Given that it is known that a graph is locatable if and only if it is 4-colourable (see [6, 8]), in this section we consider minimizing the number of lines in such a representation. In other words, given a concrete 4-colorable graph, how many lines are needed to represent it?
Although Theorem 0.1 gives an upper bound for the number of lines needed to rep-resent a graph, that bound is of course not always optimal. For instance,
Proposition 2.1. Any outerplanar graph can be located in two lines.
Proof (sketch). The proof can be obtained by induction, taking into account that any maximal outerplanar graph has always an ear. Figure 1 shows how to add an ear to an edge that is in the outerface (the tiny triangles in the central image mean that one of those points is in the outerface; then we move —if needed— all the drawing in order to
leave room for the new vertex).
In fact, we can describe the structure of any graph locatable in two or three lines.
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Figure 1. How to add a vertex of degree 2.
Theorem 2.2. Let G= (V, E) be a graph such that χ(G)≤4. Then,
(1) G is locatable in two lines if and only if V can be partitioned into two subsets V1,V2, such that the subgraph of G induced byVi is a disjoint union of paths.
(2) Gis locatable in three lines if and only if V can be partitioned into three subsets V1,V2,V3, such that the subgraph ofGinduced byV1 is a disjoint union of paths, andV2 andV3 are independent sets.
We can use Theorem 2.2 to obtain examples of planar graphs that need the four lines of Theorem 0.1 to be located in the plane. For instance, the graph depicted in Figure 2.
Basically there is only one 4-coloring and there is no possible assignment of the colors avoiding vertices of degree 3 in the subgraph induced by two colors.
Figure 2. A planar graph that needs four lines to be located.
Additionally, Theorem 2.2 is one of the main tools used to prove the following results.
Theorem 2.3. If∆(G)≤3(the maximum degree ofG), thenGis locatable in two lines.
Proof (sketch). We have to obtain a partition of the vertices inGverifying the hypothesis of Theorem 2.2 (1). In a first step, ifG is notK4, that can be located trivially in two lines, we colorGwith 3 colors. Then, if there are vertices of degree 3 in the subgraph of G induced by colors 1 and 2, this means that there is a vertex v with the color 1 with three neighbors with the color 2 (or the other way around). In that case, we can change the color ofv to 3.
So, if we assume that there is not vertex of degree 3 in the subgraph ofGinduced by colors 1 and 2, we have to check that there is no cycle in such a subgraph. If such a cycle exists, we can change the color of one of the vertices to a new color 4. The only problem could be now to find a vertex of degree 3 in the subgraph ofGinduced by colors 3 and 4,
124 Compact grid representation of graphs
but that case can be avoided again by changing the color of that vertex to 1. After all this process,V1 will have all the vertices with color 1 or 2 and V2 the vertices with color
2 or 4.
Observe that, in the previous theorem,Gcould be non-planar. It is possible to relax the condition on the maximum degree of G, but adding the additional hypothesis of planarity.
Theorem 2.4. Let G be a planar graph with ∆(G) ≤ 4. Then G is locatable in three lines.
Proof (sketch). The proof is in some way similar to the proof of Theorem 2.3. Starting with an actual embedding of G= (V, E), we complete it by adding extra dummy edges to the faces of even length. In this way, we are sure that in a 4-coloring ofGwe have no face with only two colors. We are going to change this initial coloring in order to obtain a partition ofV fulfilling the conditions of Theorem 2.2 (2). We assign vertices with colors 1 or 2 toV1, vertices with color 3 toV2 and vertices with color 4 toV3. As in Theorem 2.3, with a simple change, we can avoid vertices of degree 3 or 4 in the subgraphhV1iinduced byV1. In the case of a cycle, we can change the colors using Kempe’s chains starting in the cycles containing the most points in their interior. Now, some other cycles can be obtained, but those cycles are going to have less points in their interior, so we can apply
again the same method, if needed.
References
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[5] H. Fraysseix, J. Pach, and R. Pollack. How to draw a planar graph on a grid.Combinatorica10(1) (1990), 41–51.
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