PROBLEM 7.5. In this problem, you will prove that the Petersen graph is not planar. Let P denote the Petersen graph. The proof will be a proof by contradiction. We will assume that P can be drawn without edge crossings and then arrive at a contradiction. a) As we are assuming P is planar, let f be the number of faces in a planar drawing of P. Let n be the number of vertices of P and e the number of edges. What are n and e? Compute f using Euler's formula. b) Let l1, l2,...,l denote the lengths of the faces of the planar drawing of P. Show that 4 +l2 + +lj = 2e. %3D ... c) Every cycle of the Petersen graph has length at least 5, hence the integers l1, l2,...,lf are all at least 5. Use this together with parts (a) and (b) to obtain the desired contradiction.
PROBLEM 7.5. In this problem, you will prove that the Petersen graph is not planar. Let P denote the Petersen graph. The proof will be a proof by contradiction. We will assume that P can be drawn without edge crossings and then arrive at a contradiction. a) As we are assuming P is planar, let f be the number of faces in a planar drawing of P. Let n be the number of vertices of P and e the number of edges. What are n and e? Compute f using Euler's formula. b) Let l1, l2,...,l denote the lengths of the faces of the planar drawing of P. Show that 4 +l2 + +lj = 2e. %3D ... c) Every cycle of the Petersen graph has length at least 5, hence the integers l1, l2,...,lf are all at least 5. Use this together with parts (a) and (b) to obtain the desired contradiction.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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