Final Exam Math 38 Graph Theory Spring 2017 Due on Friday, June 2, at 12:50 pm. Good Luck!!

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1 Final Exam Math 38 Graph Theory Spring 2017 Due on Friday, June 2, at 12:50 pm NAME: Instructions: You can use the textbook (Doug West s Introduction to Graph Theory, without solutions), your notes from class, assigned homework from class, and the solutions posted on the course webpage. However, you cannot use any other (online or offline) source for assistance, or discuss the problems with anybody other than me. Please ask me if you have any questions about the rules or about the problems. Write your solutions clearly, explaining your steps and justifying all your answers. You may turn in the exam at anytime (but no later than 12:50 pm on June 2nd), by sliding it under my office door in Kemeny 332. Good Luck!! Question Possible points Points Earned Total 100

2 1. Determine if the following statements are true or false, and explain why. (a) (4) The following graph is self-complementary. Solution. True. To prove it, one must give an explicit bijection f : V (G) V (G) such that uv E(G) if and only if f(u)f(v) / E(G). (Just arguing that the vertex degrees in G and in G are the same is not enough to prove that G = G.) (b) (4) If G is a connected graph with at least 3 vertices and it contains a cut-edge, then G contains a cut-vertex. Solution. True. By Theorem 4.1.9, κ(g) κ (G). Since G contains a cut-edge, κ (G) 1, and so κ(g) 1, which implies that G contains a cut-vertex. It is also possible to give a longer argument to show that at least one of the endpoints of the cut-edge must be a cut-vertex.

3 (c) (4) A graph G has χ(g) r if and only if G contains a subgraph isomorphic to K r. Solution. False. A counterexample is G = C 5, which has χ(g) = 3 but contains no triangle. (d) (4) The sequence (7, 7, 7, 6, 6, 6, 5, 5, 5, 4, 4, 4) is a graphic sequence. Solution. True. Apply Theorem (Havel-Hakimi) repeatedly, or directly construct a simple graph with this degree sequence. Everyone got this one right.

4 (e) (4) For every graph G containing at least one edge, the sum of the coefficients of its chromatic polynomial χ(g; k) equals 0. Solution. True. The sum of the coefficients of χ(g; k) equals χ(g; 1), which is the number of proper colorings with one color. Since the graph has an edge, there are no such colorings. Alternatively, one can give a longer proof by induction using the deletioncontraction recurrence.

5 2. (10) Find a tree whose Prüfer code is (1, 7, 2, 1, 1, 7). Solution. The tree has a path and extra edges 1 3, 1 6 and 7 4. Most students found this tree.

6 3. (10) Let G be a graph with n vertices and m edges, and suppose that m n 1. Prove that κ(g) 2m n. Solution. By Theorem 4.1.9, δ(g) κ(g). By the degree-sum formula, 2m nδ(g) nκ(g), from where the inequality follows.

7 4. (15) Find a maximum flow from s to t in the following network. Each edge is labeled by its capacity. Verify that the value of your flow is maximum using the Max-Flow Min-Cut theorem. 8 t s Solution. Everybody solved this problem correctly. The maximum flow has value 18, and there are several source-sink cuts with capacity 18 that can be used to prove that such a flow is maximum.

8 5. Let G be a graph with no isolated vertices, and let M be a maximum matching of G. For each vertex v not saturated by M, choose an edge incident to v. Let T be the set of all the chosen edges, and let L = M T. Determine if each of the following statements is true or false, and explain why. (a) (6) L is always an edge cover of G. (b) (6) L is always a minimum edge cover of G. Solution. Both statements are true. This construction of L is exactly the one in the proof of Theorem (that we did in class). L is an edge cover by construction, since all edges not saturated by M are covered by the additional edges T ; and it is a minimum one because its size is n M = n α (G) = β (G), by Theorem For part (b), several students showed that L is minimal (that is, it has no superfluous edges that can be removed from it and still have a edge cover), but this does not prove that L is minimum.

9 6. (20) For n 1, let G n be the graph having 4n 2 vertices u 1, u 2,..., u 2n 1, v 1, v 2,..., v 2n 1, and having 5n 4 edges u i u i+1 and v i v i+1 for all 1 i 2n 2, and u 2j 1 v 2j 1 for all 1 j n. As an example, the graph G 4 is drawn below: (a) Find a formula for the chromatic polynomial χ(g n ; k).

