{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2 and c 4 2 we need at least one vertex there is not

Info iconThis preview shows page 4. Sign up to view the full content.

View Full Document Right Arrow Icon
)=2a n d (C 4 ) = 2, we need at least one vertex. There is not a 3-regular graph of order 5. So 6 vertices will be the best that we can do. _________________________________________________________________ 8. (10 pts.) Prove exactly one of the following propositions. Indicate clearly which you are demonstrating. (a) If G is a non-trivial graph, then there are distinct vertices u and v in G with deg(u) = deg(v). (b) If G is a graph of order n and deg(u) + deg(v) n-1f o r each pair of non-adjacent vertices u and v, then G is connected. // Review?? (a): Theorem 2.14, page 51. (b): Theorem 2.4, page 34. Proofs were also done in class. These varied somewhat from those of the text. _________________________________________________________________ 9. (10 pts.) (a) Suppose G is a bipartite graph of order at least 5. Prove that the complement of G is not bipartite. [Hint: At least one partite set has three elements. Connect the dots?] Suppose that G is a bipartite graph of order at least 5 with partite sets U and W. At least one of U and W has at least 3 elements. Suppose without loss of generality, W 3. Label three of the members of W with u,v, and w. Since these vertices are in the same partite set of G, none of these three vertices is adjacent to any other of the three. Thus, uv , vw , uw E ( G ) G contains a 3 cycle . Thus, the complement of G is not bipartite. [Problem 1.25?] (b) Display a bipartite graph G of order 4 and its bipartite complement. Label each appropriately and give partite sets for each bipartite graph. //There are a multitude of examples. See me if you need help with this!
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online