[Computer-go] UCT parameters and application to other games

Don Dailey dailey.don at gmail.com
Tue Apr 5 10:50:07 PDT 2011

I think I just don't fully understand what your proposal was.   Assuming a 4
core machine,  it sounded to me like you wanted to take 1/4 of the (legal)
points, sampled randomly, and then assign each one to a processor and start
searching a move.   When search times out,  combine all the results and get
your move.

But it sounds like you are proposing something more recursive and fine
grained than this.   It's totally not clear to me exactly what you mean but
it doesn't matter,  I'm sorry if I misunderstood you.


On Tue, Apr 5, 2011 at 1:20 PM, steve uurtamo <uurtamo at gmail.com> wrote:

> > I think you completely miss the point.
> that's super unfortunate.
> but in case i haven't, let me make myself more clear:
> whenever you have a distribution from which you would like to
> uniformly sample, if the number of things to be sampled is small
> enough, you can simply randomly reorganize them, partition them, and
> ship them off to your N friends, each of which can do the sampling for
> you. because it is a partition, you can simply add up their partial
> results whenever they reach the (normalized) threshhold that you would
> have had you sampled the distribution yourself. they don't even all
> have to finish at the same time.
> no matter where you are in a tree, and no matter what restrictions you
> have in place on the subtree(s) to be sampled, the moment you want to
> uniformly sample, you can ask your otherwise unoccupied friends to do
> so for you, collect their results, and decide what to do next.
> the main objection (other than my complete misunderstanding of MCTS)
> that i was hearing was the concern that there would not be a uniform
> distribution of interesting node expansion among the processors.
> however, this concern is not valid, as the chernoff bound involved
> shows that the tail events under consideration almost never happen,
> especially so if you are willing to uniformly sample from a
> several-node-deep view of the tree. adding up the partial results is
> super easy, since each sub-sum will be prefixed by its location in the
> tree.
> i didn't mean anything more or less than this.
> over and out for another few months,
> s.
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