On 2/23/07, Heikki Levanto <[EMAIL PROTECTED]> wrote:
Sure, but not all such boards are equivalent anyway!
Add a stone to the board. Add another stone to one of its liberties. Add
a third stone to any (empty) liberty of the last stone. There are three
possibilities. Choose the one that maximises the liberties of the
string. You have now defined a straight line. Continue this line until
you meet a black stone (which must be part of the original line). I
guess you meet the beginning of the line, where it all started. How big
portion of the board is now filled with black stones? That can vary
depending on the properties of the grid.
In the simple case you have drawn a circle of a fairly small size (say
19). In another simple case you have filled the whole board, and used
many more stones (say 361). In some cases you have filled half the
available points, or some other fraction. How big will this fraction be
on a totally random grid?
What, exactly, do you mean by "a totally random grid"? There is no
single obvious (to me) way of distributing vertexes between nodes. I
can think of several interesting ways, but no single obvious way.
a) start with a conventional board, add random "wraps" at the edges
(makes for convenient visualization)
b) start with an empty graph with n^2 nodes and pick random pairs of
nodes and add a vertex between them if neither already has 4 vertexes
(hard to visualize, risks disjoint boards)
c) start with a conventional board, pick a random pairs of nodes and a
a random vertex in each node. Switch the end points of the two
vertexes if the result is not a disjoint graph. Repeat N times.
...
It could easily be argued that only (a) results in a grid at all...
cheers
stuart
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