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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...
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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