Re: Access the neighbors of an element

[Mike Miller]

I'm not clear on what you're doing, but does this help?
For matrix X:

[n,m] = size(X)

X(2:n,:) has "up neighbors" X(1:n-1,:)
X(1:n-1,:) has "down neighbors" X(2:n,:)
X(:,2:m) has "left neighbors" X(:,1:m-1) and
X(:,1:m-1) has "right neighbors" X(:,2:m)

Maybe that is irrelevant to your question.

Mike


On Wed, 9 Feb 2005, Joerg Sommer wrote:

> Hi,
>
> I want to compare the neighbors of an element to the element, but this is
> harder than I thought.
>
> My idea:
>
>    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
>    Y_energ = X_energ + 2*(length(neighbor) -...
>                            2*sum( X(pos(t)) ~= X(neighbor) ));
>
> But this is slower than
>    Y_energ = 2* (n*(n-1)...
>                  - sum(sum( Y(1:n-1, :) == Y(2:n, :) ))...
>                  - sum(sum( Y(:, 1:n-1) == Y(:, 2:n) )) );
>
> with n=20. In one sentence: it is faster to compare and count twice
> nearly 800 elements, than find and access the four neighbors of an
> element. That's lunatic.
>
> How can I make it better?
>
> Jörg.
>
> --
> Es ist außerdem ein weit verbreiteter Irrtum das USENET "helfen" soll.
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> Jörg Klemenz <address@hidden>, <address@hidden>
>
>
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