Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.