Curve Curve: The image of a continous map from [0,1] to R 2. Polygonal curve: A curve composed of finitely many line segments. Polygonal u,v-curve: A polygonal curve that starts at u and ends at v.

Drawing of G Drawing of a graph G: A function f defined on V(G)  E(G) that assigns each vertex v a point f(v) in the plane and assigns each edge with endpoints u,v a polygonal f(u),f(v)-curve. Crossing: A point in f(e)  f(e’) that is not a common endpoint.

Planar Embedding of G Planar graph: A graph has a drawing without crossings. Planar embedding of planar graph G: A drawing of G that has no crossings. Plane graph: A particular planar embedding of a planar graph. Closed curve: A curve whose first and last points are the same. Simple closed curve: A closed curve that has no repeated points except possibly first=last.

Chord Chord of cycle C: An edge not in C whose endpoints lie in C

Proposition K5 and K3,3 cannot be drawn without crossings. Proof. 1. Let C be a spanning cycle in a drawing of K 5 or K 3,3. 2. If the drawing does not have crossing edges, C is drawn as a closed curve.

Proposition Two chords conflict if their endpoints on C occur in alternating order. 4. When two chords conflict, we can draw only one inside C and one outside C. Two Chords Conflict Two Chords do not Conflict

Proposition K 3,3 has three pairwise conflict chords. We can put at most one inside and one outside, so it is not possible to complete the embedding.

Proposition In K 5, at most two chords can go outside or inside. Since there are five chords, it is not possible to complete the embedding. Red Line : chord

Face Open set: A set U that belongs to R 2. Region: An open set U that contains a polygonal u, v- curve for every pair u, v belongs to U. Face: A maximal region of the plane that contains no point used in the embedding.

Dual Graph Dual graph G* of a plane graph G: –1. The vertices of G* correspond to the faces of G. –2. The edges of G* correspond to the edges of G: If e is an edge of G with face X on one side and face Y on the other side, then the endpoints of the dual edge e*  E(G*) are the vertices x, y of G* that represent the faces X, Y of G. The order in the plane of the edges incident to x  V(G*) is the order of the edges bounding the face X of G in a walk around its boundary. G* G

Dual Graph Two embeddings of a planar graph may have non- isomorphic duals: 1. Each embedding shown below has 3 faces, so in each case the dual has 3 vertices. 2. In the embedding on the right, the dual vertex corresponding to the outside face has degree In the embedding on the left, no dual vertex has degree 4, so the duals are not isomorphic.

Face Length Length of a face in a plane graph G: The total length of the closed walk in G bounding the face.

Example L(F 2 )=6L(F 0 )=7L(F 1 )=3 L(F 2 )=9L(F 0 )=4L(F 1 )=3 Cut edge F0F0 F1F1 F2F2 F0F0 F1F1 F2F2 1. A cut-edge belongs to the boundary of only one face, and it contributes twice to its length. 2. In the embedding on the left the lengths of three faces are 3, 6, 7; on the right they are 3, 4, The sum of the lengths is 16 in each case, which is twice the number of edges.

Proposition If L(F i ) denotes the length of face F i in a plane graph G, then 2e(G) =  L(F i ).

Euler’s Formula If a connected plane graph G has exactly n vertices, e edges, and f faces, then n – e + f = 2. n = 7 e = 8 f = 3 n – e + f = 2

Euler’s Formula Proof. 1. Use induction on n (number of vertices). 2. Basis (n = 1): –G is a “bouquet” of loops, each a closed curve in the embedding. If e = 0, then f = 1, and the formula holds. –Each added loop passes through a face and cuts it into 2 faces. This augments the edge count and the face count each by 1. Thus the formula holds when n = 1 for any number of edges.

Euler’s Formula 3. Induction step (n>1): –There exists an edge e that is not a loop because G is connected. –Obtain a plane graph G’ with n’ vertices, e’ edges, and f’ faces by contracting e. –Clearly, n’=n–1, e’=e–1, and f’=f. –n’– e’+ f’ = 2 –Therefore, n-e+f=2. e (induction hypothesis)

Theorem If G is a simple planar graph with at least three vertices, then e(G)≤3n(G)–6. If also G is triangle-free, then e(G) ≤ 2n(G)–4. Proof. 1. It suffices to consider connected graphs; otherwise, we could add edges. 2. If n(G)  3, every face boundary in a simple graph contains at least three edges (  L(F i )  3f). 3. By Proposition , 2e=  L(F i ), implying 2e  3f. (2e  4f) 4. By Euler’s Formula, n–e+f=2, implying e≤ 3n– 6. (e≤ 2n–4) (  L(F i )  4f)

Nonplanarity of K 5 and K 3,3 K 5 (e = 10, n = 5) K 3,3 (e = 9, n = 6) These graphs have too many edges to be planar. –For K 5, we have e = 10>9 = 3n-6. –Since K 3,3 is triangle-free, we have e = 9>8 = 2n-4.