that the graph has thickness *t*, completing the proof.

We refer to a vertex in \(\mathcal{G}_C\) by the notation \(\rho _{a,\ell ,c}\) where \(0 \le a < 48\), \(0 \le \ell < t\), \(0 \le c < C\). We conceptually partition the \(16(t-1)\) nested triangles of \(\mathcal{G}\) into \(t-1\) groups of 16 triangles, called *levels*. In addition, we refer to level 0 as the group formed by the outer and inner (degree-5) vertices. The index \(\ell \) corresponds to vertices within level \(\ell \). The index *c* represents one of the *C* disjoint copies of \(\mathcal{G}\). Consequently, if two vertices share an edge, they must have the same *c*-index. For any given level \(\ell \) and copy *c*, there are exactly 48 vertices of \(\mathcal{G}_C\). The specific ordering of these 48 vertices does not particularly matter so long as there is no edge connecting \(\rho _{a,\ell ,c}\) and \(\rho _{a,\ell ',c}\). That is, two vertices with the same indices *a* and *c* cannot share an edge. By ensuring that the innermost three vertices of one level do not share an *a*-index with the outermost three vertices of the next level, we can easily construct such an ordering.

*G*in the same manner such that \(\pi _0(v_{a,\ell ,c}) = \rho _{a,\ell ,c}\). We define the permutations as follows:

*i*ranges from 0 to \(t-1\), the permutation guarantees that \(\ell + i \equiv 0\mod t\) for exactly one value of

*i*.

Suppose now that condition 2 does not hold. That is, there exist two vertices \(v=v_{a,\ell ,c}\) and \(w=v_{a',\ell ',c'}\) in *V*(*G*) and two layers \(i \ne j\) such that \((\pi _i(v), \pi _i(w)) \in E(\mathcal{G}_C)\) and \((\pi _j(v), \pi _j(w)) \in E(\mathcal{G}_C)\). Assume, without loss of generality, that \(i < j\). Since two vertices with the same *a*-index do not share an edge, we know that \(a \ne a'\). In addition, since the *c* copies are disjoint in \(\mathcal{G}_C\), we know that \(c+ai \equiv c'+a'i\mod C\) (or else there would be no edge in the *i*-th permutation). Similarly, we know that \(c+aj \equiv c'+a'j\mod C\). Subtracting the two values, we see that \(a(j-i) \equiv a'(j-i)\mod C\). However, since \(0 \le a,a' < 48\) and \(0 < j-i < t\) and because we chose to use \(C=48t\) copies, we know that \(0 \le a(j-i), a'(j-i) < C\). Thus, since \(a \ne a'\), the equivalence only holds when \(j=i\), a contradiction. \(\square \)

## 3 5*t*-Regular Geometric Thickness-*t* Graphs

Although we have shown an optimal example for the thickness problem, when we look at the same problem with the restriction that the graph have *geometric* thickness *t*, the optimal solution appears more challenging. Before we prove our main theorem for this section, we first discuss a relationship between the *Cartesian product* of two graphs, \(G_1 \square G_2\), and geometric thickness.

### **Lemma 1**

The Cartesian product of two graphs, \(G = G_1 \square G_2\), has geometric thickness at most \(t_1 + t_2\) where \(t_1\) and \(t_2\) are the geometric thicknesses of \(G_1\) and \(G_2\), respectively. Furthermore, if \(G_1\) (resp., \(G_2\)) can be drawn on an integer grid of dimension \(h_1 \times w_1\) (resp., \(h_2 \times w_2\)) such that each row and column contains at most one vertex, then *G* can be drawn on an integer grid of dimension \(h_1h_2 \times w_1w_2\) such that each row and column contains at most one vertex.

