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Unformatted text preview: MATH 682 Exam #2 Solutions 1. (8 students attempted this problem) Suppose that G is a simple graph that con tains two edges whose removal destroys all cycles in G . Prove that G is planar. The easiest approach is to prove the contrapositive via Kuratowskis Theorem: if G is nonplanar, then G still contains cycles after removal of any two edges. We know G is nonplanar only if G contains a subdivision of K 3 , 3 or K 5 . A subdivision of K 3 , 3 with n vertices will contain n + 3 edges; a subdivision of K 5 with n vertices will contain n + 5 edges. In order for a graph to be cyclefree, it must have fewer edges than vertices (a tree being such a graph with as many edges as possible), so one would have to remove at least 4 edges from a nonplanar graph to bring the edge count on its K 3 , 3 or K 5 subdivision down low enough. Another approach may be by explicit treemanipulation. If G e 1 e 2 is cyclefree, this means that F = G e 1 e 2 is a forest. Let us consider the reinsetion of these edges into the forest. One edge can be added easily to any planar representation of a forest, since there are no cycles to serve as face boundaries and separate one vertex from another. If this edge does not create a cycle, then the second edge can be trivially added by the same rule. If, on the other hand, the first edge adds a cycle, then we have a planar representation with exactly one cycle (since a path between the two endpoints of a putative added edge in a tree is unique, adding this edge produces a unique cycle). Thus, in this case F + e 1 consists of one or more trees as well as a component with a single cycle. We may orient all other features of this graph to point out of the cycle; now the edges subdivide the plane into two parts the interior of the cycle and the exterior. By our edgeorientation, every vertex lies on the exterior or boundary of the cycle, so the second edge can be added in accordance with planarity, by tracing it through the cycles exterior....
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 Spring '09
 WILDSTROM
 Math

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