Periodic indexing of a matrix

I need some help, i have a matrix rapresenting points on a grid and when given a element i would like to find the indices of its nearest neighbors keeping in mind that i have periodic boundary conditions, so that if i have the element A(1,1) its nearest neighbors are

• A(1,N)
• A(2,1)
• A(1,2)
• A(N, 1)

Where A is my matrix and N is the dimension, and i need a code which will find the indices of n.n of a given element. Thanks in advance

[Image Analyst]

Is A all zeros except at certain points (the "grid" points) it has value 1??? Like A is 5x5 and columns 1, 3, and 5 and rows 1, 3, and 5 all ones?

Please give a full example for A for a small N, like 4 or 5.

[Guillaume]

Subsequent to you finding these indices, you're going to do something with them. What is that going to be? It may be significantly easier to do that subsequent thing by just constructing a matrix of neighbouring values without ever figuring their indices.

[andrea]

So what I'm doing is an ising model, my matrix is a matrix of 1 and -1 randomly chosen. I choose a random element of my matrix then I need to choose another element among the nearest neighbors randomly and i want its index.

[Guillaume]

I don't know what's an ising model.

"and i want its index."

What for?

Answer 1

[David Goodmanson]

Hi Andrea,

for N points, index j, in 1 dimension,

fnext largest index = mod(j,N)+1

next smallest index = mod(j-2,N)+1

then just include the non-changing index for 2 dimensions.

As mentioned by Guillaume, it might be convenient to have entire matrices of new indices. For example, all the indices for the next right neighbor is

indright = (mod(1:N)+1,1:N)  

and so forth.

[Alexander]

This worked like a charm! Though I believe it should be:

indright = (mod(1:N,N)+1,1:N)

so that circularly shifting a vector's components by one index should look something like this:

x = [2 4 5]
N=length(x)
x_shift = x(mod(1:N,N)+1)