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Unformatted text preview: Discrete Mathematics, Spring 2004 Homework 9 Sample Solutions 7.1 #36. A vertex v in a tree T is a center for T if the eccentricity of v is minimal; that is, if the maximum length of a simple path starting from v is less than or equal to the maximum length of a simple path starting from any other vertex w . Show that a tree has either one or two centers. Solution . Let P be a simple path of maximal length in T . (Such a path surely exists; in fact, we have seen this construction before, when establishing equivalent characterizations of trees.) The length of this path is either even or odd. In the first case, denote this length by 2 m ; then we have P = ( v ,v 1 ,... ,v m − 1 ,v m ,v m +1 ,... ,v 2 m − 1 ,v 2 m ) , where all of the v i are distinct. We claim that v m is the unique center for T . To see this, notice first that the eccentricity of v m is actually equal to m (clearly it must be at least m ). If it were not, then there would exist a simple path ( v m = w ,w 1 ,...,w k = w ) with k > m . Since trees are acyclic, there are only two possibilities: • w j = v m + j for all j ≤ some nonnegative integer j , and w j / ∈ { v ,... ,v 2 m } for all j > j , or • w j = v m − j for all j ≤ j , and w j / ∈ { v ,... ,v 2 m } for all j > j . In the first case, the path P ′ = ( v ,v 1 ,...,v m = w ,w 1 ,... ,w k = w ) would be a simple path of length greater than that of P , contradicting the way in which we chose P . In the second case, the path P ′′ = ( w = w k ,... ,w 1 ,w = v m ,v m − 1 ,...,v ) would be a simple path of length greater than that of P , again a contradiction. Hence the eccentricity of v m is equal to m . Now consider any of the remaining vertices u in T . If u is one of the v i , then its eccentricity must be > m , for if u = v i with i ≤ m then ( v i ,v i +1 ,... ,v,....
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.
 Spring '08
 WOUTERS
 Math, Algebra

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