Figure 1: A Planar Drawing of K 4. Figure 2: A Planar Drawing of K 5 minus one edge

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1 1 Planar Graphs 1.1 Plane Drawings A plane drawing of a graph is a drawing of the graph in the plane such that edges only meet and at their endpoints. A graph is planar if it admits a plane drawing. For example, K 4 is planar since it has a plane drawing: Figure 1: A Planar Drawing of K 4 The graph K 5 is non-planar, but if we delete one edge, it becomes planar: Figure 2: A Planar Drawing of K 5 minus one edge The graph K 33 is non-planar. To see this, K 33 has a Hamilton circuit. When drawn in a plane drawing, there would be three edges not on the circuit, and at least two of them would have to be drawn inside the circuit, or two of them must be drawn outside the circuit. Either way, both of the edges must cross. Thus K 33 has no plane drawing and therefore is non-planar. 1

2 Figure 3: K 3,3 has no planar drawing 1.2 Euler s Formula A planar drawing of a graph divides the plane into regions, called faces. F 1 F 2 F 4 F 3 Figure 4: The faces of a planar drawing For a face F i, let f i denote the number of edges bordering F i, where edges which border F i on both sides are counted twice. f i is called the degree of F i. Observation: If a planar drawing has faces F 1, F 2,..., F k, then f 1 + f f k = 2ε where ε is the number of edges in the graph. 1.1 Theorem ( Euler s Formula ) If G is a connected graph having a plane drawing with ν vertices, ε edges, and f faces, then ν ε + f = 2. 2

3 F 4 F 2 F 3 F 1 F 5 ν = 7, ε = 10, f = 5 ν ε + f = = 2 Figure 5: An example of Euler s formula 1.3 Proof of Euler s Formula A tree is a connected graph which has no cycles (simple circuits). Figure 6: A tree Fact: If a tree has ν vertices, then it has ν 1 edges. We shall start by proving Euler s formula for trees. Let T be a tree with ν vertices. Then it has ε = ν 1 edges. Moreover, T is clearly planar and any plane drawing of T has only one face (thus f = 1). Now ν ε + f = ν (ν 1)+ = 2. Thus Euler s formula holds for T. To prove Euler s formula for all planar graphs, we shall use induction on ε. Let G be a planar graph having ν vertices, ε edges, and a plane drawing with f faces. Suppose ε = 1. Then there are only two possibilities for G, shown below. In the first case where G is an edge, we have ν = 2, ε = 1, and f = 1. In the second case, where G is a loop, we have ν = 1, ε = 1, and f = 2. In both cases, we see that Euler s formula holds. Suppose now that ε > 1 and Euler s formula holds for any planar graph G having fewer than ε edges. If G is a tree, then we know Euler s formula holds by our previous argument. So we may assume that G is not a tree. 3

4 Therefore it has a cycle C. Let e be an edges on C. Then e belongs to exactly two faces, say F and F the face F lies inside C, and F lies outside. F e F e F Figure 7: The faces F and F bordering e Let G be the graph obtained from G by deleting e. Then G is planar and inherits a plane drawing from G. Note that the faces F and F become one face in G. The graph G has ν = ν vertices, ε = ε 1 edges, and f = f 1 faces. By assumption, Euler s formula holds for G and therefore ν ε + f = 2 ν (ε 1) + (f 1) = 2 ν ε + f = 2. Thus Euler s formula holds for G. The proof now follows by induction. 1.2 Corollary K 33 is non-planar. Proof. Proof by contradiction. Suppose K 33 can be drawn in the plane with f faces. Then ν ε + f = 2. However, we have ν = 6 and ε = 9, and as such f = 2 ν+ε = = 5. Thus there are f = 5 faces, say F 1,..., F 5. Since all cycles of K 33 have at least 4 edges, each face F i in K 33 must have degree at least 4 (that is, f i 4). Now since 5 i=1 f i = 2ε, and f i 4, i = 1,..., 5, we see that 20 = 5 4 2ε = 18. This gives a contradiction. Thus K 33 is non-planar. 4

5 1.4 Kuratowski s Theorem How do we know when a graph is planar? To start with, we shall give a few simple observations. Observation 1: If a graph is planar, then any subgraph of it will also be planar. A subdivision of a graph G is a graph obtained from G by inserting vertices of degree two into some edges of G. Figure 8: Subdivision of K 3,3 Observation 2: A subdivision of a planar (resp. non-planar) graph is planar (resp. non-planar). Two graphs H and G are homeomorphic if one graph can be obtained from the other by either suppressing vertices of degree two and/or subdividing edges. Figure 9: Homeomorphic Graphs Observation 3: If two graphs are homeomorphic, then one graph is planar iff the other graph is planar. 1.3 Theorem ( Kuratowski s Theorem ) A graph is planar iff it contains no subgraph which is homeomorphic to K 33 or K 5. 5

6 2 Exercises 1. A dodecahedron has 12 faces and 20 vertices. Each face has the same number of edges. Use Euler s Formula to determine what the number must be. 2. Show that the following graph is non-planar. 3. Suppose f is the number of faces in a planar graph G drawn in the plane having ν vertices and ε edges. Find ν ε + f if G has k components. 4. A soccer ball has 32 faces, each of which is a regular pentagon or hexagon. Because of the angles involved, exactly three faces meet at each corner. Without looking at the ball, determine how many of each type of face there are. 5. Let G be a connected planar graph with 3 or more vertices which is drawn in the plane. Let ν, ε, and f be as usual. a) Use i f i = 2ε to show that f 2ε 3. b) Prove that ε 3ν 6. c) Use b) to show that K 5 is not planar. 6. Show that there cannot exist a bipartite planar graph where each vertex has degree four. 7. Show that the graph below (called the Petersen Graph) is nonplanar by finding a a subgraph which is a subdivision of K 3,3 of K 5. 6

7 Figure 10: The Petersen Graph 7

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