Fall 2015 Midterm 1 24/09/15 Time Limit: 80 Minutes


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1 Math 340 Fall 2015 Midterm 1 24/09/15 Time Limit: 80 Minutes Name (Print): This exam contains 6 pages (including this cover page) and 5 problems. Enter all requested information on the top of this page, and put your name on the top of every page, in case the pages become separated. You are required to show your work on each problem on this exam. If you need more space, use the back of the pages; clearly indicate when you have done this. Do not write in the table to the right. Problem Points Score Total: 80
2 Math 340 Midterm 1  Page 2 of 6 Name: 1. (a) (5 points) Draw a planar, bipartite graph with 6 vertices and 6 edges. (b) (5 points) Draw a graph with 7 vertices such that it has both an Euler circuit, and a Hamiltonian circuit. Solution: For part (a), we have for example For part (b), the graph below works:
3 Math 340 Midterm 1  Page 3 of 6 Name: 2. (10 points) Determine if the following pair of graphs are isomorphic or not. If they are, provide the vertex correspondence. If not, provide a proof d f e g b h a 8 1 c Solution: In the second graph, vertex g is connected to three vertices with degree 4. In the first graph, there is no vertex that is connected to three out of the four vertices {3, 4, 5, 6}. Hence, the graphs are not isomorphic.
4 Math 340 Midterm 1  Page 4 of 6 Name: 3. A certain planar graph G has seven vertices with degrees 3, two vertices with degree 4, and one vertex with degree 5. (a) (5 points) How many edges does G have? (b) (5 points) How many regions does G have, including the outer region? (c) (10 points) Is there region with exactly three edges (a triangular region) in G? Hint: Use edge counting with regions to get an inequality. Solution: (a) Using v G deg(v) = 2 E, we have that = 2 E. Solving for E, we find that the graph has 17 edges. (b) Using Euler s formula for planar graphs, r = e v + 2, with v = , e = 17, we find that r = 9. (c) If there are no triangles, each region has at least 4 edges. Then the number of edges times two, is at least 9 4 = 36, by double counting. So, any planar graph with 9 regions and no triangles, must have at least 36 edges. But we know G has only 17 2 edges, so G must have at least one triangle. Note: part (b) requires the extra assumption that G is connected, which should have been part of the formulation of the question. I will not subtract any points for not assuming that G is connected. As a side note, there are a few possible nonconnected graphs with two components, (one of these components must be K 4 ), and all these contains at least one triangle.
5 Math 340 Midterm 1  Page 5 of 6 Name: 4. (a) (10 points) Compute the chromatic number for the graph G with vertices and edges given by V = {a, b, c, d}, E = {(a, b), (b, c), (b, d), (c, d)}. You need to show that this number is the smallest possible! (b) (10 points) Compute the chromatic polynomial for the graph G in part (a). Make sure that your polynomial does not contradict the answer you got in (a). Solution: (a) The triangle (b, c, d) is a complete graph, so we need at least 3 colors. We can then pick a to have the same color as c. Thus, the chromatic number is 3. (b) If we have k colors, the triangle can be colored in k(k 1)(k 2) ways, and the color of a can be any of the (k 1) colors different from that of b. The chromatic polynomial of G is therefore p(k) = k(k 1) 2 (k 2). We note that p(1) = p(2) = 0, but p(3) 0. This means that the chromatic number is 3, which agrees with the answer from (a).
6 Math 340 Midterm 1  Page 6 of 6 Name: 5. Given a graph G, we construct its bunkbed graph H as follows: For every vertex v G, there are two corresponding vertices v 1 and v 2 in H, which are connected with an edge. Furthermore, if (u, v) is an edge in G, then (u 1, v 1 ) and (u 2, v 2 ) are edges in H. We can imagine this as H being two identical copies of G, where corresponding vertices in the copies are connected with edges. (a) (10 points) Draw the bunked graph corresponding to the complete graph K 3. (b) (10 points) If G has a Hamiltonian path, show that the bunkbed graph of G has a Hamiltonian circuit. Solution: (a) (b) If G has a Hamiltonian path p visiting every vertex v G, then this path corresponds to two paths, p 1, and p 2 in the bunkbed graph H, such that if (u, v) is an edge in p, then (u 1, v 1 ) is an edge in p 1 and (u 2, v 2 ) is an edge in p 2. Here we used the same notation as in the problem statement. Note that p 1 and p 2 together contains all vertices of H, and that they do not share any vertex. Finally, we see that if u and v are the start and end of the path p, respectively, then (v 1, v 2 ) and (u 1, u 2 ) are edges in H. Thus, we can construct a Hamiltonian circuit in H by first following p 1, use the edge (u 1, u 2 ), then follow the path p 2 backwards compared to p 1, and finish by using the edge (v 1, v 2 ).
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PRIMARY CONTENT MODULE Algebra  Linear Equations & Inequalities T37/H37 What does the number m in y = mx + b measure? To find out, suppose (x 1, y 1 ) and (x 2, y 2 ) are two points on the graph of
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