Hi Hiroshi, I think the problem with adding moves of the opponent is that they don't come from the same distribution. E.g., imagine if Black is winning 90% then it's normal rave moves are also winning 90% on average (so mixing with the direct statistics doesn't cause a big change on average regardless of the mixing ratio). However, once you start to add White's moves the distribution is reversed (on average only winning 10%) and together those statistics will average roughly on 50%, which causes a systematic bias.
The intersections where both sides want to play first will of course still stand out, so that's why reversing the sign helps at least a bit. However, to get this to work really well you need to find a clever way to get the bias down... Expand on this a bit and you have yourself a new rave paper ;-) Best, Erik On Fri, Aug 12, 2011 at 7:18 PM, Hiroshi Yamashita <[email protected]> wrote: > Hi Erik, > >> I think the usual procedure is to update only moves by the same >> player. However, if you also want to use the opponent's moves (which >> is not an unreasonable idea), then you should probably reverse their >> sign (because those moves lost), so you only get something like: > > I tried this idea. > > 9x9 (1000games 6000po against Fuego 10000po) > 0.516 RAVE normal (same player only) > 0.211 RAVE Black and White > 0.264 RAVE Black and White, reverse their sign > 0.092 RAVE not same player only, reverse their sign > > 19x19 (1000games 700po agaist GnuGo Lv0) > 0.570 RAVE normal (same player only) > 0.013 RAVE Black and White > 0.078 RAVE Black and White, reverse their sign > 0.141 RAVE not same player only, reverse their sign > > Normal is clearly better, but "reverse their sign" looks a bit meaningful. > >> In a B node, update only B moves - and only if there's no W move at >> same location before B's move. > > I use this in 9x9, but I ignore all W moves in 19x19. I got a little > better result from this. > > Regards, > Hiroshi Yamashita > > > > ----- Original Message ----- From: "Erik van der Werf" > <[email protected]> > To: <[email protected]> > Sent: Thursday, August 11, 2011 8:42 PM > Subject: Re: [Computer-go] RAVE uses all moves in playout? > > > On Thu, Aug 11, 2011 at 3:43 AM, Hiroshi Yamashita <[email protected]> wrote: >> >> Hi, >> >> I have a question about RAVE. >> >> For example, there are 5 moves a,b,c,d,e. Black to play. >> In playout, Black plays "a", White plays "b", Black plays "c", >> W plays "d" and B plays "e". >> Then game is over, result is B win. >> >> B:a W:b B:c W:d B:e ... B win, result = +1 >> >> I update only B moves RAVE, like this. >> >> RAVEcount(a) += 1, RAVEwins(a) += 1 >> RAVEcount(c) += 1, RAVEwins(c) += 1 >> RAVEcount(e) += 1, RAVEwins(e) += 1 >> >> But on Sylvain's paper Figure4, >> It looks like updating all moves Black and White, like this. >> >> RAVEcount(a) += 1, RAVEwins(a) += 1 >> RAVEcount(c) += 1, RAVEwins(c) += 1 >> RAVEcount(e) += 1, RAVEwins(e) += 1 >> RAVEcount(b) += 1, RAVEwins(b) += 1 >> RAVEcount(d) += 1, RAVEwins(d) += 1 >> >> Is this right RAVE? >> I have misunderstood long time. > > Hi Hiroshi, > > I think the usual procedure is to update only moves by the same > player. However, if you also want to use the opponent's moves (which > is not an unreasonable idea), then you should probably reverse their > sign (because those moves lost), so you only get something like: > >> RAVEcount(b) += 1 >> RAVEcount(d) += 1 > > Best, > Erik > _______________________________________________ > Computer-go mailing list > [email protected] > http://dvandva.org/cgi-bin/mailman/listinfo/computer-go > _______________________________________________ > Computer-go mailing list > [email protected] > http://dvandva.org/cgi-bin/mailman/listinfo/computer-go > _______________________________________________ Computer-go mailing list [email protected] http://dvandva.org/cgi-bin/mailman/listinfo/computer-go
