On Nov 14, 2007 3:19 PM, John Tromp <[EMAIL PROTECTED]> wrote:
> On Nov 14, 2007 2:00 PM, John Tromp <[EMAIL PROTECTED]> wrote:
>
> > My solution doesn't make use of that, and satisfies the stronger property:
>
> > 0 <= a_i <= 4 and sum a_i * n_i is in 1*nums union 2*nums union 3*nums
> > union 4*nums
> > => only one a_i is nonzero.
>
> that was not quite correct. it should say:
>
> let a_i be the number of adjacencies to a liberty at point i.
>
> if sum a_i <= 4, and sum (a_i * n_i) is in {1,2,3,4} * nums,
> then only one a_i is nonzero.
I'm really lost now. a_i is the number of stones we have adjacent to a
liberty at intersection i? Do we need to know the location of our
liberties to update sum a_i? How is this easier than just remembering
the locations of all real liberties you saw? How do we know that the
stones around i are from the same group? What are the n_i in
sum(a_i*n_i)? is {1,2,3,4}*nums supposed to be a cartesian product of
two sets? Is this related to what you said about encoding x and y
separately?
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