10 Solution. χ(g n ; k) = k(k 1)(k 4 5k k 2 10k + 5) n 1. We can prove this formula by induction on n and repeatedly applying the deletion/contraction recurrence. For n = 1, G 1 has one edge, and χ(g 1 ; k) = k(k 1). For the induction step, let n 2 and assume that χ(g n 1 ; k) = k(k 1)(k 4 5k k 2 10k + 5) n 2. Let e be the rightmost vertical edge of G, that is, e = u 2n 1 v 2n 1. Then χ(g n e; k) = χ(g n 1 ; k)(k 1) 4, since we can color the graph G n e by first coloring G n 1, and then coloring the vertices u 2n 2, u 2n 1, v 2n 2, v 2n 1 in this order, having k 1 color choices for each one. To compute χ(g n e; k), let e be any of edges in the 5-cycle at the rightmost end of this graph (except the edge u 2n 3 v 2n 3 ), and apply the deletion/contraction recurrence again on this edge. Then χ((g n e) e ; k) = χ(g n 1 ; k)(k 1) 3, and (G n e) e has a 4-cycle on the right. We can apply the deletion/contraction recurrence to this graph by considering an edge e from the 4-cycle (other than the edge u 2n 3 v 2n 3 ). Deleting e results in a graph with chromatic polynomial χ(g n 1 ; k)(k 1) 2. Contracting e results in a graph with a triangle on the right, whose chromatic polynomial is χ(g n 1 ; k)(k 2). Putting all the steps together, we obtain as desired. χ(g n ; k) = χ(g n 1 ; k)[(k 1) 4 (k 1) 3 + (k 1) 2 (k 2)] = χ(g n 1 ; k)(k 4 5k k 2 10k + 5) = k(k 1)(k 4 5k k 2 10k + 5) n 1 An alternative for the induction step would be to use the deletion/contraction recurrence in the backwards direction, by adding the edge h = u 2n 2 v 2n 2. Then χ(g n ; k) = χ(g n + h; k) + χ((g n + h) h; k), where (G n + h) h is simply the graph obtained from G n by identifying the vertices u 2n 2 and v 2n 2. Then χ((g n + h) h; k) = χ(g n 1 ; k)(k 2)(k 1)(k 2) since, after coloring G n 1, there are k 2 ways to color the identified vertex u 2n 2 = v 2n 2, then k 1 ways to color u 2n 1, and k 2 ways to color v 2n 1. On the other hand, one can show that χ(g n + h; k) = χ(g n 1 ; k)(k 2 3k + 3) 2 as in the solution to homework problem Adding these two formulas we get χ(g n ; k) = χ(g n 1 ; k)(k 4 5k k 2 10k + 5) as before. There are also a few other ways to solve this problem. However, some students came up with formulas like χ(g n ; k) = k(k 1)(k 2 3k + 3) n 1, χ(g n ; k) = k(k 1) 2n 1 (k 2 3k + 3) n 1, χ(g n ; k) = k(k 1) 3 (k 2 3k + 3) 2n 3 and similar variations. These formulas are wrong, and they fail already for G 2 = C 6.

11 (b) Use the formula from part (a) to compute χ(g 3 ; 3). Solution. χ(g 3 ; 3) = 726.

12 7. Let G be a connected planar graph with n 5 and e edges, and suppose that the length of the smallest cycle in G is 5. (a) (6) Prove that e 5 (n 2). 3 Solution. Let f be the number of faces in an embedding of G. Since the sum of the length of the faces equals 2e and every face has length at least 5, we have 5f 2e. Plugging it in into Euler s formula, and so e 5 (n 2). 3 2 = n e + f n e e = n 3 5 e, (b) (3) Use part (a) to show that the Petersen graph is not planar. Solution. The Petersen graph has n = 10 vertices and e = 15 edges, and so it doesn t satisfy the inequality e 5 (n 2). 3 (c) (4) Use Kuratowski s theorem to show that the Petersen graph is not planar. Solution. To apply Kuratowski s theorem, we have to find a subgraph of the Petersen graph which is a subdivision of K 5 or K 3,3. While it contains no subdivision of K 5, one can easily find a subdivision of K 3,3 :