exam-2-100407-solutions

# exam-2-100407-solutions - MATH 682 Exam#2 Solutions 1(8...

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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 Kuratowski’s 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 cycle-free, 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 tree-manipulation. If G- e 1- e 2 is cycle-free, 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 edge-orientation, 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 cycle’s exterior....
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exam-2-100407-solutions - MATH 682 Exam#2 Solutions 1(8...

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