# 1 CIS260-201/204—Spring 2008 Recitation 10 Friday, April 4, 2008.

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1 CIS260-201/204—Spring 2008 Recitation 10 Friday, April 4, 2008

2 Recap: Proof of Euler Tour Let G be an Eulerian graph. Find a cycle C in G. (A cycle exists.) Remove C. Get a smaller graph. By inductive hypothesis, each connected component has an Euler tour. The Euler tour of G is obtained by traversing C and diverge to the Euler tour of each component when we encounter it. Example will help.

3 Euler Tour: Example First, verify that this graph is Eulerian. How? Every vertex has even degree. So, it has an Euler tour.

4 Euler Tour: Example (cont.) Find a cycle. Remove this cycle. The remaining graph is still Eulerian. So, we can find an Euler tour in the remaining graph.

5 Euler Tour: Example (cont.) If the remaining graph is still complicated (like this), repeat the procedure. Note: Now there are two connected components.

6 Euler Tour: Example (cont.) Find a cycle. Remove this cycle.

7 Euler Tour: Example (cont.) The resulting graph is still Eulerian. Now 3 components. Again, find another cycle. Remove it.

8 Euler Tour: Example (cont.) The resulting graph is still Eulerian. Many components now. Again, find another cycle. Remove it.

9 Euler Tour: Example (cont.) The resulting graph is still Eulerian. Now simple enough to see all the Euler tours of each nontrivial component. So, ready to construct the tour for the whole graph.

10 Euler Tour: Example (cont.) Start at the vertex on a cycle we removed. Once we encounter a vertex having an Euler tour attached to it, traverse that tour.

11 Euler Tour: Example (cont.) When done, go back to previous cycles… … and do the same. In this case we don’t encounter any other tour.

12 Euler Tour: Example (cont.) Two more cycles to go.

13 Euler Tour: Example (cont.) Last cycle… And we are done!

14 Hamiltonian Cycle: Example Prove that this graph does not have a Hamiltonian cycle.

15 Hamiltonian Cycle: Example (cont.) This graph is bipartite! Well, let’s color it.

16 Hamiltonian Cycle: Example (cont.) 13 yellow vertices 12 blue vertices Any cycle must be yellow, blue, yellow, blue, …, blue, yellow (first vertex). Is it possible to traverse every vertex and come back to the first vertex?

17 Hamiltonian Cycle: Example (cont.) Start with a blue vertex. blue, yellow, blue, yellow, …, blue, yellow. One yellow vertex left! Can’t start with a blue.

18 Hamiltonian Cycle: Example (cont.) Start with a yellow vertex. yellow, blue, yellow, blue, …, yellow, blue, yellow. But the first and last vertices are yellow. Can’t get back to the first vertex. Can’t start with a yellow. Can’t start with anything! No Hamiltonian cycle.

19 Hamiltonian Path: Example Prove that this graph does not have a Hamiltonian path.

20 Hamiltonian Path: Example (cont.) Again, this graph is bipartite. Let’s color it.

21 Hamiltonian Path: Example (cont.) 32 blue vertices 30 yellow vertices Any path must be blue, yellow, …, blue. Is it possible to traverse every vertex?

22 Hamiltonian Path: Example (cont.) Even starting with a blue vertex, we can’t get to all blue vertices because there are not enough yellow vertices. Starting with a yellow vertex is out of question; we can’t visit every blue vertex. No Hamiltonian path.