Discrete Mathematics with Graph Theory (3rd Edition) 303

Discrete Mathematics with Graph Theory (3rd Edition) 303 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 11.2 301 14. 4 3 1 2 (c) []>5 []>5 1 2 4 3 o 1 0 0 0 0 o 0 1 000 000 1 0 0 o 0 0 0 1 0 o 0 0 0 0 1 1 0 0 0 0 0 H n:1 10000 o 0 0 0 0 1 2 1 2 (d) With the graphs labeled as shown, the adjacency ma- trices are 0>5 0>5 4 3 000 1 0 1 011 000 000 4 3 1 000 o 100 o 0 1 0 o 000 1 100 The row of O's in Al and the lack of such a row in A2 is a further indication of the fact that the corresponding graphs are not isomorphic. (One has a vertex of outdegree 0 and the other does not.) [ ~ 1 0 ~ 1 [ ~ 0 0 (a) [BB] Al = 0 1 and A2 = 0 1 o 0 1 0 1 0 0 1 (b) [BB] With the vertices of 91 relabeled according to
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cp, its adjacency matrix becomes that of 92. [ ~ 1 0 n (c) [BB] P = o 0 o 0 0 1 (d) [BB] The digraphs are strongly connected: In 91, for instance, VI V2V3 V4 VI is a circuit which respects arrows and 92 is isomorphic to 91 hence, also strongly connected. (e) [BB] The digraphs are not Eulerian. In 910 for instance, vertex V2 has indegree 2 but outdegree 1. 15. (a) Al = [H ~ H 1 0 1 0 1 1 1 1 0 0 and A 2 = [~ H H o 1 000 1 1 1 0 0 (b) With the vertices of 91 relabeled according to cp, the adjacency matrix of 91 becomes that of 92. ----------------------...
View Full Document

Ask a homework question - tutors are online