# eccentricity of all the vertices in the strong product of two given graphs

I am trying to prove a result, where I have to show that the strong product of $K_{2}$ and $P_n$, n>2, is not self-centered graph, that is, eccentricity is not equal for all the vertices. the idea which i am having is that in the product of graphs, distance between two vertices $g_{hi}$ and $g_{jk}$ is equal to the sum of $d_V$($u_h$,$u_h$) and $d_V$($v_i$,$v_k$), and thus $e_{U \times V}$($g_{ij}$) $=$ $e_Uu_i$ + $e_Vv_j$ where V and U are given graphs. i am using this idea that eccentricty of vertices in a path keep changing as we increase the number of vertices.

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What's "the product of graphs"? I'm not aware that any of the various products of graphs is referred to without further qualification simply as "the product". Your assertion appears to be true for the Cartesian product but not for the strong product; but in the title and the first part of the question you refer to the strong product. –  joriki Oct 10 '12 at 5:39

in the strong product, the distance $d(x,y)$ for vertices with coordinates $x=(x_1,..x_n)$ and $y=(y_1,..y_n)$ is the maximum of the distances $d(x_i,y_i)$ over each factor. The sum holds for Cartesian products only.