{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 12 Solutions W11 - EECS 203 Homework 12 Solutions...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
EECS 203: Homework 12 Solutions Section 9.3 1. (E) 6,8 (6) a b c d e a 0 1 0 1 0 b 1 0 0 1 1 c 0 0 0 1 1 d 1 1 1 0 0 e 0 1 1 0 0 (8) a b c d e a 0 1 0 1 0 b 1 0 1 1 1 c 0 1 1 0 0 d 1 0 0 0 1 e 0 0 1 0 1 2. (E) 14 a b c d a 0 3 0 1 b 3 0 1 0 c 0 1 0 3 d 1 0 3 0 3. (E) 36,38 (36) The graphs are not isomorphic because the right graph’s v 2 has a degree of 4 whereas none of the left graph’s nodes have any nodes with a degree of 4. (38) The graphs are isomorphic. An example of corresponding vertices are { v 2 u 3 } , { v 3 u 4 } , { v 5 u 2 } , { v 1 u 1 } and { v 4 u 5 } . 4. (C) 52 In order for graph G to be self-complementary, the number of edges in G and G must be equal. Because the total number of distinct connections in both G and G is n ( n - 1) 2 , meaning that there can be n ( n - 1) 4 edges in each of G and G . It means that either n or n - 1 must be divisible by 4 in order for this quantity to be an integer. Therefore, if G is self-complementary, n is either 0 or 1 modulo 4 . 5. (M) 66 Let graphs G and H to be isomorphic graphs. If G is a bipartite graph, then the nodes in G can be divided into two groups with no edges between any pair of nodes that belong to the same group. Because of the mapping (due to isomorphism), one can replace nodes in G with equivalent ones from H . Therefore, the new nodes from H must have the same relationship with other nodes in the same way as the original nodes from G , which means H must also be a bipartite graph.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Section 9.4 6. (E) 12bc (12b) The graph is strongly connected because there are paths from one node to any other nodes. (Look at the cycle a b c d e f a ). (12c) The graph is neither strongly nor weakly connected because the vertices { a, g, c, d } and vertices { b, e, f } are completely disconnected. 7. (E) 20 The graphs G and H are isomorphic. Possible steps to find corresponding vertices with paths are as followed: (a) There are two cycles with length of four and two cycles with length of three in H : { v 1 , v 3 , v 4 , v 8 } , { v 8 , v 4 , v 5 , v 7 } , { v 2 , v 1 , v 3 } and { v 5 , v 7 , v 6 } . (b) There are two cycles with length of four and two cycles with length of three in G : { u 2 , u 3 , u 4 , u 5 } , { u 7 , u 4 , u 5 , u 6 } , { u 1 , u 2 , u 3 } and { u 6 , u 7 , u 8 } . (c) Note that the two cycles with length of four share two nodes. Therefore, { v 4 , v 8 } are associated with { u 4 , u 5 } .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern