Cyclic Graph Dr. I.Pradeepa Deparment of

Theorem 1  (a) Let u and v be distinct vertices of a tree T. Then there is precisely one path from u to v . (b) Let G be a graph without any loops. If for every pair of distinct vertices u and v of G there is precisely one path from u to v, then G is a tree. 4 Proof:
Suppose the result is false
Then there exist two vertices
P=u1u2…..umv PI=uv1v2……vnv

w - common vertex
This produces a cycle, since no two of the “u” terms are repeated.
no two the “v” terms are repeated.

Since T is a tree, T has no cycles.
This contradiction means that our initial assumption must be false.
Thus there is precisely one path from u to v. p pI<br>

5 Theorem 2
Let T be a tree with at least two vertices and P=u0u1…..un be a
Longest path in T (so that there is no path in T of length greater than n).
Then both u0 and un have degree 1. (i.e)., d(u0)=1=d(un).<br>

Corollary  Any tree T with at least two vertices has more than one vertex degree 1.Proof: In such a tree T there is a longest path P ( of length greater than 0 ) and so the theorem produces at least two vertices of degree 1. 6<br>