On Nov 14, 2007 5:03 PM, Imran Hendley <[EMAIL PROTECTED]> wrote:
>
> 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
For every string, you can keep track of this sum incrementally.
When the string establishes a new adjacency to an empty point i,
you add code[i] into the sum.
> 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
n_i was the name that Tom gave to my code[i].
> sum(a_i*n_i)? is {1,2,3,4}*nums supposed to be a cartesian product of
no, it's just my shorthand for the union of the 4 sets nums, 2*nums,
3*nums and 4*nums.
regards,
-John
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