Solved Problem on First-Year Interest Group Seminar | N 1, Exams of Health sciences

Download Solved Problem on First-Year Interest Group Seminar | N 1 and more Exams Health sciences in PDF only on Docsity! (Problem 11) A tree is a maple if and only the degree of every vertex is either one or five. Given a maple on n vertices, find a formula to express the maximum and minimum number of pendants it contains. Prove your result is correct. Let’s start from an easy case: if n = 2, then the number of pendants vertices is 2. Now consider n > 2. As the maple is a tree, the number of edges is n − 1. Let i be the number of pendant vertices of the maple. This is also the number of edges that connect a pendant to a non-pendant vertex. So the number of non-pendant vertices is n − i, and the number of edges that connect exclusively non-pendant vertices are n− 1− i. If we count the number of edges that come out from all the non-pendant vertices, as they have degree five, the sum is 5(n− i). In this sum each edge that connect two non-pendant vertices is counted twice, while those that connect a pendant to a non-pendant one only once. Consequently 5(n− i) = n− 1 + (n− 1− i). This contraint, that can be simplified to i = 3n+2 4 , indicates that there only one possible value for i given n. In addition to this, as i has to be an integer, n should be divisible by 2 but not by four. 1