AMS301 Exam Solutions

8/22/2019 AMS301 Exam Solutions

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AMS 301.03 - Finite Mathematical Structures - Midterm

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There are 6 questions for a total of 100 points.

1. (17 points) Are the two graphs below isomorphic? If so, give an isomorphism and if not, give carefulreasons why not.

Solution: No, they are not isomorphic. The first graph is bipartite while the second is not (itcontains triangles).

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2. (17 points) Is the following graph planar? If so, give a planar embedding and if not, give a K3,3 or aK5 configuration.

Solution: No, it is not planar. a,c,f and b,e,g form the two parts of a K3,3 configuration.

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3. (16 points) Prove that if G is graph with at least two vertices then G has at least two vertices with thesame degree.

Solution: Let G have n > 1 vertices. The possible degree values for each vertex are 0, 1, . . . , n 1,for a total of n possibilities. Note that G cannot contain both a vertex of degree n 1 and a vertex

of degree 0. This implies that there are at most n

1 different degrees in G. Since there are nvertices, by the pigeonhole principle, there exists a pair of vertices with equal degree.

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4. (17 points) Does the following graph have a Hamilton circuit? If so, give a circuit and if not, explaincarefully why not.

Solution: Yes, it has a Hamilton circuit. Since 12 has degree 2, force in edges (6, 12) and (11, 12).Then, we split into cases.

Case 1: The path visits 9 through 6. Then we put in edge (6, 9) and erase edges (2, 6), (5, 6), and(9, 11). Then, edges (9, 5), (2, 1), and (2, 5) are forced and we erase (5, 8). Now we condition onwhether (1, 3) or (1, 4) is used.

Case 1a: The path uses edge (1, 3). Then, erase (1, 4) and force in (3, 4) and (4, 8). This causes usto erase (3, 7) and force in (7, 8) and (7, 10). This yields a Hamilton circuit, 1 2 5 9 6 1211 10 7 8 4 3 1.

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5. (17 points) Find the chromatic number of the following graph. Give a coloring with that many colorsand carefully explain why the graph cannot be colored with fewer colors.

Solution: The chromatic number of this graph is 3. Three colors are clearly necessary, as there aretriangles in the graph. A 3-coloring is 1, 5, 10 blue, 2, 3, 8, 9, 12 red and 4, 6, 7, 11 yellow.

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6. (a) (12 points) For what values ofn does there exist a tree with n vertices whose complement is alsoa tree? Draw the largest such graph and its complement.

Solution: We have that for trees e = n 1. Since a graph and its complement have the same

number of edges, and those edges total n(n1)2 , we have that 2(n1) =n(n1)

2 . Equality occursexactly if n = 1 or n = 4. The 4 vertex path graphs give the largest such graphs.

(b) (4 points) Use a BFS to determine if the graph given by the following adjacency matrix is connected.

0 1 0 0 1 0 0 0 1 11 0 0 1 0 1 0 0 0 00 0 0 0 0 0 0 1 0 00 1 0 0 1 0 0 0 0 11 0 0 1 0 1 0 0 1 00 1 0 0 1 0 1 0 1 00 0 0 0 0 1 0 0 0 10 0 1 0 0 0 0 0 0 01 0 0 0 1 1 0 0 0 11 0 0 1 0 0 1 0 1 0

Solution: The graph is not connected as a BFS started at 1 does not contain 3 or 8.

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