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CPTS553- Graph Theory: Assignment 5 Solved

Questions with a (โ‹†) are each worth 1 bonus point for 453 students.

The graph ๐บ = ๐ถ3 × ๐ถ3 is drawn below:Show that ๐บ is Eulerian. The suggested way is to use the characterization we discussed in class.  The not-so-easy way is to actually produce an Euler circuit; if you choose this latter route, it suffices to list the vertices in order as they’d be encountered in the circuit.
Show that ๐บ is Hamiltonian. Here, you’ll want to produce a Hamilton cycle.  Again, it suffices to list the vertices in order as they’d be encountered in the cycle.         
Now, for ๐‘›, ๐‘ž ≥ 2, consider the grid graph ๐‘ƒ๐‘›,๐‘ž = ๐‘ƒ๐‘› × ๐‘ƒ๐‘ž.Show that if either ๐‘› or ๐‘ž is at least 3, then ๐‘ƒ๐‘›,๐‘ž is not Eulerian. The easy way is to consider what happens along the boundary.
Show that for any ๐‘ž ≥ 2, ๐‘ƒ2,๐‘ž is Hamiltonian
Show that ๐‘ƒ4,4 is Hamiltonian. The most straightforward way is to produce a Hamilton cycle.
Show that ๐‘ƒ3,3 is not Hamiltonian. The drawing below may suggest a clever approach:
                                                                                                                 02

                                                                                                         12

If ๐‘› and ๐‘ž are both odd, show that ๐‘ƒ๐‘›,๐‘ž is not Hamiltonian. (Use the same clever argument from part d).
(โ‹†) If either ๐‘› or ๐‘ž is even, show that ๐‘ƒ๐‘›,๐‘ž is Hamiltonian.             
Recall that if a graph ๐บ has no links (this means every edge of ๐บ is a bridge), then there is a relationship among ๐‘˜, ๐‘š, and ๐‘›. Here, ๐‘˜ is the number of components, ๐‘š is the number of edges, and ๐‘› is the number of vertices in the graph ๐บ.  
We now consider when ๐บ has links.

What is the relationship among ๐‘˜, ๐‘š, ๐‘› if deleting any 1 link of ๐บ turns every other link into a bridge (Example: ๐บ is a cycle)? Consider the effect of deleting a link.
Now, suppose ๐บ is a graph where you can arrive at an all-bridge graph by deleting โ„“ links, one at a time, taking care not to delete any edges that become bridges in this process. What is the relationship among ๐‘˜, ๐‘š, ๐‘› now?
(โ‹†) Show that for any graph ๐บ, there is a unique value โ„“ that works in part ๐‘.  (What would happen if there were two such values โ„“1 and โ„“2?)
What is the value of โ„“ in the “theta” graph (๐ถ8 with an extra edge) below?
What is the value of โ„“ for ๐พ5, the complete graph on 5 vertices?

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