Planar graphs. Math Prof. Kindred - Lecture 16 Page 1

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1 Planar graphs Typically a drawing of a graph is simply a notational shorthand or a more visual way to capture the structure of the graph. Now we focus on the drawings themselves. Definition A drawing of a graph G = (V, E) in the plane is a pair of functions f and g such that 1. f : V R 2 is one-to-one; (no two vertices are drawn on top of one another) 2. for an edge uv, g(uv) is a simple curve whose endpoints are f(u) and f(v), but for any other vertex w, f(w) is not on g(uv). (no edge crosses itself and the only vtcs on it are its endpts) Furthermore, a drawing is crossing-free if g(uv) g(xy) = and g(uv) g(uw) = {f(u)} for distinct vtcs u, v, w, x, y. (distinct edges do not intersect except at a shared endpoint) A crossing-free drawing of G in the plane is also referred to as a planar embedding of G. The study of planar graphs necessarily involves topology of the plane, but we do not attempt to be strictly rigorous in topological matters, so that we do not obscure the combinatorial aspects of the topic. Definition A curve is the image of a continuous map from [0, 1] to plane R 2. A curve is closed if its first and last points (endpts) are the same. It is simple if it has no repeated points (mapping is one-to-one) except possibly the first and last pts. Math Prof. Kindred - Lecture 16 Page 1

2 Theorem (Jordan curve theorem). Let C be a simple closed curve in the plane, R 2. Then R 2 C contains exactly two regions, each having C is a boundary. The regions are classified as inside (bounded) and outside (unbounded). The statement of the Jordan curve thm seems obvious but is hard to prove. Easy to establish the result for simple curves such as polygonal lines the problem came in generalizing it to all kind of curves, which included nowhere differentiable curves such as the Koch snowflake. To give you some idea of the difficulty, consider the simple, closed curve below is the house in the interior or exterior of the curve? Definition A graph is planar if it has a planar embedding. Primary examples of nonplanar graphs K 5 and K 3,3 We can consider embedding, or drawing, graphs on other surfaces besides the plane. Math Prof. Kindred - Lecture 16 Page 2

3 Question What kinds of graphs can be drawn without crossings on the surface of a sphere? Theorem. A graph can be embedded on the surface of a sphere without crossings if and only if it can be embedded in the plane without crossings. n Proof. Stereographic projection of a sphere on a plane. The projection maps each point x n of the sphere to a point x in the plane. (For point n (the North pole), projection is undefined.) The closer x is to n, the more distant its image is from (0, 0) in the plane. For this reason, it is common to speak of n as mapping to infinity in the plane, and of the sphere as completing the plane by adding a point at infinity. There is a one-to-one correspondence between points of the sphere and the finite points on the plane; points at infinity in the plane correspond to the point n on the sphere. Given a spherical embedding of G (no crossings), where the point n lies on no curve/edge of the drawing, the projection yields a planar embedding of G. Conversely, a planar embedding yields a crossing-free drawing on the sphere by the inverse projection. Math Prof. Kindred - Lecture 16 Page 3

4 Definition A crossing-free drawing separates the plane R 2 into faces. (Cut out the graph, and pick up the pieces.) Both planar embeddings above have four faces. Question Given a planar graph, does the number of faces vary based on the choice of planar embedding? Theorem (Euler s formula). If a connected planar graph G with n vtcs and e edges has a planar embedding with f faces, it follows that n e + f = 2. We give a proof by induction on # of edges, but there are several other possible proofs, including a proof by induction on # of faces. Proof. Proof by induction on e = E(G). Fix the number of vtcs in G as n. Base case: Since G is connected, it must have at least n 1 edges, so base case is e = n 1. Then G is a tree. Since it is acyclic, it has exactly one face in any planar embedding. So as desired. n e + f = n (n 1) + 1 = 2 Induction hypothesis: Suppose the result holds for any connected, planar graph with n vtcs and e n 1 edges. Math Prof. Kindred - Lecture 16 Page 4

5 Let G be a connected planar graph with n vtcs and e + 1 edges. Consider a planar embedding of G with f faces. Since G is not a tree (it has at least n edges), there exists an edge uv of G that is in a cycle and hence is not a cut edge. Let G = G uv. Then G has n = n vtcs and e = e 1 edges. As a consequence of the Jordan curve thm, there are two distinct faces in the planar embedding of G which have edge uv as a boundary; when uv is removed, these two faces become one face, so this embedding of G has one less face that that of G, i.e., f = f 1. We have n e + f = n (e + 1) + (f + 1) = n e + f = 2 by applying the induction hypothesis for the planar embedding of G. Corollary. Let G be a planar graph with n vtcs, e edges, and c components. Suppose a planar embedding of G has f faces. Then n e + f c = 1. Proof. For a planar embedding G of a planar graph, let ψ(g) denote the number of faces in the embedding. If H 1,..., H c are components of a planar embedding of G, then n = V (H 1 ) + + V (H c ) e = E(H 1 ) + + E(H c ) f = ψ(h 1 ) + (ψ(h 2 ) 1) + + (ψ(h c ) 1) = ψ(h 1 ) + + ψ(h c ) (c 1). Math Prof. Kindred - Lecture 16 Page 5

6 Thus, n e + f = i [ V (H i ) E(H i ) + φ(h i )] (c 1) = 2c (c 1) = c + 1 and so the result follows. Corollary. Let G be a planar graph. Then the number of faces in any planar embedding of G is the same. Definition The degree of a face in a planar embedding is the number of edges on the boundary of the face. If both sides of an edge are on the face, we count it twice in the degree. The degree of a face can also be thought of as the total length of the closed walk in G bounding the face. Examples Proposition. Any planar embedding of a planar graph G has the sum of the degrees of its faces equal to 2 E(G). Remark For a simple planar graph, the degree of a face is always 3 unless the embedding has at most one edge. Math Prof. Kindred - Lecture 16 Page 6

7 We can use Euler s formula to derive upper bounds on the number of edges in planar graphs. Theorem. Let G be a simple planar graph with n 3 vertices and e edges. Then e 3n 6. If we know further that G is triangle-free, then e 2n 4. Exercise for students Construct a proof of this result. Remark Notice that this necessary condition is not a sufficient condition for a simple graph to be planar. For example, the triangle-free graph G resulting from adding one vertex and incident edge to K 3,3 has 7 vtcs and 10 edges. Then E(G) = = 2 V (G) 4 but this graph contains K 3,3, which is nonplanar by a previous remark. An immediate consequence of this result is that a planar graph cannot have all vertices be of high degree. Corollary. If G is a planar graph, then δ(g) 5. Proof. If, for sake of contradiction, δ(g) 6, then 2 E(G) = d(v) 6 V (G). v V (G) Thus, E(G) 3 V (G) but by previous theorem, it must be that E(G) 3 V (G) 6. Question What do the upper bounds on numbers of edges in planar graphs imply about K 3,3 and K 5? Theorem. K 5 and K 3,3 are nonplanar. Math Prof. Kindred - Lecture 16 Page 7

8 Proof. For K 5, we have E(K 5 ) = 10 > 9 = = 3 V (K 5 ) 6 and for the triangle-free graph K 3,3, we have E(K 3,3 ) = 9 > 8 = = 2 V (K 3,3 ) 4 so by the contrapositive statements of previous theorem, neither graph is planar. Math Prof. Kindred - Lecture 16 Page 8