From help-request at octave dot org Fri Feb 11 04:31:05 2005
Subject: Re: Access the neighbors of an element
From: =?iso-8859-1?Q?J=F6rg?= Sommer <joerg at alea dot gnuu dot de>
To: Przemek Klosowski <przemek at jazz dot ncnr dot nist dot gov>
Cc: help at octave dot com
Date: Thu, 10 Feb 2005 14:24:26 +0100

Przemek Klosowski schrieb am Wed 09. Feb, 15:57 (-0500) :
> I can't figure out the intent of your code (I don't know what pos(t) is,
> and what's down/up and n). On the general description, I am guessing that
> you are simulating something like a spin system on a rectangular lattice,

Yes, that's right. I have a n*n matrix with -1 and 1 and simulate their
behaviour -- I know it as the ising modell. I energie of the matrix is
the number of unequal neighbors minus the number of equal neighbor, where
a neighbor is the right M(i,j+1) or the lower M(i+1,j) element in the
matrix.

For the whole matrix I can determine the energie with:

X_energ = 2* (n*(n-1)...
            - sum(sum( X(1:n-1, :) == X(2:n, :) ))... 
            - sum(sum( X(:, 1:n-1) == X(:, 2:n) )) );

Over the time, the matrix changes only at one position. By hand it is
much easier to calculate the passage of the energie of one matrix to the
other, than calculate the whole energie again. The transit is in O(1),
the energie is in O(n^2).

My idea was to do the same in octave, but it doesn't work. It's faster to
compare 800 elements than find and access 4 elements.

> and that neighbors of X(m,n) refer to X(m-1,n), X(m+1,n), X(m,n-1) and X(m,n+1).

Yes.

> Maybe you want a more arbitrary neighbor configuration...
> 
> If that's your case, then you can do it the way I said, using terms of the form
> (X(m,n) == X(m-1,n)).

But n and m are chosen randomly and I don't know if m+1, m-1, n-1, n+1
exist.

My current way is:
    % upper and lower bounds of the matrix
    up = 0:n:n*n;
    down = 1 + n + up;

    % calculate all possible neighbors
    neighbor = [-n, n, -1, 1] + pos(t);
    
    if any(neighbor(4) == down)
        neighbor(4) = [];
    end
    if any(neighbor(3) == up)
        neighbor(3) = [];
    end
    if neighbor(2) > n*n
        neighbor(2) = [];
    end
    if neighbor(1) < 1
        neighbor(1) = [];
    end
    
    % calculate the energie the "simple" way
    Y_energ = X_energ + 2*(length(neighbor) - 2*sum( X(pos(t)) ~= X(neighbor) ));

note: I found out generating n random number at one time is _much_ faster
then generating one random number n times. Because this I generate all my
positions before entering the loop.

> In general, the linear (or linearized) hamiltonian usually looks like
> a 2D convolution:  E = sum_i=allNeighbors (a*Si*S0)
> 
> [...] = conv2 (...)

What's that?
octave:33> conv2
error: `conv2' undefined near line 33 column 1


-- 
Viele Leute glauben, daß sie denken, wenn sie lediglich
ihre Vorurteile neu ordnen.



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