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If $G$ is 2 - self centered graph. then how to prove that $G$ has at least $2n - 5$ edges? where $n\geq 5$.

I started by assuming if number of edges $\mid E\mid\leq 2n-6$ then there exist a vertex say $u$ such that $deg = 2$ otherwise if no such vertex exists then

$\mid E\mid\geq \frac{3n}{2}>2n-5$ (I am stucked here. How to prove this. Sincerely thanks for giving me time.)

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    Since there are 2-self-centered graphs of arbitrarily large n, you will not be able to prove $\frac{3n}{2} \gt 2n-5$.2012-05-16

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I take it $n$ is the number of vertices. You might be interested in https://mathoverflow.net/questions/32301/examples-of-self-centered-graphs-with-large-radius. Among the things you'll find there is a link to a complete list of all self-centered graphs with diameter 2 that have minimum number of edges, and a link to a survey paper by Fred Buckley.

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i strongly recommend you to refere the following paper: Akiyama. j., Ando.K and Davis.D., Miscellaneous properties of equi eccentric graph in Convexity and Graph Theory. Proc. Conf. in Haifa Isreal-1981.

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    Would you ex$p$lain why this $p$aper is useful for the Question? Answers are generally expected to address the Question being asked.2012-05-15