### *Proof*

Let \(G_1\) and \(G_2\) be two graphs with geometric thicknesses \(t_1\) and \(t_2\) respectively. In addition, for \(i \in \{1,2\}\), let \(\Gamma _i\) be a \(t_i\)-layered planar drawing of \(G_i\). We treat the \(t_1\) colors used to partition the edges of \(\Gamma _1\) as distinct from the \(t_2\) colors of \(\Gamma _2\). Note, if the drawings do not have the property that each row and column contains at most one vertex (that is, if two vertices share the same *y*-coordinate or *x*-coordinate), we can slightly rotate the drawing so that this property holds, although the grid dimension would increase significantly. Therefore, we assume that the drawings are given on an integer grid of dimension \(h_i \times w_i\), for \(i \in \{1,2\}\) such that no two vertices share the same *x*-coordinate or *y*-coordinate.

We now argue that no two edges in the same layer (with same color) cross. Suppose for the sake of contradiction that two edges with color *c* cross. First, assume that *c* is one of the \(t_1\) colors from \(\Gamma _1\). Since \(\Gamma '_1\) is a \(t_1\)-layered planar drawing, we know that the two crossing edges cannot be from the same copy of \(\Gamma '_1\). However, the origins of the \(\Gamma '_1\) copies are separated by at least \(w_1\) units, because they are placed at vertices of \(\Gamma '_2\), which have been scaled horizontally by a factor of \(w_1\), and because no two vertices in \(\Gamma '_2\) share the same column. Since \(w_1\) is the width of \(\Gamma '_1\), no two copies of \(\Gamma '_1\) can intersect each other. So, *c* cannot be one of the \(t_1\) colors. Similarly, *c* cannot be one of the \(t_2\) colors. Since these are the only colors used, we have a contradiction, and the drawing is a \((t_1 + t_2)\)-layered planar drawing. \(\square \)

### **Theorem 2**

For any \(t \in \mathbb {Z}^+\), there exist 5*t*-regular graphs with geometric thickness at most *t*.

### *Proof*

Our construction for such a graph starts with any 5-regular graph *G*. To use the integer grid property of Lemma 1, we create a drawing \(\Gamma \) of *G* such that every vertex has a unique column and row; see Fig. 2d.

We now simply compute \(\mathbb {G}= G \square G \square \cdots \square G\), applying the Cartesian product \(t-1\) times. This results in a 5*t*-regular graph and by Lemma 1 we know that \(\mathbb {G}\) has thickness at most *t*. Using the example drawing from Fig. 2d, we observe that the resulting drawing has area \(18^t \times 18^t\) and that \(\mathbb {G}\) has order \(12^t\). \(\square \)

## 4 Conclusion and Open Questions

Ideally, to create a thickness-*t* regular graph from \(\mathcal{G}\), we would simply create *t* permutations of the vertex order with each permutation corresponding to a separate layer to ensure that every vertex is assigned a degree-5 role exactly once and that two vertices never share an edge in more than one permutation layer. Although this is certainly plausible, we did not see a simple description for such a permutation that would not result in a large case analysis. In addition, our aim was not to construct the smallest order of such a regular graph but simply to maximize the degree.

### *Question 1*

What is the smallest \((6t-1)\)-regular graph of thickness *t*?

Our solution for geometric thickness *t* is very generalized and has plenty of room to add extra edges. It seems possible that one might be able to create regular graphs with larger degree while still maintaining the same thickness.

### *Question 2*

What is the largest *k* such that there exists a *k*-regular graph of geometric thickness *t*? Or is 5*t* optimal?

In Lemma 1, we were careful to state that our Cartesian product produced graphs with geometric thickness *at most* \(t_1 + t_2\), because we did not prove that the geometric thickness was not less than this value. This is clearly possible as the Cartesian product of two line segments yields a planar graph (a 4-cycle). Nonetheless, \(\mathbb {G}\) has geometric thickness *exactly* *t* for \(t < 7\).

### *Question 3*

Does the graph \(\mathbb {G}\) have geometric thickness *exactly* *t* for all \(t \in \mathbb {Z}^+\)?

*We thank the anonymous reviewers for their helpful comments and suggestions.*

## Footnotes

- 1.
Although this means there are \((48t)^2\) vertices in \(\mathcal{G}_C\), it is more convenient to refer to

*C*distinctly for now.

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