[computer-go] 10k UCT bots

[Don Dailey]

dhillismail at netscape.net wrote:
> For those currently coding this up, I think the most important thing about this playout algorithm is that it is *temporary*. You will almost certainly be?replacing it with something different and better just a little bit down the road.
>
> Creating an MC-UCT bot has a well worn path and its kind of an ontology recapitulates phylogeny deal. First you implement light playouts (the random non-eyefilling moves people are describing), then you implement UCT, then you throw away the light playouts and replace them?with heavy playouts, then you start extreme modifications to UCT...
>
> So you probably don't want to worry about hair-splitting tweaks except as an academic exercise.
>   
Yes,  I agree.   Also  my hair brained scheme of pre-generated tables of 
list traversal orderings was just an academic exercise as you say.  

- Don



> - Dave Hillis
>
>
> -----Original Message-----
> From: Christoph Birk <birk at ociw.edu>
> To: computer-go <computer-go at computer-go.org>
> Sent: Tue, 13 May 2008 3:40 pm
> Subject: Re: [computer-go] 10k UCT bots
>
>
> On Tue, 13 May 2008, Mark Boon wrote:?
>   
>>>> If this asymmetry really bothers you, you could very easily fix this by?
>>>> wrapping the search around. There's no asymmetry in a circle.?
>>>>         
>>>>> That doesn't fix anything.?
>>>>>           
>> ?
>> Why not? The whole argument is about a bias against points towards the end. > In a circular list there is no 'end'.?
>>     
> ?
> No, it was a bias towards moves "behind" illegal moves.?
> Those moves are twice as likely to be played than other moves. Consider a list with 5 moves:?
> ?
> [Move1] [Move2] [Move3] [Move4] [Move5]?
> ?
> You create a random number between 1 and 5. If Move2 is illegeal?
> for example, then you will play?
> ?Move1 if random#=1?
> ?Move3 if random#=2 or 3,?
> ?Move4 =4?
> ?Move5 =5?
> ?
> Move3 is twice as likely to be played. Even if you make a circular?
> list.?
> ?
> Christoph?
> ?
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