Math 575 Problem Set 8 1. A graph G is outerplanar if it is possible to draw it in the plane so that every vertex lies in the boundary of the exterior region. Show that if G is an outerplanar graph having m edges and n vertices, then m ≤ 2n - 3. Hint: Form a new graph G* by adding a single new vertex and joining it to all the vertices of G. 2. Prove Euler’s Theorem below by strong induction. Theorem. Let G be a connected plane graph having n vertices, m edges and f faces, then n ! m + f = 2 . Solution: We argue by induction on the number of cycles in G. Suppose that G is a connected plane graph having n vertices, m edges and f faces nd having 0 cycles. Then G is a tree and so m = n – 1 and f = 1. So in this case, n ! m + 1 = n ! (n ! 1) + 1 = 2 . Now suppose that G is a connected plane graph having n vertices, m edges, f faces and k cycles, and that the theorem holds for all graphs with fewer than k cycles. Now let e be any edge of G that is not a bridge; i.e., e is any edge of G that lies in a cycle of G. Then G – e is still connected and has n vertices, m – 1 edges, f – 1 faces and fewer that k cycles. Thus by our inductive assumption, 2 = n – (m – 1) + ( f – 1 ) = n – m + f. And so the result follows by induction. 3. State and prove a generalization of Euler’s Theorem for graphs that are not connected; i.e., generalize Euler’s Theorem to planar graphs having k components. Solution: We will show that if G is a plane (or planar) graph having n vertices, m edges, f faces and k components, then n – m + f = k + 1. Suppose that G is a plane graph having n vertices, m edges, f faces and k components. * Say the components of G are A1,!A2 ,!…,!Ak . Then form a new graph G by adding an edge between two vertices of Ai and Ai-1 for each i = 1, 2, …, k – 1. * Then G is a connected graph having n* = n,!m* = m + k ! 1,! f * = f . So, by Euler’s Theorem, we have 2 = n* ! m* + f * = n ! (m + k ! 1) + f = n ! m + f + 1! k . But then this reduces to n ! m + f = k + 1. Alternate Argument. For each i = 1,!2,…,!k let ni ,!mi ,!and fi denote the number of vertices, edges, and faces respectively for each Ai . Then clearly, k k k k . n = !ni ,!!m = !mi ,! f = 1+!( fi " 1) = (1" k) + !mi i=1 i=1 i=1 i=1 And so from Euler’s Theorem, k . n ! m + f = 1! k + "(ni ! mi + fi ) = 1! k + 2k = k + 1 i=1 4. Show that if G is a planar graph on n ≥ 11 vertices, then its complement is not planar. Hint: Consider the number of edges in the complement of G. 5. If a graph G on n vertices has fewer than 3n – 6 edges, must G be planar? Solution: No, the graph K3,3 is an example of a non-planar graph on 6 vertices which has 9 ≤ 3 ! 6 " 6 = 12 edges. 6. Show that if G is a bipartite planar graph with n vertices and m edges, then m ! 2n " 4 . Hint: You can assume that every edge of G is in a cycle and that every face has at least four edges in its boundary. Emulate the proof that m = 3n – 6 for maximal planar graphs. 7. Suppose that G is a planar graph having no bridges, and every face of G contains at least 5 edges in its boundary. If G has m edges and n vertices, then m ≤ an + b for some a and b. Find the best values for a and b that you can. Solution: Emulating the argument from the solution to problem 6, we get 5 10 m ! n " . 3 3 8. How many distinct trees on the vertex set V = {1,2,3,4…,n} , n ≥ 10 are there in which both vertices 1 and 2 have degree 5? " n ! 2% " n ! 6% n!10 Solution: (n ! 2) . #$ 4 &' #$ 4 &' 9. If a tree on the vertex set V = {1,2,3,4…,n} is picked at random, let pn be the probability that both vertex 1 and vertex 2 are end-vertices. What is the value of lim!pn ? n!" n#2 n #2 n # 2 $ 2' $ 2' $ 2' 1 Solution: ( ) . lim!pn = lim! n#2 = lim!&1# ) *&1# ) = lim!&1# ) = 2 n!" n!" n n!" % n( % n( n!" % n( e .