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Let $L\geq 3, d\geq 2, B_L=\left\{0,1,...,L-1\right\}^d$. With $I$ denote the set of pairs of adjacent points in $B_L$.

I think by adjacent the following is meant: http://en.wikipedia.org/wiki/Von_Neumann_neighborhood

I do not understand why the cardinality of $I$ is $d\cdot L^d$.

Can you explain that to me, please?

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    $\begingroup$ What does it mean to be adjacent? $\endgroup$ – Andrés E. Caicedo Dec 9 '14 at 18:27
  • $\begingroup$ On the whole $Z^2$ lattice for example each point has 4 neighbors, the point above, below, to the left and to the right. $\endgroup$ – Salamo Dec 9 '14 at 18:28
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    $\begingroup$ That's an example. What is the definition? $\endgroup$ – Andrés E. Caicedo Dec 9 '14 at 18:28
  • $\begingroup$ This is the problem: I do not know. Only have the result, that the cardinality is $dL^d$. $\endgroup$ – Salamo Dec 9 '14 at 18:29
  • $\begingroup$ @AndresCaicedo I think this is meant: en.wikipedia.org/wiki/Von_Neumann_neighborhood $\endgroup$ – Salamo Dec 9 '14 at 20:32
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Two points in $B_L$ are adjacent if and only if they differ in exactly one coordinate, and in that coordinate they differ by exactly $1$ modulo $L$. Thus, each point is adjacent to $2d$ other points, which are obtained by adding $\pm1\pmod L$ to each coordinate. There are $L^d$ points in $B_L$, and each is adjacent to $2d$ other points, so that’s a total of $2dL^d$ adjacencies. However, each adjacency is counted twice, once for each of the adjacent points, so there are only $dL^d$ pairs of adjacent points.

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  • $\begingroup$ Merci bien! I got it now. 😀 $\endgroup$ – Salamo Dec 9 '14 at 22:44
  • $\begingroup$ @Salamo: You’re welcome! $\endgroup$ – Brian M. Scott Dec 9 '14 at 22:45

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