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# Euler Digraph: Characterization

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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.

https://brainmass.com/math/graphs-and-functions/euler-digraph-characterization-122201

#### Solution Preview

Problem 4.1 and 4.5, I have attached figures.

Problem 4.1:
-------------
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 4.6
-------------
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).

Proof:
A trail in a graph is a path without repeated edges.

Suppose D is Eulerian. Consider any ...

#### Solution Summary

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.

\$2.49
The provided text does not contain any mathematical questions or answers. It appears to be a description of problems and solutions related to graph theory, specifically dealing with Eulerian and Hamiltonian graphs.
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Explore BrainMass

# Euler Digraph: Characterization

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

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.

https://brainmass.com/math/graphs-and-functions/euler-digraph-characterization-122201

#### Solution Preview

Problem 4.1 and 4.5, I have attached figures.

Problem 4.1:
-------------
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 4.6
-------------
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).

Proof:
A trail in a graph is a path without repeated edges.

Suppose D is Eulerian. Consider any ...

#### Solution Summary

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.

\$2.49
user
The provided text does not contain any mathematical questions or answers. It appears to be a description of problems and solutions related to graph theory, specifically dealing with Eulerian and Hamiltonian graphs.

TYPE: NO_QUESTION