Page 90, Theorem 4.6 ( To show that a digraph is hamiltonian if and only if for each vertex v, indegree (v) =outdegree (v)
Page 92 Exercise problem 4.1 ( A modified version of konigsberg problem where, two extra bridges are built. )
Page 93, Figure 4.5 need to show as Hamilton. (Show that dodecahedron is hamiltonian)
See attached file for full problem description.© BrainMass Inc. brainmass.com October 24, 2018, 9:17 pm ad1c9bdddf
Problem 4.1 and 4.5, I have attached figures.
We construct a graph of the present day Konigsberg bridges and see that there are exactly two vertices (A and B) have odd degree. We know that every connected graph with exactly 2 odd vertices have an Eulerian tour starting and ending at these vertices. Hence we can device such a route passing through all the edges. The attached figure shows such a route.
Problem of figure 4.5
We need to show the graph of dodecahedron is Hamiltonian. We do this by providing such a cycle explicitly. In the attached figure, follow the vertex sequence 1,2,3,...,19,20,1 which is such a Hamiltonian cycle.
Theorem: Let D be a connected non-trivial digraph. D is Eulerian if and
only if id(v) = od(v) for every vertex. (id:in degree, od:out degree).
A trail in a graph is a path without repeated edges.
Suppose D is Eulerian. Consider any ...
Here we prove the Necessary and sufficient condition for a digraph to be Eulerian. It also solves three problems from Graphs and Digraphs concerning a modified Konigsberg problem and provide a Hamiltonian cycle for dodecahedron.