Difference between revisions of "GATE2010 q1"

Revision as of 22:50, 13 July 2014

Let $G=(V, E)$ be a graph. Define $\xi(G) = \sum\limits_d i_d*d$, where $i_d$ is the number of vertices of degree d in G. If S and T are two different trees with $\xi(S) = \xi(T)$, then

(A) $|S| = 2|T|$

(B) $|S| = |T| - 1$

(C)$ |S| = |T| $

(D) $|S| = |T| + 1$

Solution by Happy Mittal

$\xi(G)$ basically is the sum of degrees of all vertices in a graph. If sum of degrees of two different trees are same, then number of nodes in the trees has to be same. We prove this by induction on sum of degree of all vertices in the two trees.

• Base case : When $\xi(S) = \xi(T) = 1$, then we have only single node in both trees and hence $|S| = |T|$.
• Induction hypothesis : Assume, for $\xi(S) = \xi(T) = k$, we have $|S| = |T|$.
• Induction step : For $\xi(S) = \xi(T) = k+1$, we must add a leaf vertex in both trees, and hence $|S| = |T|$.

So $|S| = |T|$. So option (C) is correct.