English

Relation between the number of leaves of a tree and its diameter

Combinatorics 2019-04-30 v1

Abstract

Let L(n,d)L(n,d) denote the minimum possible number of leaves in a tree of order nn and diameter d.d. In 1975 Lesniak gave the lower bound B(n,d)=2(n1)/dB(n,d)=\lceil 2(n-1)/d\rceil for L(n,d).L(n,d). When dd is even, B(n,d)=L(n,d).B(n,d)=L(n,d). But when dd is odd, B(n,d)B(n,d) is smaller than L(n,d)L(n,d) in general. For example, B(21,3)=14B(21,3)=14 while L(21,3)=19.L(21,3)=19. We prove that for d2,d\ge 2, L(n,d)=2(n1)d L(n,d)=\left\lceil \frac{2(n-1)}{d}\right\rceil if dd is even and L(n,d)=2(n2)d1L(n,d)=\left\lceil \frac{2(n-2)}{d-1}\right\rceil if dd is odd. The converse problem is also considered. Let D(n,f)D(n,f) be the minimum possible diameter of a tree of order nn with exactly ff leaves. We prove that D(n,f)=2D(n,f)=2 if n=f+1,n=f+1, D(n,f)=2k+1D(n,f)=2k+1 if n=kf+2,n=kf+2, and D(n,f)=2k+2D(n,f)=2k+2 if kf+3n(k+1)f+1.kf+3\le n\le (k+1)f+1.

Keywords

Cite

@article{arxiv.1904.12150,
  title  = {Relation between the number of leaves of a tree and its diameter},
  author = {Pu Qiao and Xingzhi Zhan},
  journal= {arXiv preprint arXiv:1904.12150},
  year   = {2019}
}