The UCT heuristic of trying every child of a node once before trying
any child twice is reasonable when the payoff distribution is unknown.
Why try the lever that paid $5 a second time if there might be another
lever that pays $1,000,000? But when the set of possible payoffs is
known to be {1,
With repeat-winners, if there is a move is seems flawless at first but
some flaw is eventually found, there might be a rough transition once
the flaw is identified, since there is no backup plan. It might make
more sense to study two apparently flawless children equally until a
flaw is found in
Just one comment. The statement that 'it (CrazySton) has a 2k rating on KGS...'
could be misleading. Crazystone never achived 2k rating against human player in
real gemes. It achieved the ranking by basically winning on time.
Daniel Liu
-Original Message-
From: Nick Wedd [EMAIL
On Sun, 2007-06-10 at 21:40 -0400, [EMAIL PROTECTED] wrote:
Just one comment. The statement that 'it (CrazySton) has a 2k rating
on KGS...' could be misleading. Crazystone never achived 2k rating
against human player in real gemes. It achieved the ranking by
basically winning on time.
On Sun, 2007-06-10 at 22:19 -0400, Don Dailey wrote:
Did something happen that unfairly caused the player to lose on time?
No, but the games were absolute time games where CrazyStone was often in
a losing position but ended up winning on time. The endgame in Go takes
a long time but is mostly
On Sun, 2007-06-10 at 22:42 -0400, Jeff Nowakowski wrote:
On Sun, 2007-06-10 at 22:19 -0400, Don Dailey wrote:
Did something happen that unfairly caused the player to lose on time?
No, but the games were absolute time games where CrazyStone was often in
a losing position but ended up
byo-yomi is important for go, or at the very least,
canadian time standards.
s.
- Original Message
From: Jeff Nowakowski [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; computer-go computer-go@computer-go.org
Sent: Sunday, June 10, 2007 10:42:50 PM
Subject: Re: [computer-go] Congratulations
