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Planar Graphs and Coloring Math 443/543 Graph Theory Notes 5: Planar graphs and coloring David Glickenstein October 10, 2014 1 Planar graphs The Three Houses and Three Utilities Problem: Given three houses and three utilities, can we connect each house to all three utilities so that the utility lines do not cross. We can represent this problem with a graph, connecting each house to each utility. We notice that this graph is bipartite: De…nition 1 A bipartite graph is one in which the vertices can be partitioned into two sets X and Y such that every edge joins one vertex in X with one vertex in Y: The partition (X; Y ) is called a bipartition.A complete bipartite graph is one such that each vertex of X is joined with every vertex of Y: The complete bipartite graph such that the order of X is m and the order of Y is n is denoted Km;n or K (m; n) : It is easy to see that the relevant graph in the problem above is K3;3: Now, we wish to embed this graph in the plane such that no two edges cross except at a vertex. De…nition 2 A planar graph is a graph that can be drawn in the plane such that no two edges cross except at a vertex. A planar graph drawn in the plane in a way such that no two edges cross except at a vertex is called a plane graph. Note that it says “can be drawn”not “is drawn.”The problem is that even if a graph is drawn with crossings, it could potentially be drawn in another way so that there are no crossings. Thus the Three Houses and Three Utilities Problem is whether or not K3;3 is a planar graph. A planar graph divides the plane into regions. That is, if we remove the vertices and edges from the plane, there are a number of disconnected pieces, each of which we call a region. The boundary of a given region is all of the edges and vertices incident on the region. Notice that there is always one exterior region which contains all of the unbounded parts of the plane. 1 Look at examples. Notice that p q + r = 2: A theorem of Euler says that this is always true. For the following theorem, we will need the de…nition of a tree. De…nition 3 A connected graph with no cycles is called a tree. Note the following fact about trees. Proposition 4 A tree with p vertices and q edges satis…es p = q + 1: Proof. This can be proven by induction on the number of edges, since each time a new edge is attached, it must come with a new vertex. This is because if we attach two vertices with an edge, since the tree in the previous step is connected, we had a path between the two vertices and are hence introducing a cycle. Theorem 5 (Euler’sTheorem) Let G be a connected plane graph with p ver- tices, q edges, and r regions. Then p q + r = 2: We remark that the number 2 has to do with the plane, and will become important when we look at topology. Proof. We induct on q: If q = 0; then we must have p = 1 and r = 1: Thus p q + r = 1 0 + 1 = 2: Now suppose it is true for all graphs with k or fewer edges. Consider a connected graph G with k + 1 edges. If G is a tree, then we know that p = q + 1 = k + 2: Furthermore, since there are no cycles in a tree, r = 1: Thus p q + r = (k + 2) (k + 1) + 1 = 2: If G is not a tree, then it contains a cycle. Let e G be an edge on a cycle. If 2 we look at G0 = G e; it has k edges, and p0 = p; q0 = k = q 1; and by the inductive hypothesis, p0 q0 + r0 = 2: We see that r0 = r 1 since removing the edge e joins the regions on either side of the e: Thus p q + r = p0 (q0 + 1) + (r0 + 1) = 2: Notice the following corollary: Corollary 6 Any representation of a planar graph as a plane graph has the same number of regions. We can now solve the Problem of Three Houses and Three Utilities. Theorem 7 The graph K3;3 is not planar. 2 Proof. Suppose we can represent K3;3 as a plane graph. We know that p = 6 and q = 9; so we need to understand something about r: Since K3;3 is bipartite, we see that all regions have boundaries with at least 4 edges. Suppose the graph divides the plane into regions R1;R2;:::;Rr: Let B (R) be the number of edges in the boundary of region R: Consider the number r N = B (Ri) : i=1 X Since all regions have boundaries with at least 4 edges, we have N 4r: However, since each edge is counted twice, we have that N = 2q = 18: Thus 4r 18 or r 4:5: However, that would mean that p q + r = 6 9 + r 1:5; which is impossible if the graph is a plane graph. One might ask about other non-planar graphs. Another important one is K5: Here is a theorem that allows us to show this. Theorem 8 Let G be a connected, planar graph with p vertices and q edges, with p 3: Then q 3p 6: Proof. The proof is quite similar to that of the previous theorem. For p = 3; we certainly have q 3: For p 4; we consider a representation of G as a plane graph, which gives us r regions. Denote the regions as R1;R2;:::;Rr: We compute again r N = B (Ri) : i=1 X This time, we note that B (R) 3 for each region, and a similar argument give us that 3r N 2q: Using Euler Theorem, we have that 2 = p q + r 2 1 p q + q = p q 3 3 3 or q 3p 6: Corollary 9 The complete graph K5 is not planar. 5 Proof. We see that K5 has p = 5 and q = 2 = 10; and so q = 10 > 15 6 = 3p 6: 2 Kuratowski’stheorem We saw that K5 and K3;3 are not planar. It easily follows that any graph containing one of these as a subgraph (technically, any graph with a subgraph isomorphic to one of these two graphs) is not planar either. In fact, we can do slightly better by seeing that any graph with a subgraph isomorphic to a subdivision of K5 or K3;3 is not planar. De…nition 10 A subdivision of a graph G is a new graph obtained by adding vertices inside the edges. (The new vertices are all of degree 2.) It is then clear that : Proposition 11 If a subdivision of G is planar, then G is planar. Proof. Given a representation of the subdivision of G as a plane graph, simply remove the vertices which form the subdivision and we arrive at a plane graph representation of G: Stated another way, if G is not planar, then any subdivision is not planar. Thus any subdivision of K5 and K3;3 is nonplanar. Proposition 12 If G is planar, then every subgraph is planar. Proof. Represent G as a plane graph, then the subgraphs are also plane graphs. Thus if a subgraph of G is a subdivision of K5 or K3;3; then it is not planar, and thus G is not planar. Kuratowski’stheorem is the converse: Theorem 13 (Kuratowski) If a graph is not planar, then it contains a sub- graph isomorphic to a subdivision of K5 or K3;3: Proof is in BM1, BM2, or D. We may do it later in the semester. 4 3 Topology comments Plane graphs are also graphs on spheres. Using stereographic projection, one can convert any plane graph to a graph on the sphere with no vertex at the north pole. Stereographic projection is a map from the sphere except the north pole to the plane which is a homeomorphism (i.e., a bijection which is continuous with continuous inverse). The map is de…ned as follows. Consider the plane in R3 and the xy-plane, R2: For any point P on the sphere, draw a line in R3 2 from the north pole to that point. It will intersect the plane R at a point P 0: Stereographic projection is the map P P 0: With some basic geometry, one can explicitly write down the map, which! is x y (x; y; z) = ; : 1 z 1 z The inverse map can also be written explicitly, as 2 2 1 2x 2y x + y 1 (x; y) = ; ; : x2 + y2 + 1 x2 + y2 + 1 x2 + y2 + 1 We have the following: Proposition 14 Let v V (G) for a planar graph G: There is a plane graph representing G such that2v is in the boundary of the exterior region. Proof. Take a plane graph representing G: Map it by inverse stereographic projection to the sphere. Some region contains v in its boundary. Rotate the sphere so that a point in that region is at the north pole, then use stereographic projection to project it back to the plane. Recall that our proof of nonembeddability of K3;3 and K5 used Euler’sthe- orem, which is essentially a theorem about topology. We can replace the use of Euler’stheorem with a di¤erent topological theorem, the Jordan curve theorem.
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