Rémi:
John and I are planning a similar usage of patterns and features in
the MC simulations, although we were thinking of a very different
method to update "strengths". Although we derived our formulas from an
extension of the logistic regression instead of Bradley-Terry models,
we arrived at very similar expressions. The main difference is that
what we have been calling "score" seems to be the log of your
"strength", and instead of multiplying strengths when considering
"teams", we just add the scores of different features. We use an
online learning algorithm to adjust the scores during the game, which
severely limits the kind of algorithms we can use, but also allows us
to learn things that apply specifically to the situations that are
appearing on the board in this game.
Now we only have to make dimwit 400 points stronger to show the world
that our approach works. :)
There are many things in the paper that we had never thought of, like
considering the distance to the penultimate move. Most of those things
won't be directly applicable to dimwit, but it gives us many
suggestions of things to try.
Thanks for this important contribution to computer go.
Álvaro.
On 5/17/07, Rémi Coulom <[EMAIL PROTECTED]> wrote:
Hi,
It seems that e-mail at my university does not work any more. I have
received none of the replies to my message of yesterday, but I could
read them on the web archives of the list. So I have registered from
another address, and will answer to the questions I have read on the web.
to Matt: In Crazy Stone, UCT is still used as before to select moves.
Patterns are used to shorten the list of moves that UCT considers.
to Sylvain: Here are tests of Crazy Stone at 90s/game 1CPU against GNU
3.6 level 10, measured over about 200 games:
no pattern: 38.2% (version available on my web page)
patterns in simulations:68.2%
patterns for pruning and simulations:90.6%
I found these number in my history of Crazy Stone versions, so they may
not be extremely accurate, because other things have changed in between.
I will run cleaner tests for the final version of the paper. I will also
try to test what each feature brings.
to Magnus: If you consider the example of section 2.2: 1,2,3 wins
against 4,2 and 1,5,6,7. The probability is
P=c1c2c3/(c1c2c3+c4c2+c1c5c6c7). For this example:
N1j=c2c3,B1j=0,M1j=c2c3+c5c6c7,A1j=c4c2
N2j=c1c3,B2j=0
N3j=c1c2,B3j=0
N4j=0,B4j=c1c2c3
I will add this example to the paper if it makes things clearer.
Regarding the definition of Wi, |{Nij|Nij!=0}| means "number of elements
of the set of all the Nij that verify Nij!=0". It is incorrect. It
should be |{j|Nij!=0}|. I'll fix it in the final version. As I wrote in
the next sentence, it is equal to the number of wins of i.
Thanks to all for the kind words, and the interesting remarks and questions.
Rémi
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