{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Discrete Mathematics with Graph Theory (3rd Edition) 297

# Discrete Mathematics with Graph Theory (3rd Edition) 297 -...

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

Section 11.1 295 Partition into pairs Sum of lengths of shortest paths Unweighted graph Weighted graph {A, B}, {C,D}, {E, F} 2+1+1=4 4+1+2=7 {A,B}, {C,E}, {D,F} 2+2+2=6 4+3+2=9 {A, B}, {C, F}, {D, E} 2+1+1=4 4+1+3=8 {A, C}, {B, D}, {E, F} 3+2+1=6 3+2+2=7 {A,C},{B,E}, {D,F} 3+3+2=8 3+4+2=9 {A, C}, {B, F}, {D, E} 3+2+1=6 3+2+3=8 {A, D}, {B, C}, {E, F} 4+1+1=6 4+1+2=7 {A,D},{B,E}, {C,F} 4+3+1=8 4+4+1=9 {A,D}, {B,F}, {C,E} 4+2+2=8 4+2+3=9 {A, E}, {B, C}, {D, F} 3+1+2=6 4+1+2=7 {A,E}, {B, D}, {C, F} 3+2+1=6 4+2+1=7 {A, E}, {B, F}, {C, D} 3+2+1=6 4+2+1=7 {A,F},{B,C}, {D,E}
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2+1+1=4 2+1+3=6 {A,F}, {B, D}, {C,E} 2+2+2=6 2+2+3=7 {A, F}, {B, E}, {C, D} 2+3+1=6 2+4+1=7 5. In graphs (a), (b), and (c), there are six odd vertices, labeled A, ... , F. In case (a), we must add at least three edges. A solution with just three edges added is shown. In case (b), edges of total weight 7 must be added. In case (c), edges of total weight 11 must be added. In case (d), there are only four odd vertices, labeled A, B, C, D. The solution is to duplicate four edges, with a total weight of 8, as shown. A A (a) (b) 4 6 C E D D A (c) (d) D C...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online