A net pattern whose line-of-sight graph is Eulerian, but not Hamiltonian

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Chapter 13, Problem 18RE

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A net pattern whose line-of-sight graph is Eulerian, but not Hamiltonian

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7. [10 marks] Let G = (V,E) be a 3-connected graph. We prove that for every x, y, z Є V, there is a cycle in G on which x, y, and z all lie. (a) First prove that there are two internally disjoint xy-paths Po and P₁. (b) If z is on either Po or P₁, then combining Po and P₁ produces a cycle on which x, y, and z all lie. So assume that z is not on Po and not on P₁. Now prove that there are three paths Qo, Q1, and Q2 such that: ⚫each Qi starts at z; • each Qi ends at a vertex w; that is on Po or on P₁, where wo, w₁, and w₂ are distinct; the paths Qo, Q1, Q2 are disjoint from each other (except at the start vertex 2) and are disjoint from the paths Po and P₁ (except at the end vertices wo, W1, and w₂). (c) Use paths Po, P₁, Qo, Q1, and Q2 to prove that there is a cycle on which x, y, and z all lie. (To do this, notice that two of the w; must be on the same Pj.)