Re: Access the neighbors of an element

[Francesco Potorti`]

>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.

sum is a very quick operation, while evaluating an if is not, and you do
it four times.  Generally speaking, builtin functions are very quick, as
long as your data fit in real memory, while interpreting code is not.

>How can I make it better?

I am not sure I understand what you want to do.

-- 
Francesco Potortì (ricercatore)        Voice: +39 050 315 3058 (op.2111)
ISTI - Area della ricerca CNR          Fax:   +39 050 313 8091
via G. Moruzzi 1, I-56124 Pisa         Email: address@hidden
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