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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...
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 Summer '10
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 Graph Theory

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