{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Discrete Mathematics with Graph Theory (3rd Edition) 298

# Discrete Mathematics with Graph Theory (3rd Edition) 298 -...

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

296 Solutions to Exercises Partition into pairs Sum of lengths of shortest paths Graph (a) Graph (b) Graph (c) {A,B}, {C,D}, {E,F} 1+1+1=3 2+2+3=7 5 + 2 + 5 = 12 {A, B}, {C, E}, {D, F} 1+2+2=5 2+2+4=8 5 + 5 + 6 = 16 {A,B}, {C, F}, {D, E} 1+3+1=5 2+3+2=7 5 + 8+3 = 16 {A,C}, {B,D}, {E,F} 2+2+1=5 3+4+3=10 6+6+5=17 {A,C}, {B,E}, {D,F} 2+3+2=7 3+3+4=10 6 + 6 +6 = 18 {A, C}, {B,F}, {D,E} 2+2+1=5 3+2+2=7 6+7+3 = 16 {A,D}, {B,C}, {E,F} 3+1+1=5 4+3+3=10 4+4+ 5 = 13 {A,D}, {B,E}, {C,F} 3+3+3=9 4+3+3=10 4+6+8 = 18 {A, D}, {B, F}, {C, E} 3+2+2=7 4+2+2=8 4+ 7+ 5 = 16 {A, E}, {B, C}, {D, F} 2+1+2=5 3+3+4=10 3 +4 + 6 = 13 {A, E}, {B, D}, {C, F} 2+2+3=7 3+4+3=10 3+6+8=17 {A,E}, {B,F}, {C,D} 2+2+1=5 3+2+2=7 3 + 7 + 2 = 12 {A,F}, {B, C}, {D,E} 1+1+1=3 2+3+2=7 4+4+3=11 {A,F}, {B,D}, {C,E} 1+2+2=5 2+4+2=8 4+ 6 +5 = 15 {A, F}, {B, E}, {C, D} 1+3+1=5 2+3+2=7 4+ 6+2 = 12 Graph (d) Partition into pairs Sum of lengths of shortest paths {A,B}, {C,D}
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: {A, C}, {B, D} {A,D},{B,C} 5+3=8 4+6=10 3+6=9 6. [BB] The shortest route from X to X is a circuit (perhaps in a pseudograph) passing through Y. Hence, it can also be viewed as a circuit from Y to Y. Any other route from Y to Y could also be viewed as a route from X to X. So there cannot be any shorter route from Y to Y. 7. The six vertices of odd degree are {A, B, C, D, E, F} and the shortest path totals for all partitions are given. {A, B}, {C, D}, {E, F} {A, B}, {C, F}, {D, E} {A, C}, {B, E}, {D, F} {A, D}, {B, C}, {E, F} {A, D}, {B, F}, {C, E} {A, E}, {B, D}, {C, F} {A, F}, {B, C}, {D, E} {A, F}, {B, E}, {C, D} 2+3+1=6 2+3+1=6 3+3+2=8 4+3+1=8 4+4+3=11 5+2+3=10 6+3+1=10 6+3+3 = 12 {AB,CE,DF} {AC,BD,EF} {AC,BF,DE} {AD,BE,CF} {AE,BC,DF} {AE,BF,CD} {AF,BD,CE} 2+3+2=7 3+2+1=6 3+4+1=8 4+3+3=10 5+ 3 +2 = 10 5+4+ 3 = 12 6+2+3=11 So we see that one solution is to add cBPies of edges AX, X B, C M, M N, N D and EF, as shown. D A...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online