To find maximum of the given formula I am reading about graph operations. $G$ is the product of $n$ graphs $G_i$, $1\leq i\leq n$. In particular its strong product. $x = (x_1,x_2,...,x_n)$ and $y = (y_1,y_2,...,y_n)$ are two vertices of $G$. Distance between x and y is:
$d_{G}(x,y) = \max_{1\leq i\leq n }\{d_{G_{i}} (x_i, y_i)\}$
To find the eccentricity of $x$ we have to find the maximum of $d(x,y)$, where $y \in V(G)$. I am not able to find the maximum value. Can anybody help me in this case. Any help or hint will be useful. Heartily thanks.
Here $d$ denotes the distance between vertices of the graph.
 A: Hint: You know that $d_G(x,y)=\max_{1\leq i \leq n}\{d_{G_i}(x_i,y_i)\}$ hence the maximum you are looking will at least be greater than the maximum of $d_{G_i}(x_i,y_i)$ but can it be greater than every of those maximum?
Answer:
Let $m=(m_1,...,m_n)$ be a vertex of $G$ such that $d_{G_i}(x_i,m_i)=\max_{v\in V(G_i)} d_{G_i}(x_i,v)$ (lets denote $d_i$ this maximal distance to $x_i$ in $G_i$).
Let $d=d_G(x,m)=\max_{i\in\{1..n\}}d_i=\max_{i\in\{1..n\}}\max_{v\in V(G_i)} d_{G_i}(x_i,v)$.
Now that we have a "natural" candidate for the maximal distance lets show that $d$ is the maximal distance you are looking for.
Let $y=(y_1,...,y_n)\in V(G)$ be a vertex of $G$ such that $d_G(x,m)\leq d_G(x,y)$. By definition $d_G(x,y)=\max_{1\leq i \leq n}\{d_{G_i}(x_i,y_i)\}\underbrace{=}_{def}d_{G_j}(x_j,y_j)$ for some $j$. By definition of $d$ we know that $d=d_G(x,m)=\max_{i\in\{1..n\}}d_i$ in particular $d\geq d_j$ and by definition of $d_j$ we know that $d_j\geq d_{G_j}(x_j,y_j)=d_G(x,y)$.
Hence we have $d_G(x,y)\geq d\geq d_G(x,y)$. Hence for all $y\in V(G)$ such that $d_G(x,m)\leq d_G(x,y)$ we have $d_G(x,m)= d_G(x,y)$. Hence $d$ is a maximum (reached by $m$).
As a conclusion if you want to compute the eccentricity of $x$ in $G$ you have to find for all $i$ the eccentricity of $x_i$ in $G_i$ and take the maximum.
I hope it helps, I tried to give a full detailed proof staying clear, but may be there is too much details;
A: Let, $A$ is the matrix with $$A_{ij}=d\left(x_i,y_i^{(j)}\right)$$ where $y^{(j)}=(y_1^{(j)},y_2^{(j)},\cdots,\ y_n^{(j)}) $ represents the $j$th node in the graph $G$. Clearly $A$ will be a $n \times |V(G)|$ matrix. Then the eccentricity of $x$ will be $$e_x=\max_{y\in V(G)}\max_{1\leq i\leq n}d\left(x,y^{(j)}\right)$$ So $e_x$ is equal to the maximum element(s) of the matrix $A$.
